Download - Fundamentals of Electric Circuits 3rd Sadiku
Chapter 1, Solution 1 (a) q = 6.482x1017
x [-1.602x10-19 C] = -0.10384 C
(b) q = 1. 24x1018
x [-1.602x10-19 C] = -0.19865 C
(c) q = 2.46x1019
x [-1.602x10-19 C] = -3.941 C
(d) q = 1.628x1020
x [-1.602x10-19 C] = -26.08 C
Chapter 1, Solution 2
(a) i = dq/dt = 3 mA (b) i = dq/dt = (16t + 4) A (c) i = dq/dt = (-3e-t + 10e-2t) nA (d) i=dq/dt = 1200 120π πcos t pA (e) i =dq/dt = − +−e tt4 80 50 1000 50( cos sin ) Aµt
Chapter 1, Solution 3
(a) C 1)(3t +=+= ∫ q(0)i(t)dt q(t)
(b) mC 5t)(t 2 +=++= ∫ q(v)dt s)(2tq(t)
(c) ( )q(t) 20 cos 10t / 6 q(0) (2sin(10 / 6) 1) Ctπ π µ= + + = + +∫
(d) C 40t) sin 0.12t(0.16cos40e 30t- +−=
−+
=+= ∫ t)cos 40-t40sin30(1600900
e10q(0)t40sin10eq(t)-30t
30t-
Chapter 1, Solution 4
( ) mC 4.698=−=
−=== ∫ ∫
ππ
π
06.0cos165
tπ6cos6
5dt t π 6 5sinidtq10
0
Chapter 1, Solution 5
µCmC )e1(21
e21-mC dteidtq
4
2
0
2t-2t-
490=−=
=== ∫ ∫
Chapter 1, Solution 6
(a) At t = 1ms, mA 40===2
80dtdqi
(b) At t = 6ms, mA 0==dtdqi
(c) At t = 10ms, mA 20-===4
80dtdqi
Chapter 1, Solution 7
<<<<<<
==8t6 25A,6t2 25A,-2t0 A,25
dtdqi
which is sketched below:
Chapter 1, Solution 8
C 15 µ1102
110idtq =×+×
== ∫
Chapter 1, Solution 9
(a) C 10=== ∫∫1
0dt 10idtq
(b) C 22.5=−+=
×+
×
−+×== ∫251015
152
1510110idtq3
0
(c) C 30=++== ∫ 101010idt5
0q
Chapter 1, Solution 10
q ixt x x x= = =−8 10 15 10 1203 6 Cµ Chapter 1, Solution 11
q = it = 85 x10-3 x 12 x 60 x 60 = 3,672 C E = pt = ivt = qv = 3672 x1.2 = 4406.4 J
Chapter 1, Solution 12
For 0 < t < 6s, assuming q(0) = 0,
q t idt q tdt tt t
( ) ( ) .= + = + =∫ ∫0 3 0 150
2
0
At t=6, q(6) = 1.5(6)2 = 54 For 6 < t < 10s,
q t idt q dt tt t
( ) ( )= + = + = −∫ ∫6 18 54 18 56 6
4
66
At t=10, q(10) = 180 – 54 = 126 For 10<t<15s,
q t idt q dt tt t
( ) ( ) ( )= + = − + = − +∫ ∫10 12 126 12 24610 10
At t=15, q(15) = -12x15 + 246 = 66 For 15<t<20s,
q t dt qt
( ) ( )= + =∫ 0 1515
Thus,
q t
ttt
( )
.
,
=−
− +
1518 5412 24666
2 C, 0 < t < 6s C, 6 < t < 10s C, 10 < t < 15s
C 15 < t < 20s
The plot of the charge is shown below.
0 5 10 15 200
20
40
60
80
100
120
140
t
q(t)
Chapter 1, Solution 13
kJ 2.486-=
−=
−=
==
==
∫∫∫
216sin4112008sin
821200
1)-2x cos 2xcos (since,1)dt -t8cos2(1200
dtt 4cos1200vidtw
2
0
2
0
2
2
0
22
0
tt
Chapter 1, Solution 14
(a) ( ) ( )
( ) C 2.131=−+=
+=== ∫ ∫2e2110
2et10dte-110idtq0.5-
1
0
0.5t-1
0
0.5t-
(b) p(t) = v(t)i(t) p(1) = 5cos2 ⋅ 10(1- e-0.5) = (-2.081)(3.935) = -8.188 W Chapter 1, Solution 15
(a)
( ) C 1.297=−−=
−=== ∫ ∫
1e5.1
e23dt3eidtq
2-
2
0
2t2
0
2t-
(b) We90
)(
t4−−==
−=−==
vip
e305e6dtdi5v 2t-2t
(c) J 22.5−=−−
=== ∫ ∫3
0
4t-3
0
4t- e490dt e-90pdtw
Chapter 1, Solution 16
mJ 916.7
)(
(
(((
,
=
++
+−−++=
−+
++=
−+−+++==
<<<<<<
=<<<<
=
∫
∫ ∫∫∫∫
4
3
22
4
3
2
3
2
22
1
1
0
3
4
3
3
2
2
1
1
0
3t4t-t1625028
29122503
2250
3250
dtt)42502t-t4250
2250t
3250
25t)mJ-t)(1001040t)dt2510010t)dt2510(25t)dt10v(t)i(t)dtw
4t3 V10t -403t1 V 101t0 V10t
v(t)4t2 mA25t -1002t0 mA25t
i(t)
Chapter 1, Solution 17 Σ p = 0 → -205 + 60 + 45 + 30 + p3 = 0 p3 = 205 – 135 = 70 W Thus element 3 receives 70 W. Chapter 1, Solution 18
p1 = 30(-10) = -300 W p2 = 10(10) = 100 W p3 = 20(14) = 280 W p4 = 8(-4) = -32 W p5 = 12(-4) = -48 W
Chapter 1, Solution 19
p I x x xs s= → − − − + = → =∑ 0 4 2 6 13 2 5 10 0 3 AI
Chapter 1, Solution 20
Since Σ p = 0 -30×6 + 6×12 + 3V0 + 28 + 28×2 - 3×10 = 0 72 + 84 + 3V0 = 210 or 3V0 = 54 V0 = 18 V
Chapter 1, Solution 21
nA8.
(.
C/s100.8 C/s10611084
electron) / C1061photon
electron81
secphoton104
8-1911
1911
=×=×××=
×⋅
⋅
×=
∆∆
=
−
tqi
Chapter 1, Solution 22
It should be noted that these are only typical answers.
(a) Light bulb 60 W, 100 W (b) Radio set 4 W (c) TV set 110 W (d) Refrigerator 700 W (e) PC 120 W (f) PC printer 18 W (g) Microwave oven 1000 W (h) Blender 350 W
Chapter 1, Solution 23
(a) W12.5===1201500
vpi
(b) kWh 1.125. =×=⋅×××== kWh60451.5J60451051 3ptw
(c) Cost = 1.125 × 10 = 11.25 cents
Chapter 1, Solution 24
p = vi = 110 x 8 = 880 W Chapter 1, Solution 25
cents 21.6 cents/kWh 930hr 64 kW 1.2 Cost =×××=
Chapter 1, Solution 26
(a) mA 80.=
⋅=
10hhA80i
(b) p = vi = 6 × 0.08 = 0.48 W (c) w = pt = 0.48 × 10 Wh = 0.0048 kWh
Chapter 1, Solution 27
∫ ∫ =××====
×==
kC 43.2 36004 3 T33dt idt q
36005 4 4h T Let (a)T
0
[ ]
kJ 475.2
..
.)((
=
××+×=
+=
+===
×
∫ ∫∫
36001625036004033600
250103
dt3600
t50103vidtpdtW b)
36004
0
2
0
T
0
tt
T
cents 1.188
(
=×=
==
cent 9kWh3600475.2 Cost
Ws)(J kWs, 475.2W c)
Chapter 1, Solution 28
A 0.25===12030 (a)
VPi
$31.54(
=×==××==
262.8 $0.12CostkWh 262.8Wh 2436530ptW b)
Chapter 1, Solution 29
cents 39.6
.
=×==+=
+
+++==
3.3 cents 12CostkWh 3.30.92.4
hr6030kW 1.8hr
6045)1540(20kW21ptw
Chapter 1, Solution 30
Energy = (52.75 – 5.23)/0.11 = 432 kWh Chapter 1, Solution 31
Total energy consumed = 365(4 +8) W Cost = $0.12 x 365 x 12 = $526.60
Chapter 1, Solution 32
cents 39.6
.
=×==+=
+
+++==
3.3 cents 12CostkWh 3.30.92.4
hr6030kW 1.8hr
6045)1540(20kW21ptw
Chapter 1, Solution 33
C 6=××==→= ∫ 31032000idtqdtdqi
Chapter 1, Solution 34
(b) Energy = ∑ = 200 x 6 + 800 x 2 + 200 x 10 + 1200 x 4 + 200 x 2 pt
= 10,000 kWh (c) Average power = 10,000/24 = 416.67 W
Chapter 1, Solution 35
kWh 10.4
)((
=+=
×+×××+×+×== ∫28007200
24002120012200210006400W a) dttp
W/h 433.3( =h 24kW 10.4 b)
Chapter 1, Solution 36
days 6,667,(
A 4
===
=⋅
=
day / 24hh000160
0.001A160Ah tb)
40hA160i (a)
Chapter 1, Solution 37
( )J 901.2.
..−=×−==
−=×−×= −
12180WC 180106021105q 1920
qv
Chapter 1, Solution 38
P = 10 hp = 7460 W W = pt = 7460 × 30 × 60 J = 13.43 × 106 J
Chapter 1, Solution 39
A 16.667=×
==→=120
102vpivip
3
Chapter 2, Solution 1
v = iR i = v/R = (16/5) mA = 3.2 mA
Chapter 2, Solution 2
p = v2/R → R = v2/p = 14400/60 = 240 ohms
Chapter 2, Solution 3
R = v/i = 120/(2.5x10-3) = 48k ohms Chapter 2, Solution 4
(a) i = 3/100 = 30 mA (b) i = 3/150 = 20 mA
Chapter 2, Solution 5
n = 9; l = 7; b = n + l – 1 = 15 Chapter 2, Solution 6
n = 12; l = 8; b = n + l –1 = 19 Chapter 2, Solution 7
7 elements or 7 branches and 4 nodes, as indicated. 30 V 1 20Ω 2 3 -
2A 30Ω 60Ω 40Ω 4
++++ -
10Ω
Chapter 2, Solution 8
dc
b
a
9 A
i3i2 12 A
12 A
i1
8 A At node a, 8 = 12 + i1 i1 = - 4A At node c, 9 = 8 + i2 i2 = 1A At node d, 9 = 12 + i3 i3 = -3A Chapter 2, Solution 9
Applying KCL, i1 + 1 = 10 + 2 i1 = 11A 1 + i2 = 2 + 3 i2 = 4A i2 = i3 + 3 i3 = 1A Chapter 2, Solution 10 At node 1, 4 + 3 = i1 i1 = 7A At node 3, 3 + i2 = -2 i2 = -5A
3
2-2A
3A
1
4A
i2i1
Chapter 2, Solution 11
Applying KVL to each loop gives -8 + v1 + 12 = 0 v1 = 4v -12 - v2 + 6 = 0 v2 = -6v 10 - 6 - v3 = 0 v3 = 4v -v4 + 8 - 10 = 0 v4 = -2v Chapter 2, Solution 12 For loop 1, -20 -25 +10 + v1 = 0 v1 = 35v For loop 2, -10 +15 -v2 = 0 v2 = 5v For loop 3, -v1 +v2 +v3 = 0 v3 = 30v
+ 15v -
loop 3
loop 2
loop 1 + 20v -
+ 10v - – 25v + + v2 -
+ v1 -
+ v3 -
Chapter 2, Solution 13
I2 1 I1
2A
7A I4 2 3 4
4A
3A I3
At node 2, 3 7 0 102 2+ + = → = −I I A
12
2 A
2 5 A
At node 1, I I I I A1 2 1 22 2+ = → = − =
At node 4, 2 4 2 44 4= + → = − = −I IAt node 3, 7 7 4 3 3+ = → = − =I I IHence, I A I A I A I1 2 3 412 10 5 2= = − = =, , , A−
V
11
8
Chapter 2, Solution 14 + + - 3V V1 I4 V2 - I3 - + 2V - + - + V3 - + + 4V I2 - I1 V4 + -
5V
For mesh 1, − + + = → =V V4 42 5 0 7 For mesh 2, + + + = → = − − = −4 0 4 73 4 3V V V V For mesh 3, − + − = → = + = −3 0 31 3 1 3V V V V V For mesh 4, − − − = → = − − =V V V V V1 2 2 12 0 2 6 Thus, V V V V V V V1 2 3 48 6 11= − = = − V7=, , ,
Chapter 2, Solution 15
+ +
+ 12V 1 v2 - - 8V + - v1 - 3 + 2 - v3 10V
- + For loop 1,
8 12 0 42 2− + = → =v v V
V
V
For loop 2,
− − − = → = −v v3 38 10 0 18 For loop 3,
− + + = → = −v v v1 3 112 0 6 Thus, v V v V v1 2 36 4= − = = −, , V18 Chapter 2, Solution 16
+ v1 -
+ - + -
6V -+ loop 1
loop 2 12V
10V
+ v1 -
+ v2 -
Applying KVL around loop 1,
–6 + v1 + v1 – 10 – 12 = 0 v1 = 14V
Applying KVL around loop 2,
12 + 10 – v2 = 0 v2 = 22V
Chapter 2, Solution 17 + v1 -
+
- +
-+ 10V
12V
24V
loop 2
+ v3 -
v2
-loop 1
-+
It is evident that v3 = 10V Applying KVL to loop 2, v2 + v3 + 12 = 0 v2 = -22V Applying KVL to loop 1, -24 + v1 - v2 = 0 v1 = 2V Thus, v1 = 2V, v2 = -22V, v3 = 10V Chapter 2, Solution 18
Applying KVL, -30 -10 +8 + I(3+5) = 0
8I = 32 I = 4A -Vab + 5I + 8 = 0 Vab = 28V
Chapter 2, Solution 19
Applying KVL around the loop, we obtain
-12 + 10 - (-8) + 3i = 0 i = -2A Power dissipated by the resistor: p 3Ω = i2R = 4(3) = 12W Power supplied by the sources: p12V = 12 (- -2) = 24W p10V = 10 (-2) = -20W
p8V = (- -2) = -16W Chapter 2, Solution 20
Applying KVL around the loop,
-36 + 4i0 + 5i0 = 0 i0 = 4A Chapter 2, Solution 21
10 Ω
+-45V -
+
+ v0 -
Apply KVL to obtain -45 + 10i - 3V0 + 5i = 0 But v0 = 10i, 3v0-45 + 15i - 30i = 0 i = -3A P3 = i2R = 9 x 5 = 45W 5 Ω
Chapter 2, Solution 22 Ω
+ v0 -
10A
2v0 Ω At the node, KCL re
0 v2104
v++
The current through i = 2V0 = -8. and the voltage acro
v = (6 + 4) i0
Hence, p2 vi = (-8.88 Chapter 2, Solution 8//12 = 4.8, 3//6 = 2The circuit is reduce 6A
6
quires that
0 = 0 v
the controlled
888A
ss it is
= 10 14
v0 −=
8)(-11.111) =
23
, (4 + 2)//(1.2d to that show
ix
+
2
4
0 = –4.444V
source is
111.1
98.75 W
+ 4.8) = 6//6 = 3 n below.
1Ω
vx -
Ω 3Ω
Applying current division,
i A A v ix x=+ +
= =2
2 1 36 2 1 2( ) , Vx =
The current through the 1.2- resistor is 0.5iΩ x = 1A. The voltage across the 12- resistor is 1 x 4.8 = 4.8 V. Hence the power is
Ω
p vR
W= = =2 24 8
12192. .
Chapter 2, Solution 24
(a) I0 = 21 RR
Vs
+
α−=0V I0 ( )43 RR = 43
43
21
0
RRRR
RRV
+⋅
+−
α
( )( )4321
430
RRRRRR
Vs ++V −
=α
(b) If R1 = R2 = R3
= R4 = R,
1042
RR2V
V
S
0 =α
=⋅α
= α = 40
Chapter 2, Solution 25
V0 = 5 x 10-3 x 10 x 103 = 50V
Using current division,
I20 =+
)5001.0(205
x=5 0.1 A
V20 = 20 x 0.1 kV = 2 kV p20 = I20 V20 = 0.2 kW
Chapter 2, Solution 26
V0 = 5 x 10-3 x 10 x 103 = 50V
Using current division,
I20 =+
)5001.0(205
x=5 0.1 A
V20 = 20 x 0.1 kV = 2 kV p20 = I20 V20 = 0.2 kW Chapter 2, Solution 27
Using current division,
i1 = =+
)20(64
4 8 A
i2 = =+
)20(64
6 12 A
Chapter 2, Solution 28
We first combine the two resistors in parallel =1015 6 Ω We now apply voltage division,
v1 = =+
)40(614
14 20 V
v2 = v3 = =+
)40(614
6 12 V
Hence, v1 = 28 V, v2 = 12 V, vs = 12 V
Chapter 2, Solution 29
The series combination of 6 Ω and 3 Ω resistors is shorted. Hence i2 = 0 = v2
v1 = 12, i1 = =4
12 3 A
Hence v1 = 12 V, i1 = 3 A, i2 = 0 = v2 Chapter 2, Solution 30
i +v-
8 Ω
6 Ω
i1
9A 4 Ω
By current division, =+
= )9(126
12i 6 A
4 x 3 = ===−= 11 i4v,A369i 12 V p6 = 12R = 36 x 6 = 216 W Chapter 2, Solution 31
The 5 Ω resistor is in series with the combination of Ω=+ 5)64(10 . Hence by the voltage division principle,
=+
= )V20(55
5v 10 V
by ohm's law,
=+
=+
=64
1064
vi 1 A
pp = i2R = (1)2(4) = 4 W
Chapter 2, Solution 32
We first combine resistors in parallel.
=3020 =50
30x20 12 Ω
=4010 =50
40x10 8 Ω
Using current division principle,
A12)20(2012ii,A8)20(
1288ii 4321 ==+=+
=+
== )8(5020i1 3.2 A
== )8(5030i2 4.8 A
== )12(5010i3 2.4A
== )12(5040i4 9.6 A
Chapter 2, Solution 33 Combining the conductance leads to the equivalent circuit below i
+v-
9A 1S
i
+ v -
4S
4S
1S
9A 2S
=SS 36 259
3x6= and 25 + 25 = 4 S
Using current division,
=+
= )9(
211
1i 6 A, v = 3(1) = 3 V
Chapter 2, Solution 34 By parallel and series combinations, the circuit is reduced to the one below:
-+
+ v1 -
8 Ωi1
=+ )132(10 Ω= 625
1510x
=+ )64(15 Ω= 625
1515x 28V 6 Ω Ω=+ 6)66(12
Thus i1 = =+ 6828 2 A and v1 = 6i1 = 12 V
We now work backward to get i2 and v2.
+ 6V -
1A
1A 6 Ω
-+ +
12V -
12 Ω
8 Ωi1 = 2A 28V 6 Ω 0.6A
+ 3.6V
-
4 Ω
+ 6V -
1A
1A
15 Ω
6 Ω
-+ +
12V -
12 Ω
8 Ω i1 = 2A 28V 6 Ω
Thus, v2 = ,123)63(1513
⋅=⋅ i2 = 24.013v2 =
p2 = i2R = (0.24)2 (2) = 0.1152 W i1 = 2 A, i2 = 0.24 A, v1 = 12 V, v2 = 3.12 V, p2 = 0.1152 W Chapter 2, Solution 35
i
20 Ω + V0 - i2
a b
5 Ω
30 Ω70 Ω
I0i1
+ V1 -
-+
50V
Combining the versions in parallel,
=3070 Ω= 21100
30x70 , =1520 =25
5x20 4 Ω
i = =+ 421
50 2 A
vi = 21i = 42 V, v0 = 4i = 8 V
i1 = =70v1 0.6 A, i2 = =
20v2 0.4 A
At node a, KCL must be satisfied i1 = i2 + I0 0.6 = 0.4 + I0 I0 = 0.2 A Hence v0 = 8 V and I0 = 0.2A Chapter 2, Solution 36
The 8-Ω resistor is shorted. No current flows through the 1-Ω resistor. Hence v0 is the voltage across the 6Ω resistor.
I0 = ==+ 4
41632
4 1 A
v0 = I0 ( ) == 0I263 2 V
Chapter 2, Solution 37
Let I = current through the 16Ω resistor. If 4 V is the voltage drop across the R6 combination, then 20 - 4 = 16 V in the voltage drop across the 16Ω resistor.
Hence, I = =1616 1 A.
But I = 1R616
20=
+ 4 = =R6
R6R6+
R = 12 Ω
Chapter 2, Solution 38
Let I0 = current through the 6Ω resistor. Since 6Ω and 3Ω resistors are in parallel. 6I0 = 2 x 3 R0 = 1 A The total current through the 4Ω resistor = 1 + 2 = 3 A. Hence vS = (2 + 4 + 32 ) (3 A) = 24 V
I = =v 10
S 2.4 A
Chapter 2, Solution 39
(a) Req = =0R 0
(b) Req = =+ RRRR =+2R
2R R
(c) Req = ==++ R2R2)RR()RR( R
(d) Req = )R21R(R3)RRR(R +=+3
= =+ R
23R3
R23Rx3
R
(e) Req = R3R3R2R =
⋅
R3R2R
= R3 =+
=R
32R3
R32Rx3
R32 R
116
Chapter 2, Solution 40
Req = =+=++ 23)362(43 5Ω
I = =5
10qRe=
10 2 A
Chapter 2, Solution 41
Let R0 = combination of three 12Ω resistors in parallel
121
121
121
R1
o
++= Ro = 4
)R14(6030)RR10(6030R 0eq ++=+++=
R74
)R14(6030++
+=50 74 + R = 42 + 3R
or R = 16 Ω Chapter 2, Solution 42
(a) Rab = ==+=+25
20x5)128(5)30208(5 4 Ω
(b) Rab = =++=++=++++ 5.22255442)46(1058)35(42 6.5 Ω
Chapter 2, Solution 43
(a) Rab = =+=+=+ 8450400
2520x54010205 12 Ω
(b) =302060 Ω==
++
−
10660
301
201
601 1
Rab = )1010( +80 =+
=100
2080 16 Ω
Chapter 2, Solution 44
(a) Convert T to Y and obtain
R x x x1
20 20 20 10 10 2010
80010
80=+ +
= = Ω
R R2 380020
40= = =Ω
The circuit becomes that shown below. R1 a R3 R2 5Ω b R1//0 = 0, R3//5 = 40//5 = 4.444Ω R Rab = + = =2 0 4 444 40 4 444 4/ /( . ) / / . Ω
(b) 30//(20+50) = 30//70 = 21 Ω Convert the T to Y and obtain
R x x x1
20 10 10 40 40 2040
140040
35=+ +
= = Ω
R2140020
70= = Ω , R3140010
140= = Ω
The circuit is reduced to that shown below. 11Ω R2 30Ω
15Ω
Combining the resist
R1
R3
21Ω
21Ω
ors in parallel
R1//15 =35//15=10.5, 30//R2=30//70 = 21 leads to the circuit below. 11 10.5Ω Ω 21Ω 140Ω
21Ω 21Ω Coverting the T to Y leads to the circuit below. 11Ω 10.5Ω R4 R5 R6 21Ω R x x x R4 6
21 140 140 21 21 2121
632121
301=+ +
= = =Ω
R5
6321140
4515= = .
10.5//301 = 10.15, 301//21 = 19.63 R5//(10.15 +19.63) = 45.15//29.78 = 17.94 Rab = + =11 17 94 28 94. . Ω Chapter 2, Solution 45
(a) 10//40 = 8, 20//30 = 12, 8//12 = 4.8
Rab = + + =5 50 4 8 59 8. . Ω (b) 12 and 60 ohm resistors are in parallel. Hence, 12//60 = 10 ohm. This 10 ohm and 20 ohm are in series to give 30 ohm. This is in parallel with 30 ohm to give 30//30 = 15 ohm. And 25//(15+10) = 12.5. Thus Rab = + + =5 12 8 15 32 5. . Ω
Chapter 2, Solution 46
(a) Rab = =++ 2060407030 80
206040100
70x30 +++
=++ 154021= 76 Ω
(b) The 10-Ω, 50-Ω, 70-Ω, and 80-Ω resistors are shorted.
=3020 Ω= 1250
30x20
40 =60 24100
60x=
40
Rab = 8 + 12 + 24 + 6 + 0 + 4 = 54 Ω Chapter 2, Solution 47
=205 Ω= 425
20x5
6 =3 Ω= 29
3x6
8 Ωa b
10 Ω
2 Ω 4 Ω Rab = 10 + 4 + 2 + 8 = 24 Ω
Chapter 2, Solution 48
(a) Ra = 3010
100100100R
RRRRRR
3
133221 =++
=++
Ra = Rb = Rc = 30 Ω
(b) Ω==++
= 3.10330
310030
50x2050x3020x30R a
,15520
3100R b Ω== Ω== 6250
3100R c
Ra = 103.3 Ω, Rb = 155 Ω, Rc = 62 Ω
Chapter 2, Solution 49
(a) R1 = Ω=+
=++
436
1212RRR
RR
cba
ca
R1 = R2 = R3 = 4 Ω
(b) Ω=++
= 18103060
30x60R1
Ω== 6100
10x60R 2
Ω== 3100
10x30R 3
R1 = 18Ω, R2 = 6Ω, R3 = 3Ω Chapter 2, Solution 50 Using = 3R∆R Y = 3R, we obtain the equivalent circuit shown below:
3R 30mA
R
R 3R
3R 3R
30mA 3R/2
=RR3 R43
R4RxR3
=
)R4/(3)R4/()RxR3(R =3
RR
23R3
R23Rx3
R23R3R
43R
43R3
=+==
+
P = I2R 800 x 10-3 = (30 x 10-3)2 R R = 889 Ω Chapter 2, Solution 51
(a) Ω= 153030 and Ω== 12)50/(20x302030
Rab = ==+ )39/(24x15)1212(15 9.31 Ω
b
15 Ω
a
20 Ω
30 Ω
b
a
20 Ω
30 Ω 30 Ω 30 Ω
12 Ω
12 Ω
(b) Converting the T-subnetwork into its equivalent network gives ∆ Ra'b' = 10x20 + 20x5 + 5x10/(5) = 350/(5) = 70 Ω Rb'c' = 350/(10) = 35Ω, Ra'c' = 350/(20) = 17.5 Ω
Also Ω== 21)100/(70x307030 and 35/(15) = 35x15/(50) = 10.5
Rab = 25 + 5.315.1725)5.1021(5. +=+17 Rab = 36.25 Ω
b’
c’ c’ b
a
15 Ω 35 Ω 17.5 Ω
25 Ω 70 Ω a’
b
a
30 Ω
25 Ω
5 Ω
10 Ω
15 Ω
20 Ω
30 Ω
Chapter 2, Solution 52 (a) We first convert from T to ∆ .
R3 R2
b
a
100 Ω
100 Ω 100 Ω
100 Ω
100 Ω
100 Ω
100 Ω
b
a
200 Ω
200 Ω
100 Ω
100 Ω 100 Ω
100 Ω
100 Ω
100 Ω
100 Ω
100 Ω R1
Ω==++
= 800100
80000100
100x200200x200200x1001R
R2 = R3 = 80000/(200) = 400
But Ω== 80500
400x100400100
We connect the ∆ to Y.
Ra = Rc = Ω==++ 3
400960
000,648008080
800x80
Rc
Rb
b
aRa
100 Ω
100 Ω
100 Ω
100 Ω
100 Ω
b
a
800 Ω
80 Ω
80 Ω 100 Ω
100 Ω
100 Ω
100 Ω
100 Ω
Rb = Ω=320
96080x80
We convert T to . ∆
R3’
R2’
b
a
500/3 Ω
500/3 Ω
R1’
b
a
500/3 Ω
500/3 Ω
320/3 Ω
100 Ω
100 Ω
Ω==++
= 75.293)3/(320
)3/(000,94
3320
3320x100
3320x100100x100
'1R
33.313100
)3/(000,94RR 13
'2 ===
796.108)3/(1440
)3/(500x)3/(940)3/(500)30/(940 ==
Rab = ==36.511
6.217x75.293)796.108x2(75.293 125 Ω
(b) Converting the Ts to ∆ s, we have the equivalent circuit below.
100 Ω a
b
300 Ω
300 Ω 100 Ω
100 Ω
300 Ω
100 Ω 300 Ω
300 Ω
100 Ω
300 Ω
100 Ω
100 Ω
100 Ω
b
a 100 Ω
100 Ω
100 Ω
100 Ω
100 Ω
100 Ω ,75)400/(100x300100300 == 100)450/(150x300)7575( ==+300
Rab = 100 + )400/(300x100200100300 +=+100 Rab = 2.75 Ω
100 Ω
300 Ω
300 Ω
100 Ω
300 Ω
100 Ω
100 Ω
Chapter 2, Solution 53 (a) Converting one ∆ to T yields the equivalent circuit below:
20 Ω
b’
c’
a’
b 80 Ω
a 20 Ω
5 Ω
4 Ω
60 Ω
30 Ω
Ra'n = ,4501040
10x40Ω=
++ ,5
10050x10R n'b Ω== Ω== 20
10050x40R n'c
Rab = 20 + 80 + 20 + 6534120)560()430( +=++ Rab = 142.32 Ω
(a) We combine the resistor in series and in parallel.
Ω==+ 2090
60x30)3030(30
We convert the balanced ∆ s to Ts as shown below:
b
10 Ω
10 Ω 10 Ω
10 Ω
10 Ω
30 Ω30 Ω
30 Ω
30 Ω
30 Ω
30 Ω b
a
20 Ω
20 Ω 10 Ω
a
Rab = 10 + 40202010)102010()1010 +=++++( Rab = 33.33 Ω
Chapter 2, Solution 54
(a) Rab = + + + = + =50 100 150 100 150 50 100 400 130/ /( ) / / Ω (b) Rab = + + + = + =60 100 150 100 150 60 100 400 140/ /( ) / / Ω
Chapter 2, Solution 55 We convert the T to . ∆
50 Ω
I0
-+
24 V
b
a
35 Ω
70 Ω
140 Ω
Req
-+
24 V 60 Ω
b
a
20 Ω
40 Ω
10 Ω
20 Ω
I0
60 Ω
70 Ω
Req
Rab = Ω==++
=++
3540
140040
20x1010x4040x20R
RRRRRR
3
133221
Rac = 1400/(10) = 140Ω, Rbc = 1400/(40) = 35Ω 357070 = and =160140 140x60/(200) = 42
Req = Ω=+ 0625.24)4235(35 I0 = 24/(Rab) = 0.9774A Chapter 2, Solution 56 We need to find Req and apply voltage division. We first tranform the Y network to ∆ .
c
35 Ω
16 Ω
30 Ω
37.5 Ω a b
30 Ω 45 Ω
Req
+100 V
-20 Ω
35 Ω
16 Ω
30 Ω
10 Ω
12 Ω
15 Ω
Req
+ 100 V
-
20 Ω
Rab = Ω==++ 5.37
12450
1215x1212x1010x15
Rac = 450/(10) = 45Ω, Rbc = 450/(15) = 30Ω Combining the resistors in parallel,
30||20 = (600/50) = 12 Ω,
37.5||30 = (37.5x30/67.5) = 16.667 Ω
35||45 = (35x45/80) = 19.688 Ω
Req = 19.688||(12 + 16.667) = 11.672Ω
By voltage division,
v = 10016672.11
672.11+
= 42.18 V
Chapter 2, Solution 57
4 Ω
ec
f
a
bd
36 Ω
7 Ω
27 Ω
2 Ω
18 Ω
28 Ω 10 Ω
1 Ω 14 Ω
Rab = Ω==++ 18
12216
126x88x1212x6
Rac = 216/(8) = 27Ω, Rbc = 36 Ω
Rde = Ω=++ 7
856
84x88x22x4
Ref = 56/(4) = 14Ω, Rdf = 56/(2) = 28 Ω Combining resistors in parallel,
,368.7382802810 Ω== Ω== 868.5
437x36736
Ω== 7.230
3x27327
4 Ω
0.5964 Ω 3.977 Ω1.829 Ω
4 Ω
7.568 Ω 14 Ω
2.7 Ω
14 Ω
18 Ω
5.868 Ω 7.568 Ω
Ω==++
= 829.1567.26
7.2x18867.57.218
7.2x18R an
Ω== 977.3567.26868.5x18R bn
Ω== 5904.0567.26
7.2x868.5R cn
)145964.0()368.7977.3(829.14R eq ++++=
=+ 5964.14346.11829.5= 12.21 Ω i = 20/(Req) = 1.64 A
Chapter 2, Solution 58 The resistor of the bulb is 120/(0.75) = 160Ω 2.25 A 1.5 A 40 Ω
0.75 A
80 Ω 160 Ω+ 90 V - +
120-+
VS Once the 160Ω and 80Ω resistors are in parallel, they have the same voltage 120V. Hence the current through the 40Ω resistor is 40(0.75 + 1.5) = 2.25 x 40 = 90 Thus vs = 90 + 120 = 210 V
Chapter 2, Solution 59 Total power p = 30 + 40 + 50 + 120 W = vi
or i = p/(v) = 120/(100) = 1.2 A
Chapter 2, Solution 60
p = iv i = p/(v) i30W = 30/(100) = 0.3 A i40W = 40/(100) = 0.4 A i50W = 50/(100) = 0.5 A Chapter 2, Solution 61 There are three possibilities
(a) Use R1 and R2: R = Ω== 35.429080RR 21 p = i2R i = 1.2A + 5% = 1.2 ± 0.06 = 1.26, 1.14A p = 67.23W or 55.04W, cost = $1.50
(b) Use R1 and R3:
R = Ω== 44.4410080RR 31 p = I2R = 70.52W or 57.76W, cost = $1.35
(c) Use R2 and R3: R = Ω== 37.4710090RR 32 p = I2R = 75.2W or 61.56W, cost = $1.65
Note that cases (b) and (c) give p that exceed 70W that can be supplied. Hence case (a) is the right choice, i.e. R1 and R2
Chapter 2, Solution 62
pA = 110x8 = 880 W, pB = 110x2 = 220 W Energy cost = $0.06 x 360 x10 x (880 + 220)/1000 = $237.60
Chapter 2, Solution 63
Use eq. (2.61),
Rn = Ω=−
=− −
−
04.010x25
100x10x2RII 3
3
mm
mI
In = I - Im = 4.998 A p = (I = 9992.0)04.0()998.4R 22
n = ≅ 1 W Chapter 2, Solution 64
When Rx = 0, i R = A10x = Ω=1110
110
When Rx is maximum, ix = 1A Ω==+ 1101
110R xR
i.e., Rx = 110 - R = 99 Ω Thus, R = 11 Ω, Rx = 99 Ω Chapter 2, Solution 65
=Ω−=−= k1mA1050R
IV
R mfs
fsn 4 kΩ
Chapter 2, Solution 66
20 kΩ/V = sensitivity = fsI1
i.e., Ifs = A50V/k20
µ=Ω1
The intended resistance Rm = Ω=Ω= k200)V/k20(10IV
fs
fs
(a) =Ω−µ
=−= k200A50V50R
iV
R mfs
fsn 800 kΩ
(b) p = =Ωµ= )k800()A50(RI 2n
2fs 2 mW
Chapter 2, Solution 67
(a) By current division, i0 = 5/(5 + 5) (2 mA) = 1 mA V0 = (4 kΩ) i0 = 4 x 103 x 10-3 = 4 V
(b) .k4.2k6k4 Ω= By current division,
mA19.1)mA2(54.21
5i '0 =
++=
=Ω= )mA19.1)(k4.2(v'0 2.857 V
(c) % error = 0
'00
vvv −
x 100% = =100x4143.1 28.57%
(d) .k6.3k30k4 Ω=Ω By current division,
mA042.1)mA2(56.31
5i '0 =
++=
V75.3)mA042.1)(k6.3(v'0 =Ω
% error = ==−
4100x25.0%100x
vvv
0
'0 6.25%
Chapter 2, Solution 68
(a) Ω= 602440
i = =+ 24164 0.1 A
(b) =++
=24116
4i ' 0.09756 A
(c) % error = =− %100x
1.009756.01.0 2.44%
Chapter 2, Solution 69
With the voltmeter in place,
Sm2S1
m20 V
RRRRRR++
=V
where Rm = 100 kΩ without the voltmeter,
SS21
20 V
RRRR
++=V
(a) When R2 = 1 kΩ, Ω= k101100RR 2m
V0 = =
+
)40(
30101100101100
1.278 V (with)
V0 = =+
)40(301
1 1.29 V (without)
(b) When R2 = 10 kΩ, Ω== k091.91101000RR m2
V0 = =+
)40(30091.9
091.9 9.30 V (with)
V0 = =+
)40(3010
10 10 V (without)
(c) When R2 = 100 kΩ, Ω= k50RR m2
=+
= )40(3050
50V0 25 V (with)
V0 = =+
)40(30100
100 30.77 V (without)
Chapter 2, Solution 70
(a) Using voltage division, v Va = +
=1212 8
25 15( )
v Vb = +=
1010 15
25 10( )
v v vab a b= − = V− =15 10 5
(b) + 8kΩ 15kΩ 25 V - a b
12kΩ 10kΩ o
v v V v v va b ab a b= = = V− = − = −0 10 0 10 1, , 0 Chapter 2, Solution 71
Vs RL
R1
+−
iL
Given that vs = 30 V, R1 = 20 Ω, IL = 1 A, find RL.
v i R R Rvi
Rs L L Ls
L= + → = − = − =( )1 1
301
20 10Ω
Chapter 2, Solution 72
The system can be modeled as shown. +
9V - The n pa
9 12= x
Chapter 2,
Chapter 2,
12A
R R ••• R
rallel resistors R give a combined resistance of R/n. Thus,
129
12 159
20 → = = =Rn
n xR x
Solution 73
By the current division principle, the current through the ammeter will be one-half its previous value when
R = 20 + Rx 65 = 20 + Rx Rx = 45 Ω
Solution 74
With the switch in high position,
6 = (0.01 + R3 + 0.02) x 5 R3 = 1.17 Ω
At the medium position,
6 = (0.01 + R2 + R3 + 0.02) x 3 R2 + R3 = 1.97
or R2 = 1.97 - 1.17 = 0.8 Ω
At the low position, 6 = (0.01 + R1 + R2 + R3 + 0.02) x 1 R1 + R2 + R3 = 5.97 R1 = 5.97 - 1.97 = 4 Ω
Chapter 2, Solution 75
MR 100 Ω
VS -+
12 Ω
(a) When Rx = 0, then
Im = Ifs = mRR
t+
R2 = mfs
2
RIE
− = Ω=− k9.1910010x1.0
23
(b) For half-scale deflection, Im = mA05.02Ifs =
Im = xm RRR
E++
Rx = =Ω−=+− − k2010x05.0
2)RR(IE
3mm
20 kΩ
Chapter 2, Solution 76
For series connection, R = 2 x 0.4Ω = 0.8Ω
===8.0)120(
RV 22
p 18 kΩ (low)
For parallel connection, R = 1/2 x 0.4Ω = 0.2Ω
p = =2.0)120
=(
RV
22
72 kW (high)
Chapter 2, Solution 77
(a) 5 Ω = 2020202010 =10
i.e., four 20 Ω resistors in parallel.
(b) 311.8 = 300 + 10 + 1.8 = 300 + 8.12020 + i.e., one 300Ω resistor in series with 1.8Ω resistor and a parallel combination of two 20Ω resistors.
(c) 40kΩ = 12kΩ + 28kΩ = k50k56k2424 + i.e., Two 24kΩ resistors in parallel connected in series with two 50kΩ resistors in parallel.
(d) 42.32kΩ = 42l + 320
= 24k + 28k = 320 = 24k = 20300k56k ++56
i.e., A series combination of 20Ω resistor, 300Ω resistor, 24kΩ resistor and a parallel combination of two 56kΩ resistors.
Chapter 2, Solution 78
The equivalent circuit is shown below:
R
+ V0
--+
VS (1-α)R
V0 = S0S VR)1(VR)1(R
R)1α−=
α−+α−(
R)1(VV
S
0 α−=
Chapter 2, Solution 79
Since p = v2/R, the resistance of the sharpener is R = v2/(p) = 62/(240 x 10-3) = 150Ω I = p/(v) = 240 mW/(6V) = 40 mA Since R and Rx are in series, I flows through both. IRx = Vx = 9 - 6 = 3 V Rx = 3/(I) = 3/(40 mA) = 3000/(40) = 75 Ω
Chapter 2, Solution 80 The amplifier can be modeled as a voltage source and the loudspeaker as a resistor:
V -+
Case 1
R1 -+ V R2
Hence ,R
Vp2
= 2
1
1
2
RR
pp
= === )12(4
10pRR
12
12p 30 W
Case 2
Chapter 2, Solution 81
Let R1 and R2 be in kΩ. 5RR 21eq +=R (1)
12
2
S
0
RR5R5
VV
+= (2)
From (1) and (2), 40R5
05. 1=0 2 = 2
22 R5
R5R5
+= or R2 = 3.33 kΩ
From (1), 40 = R1 + 2 R1 = 38 kΩ Thus R1 = 38 kΩ, R2 = 3.33 kΩ
Chapter 2, Solution 82 (a) 10 Ω
1 2
10 Ω
40 Ω 80 Ω R12
R12 = 80 + =+=+6
5080)4010(10 88.33 Ω
(b)
3 20 Ω 10 Ω
10 Ω
80 Ω
40 Ω
R13 1
R13 = 80 + =+=++ 501010020)4010(10 108.33 Ω
20 Ω 10 Ω
10 Ω
1
4
80 Ω
(c) R14 40 Ω
R14 = =++=++++ 2008020)104010(080 100 Ω Chapter 2, Solution 83 The voltage across the tube is 2 x 60 mV = 0.06 V, which is negligible
compared with 24 V. Ignoring this voltage amp, we can calculate the current through the devices.
I1 = mA5V9mW
V1
1 =45p
=
I2 = mA2024mW
V2
2 =480p
=
60 mA i2 = 20 mA
R1
iR2
iR1
i1 = 5 mA
R2 -+
24 V By applying KCL, we obtain and mA402060I
1R =−= mA35540I2R =−=
Hence, R1RI 1 = 24 - 9 = 15 V ==
mA40V15R1 375 Ω
V9RI 2R 2= ==
mA35V9
2R 257.14 Ω
Chapter 3, Solution 1.
8 Ω
v2
2 Ω
40 Ω v1
6 A 10 A
At node 1, 6 = v1/(8) + (v1 - v2)/4 48 = 3v1 - 2v2 (1) At node 2, v1 - v2/4 = v2/2 + 10 40 = v1 - 3v2 (2) Solving (1) and (2), v1 = 9.143V, v2 = -10.286 V
P8Ω = ( )==
8143.9
8v 22
1 10.45 W
P4Ω = ( )
=−4vv 2
21 94.37 W
P2Ω = ( )=
==
2286.10
2v 21
2 52.9 W
Chapter 3, Solution 2
At node 1,
2
vv6
5v
10v 2111 −
+=−− 60 = - 8v1 + 5v2 (1)
At node 2,
2
vv634
v 212 −++= 36 = - 2v1 + 3v2 (2)
Solving (1) and (2), v1 = 0 V, v2 = 12 V
Chapter 3, Solution 3 Applying KCL to the upper node,
10 = 60v
230v
20v
10v 0oo +++0 + v0 = 40 V
i1 = =10
0v 4 A , i2 = =
20v0 2 A, i3 = =
30v0 1.33 A, i4 = =
60v0 67 mA
Chapter 3, Solution 4 At node 1, 4 + 2 = v1/(5) + v1/(10) v1 = 20 At node 2, 5 - 2 = v2/(10) + v2/(5) v2 = 10 i1 = v1/(5) = 4 A, i2 = v1/(10) = 2 A, i3 = v2/(10) = 1 A, i4 = v2/(5) = 2 A Chapter 3, Solution 5 Apply KCL to the top node.
k4
vk6v20
k2v30 000 =
−+
− v0 = 20 V
i1
10 Ω 10 Ω
2A
i4 i3
v1 v2
i2
4 A 5 Ω 5 Ω 5 A
Chapter 3, Solution 6
i1 + i2 + i3 = 0 02
10v6v
412v 002 =
−++
−
or v0 = 8.727 V
Chapter 3, Solution 7 At node a,
babaaa VV
VVVV3610
10153010
−=→−
+=−
(1)
At node b,
babbba VV
VVVV72240
59
2012
10−=→=
−−+
−+
− (2)
Solving (1) and (2) leads to Va = -0.556 V, Vb = -3.444V Chapter 3, Solution 8
i2
5 Ω
2 Ω
i13 Ω v1 i3
1 Ω
+–
4V0
3V–+
+
V0
–
i1 + i2 + i3 = 0 05
v4v1
3v5v 0111 =
−+
−+
But 10 v52v = so that v1 + 5v1 - 15 + v1 - 0v
58
1 =
or v1 = 15x5/(27) = 2.778 V, therefore vo = 2v1/5 = 1.1111 V
Chapter 3, Solution 9
+ v1
–
i2
6 Ω i13 Ω v1 i3
8 Ω +–
+ v0 –
2v0 12V
–+
At the non-reference node,
6
v2v8v
3v 011112 −
+=− (1)
But -12 + v0 + v1 = 0 v0 = 12 - v1 (2) Substituting (2) into (1),
6
24v38v
3v 111 −
+=−12 v0 = 3.652 V
Chapter 3, Solution 10 At node 1,
8v4
1vv 112 +=
− 32 = -v1 + 8v2 - 8v0 (1)
1 Ω
4A 2i0
i0
v1 v0
8 Ω 2 Ω
v2 4 Ω
At node 0,
00 I2
2v
+=4 and 8vI 1
0 = 16 = 2v0 + v1 (2)
At node 2,
2I0 = 4
v1
v 212 +−v and
8v1
0 =I v2 = v1 (3)
From (1), (2) and (3), v0 = 24 V, but from (2) we get
io = 624242
22v4 o
−=−=−
= - 4 A
Chapter 3, Solution 11
i3
6 Ω
vi1 i24 Ω 3 Ω
–+
10 V 5 A Note that i2 = -5A. At the non-reference node
6v5
4v10
=+− v = 18
i1 = =−4
v10 -2 A, i2 = -5 A
Chapter 3, Solution 12
i3
40 Ω
v1 v210 Ω 20 Ω
–+ 5 A
50 Ω 24 V
At node 1, 40
0v20
vv10
v24 1211 −+
−=
− 96 = 7v1 - 2v2 (1)
At node 2, 50v
20vv 221 =
−+5 500 = -5v1 + 7v2 (2)
Solving (1) and (2) gives, v1 = 42.87 V, v2 = 102.05 V
==40vi 1
1 1.072 A, v2 = =50v2 2.041 A
Chapter 3, Solution 13
At node number 2, [(v2 + 2) – 0]/10 + v2/4 = 3 or v2 = 8 volts But, I = [(v2 + 2) – 0]/10 = (8 + 2)/10 = 1 amp and v1 = 8x1 = 8volts
Chapter 3, Solution 14
1 Ω 2 Ω
4 Ω
v0 v1
5 A
8 Ω
20 V –+
40 V –+
At node 1, 1
v405
2vv 001 −
=+−
v1 + v0 = 70 (1)
At node 0, 8
20v4v
52
vv 0001 ++=+
− 4v1 - 7v0 = -20 (2)
Solving (1) and (2), v0 = 20 V
Chapter 3, Solution 15
1 Ω 2 Ω
4 Ω
v0 v1
5 A
8 Ω
20 V –+
40 V –+
Nodes 1 and 2 form a supernode so that v1 = v2 + 10 (1) At the supernode, 2 + 6v1 + 5v2 = 3 (v3 - v2) 2 + 6v1 + 8v2 = 3v3 (2) At node 3, 2 + 4 = 3 (v3 - v2) v3 = v2 + 2 (3) Substituting (1) and (3) into (2),
2 + 6v2 + 60 + 8v2 = 3v2 + 6 v2 = 1156−
v1 = v2 + 10 = 1154
i0 = 6vi = 29.45 A
P65 = =
== 6
1154Gv
R
221
21v
144.6 W
P55 = =
−= 5
1156Gv
222 129.6 W
P35 = ( ) ==− 3)2(Gvv 22
3L 12 W
Chapter 3, Solution 16 2 S
+ v0 –
13 V –+
i0
1 S 4 S
8 Sv1 v2 v3
2 A At the supernode, 2 = v1 + 2 (v1 - v3) + 8(v2 – v3) + 4v2, which leads to 2 = 3v1 + 12v2 - 10v3 (1) But
v1 = v2 + 2v0 and v0 = v2. Hence
v1 = 3v2 (2) v3 = 13V (3)
Substituting (2) and (3) with (1) gives, v1 = 18.858 V, v2 = 6.286 V, v3 = 13 V Chapter 3, Solution 17
60 V
i0
3i0
2 Ω
10 Ω
4 Ω
60 V –+
8 Ω
At node 1, 2
vv8v
4v60 2111 −
+=− 120 = 7v1 - 4v2 (1)
At node 2, 3i0 + 02
vv10
v60 212 =−
+−
But i0 = .4
v60 1−
Hence
( )
02
vv10
v604
v603 2121 =−
+−
+− 1020 = 5v1 - 12v2 (2)
Solving (1) and (2) gives v1 = 53.08 V. Hence i0 = =−4
v60 1 1.73 A
Chapter 3, Solution 18
+ v3 –
+ v1 –
10 V
– + v1
v2
2 Ω 2 Ω
4 Ω 8 Ω
v3
5 A
(a) (b)
At node 2, in Fig. (a), 5 = 2
vv2
vv 3212 −+
− 10 = - v1 + 2v2 - v3 (1)
At the supernode, 8v
4v
2vv
2vv 313212 +=
−+
− 40 = 2v1 + v3 (2)
From Fig. (b), - v1 - 10 + v3 = 0 v3 = v1 + 10 (3) Solving (1) to (3), we obtain v1 = 10 V, v2 = 20 V = v3
Chapter 3, Solution 19
At node 1,
32112131 4716
48235 VVVVVVVV
−−=→+−
+−
+= (1)
At node 2,
32132221 270
428VVV
VVVVV−+−=→
−+=
− (2)
At node 3,
32132313 724360
42812
3 VVVVVVVV−+=−→=
−+
−+
−+ (3)
From (1) to (3),
BAVVVV
=→
−=
−−−−−
360
16
724271417
3
2
1
Using MATLAB,
V 267.12 V, 933.4 V, 10267.12933.410
3211 ===→
== − VVVBAV
Chapter 3, Solution 20 Nodes 1 and 2 form a supernode; so do nodes 1 and 3. Hence
040414 321
321 =++→=++ VVVVVV (1)
. V1 . V2 2Ω V3 4Ω 1Ω 4Ω
Between nodes 1 and 3,
12012 1331 −=→=++− VVVV (2) Similarly, between nodes 1 and 2, (3) iVV 221 +=But i . Combining this with (2) and (3) gives 4/3V=
.V (4) 2/6 12 V+= Solving (1), (2), and (4) leads to
V15 V,5.4 V,3 321 −==−= VVV Chapter 3, Solution 21
4 kΩ
2 kΩ
+ v0 –
3v0v1 v2v3
1 kΩ
+
3 mA
3v0
+ v2 –
+ v3 –
+
(b) (a) Let v3 be the voltage between the 2kΩ resistor and the voltage-controlled voltage source. At node 1,
2000
vv4000
vv10x3 31213 −
+−
=− 12 = 3v1 - v2 - 2v3 (1)
At node 2,
1v
2vv
4vv 23121 =
−+
− 3v1 - 5v2 - 2v3 = 0 (2)
Note that v0 = v2. We now apply KVL in Fig. (b) - v3 - 3v2 + v2 = 0 v3 = - 2v2 (3) From (1) to (3), v1 = 1 V, v2 = 3 V
Chapter 3, Solution 22
At node 1, 8
vv3
4v
2v 011012 −
++=−
24 = 7v1 - v2 (1)
At node 2, 3 + 1
v5v8
vv 2221 +=
−
But, v1 = 12 - v1 Hence, 24 + v1 - v2 = 8 (v2 + 60 + 5v1) = 4 V 456 = 41v1 - 9v2 (2) Solving (1) and (2), v1 = - 10.91 V, v2 = - 100.36 V
Chapter 3, Solution 23
At the supernode, 5 + 2 = 5v
10v 21 + 70 = v1 + 2v2 (1)
Considering Fig. (b), - v1 - 8 + v2 = 0 v2 = v1 + 8 (2) Solving (1) and (2), v1 = 18 V, v2 = 26 V
+ v1
–5 Ω 10 Ω
v1 v2
8 V
– +
5 A 2 A
+
v2
– (b)(a)
Chapter 3, Solution 24
6mA 1 kΩ 2 kΩ 3 kΩ V1 V2 + io - 30V 15V - 4 kΩ 5 kΩ + At node 1,
212111 2796
246
130 VVVVVV
−=→−
++=− (1)
At node 2,
211222 311530
253)15(
6 VVVVVV+−=→
−+=
−−+ (2)
Solving (1) and (2) gives V1=16.24. Hence io = V1/4 = 4.06 mA Chapter 3, Solution 25 Using nodal analysis,
1 Ω
v0
2 Ω
4 Ω
2 Ω
10V –+
–+
40V –+
i0
20V
40v
2v10
2v40
1v20 0000 −
=−
+−
+−
v0 = 20V
i0 = 1
v20 0− = 0 A
Chapter 3, Solution 26 At node 1,
32121311 24745
5103
2015 VVVVVVVV
−−=−→−
+−
+=− (1)
At node 2,
554
532221 VVVIVV o −
=−
+− (2)
But 10
31 VVIo
−= . Hence, (2) becomes
321 31570 VVV +−= (3) At node 3,
32132331 52100
5510
103 VVV
VVVVV−+=−→=
−+
−−+
−+ (4)
Putting (1), (3), and (4) in matrix form produces
BAVVVV
=→
−
−=
−−
−−
10045
5213157247
3
2
1
Using MATLAB leads to
−−−
== −
96.1982.4835.9
1BAV
Thus, V 95.1 V, 982.4 V, 835.9 321 −=−=−= VVV Chapter 3, Solution 27 At node 1, 2 = 2v1 + v1 – v2 + (v1 – v3)4 + 3i0, i0 = 4v2. Hence, 2 = 7v1 + 11v2 – 4v3 (1) At node 2, v1 – v2 = 4v2 + v2 – v3 0 = – v1 + 6v2 – v3 (2) At node 3,
2v3 = 4 + v2 – v3 + 12v2 + 4(v1 – v3)
or – 4 = 4v1 + 13v2 – 7v3 (3) In matrix form,
−=
−−
−
402
vvv
71341614117
3
2
1
,1767134
1614117=
−−
−=∆ 110
71341604112
1 =−−
−−
=∆
,66744
101427
2 =−−
−=∆ 286
41340612117
3 =−
−=∆
v1 = ,V625.01761101 ==
∆∆
v2 = V375.0176662 ==
∆∆
v3 = .V625.11762863 ==
∆∆
v1 = 625 mV, v2 = 375 mV, v3 = 1.625 V. Chapter 3, Solution 28 At node c,
dcbcbccd VVVVVVVV
211505410
−+−=→+−
=−
(1)
At node b,
cbabbcba VVVVVVVV
2445848
45+−=−→=
−+
−+ (2)
At node a,
dbabaada VVVVVVVV
427300845
16430
−−=→=−+
++−−
(3)
At node d,
dcacddda VVVVVVVV
72515010204
30−+=→
−+=
−− (4)
In matrix form, (1) to (4) become
BAV
VVVV
d
c
b
a
=→
−
=
−−−
−−−
15030450
72054027
024121150
We use MATLAB to invert A and obtain
−−
−
== −
17.29736.1
847.714.10
1BAV
Thus, V 17.29 V, 736.1 V, 847.7 V, 14.10 −=−==−= dcba VVVV Chapter 3, Solution 29 At node 1,
42121141 45025 VVVVVVVV −−=−→=−++−+ (1) At node 2,
32132221 4700)(42 VVVVVVVV −+−=→=−+=− (2) At node 3,
4324332 546)(46 VVVVVVV −+−=→−=−+ (3) At node 4,
43144143 5232 VVVVVVVV +−−=→=−+−+ (4) In matrix form, (1) to (4) become
BAV
VVVV
=→
−
=
−−−−
−−−−
2605
51011540
04711014
4
3
2
1
Using MATLAB,
−
== −
7076.0309.2209.17708.0
1BAV
i.e. V 7076.0 V, 309.2 V, 209.1 V, 7708.0 4321 ===−= VVVV
Chapter 3, Solution 30
v1
10 Ω
20 Ω
80 Ω
40 Ω
1
v0
I0
2I0
2
v2
4v0
120 V
+––
+
– +
100 V At node 1,
20
vv410
v10040
vv 1o121 −+
−=
− (1)
But, vo = 120 + v2 v2 = vo – 120. Hence (1) becomes 7v1 – 9vo = 280 (2) At node 2,
Io + 2Io = 80
0v o −
80v
40v120v
3 oo1 =
−+
or 6v1 – 7vo = -720 (3)
from (2) and (3),
−
=
−−
720280
vv
7697
o
1
554497697
=+−=−−
=∆
844077209280
1 −=−−−
=∆ , 67207206
28072 −=
−=∆
v1 = ,16885
84401 −=−
=∆∆
vo = V13445
67202 −−
=∆∆
Io = -5.6 A
Chapter 3, Solution 31 1 Ω
i0
2v0 v3
1 Ω
2 Ω
4 Ω 10 V –+
v1 v2
+ v0 –
4 Ω
1 A At the supernode,
1 + 2v0 = 1
vv1
v4v 3121 −
++ (1)
But vo = v1 – v3. Hence (1) becomes, 4 = -3v1 + 4v2 +4v3 (2) At node 3,
2vo + 2
v10vv
4v 3
312 −
+−=
or 20 = 4v1 + v2 – 2v3 (3)
At the supernode, v2 = v1 + 4io. But io = 4
v 3 . Hence,
v2 = v1 + v3 (4) Solving (2) to (4) leads to, v1 = 4 V, v2 = 4 V, v3 = 0 V.
Chapter 3, Solution 32
v3
v1 v2
10 V
loop 2
– +
20 V
+ –
12 V
loop 1
–+
+ v3 –
+ v1 –
10 kΩ
4 mA
5 kΩ
(b)(a) We have a supernode as shown in figure (a). It is evident that v2 = 12 V, Applying KVL to loops 1and 2 in figure (b), we obtain,
-v1 – 10 + 12 = 0 or v1 = 2 and -12 + 20 + v3 = 0 or v3 = -8 V Thus, v1 = 2 V, v2 = 12 V, v3 = -8V. Chapter 3, Solution 33
(a) This is a non-planar circuit because there is no way of redrawing the circuit with no crossing branches.
(b) This is a planar circuit. It can be redrawn as shown below.
1 Ω
2 Ω
3 Ω
4 Ω
5 Ω 12 V
–+
Chapter 3, Solution 34 (a) This is a planar circuit because it can be redrawn as shown below,
7 Ω
10 V –+
1 Ω 2 Ω
3 Ω
4 Ω
6 Ω
5 Ω (b) This is a non-planar circuit. Chapter 3, Solution 35
5 kΩ i1
i2
+ v0 –2 kΩ
30 V –+
–+20 V
4 kΩ Assume that i1 and i2 are in mA. We apply mesh analysis. For mesh 1,
-30 + 20 + 7i1 – 5i2 = 0 or 7i1 – 5i2 = 10 (1) For mesh 2, -20 + 9i2 – 5i1 = 0 or -5i1 + 9i2 = 20 (2) Solving (1) and (2), we obtain, i2 = 5. v0 = 4i2 = 20 volts.
Chapter 3, Solution 36
i1 i2
2 Ω
4 Ω 10 V
+ –
12 V –+
I2
6 Ω I1
i3 Applying mesh analysis gives,
12 = 10I1 – 6I2
-10 = -6I1 + 8I2
or
−
−=
− 2
1
II
4335
56
,114335=
−−
=∆ ,94536
1 =−
−=∆ 7
5365
2 −=−−
=∆
,119I 1
1 =∆∆
= 11
7I 22
−=
∆∆
=
i1 = -I1 = -9/11 = -0.8181 A, i2 = I1 – I2 = 10/11 = 1.4545 A.
vo = 6i2 = 6x1.4545 = 8.727 V.
Chapter 3, Solution 37
5 Ω
4v0 2 Ω
1 Ω
3 Ω
3 V –+
+–
+ v0 –
i2
i1
Applying mesh analysis to loops 1 and 2, we get,
6i1 – 1i2 + 3 = 0 which leads to i2 = 6i1 + 3 (1) -1i1 + 6i2 – 3 + 4v0 = 0 (2) But, v0 = -2i1 (3)
Using (1), (2), and (3) we get i1 = -5/9. Therefore, we get v0 = -2i1 = -2(-5/9) = 1.111 volts Chapter 3, Solution 38
6 Ω
2v0 8 Ω
3 Ω
12 V –+ +
–
+ v0 –
i2
i1
We apply mesh analysis.
12 = 3 i1 + 8(i1 – i2) which leads to 12 = 11 i1 – 8 i2 (1) -2 v0 = 6 i2 + 8(i2 – i1) and v0 = 3 i1 or i1 = 7 i2 (2)
From (1) and (2), i1 = 84/69 and v0 = 3 i1 = 3x89/69 v0 = 3.652 volts Chapter 3, Solution 39 For mesh 1, 0610210 21 =−+− III x− But . Hence, 21 III x −=
212121 245610121210 IIIIII −=→−++−= (1) For mesh 2,
2112 43606812 IIII −=→=−+ (2) Solving (1) and (2) leads to -0.9A A, 8.0 21 == II
Chapter 3, Solution 40
2 kΩ
i2
6 kΩ
4 kΩ
2 kΩ 6 kΩ
i3
i1
–+
4 kΩ
30V Assume all currents are in mA and apply mesh analysis for mesh 1.
30 = 12i1 – 6i2 – 4i3 15 = 6i1 – 3i2 – 2i3 (1) for mesh 2,
0 = - 6i1 + 14i2 – 2i3 0 = -3i1 + 7i2 – i3 (2) for mesh 2,
0 = -4i1 – 2i2 + 10i3 0 = -2i1 – i2 + 5i3 (3) Solving (1), (2), and (3), we obtain, io = i1 = 4.286 mA. Chapter 3, Solution 41 10 Ω
i3
ii2
2 Ω
1 Ω
8 V –+
6 V
+ –
i3
i2
i1
5 Ω 4 Ω
0
For loop 1, 6 = 12i1 – 2i2 3 = 6i1 – i2 (1) For loop 2, -8 = 7i2 – 2i1 – i3 (2) For loop 3,
-8 + 6 + 6i3 – i2 = 0 2 = 6i3 – i2 (3) We put (1), (2), and (3) in matrix form,
=
−−−
283
iii
610172016
3
2
1
,234610172016
−=−−−
=∆ 240620182036
2 −==∆
38210872316
3 −=−−−
=∆
At node 0, i + i2 = i3 or i = i3 – i2 = 234
2403823
−−−
=∆∆−∆ = 1.188 A
Chapter 3, Solution 42 For mesh 1, (1) 2121 3050120305012 IIII −=→=−+−For mesh 2, (2) 321312 40100308040301008 IIIIII −+−=→=−−+−For mesh 3, (3) 3223 50406040506 IIII +−=→=−+−Putting eqs. (1) to (3) in matrix form, we get
BAIIII
=→
=
−−−
−
68
12
50400401003003050
3
2
1
Using Matlab,
== −
44.040.048.0
1BAI
i.e. I1 = 0.48 A, I2 = 0.4 A, I3 = 0.44 A Chapter 3, Solution 43
30 Ω
30 Ω
20 Ω
20 Ω
20 Ω
80 V –+
80 V –+
i3
i2
i1
a
+
Vab –
30 Ω b For loop 1, 80 = 70i1 – 20i2 – 30i3 8 = 7i1 – 2i2 – 3i3 (1)
For loop 2, 80 = 70i2 – 20i1 – 30i3 8 = -2i1 + 7i2 – 3i3 (2) For loop 3, 0 = -30i1 – 30i2 + 90i3 0 = i1 + i2 – 3i3 (3) Solving (1) to (3), we obtain i3 = 16/9
Io = i3 = 16/9 = 1.778 A Vab = 30i3 = 53.33 V. Chapter 3, Solution 44 6 V
1 Ω
4 Ω i3 i2
i1
3 A
+
6 V –+
2 Ω
5 Ω i2 i1 Loop 1 and 2 form a supermesh. For the supermesh, 6i1 + 4i2 - 5i3 + 12 = 0 (1) For loop 3, -i1 – 4i2 + 7i3 + 6 = 0 (2) Also, i2 = 3 + i1 (3) Solving (1) to (3), i1 = -3.067, i3 = -1.3333; io = i1 – i3 = -1.7333 A
Chapter 3, Solution 45 4 Ω 8 Ω
1 Ω i1
2 Ω 6 Ω
3 Ω –+
i4i3
i2
30V For loop 1, 30 = 5i1 – 3i2 – 2i3 (1) For loop 2, 10i2 - 3i1 – 6i4 = 0 (2) For the supermesh, 6i3 + 14i4 – 2i1 – 6i2 = 0 (3) But i4 – i3 = 4 which leads to i4 = i3 + 4 (4) Solving (1) to (4) by elimination gives i = i1 = 8.561 A. Chapter 3, Solution 46
For loop 1, 12811081112 2121 =−→=−+− iiii (1)
For loop 2, 02148 21 =++− ovii
But v , 13io =
21121 706148 iiiii =→=++− (2) Substituting (2) into (1),
1739.012877 222 =→=− iii A and 217.17 21 == ii A
Chapter 3, Solution 47 First, transform the current sources as shown below.
- 6V + 2Ω I3 V1 8Ω V2 4Ω V3 4Ω 8Ω I1 2Ω I2 + + 20V 12V - - For mesh 1,
321321 47100821420 IIIIII −−=→=−−+− (1) For mesh 2,
321312 2760421412 IIIIII −+−=−→=−−+ (2) For mesh 3,
321123 7243084146 IIIIII +−−=→=−−+− (3) Putting (1) to (3) in matrix form, we obtain
BAIIII
=→
−=
−−−−−−
36
10
724271417
3
2
1
Using MATLAB,
8667.1,0333.0 ,5.28667.10333.02
3211 ===→
== − IIIBAI
But
1 120 20 4 10 V
4VI V−
= → = − =1I
2 1 22( ) 4.933 VV I I= − = Also,
32 3 2
12 12 8 12.267V8
VI V I−= → = + =
Chapter 3, Solution 48 We apply mesh analysis and let the mesh currents be in mA.
3kΩ I4 4kΩ 2kΩ 5kΩ Io 1kΩ I3 - I1 I2 6V + + 12 V + 10kΩ - 8V - For mesh 1,
421421 454045812 IIIIII −−=→=−−++− (1) For mesh 2,
43214312 2101380210138 IIIIIIII −−+−=→=−−−+− (2) For mesh 3,
432423 5151060510156 IIIIII −+−=→=−−+− (3) For mesh 4,
014524 4321 =+−−− IIII (4) Putting (1) to (4) in matrix form gives
BAI
IIII
=→
=
−−−−−−−−−−
0684
145245151002101314015
4
3
2
1
Using MATLAB,
== −
6791.7087.8217.7
1BAI
The current through the 10k resistor is IΩ o= I2 – I3 = 0.2957 mA
Chapter 3, Solution 49
3 Ω
(a)
2i0
i20
16 V –+
1 Ω 2 Ω
i3
i2i1
i1
2 Ω
16V 2 Ω
i2
+ v0 –
+ v0 or–
i1
1 Ω
–+
2 Ω (b)
For the supermesh in figure (a), 3i1 + 2i2 – 3i3 + 16 = 0 (1) At node 0, i2 – i1 = 2i0 and i0 = -i1 which leads to i2 = -i1 (2) For loop 3, -i1 –2i2 + 6i3 = 0 which leads to 6i3 = -i1 (3) Solving (1) to (3), i1 = (-32/3)A, i2 = (32/3)A, i3 = (16/9)A i0 = -i1 = 10.667 A, from fig. (b), v0 = i3-3i1 = (16/9) + 32 = 33.78 V.
Chapter 3, Solution 50
10 Ω
4 Ω 2 Ω i3
i2
i1
3i0
60 V –+
8 Ω i3 i2
For loop 1, 16i1 – 10i2 – 2i3 = 0 which leads to 8i1 – 5i2 – i3 = 0 (1) For the supermesh, -60 + 10i2 – 10i1 + 10i3 – 2i1 = 0 or -6i1 + 5i2 + 5i3 = 30 (2) Also, 3i0 = i3 – i2 and i0 = i1 which leads to 3i1 = i3 – i2 (3) Solving (1), (2), and (3), we obtain i1 = 1.731 and i0 = i1 = 1.731 A Chapter 3, Solution 51
5 A
+
v0
1 Ω
4 Ω
2 Ω
i3
i2
i1
20V –+40 V
–+
8 Ω
For loop 1, i1 = 5A (1) For loop 2, -40 + 7i2 – 2i1 – 4i3 = 0 which leads to 50 = 7i2 – 4i3 (2) For loop 3, -20 + 12i3 – 4i2 = 0 which leads to 5 = - i2 + 3 i3 (3) Solving with (2) and (3), i2 = 10 A, i3 = 5 A And, v0 = 4(i2 – i3) = 4(10 – 5) = 20 V. Chapter 3, Solution 52
i3
i2
2 Ω
4 Ω
8 Ω
i3
i2
i1
3A
+ v0 –
2V0 +–
–+
VS For mesh 1, 2(i1 – i2) + 4(i1 – i3) – 12 = 0 which leads to 3i1 – i2 – 2i3 = 6 (1) For the supermesh, 2(i2 – i1) + 8i2 + 2v0 + 4(i3 – i1) = 0 But v0 = 2(i1 – i2) which leads to -i1 + 3i2 + 2i3 = 0 (2) For the independent current source, i3 = 3 + i2 (3) Solving (1), (2), and (3), we obtain,
i1 = 3.5 A, i2 = -0.5 A, i3 = 2.5 A.
Chapter 3, Solution 53
i3
i2
2 Ω
4 Ω
8 Ω
i3
i2
i1
3A
+ v0 –
2V0 +–
–+
VS For mesh 1, 2(i1 – i2) + 4(i1 – i3) – 12 = 0 which leads to 3i1 – i2 – 2i3 = 6 (1) For the supermesh, 2(i2 – i1) + 8i2 + 2v0 + 4(i3 – i1) = 0 But v0 = 2(i1 – i2) which leads to -i1 + 3i2 + 2i3 = 0 (2) For the independent current source, i3 = 3 + i2 (3) Solving (1), (2), and (3), we obtain,
i1 = 3.5 A, i2 = -0.5 A, i3 = 2.5 A.
Chapter 3, Solution 54
Let the mesh currents be in mA. For mesh 1, 2121 22021012 IIII −=→=−++− (1)
For mesh 2, (2) 321312 3100310 IIIIII −+−=→=−−+−
For mesh 3, 3223 2120212 IIII +−=→=−+− (3)
Putting (1) to (3) in matrix form leads to
BAIIII
=→
=
−−−
−
12102
210131
012
3
2
1
Using MATLAB,
mA 25.10,mA 5.8 ,mA 25.525.105.825.5
3211 ===→
== − IIIBAI
Chapter 3, Solution 55
dI1 i2
a 0
b c
1A
4A
i1
i3
I4
1A
I2
I3
8 V
12 Ω 4 Ω
2 Ω
6 Ω
10 V
+
+ –
4 A
I4
I3
I2
It is evident that I1 = 4 (1) For mesh 4, 12(I4 – I1) + 4(I4 – I3) – 8 = 0 (2) For the supermesh 6(I2 – I1) + 10 + 2I3 + 4(I3 – I4) = 0 or -3I1 + 3I2 + 3I3 – 2I4 = -5 (3) At node c, I2 = I3 + 1 (4) Solving (1), (2), (3), and (4) yields, I1 = 4A, I2 = 3A, I3 = 2A, and I4 = 4A At node b, i1 = I2 – I1 = -1A At node a, i2 = 4 – I4 = 0A At node 0, i3 = I4 – I3 = 2A Chapter 3, Solution 56 + v1 –
2 Ω
2 Ω
2 Ω
2 Ω
2 Ω
i3
i2
i1 12 V
–+
+
v2 –
For loop 1, 12 = 4i1 – 2i2 – 2i3 which leads to 6 = 2i1 – i2 – i3 (1) For loop 2, 0 = 6i2 –2i1 – 2 i3 which leads to 0 = -i1 + 3i2 – i3 (2) For loop 3, 0 = 6i3 – 2i1 – 2i2 which leads to 0 = -i1 – i2 + 3i3 (3)
In matrix form (1), (2), and (3) become,
=
−−−−−−
006
iii
311131112
3
2
1
∆ = ,8311131112=
−−−−−−
∆2 = 24301131162=
−−−−
∆3 = 24011031612=
−−−
−, therefore i2 = i3 = 24/8 = 3A,
v1 = 2i2 = 6 volts, v = 2i3 = 6 volts Chapter 3, Solution 57
Assume R is in kilo-ohms. VVVVmAxkV 2872100100,72184 212 =−=−==Ω=
Current through R is
RR
RiViR
i RoR )18(3
3283
31, +
=→=+
=
This leads to R = 84/26 = 3.23 kΩ Chapter 3, Solution 58 30 Ω
10 Ω 30 Ω 10 Ω
i3
i2
i1
120 V
–+
30 Ω
For loop 1, 120 + 40i1 – 10i2 = 0, which leads to -12 = 4i1 – i2 (1) For loop 2, 50i2 – 10i1 – 10i3 = 0, which leads to -i1 + 5i2 – i3 = 0 (2) For loop 3, -120 – 10i2 + 40i3 = 0, which leads to 12 = -i2 + 4i3 (3) Solving (1), (2), and (3), we get, i1 = -3A, i2 = 0, and i3 = 3A Chapter 3, Solution 59
i1
2I0
I0
10 Ω
20 Ω
40 Ω
+ v0 –
100V –+
120 V
– +
+–
4v0 i3
i2
80 Ω
i2 i3 For loop 1, -100 + 30i1 – 20i2 + 4v0 = 0, where v0 = 80i3 or 5 = 1.5i1 – i2 + 16i3 (1) For the supermesh, 60i2 – 20i1 – 120 + 80i3 – 4 v0 = 0, where v0 = 80i3 or 6 = -i1 + 3i2 – 12i3 (2) Also, 2I0 = i3 – i2 and I0 = i2, hence, 3i2 = i3 (3)
From (1), (2), and (3),
−−−
−
1301231
3223
=
06
10
iii
3
2
1
∆ = ,5130
12313223
=−−−
− ∆2 = ,28
1001261
32103−=
−−− ∆3 = 84
030631
1023−=−
−
I0 = i2 = ∆2/∆ = -28/5 = -5.6 A v0 = 8i3 = (-84/5)80 = -1344 volts
Chapter 3, Solution 60 0.5i0
i0
v1
1 Ω
4 Ω 8 Ω 10 V
10 V –+
v2 2 Ω At node 1, (v1/1) + (0.5v1/1) = (10 – v1)/4, which leads to v1 = 10/7 At node 2, (0.5v1/1) + ((10 – v2)/8) = v2/2 which leads to v2 = 22/7 P1Ω = (v1)2/1 = 2.041 watts, P2Ω = (v2)2/2 = 4.939 watts P4Ω = (10 – v1)2/4 = 18.38 watts, P8Ω = (10 – v2)2/8 = 5.88 watts Chapter 3, Solution 61
+ v0 –
20 Ω v2v1
+–30 Ω 5v0
40 Ω
10 Ω i0 is At node 1, is = (v1/30) + ((v1 – v2)/20) which leads to 60is = 5v1 – 3v2 (1) But v2 = -5v0 and v0 = v1 which leads to v2 = -5v1 Hence, 60is = 5v1 + 15v1 = 20v1 which leads to v1 = 3is, v2 = -15is i0 = v2/50 = -15is/50 which leads to i0/is = -15/50 = -0.3
Chapter 3, Solution 62
i1
4 kΩ A
–+
B8 kΩ
100V –+
i3i2
2 kΩ 40 V We have a supermesh. Let all R be in kΩ, i in mA, and v in volts. For the supermesh, -100 +4i1 + 8i2 + 2i3 + 40 = 0 or 30 = 2i1 + 4i2 + i3 (1) At node A, i1 + 4 = i2 (2) At node B, i2 = 2i1 + i3 (3) Solving (1), (2), and (3), we get i1 = 2 mA, i2 = 6 mA, and i3 = 2 mA. Chapter 3, Solution 63
A
5 Ω
10 Ω
i2i1
+– 4ix
50 V –+
For the supermesh, -50 + 10i1 + 5i2 + 4ix = 0, but ix = i1. Hence, 50 = 14i1 + 5i2 (1) At node A, i1 + 3 + (vx/4) = i2, but vx = 2(i1 – i2), hence, i1 + 2 = i2 (2) Solving (1) and (2) gives i1 = 2.105 A and i2 = 4.105 A vx = 2(i1 – i2) = -4 volts and ix = i2 – 2 = 4.105 amp
Chapter 3, Solution 64
i0
i1
2 A
10 Ω
50 Ω A 10 Ω i2
+ −
0.2V0
4i0 +–
100V –+
i3
i2i1
40 Ω i1 i3B For mesh 2, 20i2 – 10i1 + 4i0 = 0 (1) But at node A, io = i1 – i2 so that (1) becomes i1 = (7/12)i2 (2) For the supermesh, -100 + 50i1 + 10(i1 – i2) – 4i0 + 40i3 = 0 or 50 = 28i1 – 3i2 + 20i3 (3) At node B, i3 + 0.2v0 = 2 + i1 (4) But, v0 = 10i2 so that (4) becomes i3 = 2 – (17/12)i2 (5) Solving (1) to (5), i2 = -0.674,
v0 = 10i2 = -6.74 volts, i0 = i1 - i2 = -(5/12)i2 = 0.281 amps Chapter 3, Solution 65
For mesh 1, 421 61212 III −−= (1) For mesh 2, 54321 81660 IIIII −−−+−= (2) For mesh 3, 532 1589 III −+−= (3) For mesh 4, 5421 256 IIII −+−−= (4) For mesh 5, 5432 8210 IIII +−−−= (5)
Casting (1) to (5) in matrix form gives
BAI
IIIII
=→
=
−−−−−−−−−−−−
−
10690
12
8211025011101580118166
010612
5
4
3
2
1
Using MATLAB leads to
== −
411.2864.2733.1824.1673.1
1BAI
Thus, A 411.2 A, 864.1 A, 733.1 A, 824.1 A, 673.1 54321 ===== IIIII
Chapter 3, Solution 66 Consider the circuit below.
2 kΩ 2 kΩ
+ + 20V I1 1 kΩ I2 10V - - 1 k 1 kΩ Ω Io
1 kΩ 2 kΩ 2 kΩ I3 I4 - 12V + We use mesh analysis. Let the mesh currents be in mA. For mesh 1, (1) 321420 III −−=For mesh 2, (2) 421 410 III −+−=−For mesh 3, (3) 431 412 III −+−=For mesh 4, (4) 432 412 III +−−=−
In matrix form, (1) to (4) become
BAI
IIII
=→
−
−=
−−−−−−
−−
121210
20
411014011041
0114
4
3
2
1
Using MATLAB,
−
−== −
5.275.375.15.5
1BAI
Thus, mA 75.33 −=−= II o Chapter 3, Solution 67
G11 = (1/1) + (1/4) = 1.25, G22 = (1/1) + (1/2) = 1.5 G12 = -1 = G21, i1 = 6 – 3 = 3, i2 = 5-6 = -1
Hence, we have,
−
=
−
−1
3vv
5.11125.1
2
1
∆
=
−
− −
25.1115.11
5.11125.1 1
, where ∆ = [(1.25)(1.5)-(-1)(-1)] = 0.875
=
−−
=
−
=
24
)4286.1(1)1429.1(3)1429.1(1)7143.1(3
13
4286.11429.11429.17143.1
vv
2
1
Clearly v1 = 4 volts and v2 = 2 volts
Chapter 3, Solution 68 By inspection, G11 = 1 + 3 + 5 = 8S, G22 = 1 + 2 = 3S, G33 = 2 + 5 = 7S
G12 = -1, G13 = -5, G21 = -1, G23 = -2, G31 = -5, G32 = -2
i1 = 4, i2 = 2, i3 = -1
We can either use matrix inverse as we did in Problem 3.51 or use Cramer’s Rule. Let us use Cramer’s rule for this problem.
First, we develop the matrix relationships.
−=
−−−−−−
124
vvv
725231518
3
2
1
85721232514
,34725231518
1 =−−
−−−
=∆=−−
−−−−
=∆
87125
231418
,109715221548
32 =−−−
−−
=∆=−−
−−−
=∆
v1 = ∆1/∆ = 85/34 = 3.5 volts, v2 = ∆2/∆ = 109/34 = 3.206 volts
v3 = ∆3/∆ = 87/34 = 2.56 volts
Chapter 3, Solution 69
Assume that all conductances are in mS, all currents are in mA, and all voltages are in volts. G11 = (1/2) + (1/4) + (1/1) = 1.75, G22 = (1/4) + (1/4) + (1/2) = 1, G33 = (1/1) + (1/4) = 1.25, G12 = -1/4 = -0.25, G13 = -1/1 = -1, G21 = -0.25, G23 = -1/4 = -0.25, G31 = -1, G32 = -0.25 i1 = 20, i2 = 5, and i3 = 10 – 5 = 5 The node-voltage equations are:
=
−−−−−−
55
20
vvv
25.125.0125.0125.0125.075.1
3
2
1
Chapter 3, Solution 70
G11 = G1 + G2 + G4, G12 = -G2, G13 = 0, G22 = G2 + G3, G21 = -G2, G23 = -G3, G33 = G1 + G3 + G5, G31 = 0, G32 = -G3 i1 = -I1, i2 = I2, and i3 = I1
Then, the node-voltage equations are:
−=
++−−+−
−++
1
2
1
3
2
1
5313
3212
2421
III
vvv
GGGG0GGGG0GGGG
Chapter 3, Solution 71
R11 = 4 + 2 = 6, R22 = 2 + 8 + 2 = 12, R33 = 2 + 5 = 7, R12 = -2, R13 = 0, R21 = -2, R23 = -2, R31 = 0, R32 = -2 v1 = 12, v2 = -8, and v3 = -20
Now we can write the matrix relationships for the mesh-current equations.
−−=
−−−
−
208
12
iii
7202122
026
3
2
1
Now we can solve for i2 using Cramer’s Rule.
4087200282
0126,452
7202122
026
2 −=−
−−−=∆=−
−−−
=∆
i2 = ∆2/∆ = -0.9026, p = (i2)2R = 6.52 watts Chapter 3, Solution 72
R11 = 5 + 2 = 7, R22 = 2 + 4 = 6, R33 = 1 + 4 = 5, R44 = 1 + 4 = 5, R12 = -2, R13 = 0 = R14, R21 = -2, R23 = -4, R24 = 0, R31 = 0, R32 = -4, R34 = -1, R41 = 0 = R42, R43 = -1, we note that Rij = Rji for all i not equal to j. v1 = 8, v2 = 4, v3 = -10, and v4 = -4
Hence the mesh-current equations are:
−−
=
−−−
−−−
41048
iiii
51001540
04620027
4
3
2
1
Chapter 3, Solution 73
R11 = 2 + 3 +4 = 9, R22 = 3 + 5 = 8, R33 = 1 + 4 = 5, R44 = 1 + 1 = 2, R12 = -3, R13 = -4, R14 = 0, R23 = 0, R24 = 0, R34 = -1 v1 = 6, v2 = 4, v3 = 2, and v4 = -3
Hence,
−
=
−−−
−−−
3246
iiii
21001604
00830439
4
3
2
1
Chapter 3, Solution 74
R11 = R1 + R4 + R6, R22 = R2 + R4 + R5, R33 = R6 + R7 + R8, R44 = R3 + R5 + R8, R12 = -R4, R13 = -R6, R14 = 0, R23 = 0, R24 = -R5, R34 = -R8, again, we note that Rij = Rji for all i not equal to j.
The input voltage vector is =
−
−
4
3
2
1
VVV
V
−
−=
++−−−++−−++−
−−++
4
3
2
1
4
3
2
1
85385
88766
55424
64641
VVV
V
iiii
RRRRR0RRRR0RR0RRRR0RRRRR
Chapter 3, Solution 75 * Schematics Netlist * R_R4 $N_0002 $N_0001 30 R_R2 $N_0001 $N_0003 10 R_R1 $N_0005 $N_0004 30 R_R3 $N_0003 $N_0004 10 R_R5 $N_0006 $N_0004 30 V_V4 $N_0003 0 120V v_V3 $N_0005 $N_0001 0 v_V2 0 $N_0006 0 v_V1 0 $N_0002 0
i3
i2i1
Clearly, i1 = -3 amps, i2 = 0 amps, and i3 = 3 amps, which agrees with the answers in Problem 3.44.
Chapter 3, Solution 76 * Schematics Netlist * I_I2 0 $N_0001 DC 4A R_R1 $N_0002 $N_0001 0.25 R_R3 $N_0003 $N_0001 1 R_R2 $N_0002 $N_0003 1 F_F1 $N_0002 $N_0001 VF_F1 3 VF_F1 $N_0003 $N_0004 0V R_R4 0 $N_0002 0.5 R_R6 0 $N_0001 0.5 I_I1 0 $N_0002 DC 2A R_R5 0 $N_0004 0.25
Clearly, v1 = 625 mVolts, v2 = 375 mVolts, and v3 = 1.625 volts, which agrees with the solution obtained in Problem 3.27.
Chapter 3, Solution 77 * Schematics Netlist * R_R2 0 $N_0001 4 I_I1 $N_0001 0 DC 3A I_I3 $N_0002 $N_0001 DC 6A R_R3 0 $N_0002 2 R_R1 $N_0001 $N_0002 1 I_I2 0 $N_0002 DC 5A
Clearly, v1 = 4 volts and v2 = 2 volts, which agrees with the answer obtained in Problem 3.51.
Chapter 3, Solution 78 The schematic is shown below. When the circuit is saved and simulated the node voltages are displaced on the pseudocomponents as shown. Thus,
,V15 V,5.4 V,3 321 −==−= VVV
.
Chapter 3, Solution 79 The schematic is shown below. When the circuit is saved and simulated, we obtain the node voltages as displaced. Thus,
V 88.26V V, 6944.0V V, 28.10V V, 278.5V dcba −===−=
Chapter 3, Solution 80 * Schematics Netlist * H_H1 $N_0002 $N_0003 VH_H1 6 VH_H1 0 $N_0001 0V I_I1 $N_0004 $N_0005 DC 8A V_V1 $N_0002 0 20V R_R4 0 $N_0003 4 R_R1 $N_0005 $N_0003 10 R_R2 $N_0003 $N_0002 12 R_R5 0 $N_0004 1 R_R3 $N_0004 $N_0001 2
Clearly, v1 = 84 volts, v2 = 4 volts, v3 = 20 volts, and v4 = -5.333 volts Chapter 3, Solution 81
Clearly, v1 = 26.67 volts, v2 = 6.667 volts, v3 = 173.33 volts, and v4 = -46.67 volts which agrees with the results of Example 3.4.
This is the netlist for this circuit. * Schematics Netlist * R_R1 0 $N_0001 2 R_R2 $N_0003 $N_0002 6 R_R3 0 $N_0002 4 R_R4 0 $N_0004 1 R_R5 $N_0001 $N_0004 3 I_I1 0 $N_0003 DC 10A V_V1 $N_0001 $N_0003 20V E_E1 $N_0002 $N_0004 $N_0001 $N_0004 3 Chapter 3, Solution 82
+ v0 –
4
3 kΩ
2 kΩ
4 kΩ 8 kΩ
6 kΩ
0
1 2 33v0
2i0
100V –+
4A +
This network corresponds to the Netlist.
Chapter 3, Solution 83 The circuit is shown below.
+ v0 –
4
3 kΩ
2 kΩ
4 kΩ 8 kΩ
6 kΩ
1 3
20 V –+ 30 Ω 2 A
0
50 Ω
0
1 2 3
3v0
2i0 2
100V –+
4A +
20 Ω 70 Ω When the circuit is saved and simulated, we obtain v2 = -12.5 volts Chapter 3, Solution 84
From the output loop, v0 = 50i0x20x103 = 106i0 (1) From the input loop, 3x10-3 + 4000i0 – v0/100 = 0 (2) From (1) and (2) we get, i0 = 0.5µA and v0 = 0.5 volt. Chapter 3, Solution 85
The amplifier acts as a source. Rs + Vs RL - For maximum power transfer, Ω== 9sL RR
Chapter 3, Solution 86 Let v1 be the potential across the 2 k-ohm resistor with plus being on top. Then,
[(0.03 – v1)/1k] + 400i = v1/2k (1) Assume that i is in mA. But, i = (0.03 – v1)/1 (2) Combining (1) and (2) yields, v1 = 29.963 mVolts and i = 37.4 nA, therefore, v0 = -5000x400x37.4x10-9 = -74.8 mvolts
Chapter 3, Solution 87
v1 = 500(vs)/(500 + 2000) = vs/5 v0 = -400(60v1)/(400 + 2000) = -40v1 = -40(vs/5) = -8vs, Therefore, v0/vs = -8
Chapter 3, Solution 88 Let v1 be the potential at the top end of the 100-ohm resistor. (vs – v1)/200 = v1/100 + (v1 – 10-3v0)/2000 (1) For the right loop, v0 = -40i0(10,000) = -40(v1 – 10-3)10,000/2000, or, v0 = -200v1 + 0.2v0 = -4x10-3v0 (2)
Substituting (2) into (1) gives, (vs + 0.004v1)/2 = -0.004v0 + (-0.004v1 – 0.001v0)/20
This leads to 0.125v0 = 10vs or (v0/vs) = 10/0.125 = -80
Chapter 3, Solution 89 vi = VBE + 40k IB (1) 5 = VCE + 2k IC (2) If IC = βIB = 75IB and VCE = 2 volts, then (2) becomes 5 = 2 +2k(75IB) which leads to IB = 20 µA.
Substituting this into (1) produces, vi = 0.7 + 0.8 = 1.5 volts. 2 kΩ IB 40 kΩ
+ VBE
–
-+
-+ vi 5 v
Chapter 3, Solution 90 1 kΩ
100 kΩ
IE
IB
+ VBE
–
+ VCE –
i2
i1 -+
18V -+
500 Ω + V0 –
vs For loop 1, -vs + 10k(IB) + VBE + IE (500) = 0 = -vs + 0.7 + 10,000IB + 500(1 + β)IB which leads to vs + 0.7 = 10,000IB + 500(151)IB = 85,500IB But, v0 = 500IE = 500x151IB = 4 which leads to IB = 5.298x10-5 Therefore, vs = 0.7 + 85,500IB = 5.23 volts
Chapter 3, Solution 91 We first determine the Thevenin equivalent for the input circuit.
RTh = 6||2 = 6x2/8 = 1.5 kΩ and VTh = 2(3)/(2+6) = 0.75 volts 5 kΩ
IC
1.5 kΩ
IE
IB
+ VBE
–
+ VCE –
i2
i1 -+
9 V -+
400 Ω + V0 –
0.75 V For loop 1, -0.75 + 1.5kIB + VBE + 400IE = 0 = -0.75 + 0.7 + 1500IB + 400(1 + β)IB IB = 0.05/81,900 = 0.61 µA v0 = 400IE = 400(1 + β)IB = 49 mV For loop 2, -400IE – VCE – 5kIC + 9 = 0, but, IC = βIB and IE = (1 + β)IB VCE = 9 – 5kβIB – 400(1 + β)IB = 9 – 0.659 = 8.641 volts Chapter 3, Solution 92
IC
VC
10 kΩ
IE
IB
+ VBE
–
+ VCE –
12V -+
4 kΩ + V0 –
I1 5 kΩ
I1 = IB + IC = (1 + β)IB and IE = IB + IC = I1 Applying KVL around the outer loop, 4kIE + VBE + 10kIB + 5kI1 = 12 12 – 0.7 = 5k(1 + β)IB + 10kIB + 4k(1 + β)IB = 919kIB IB = 11.3/919k = 12.296 µA Also, 12 = 5kI1 + VC which leads to VC = 12 – 5k(101)IB = 5.791 volts Chapter 3, Solution 93
i1 i2i
4 Ω
4 Ω
2 Ω
2 Ω
1 Ω
8 Ω
3v0
+ v2 –
+ v1 –
+
–+
3v0 +
2 Ω i v2v1 i3
+ v0 –
24V
(a) (b) From (b), -v1 + 2i – 3v0 + v2 = 0 which leads to i = (v1 + 3v0 – v2)/2 At node 1 in (a), ((24 – v1)/4) = (v1/2) + ((v1 +3v0 – v2)/2) + ((v1 – v2)/1), where v0 = v2 or 24 = 9v1 which leads to v1 = 2.667 volts At node 2, ((v1 – v2)/1) + ((v1 + 3v0 – v2)/2) = (v2/8) + v2/4, v0 = v2 v2 = 4v1 = 10.66 volts Now we can solve for the currents, i1 = v1/2 = 1.333 A, i2 = 1.333 A, and i3 = 2.6667 A.
Chapter 4, Solution 1.
i io5 Ω
8 Ω
1 Ω
+ −
1 V 3 Ω
Ω=+ 4)35(8 , 51
411i =+
=
===101i
21io 0.1A
Chapter 4, Solution 2.
,3)24(6 Ω=+ A21i21 ==i
,41i
21i 1o == == oo i2v 0.5V
5 Ω 4 Ω
i2
8 Ω
i1 io
6 Ω
1 A 2 Ω If is = 1µA, then vo = 0.5µV Chapter 4, Solution 3.
R
+
vo
−
3R io
3R
3R
R
+ −
3R
+ − 1 V
1.5R Vs
(b)(a)
(a) We transform the Y sub-circuit to the equivalent ∆ .
,R43
R4R3R3R
2
== R23R
43R
43
=+
2v
v so = independent of R
io = vo/(R) When vs = 1V, vo = 0.5V, io = 0.5A (b) When vs = 10V, vo = 5V, io = 5A (c) When vs = 10V and R = 10Ω,
vo = 5V, io = 10/(10) = 500mA Chapter 4, Solution 4. If Io = 1, the voltage across the 6Ω resistor is 6V so that the current through the 3Ω resistor is 2A.
+
v1
−
3A
Is 2 Ω 4 Ω
i1 3A 1A
Is
2A
6 Ω 4 Ω3 Ω
2 Ω 2 Ω
(a) (b)
Ω= 263 , vo = 3(4) = 12V, .A34
vo1 ==i
Hence Is = 3 + 3 = 6A If Is = 6A Io = 1 Is = 9A Io = 6/(9) = 0.6667A
Chapter 4, Solution 5.
If vo = 1V, V2131V1 =+
=
3
10v322V 1s =+
=
If vs = 3
10 vo = 1
Then vs = 15 vo = =15x103 4.5V
vo3 Ω2 Ω
+ − 6 Ω 6 Ω 6 Ω
v1
Vs
Chapter 4, Solution 6
Let sT
ToT V
RRRV
RRRR
RRR132
3232 then ,//
+=
+==
133221
32
132
32
32
32
1 RRRRRRRR
RRRRRRRRR
RRR
VV
kT
T
s
o
++=
++
+=
+==
Chapter 4, Solution 7 We find the Thevenin equivalent across the 10-ohm resistor. To find VTh, consider the circuit below.
3Vx
5Ω 5Ω + + 4V 15Ω VTh - 6Ω - + Vx -
From the figure,
V3)4(515
15,0 =+
== Thx VV
To find RTh, consider the circuit below:
3Vx
5Ω 5Ω V1 V2 + 4V 15Ω 1A - 6Ω + Vx - At node 1,
122111 73258616,
5153
54 VVxVVVVVV
xx −=→==−
++=− (1)
At node 2,
9505
31 2121 −=→=
−++ VV
VVVx (2)
Solving (1) and (2) leads to V2 = 101.75 V
mW 11.2275.1014
94
,75.1011
2
max2 ===Ω==
xRV
pVRTh
ThTh
Chapter 4, Solution 8. Let i = i1 + i2, where i1 and iL are due to current and voltage sources respectively.
6 Ω
i1 + − 20V
i2
5 A 4 Ω 6 Ω
4 Ω (a) (b)
i1 = ,A3)5(46
6=
+ A2
4620
2 =+
=i
Thus i = i1 + i2 = 3 + 2 = 5A Chapter 4, Solution 9. Let i
2x1xx ii += where i is due to 15V source and i is due to 4A source,
1x 2x
12Ω
-4A
40Ω 10Ωix2
40Ω
i
10 Ω
ix1
12 Ω
+ −
15V
(a) (b)
For ix1, consider Fig. (a).
10||40 = 400/50 = 8 ohms, i = 15/(12 + 8) = 0.75
ix1 = [40/(40 + 10)]i = (4/5)0.75 = 0.6 For ix2, consider Fig. (b).
12||40 = 480/52 = 120/13
ix2 = [(120/13)/((120/13) + 10)](-4) = -1.92
ix = 0.6 – 1.92 = -1.32 A
p = vix = ix2R = (-1.32)210 = 17.43 watts
Chapter 4, Solution 10. Let vab = vab1 + vab2 where vab1 and vab2 are due to the 4-V and the 2-A sources respectively.
+
vab2
−
10 Ω+ −
3vab2
2 A
10 Ω
+ −
+
vab1
−
+ −
3vab1
4V
(a) (b) For vab1, consider Fig. (a). Applying KVL gives,
- vab1 – 3 vab1 + 10x0 + 4 = 0, which leads to vab1 = 1 V For vab2, consider Fig. (b). Applying KVL gives,
- vab2 – 3vab2 + 10x2 = 0, which leads to vab2 = 5
vab = 1 + 5 = 6 V
Chapter 4, Solution 11. Let i = i1 + i2, where i1 is due to the 12-V source and i2 is due to the 4-A source.
12V
4A
2Ω 2Ωix2
6Ω
4A
3Ω2Ω i2
3Ω
io
(a)
2 Ω
i1
6 Ω
+ −
(b) For i1, consider Fig. (a).
2||3 = 2x3/5 = 6/5, io = 12/(6 + 6/5) = 10/6
i1 = [3/(2 + 3)]io = (3/5)x(10/6) = 1 A For i2, consider Fig. (b), 6||3 = 2 ohm, i2 = 4/2 = 2 A
i = 1 + 2 = 3 A Chapter 4, Solution 12. Let vo = vo1 + vo2 + vo3, where vo1, vo2, and vo3 are due to the 2-A, 12-V, and 19-V sources respectively. For vo1, consider the circuit below.
5 Ω
5 Ω
+ vo1 −
io
2A2A
3Ω
4 Ω
6Ω 12 Ω
5 Ω
+ vo1 −
6||3 = 2 ohms, 4||12 = 3 ohms. Hence,
io = 2/2 = 1, vo1 = 5io = 5 V
For vo2, consider the circuit below. 6 Ω 5 Ω 4 Ω 6 Ω 5 Ω
+ vo2 −
3 Ω3 Ω +
v1
−
+ − 12V
+ vo2 −
12 Ω
+ −
3 Ω
12V
3||8 = 24/11, v1 = [(24/11)/(6 + 24/11)]12 = 16/5
vo2 = (5/8)v1 = (5/8)(16/5) = 2 V For vo3, consider the circuit shown below. 4 Ω 5 Ω 5 Ω 4 Ω
2Ω+ vo3 −
12 Ω+
v2
−
+ − 19V
+ −
19V 6Ω + vo3 −
12 Ω 3 Ω
7||12 = (84/19) ohms, v2 = [(84/19)/(4 + 84/19)]19 = 9.975
v = (-5/7)v2 = -7.125
vo = 5 + 2 – 7.125 = -125 mV
,
Chapter 4, Solution 13 Let iiii 321o ++= where i1, i2, and i3 are the contributions to io due to 30-V, 15-V, and 6-mA sources respectively. For i1, consider the circuit below.
1 kΩ 2 kΩ 3 kΩ + i1 30V - 4 kΩ 5 kΩ
3//5 = 15/8 = 1.875 kohm, 2 + 3//5 = 3.875 kohm, 1//3.875 = 3.875/4.875 = 0.7949 kohm. After combining the resistors except the 4-kohm resistor and transforming the voltage source, we obtain the circuit below. i1 30 mA 4 kΩ 0.7949 kΩ Using current division,
mA 4.973mA)30(7949.47949.0
1 ==i
For i2, consider the circuit below. 1 kΩ 2 kΩ 3 kΩ i2 - 15V 4 kΩ 5 kΩ +
After successive source transformation and resistance combinations, we obtain the circuit below: 2.42mA i2 4 kΩ 0.7949 kΩ
Using current division,
mA 4012.0mA)42.2(7949.47949.0
2 −=−=i
For i3, consider the circuit below.
6mA 1 kΩ 2 kΩ 3 kΩ i3 4 kΩ 5 kΩ
After successive source transformation and resistance combinations, we obtain the circuit below: 3.097mA i3 4 kΩ 0.7949 kΩ
mA 5134.0mA)097.3(7949.47949.0
3 −=−=i
Thus, mA 058.4321 =++= iiiio Chapter 4, Solution 14. Let vo = vo1 + vo2 + vo3, where vo1, vo2 , and vo3, are due to the 20-V, 1-A, and 2-A sources respectively. For vo1, consider the circuit below. 6 Ω
20V
+ −
+
vo1
−
4 Ω 2 Ω
3 Ω
6||(4 + 2) = 3 ohms, vo1 = (½)20 = 10 V
For vo2, consider the circuit below. 6 Ω 6 Ω
1A
2 Ω 4 Ω
+
vo2
−
4V
− + 2 Ω 4 Ω
3 Ω
+
vo2
−
3 Ω
3||6 = 2 ohms, vo2 = [2/(4 + 2 + 2)]4 = 1 V For vo3, consider the circuit below. 6 Ω
3 Ω
− vo3 +
3 Ω
2A
2 Ω 4 Ω
3 Ω
+
vo3
−
2A
6||(4 + 2) = 3, vo3 = (-1)3 = -3 vo = 10 + 1 – 3 = 8 V
Chapter 4, Solution 15. Let i = i1 + i2 + i3, where i1 , i2 , and i3 are due to the 20-V, 2-A, and 16-V sources. For i1, consider the circuit below.
io
4Ω
3Ω
i1
1 Ω
+ −
20V 2 Ω
4||(3 + 1) = 2 ohms, Then io = [20/(2 + 2)] = 5 A, i1 = io/2 = 2.5 A For i3, consider the circuit below.
2||(1 + 3) = 4/3, vo’ = [(4/3)/((4/3) + 4)](-16) = -4
i3 = vo’/4 = -1 For i2, consider the circuit below.
3Ω i2
1 Ω (4/3)Ω
3Ωi2
1 Ω 2A
4Ω
+
vo’
−
4Ω
3Ω
i3
1 Ω2 Ω
− +
16V
2A 2 Ω
2||4 = 4/3, 3 + 4/3 = 13/3 Using the current division principle.
i2 = [1/(1 + 13/2)]2 = 3/8 = 0.375
i = 2.5 + 0.375 - 1 = 1.875 A
p = i2R = (1.875)23 = 10.55 watts
Chapter 4, Solution 16. Let io = io1 + io2 + io3, where io1, io2, and io3 are due to the 12-V, 4-A, and 2-A sources. For io1, consider the circuit below.
5Ω
io1
10 Ω
4 Ω
+ −
3 Ω 2 Ω
12V
10||(3 + 2 + 5) = 5 ohms, io1 = 12/(5 + 4) = (12/9) A For io2, consider the circuit below.
io2
i1
4A
10Ω
2 Ω
4Ω
3 Ω
5 Ω
2 + 5 + 4||10 = 7 + 40/14 = 69/7 i1 = [3/(3 + 69/7)]4 = 84/90, io2 =[-10/(4 + 10)]i1 = -6/9
For io3, consider the circuit below.
i2
2 A
io3
10 Ω 5 Ω 4 Ω
3 Ω 2 Ω
3 + 2 + 4||10 = 5 + 20/7 = 55/7
i2 = [5/(5 + 55/7)]2 = 7/9, io3 = [-10/(10 + 4)]i2 = -5/9
io = (12/9) – (6/9) – (5/9) = 1/9 = 111.11 mA
Chapter 4, Solution 17. Let vx = vx1 + vx2 + vx3, where vx1,vx2, and vx3 are due to the 90-V, 6-A, and 40-V sources. For vx1, consider the circuit below.
30 Ω 10 Ω 20 Ω
12 Ω
+ − vx1
10 Ω
20 Ω 3 A
io
30 Ω
+ − vx1 60 Ω
+ −
90V
20||30 = 12 ohms, 60||30 = 20 ohms By using current division,
io = [20/(22 + 20)]3 = 60/42, vx1 = 10io = 600/42 = 14.286 V For vx2, consider the circuit below.
io’
20 Ω30 Ω
io’
6A 60 Ω 30 Ω
10 Ω
+ − vx2 + vx2 −
6A 20 Ω 12 Ω
10 Ω
io’ = [12/(12 + 30)]6 = 72/42, vx2 = -10io’ = -17.143 V
For vx3, consider the circuit below.
20 Ω
+ − vx3 7.5Ω
4A
io”
+ − 40V30 Ω 30 Ω
+ − vx3 60 Ω
10 Ω10 Ω 10 Ω
io” = [12/(12 + 30)]2 = 24/42, vx3 = -10io” = -5.714
vx = 14.286 – 17.143 – 5.714 = -8.571 V
Chapter 4, Solution 18. Let ix = i1 + i2, where i1 and i2 are due to the 10-V and 2-A sources respectively. To obtain i1, consider the circuit below. 2 Ω
+ −
10i1i1
2 Ω
+ − 10V
1 Ωi1
4 Ω5i1
1 Ω
+ −
4 Ω10V
-10 + 10i1 + 7i1 = 0, therefore i1 = (10/17) A For i2, consider the circuit below.
io+ −
10i2 2 Ω
+ − 2V
1 Ωio i2 1 Ω
4 Ω
2 Ω
10i2 2 A
+ − 4 Ω
-2 + 10i2 + 7io = 0, but i2 + 2 = io. Hence,
-2 + 10i2 +7i2 + 14 = 0, or i2 = (-12/17) A
vx = 1xix = 1(i1 + i2) = (10/17) – (12/17) = -2/17 = -117.6 mA Chapter 4, Solution 19. Let vx = v1 + v2, where v1 and v2 are due to the 4-A and 6-A sources respectively. ix v1 ix v2
+
v2
−
+
v1
−
8 Ω 2 Ω
4ix
6 A
− +
8Ω2 Ω
4ix
4 A
− +
(a) (b)
To find v1, consider the circuit in Fig. (a).
v1/8 = 4 + (-4ix – v1)/2 But, -ix = (-4ix – v1)/2 and we have -2ix = v1. Thus,
v1/8 = 4 + (2v1 – v1)/8, which leads to v1 = -32/3 To find v2, consider the circuit shown in Fig. (b).
v2/2 = 6 + (4ix – v2)/8 But ix = v2/2 and 2ix = v2. Therefore,
v2/2 = 6 + (2v2 – v2)/8 which leads to v2 = -16 Hence, vx = –(32/3) – 16 = -26.67 V Chapter 4, Solution 20. Transform the voltage sources and obtain the circuit in Fig. (a). Combining the 6-ohm and 3-ohm resistors produces a 2-ohm resistor (6||3 = 2). Combining the 2-A and 4-A sources gives a 6-A source. This leads to the circuit shown in Fig. (b).
i
6Ω 4A3Ω2Ω i
6A 2 Ω 2 Ω
2A (a) (b) From Fig. (b), i = 6/2 = 3 A Chapter 4, Solution 21. To get io, transform the current sources as shown in Fig. (a). 6 Ω
+
vo
−
2 A 3 Ωi
6 Ω2 A
+ − 6V
io 3 Ω
+ − 12V
(a) (b)
From Fig. (a), -12 + 9io + 6 = 0, therefore io = 666.7 mA To get vo, transform the voltage sources as shown in Fig. (b).
i = [6/(3 + 6)](2 + 2) = 8/3
vo = 3i = 8 V Chapter 4, Solution 22. We transform the two sources to get the circuit shown in Fig. (a). 5 Ω 5 Ω
10V
4Ω 10Ω 2A
i 10Ω1A
2A
(a)
10Ω4 Ω
− +
(b) We now transform only the voltage source to obtain the circuit in Fig. (b). 10||10 = 5 ohms, i = [5/(5 + 4)](2 – 1) = 5/9 = 555.5 mA
Chapter 4, Solution 23 If we transform the voltage source, we obtain the circuit below. 8Ω 10Ω 6Ω 3Ω 5A 3A 3//6 = 2-ohm. Convert the current sources to voltages sources as shown below. 10 8Ω Ω 2Ω + + 10V 30V -
- Applying KVL to the loop gives
A 10)2810(1030 =→=++++− II W82 === RIVIp
Chapter 4, Solution 24 Convert the current source to voltage source.
16Ω 1Ω
4Ω + 5Ω + 48 V 10Ω Vo
- + -
12 V - Combine the 16-ohm and 4-ohm resistors and convert both voltages sources to current Sources. We obtain the circuit below. 1Ω 2.4A 20 5Ω Ω 2.4A 10Ω Combine the resistors and current sources. 20//5 = (20x5)/25 = 4 , 2.4 + 2.4 = 4.8 A ΩConvert the current source to voltage source. We obtain the circuit below. 4Ω 1Ω + + 19.2V Vo 10Ω - - Using voltage division,
8.12)2.19(1410
10=
++=oV V
Chapter 4, Solution 25. Transforming only the current source gives the circuit below.
12V
30 V
5 Ω
9 Ω 18 V
+ − vo 4 Ω
2 Ω
i
+ −
− +
− +
+ −
30 V Applying KVL to the loop gives,
(4 + 9 + 5 + 2)i – 12 – 18 – 30 – 30 = 0
20i = 90 which leads to i = 4.5
vo = 2i = 9 V Chapter 4, Solution 26.
Transform the voltage sources to current sources. The result is shown in Fig. (a),
30||60 = 20 ohms, 30||20 = 12 ohms 10 Ω
+ + vx −
12 Ω10 Ω
− 96V i
20 Ω
+ − 60V
20Ω3A 2A
+ vx −
60Ω 30Ω6A
(a)
30Ω
(b)
Combining the resistors and transforming the current sources to voltage sources, we obtain the circuit in Fig. (b). Applying KVL to Fig. (b),
42i – 60 + 96 = 0, which leads to i = -36/42
vx = 10i = -8.571 V Chapter 4, Solution 27. Transforming the voltage sources to current sources gives the circuit in Fig. (a).
10||40 = 8 ohms Transforming the current sources to voltage sources yields the circuit in Fig. (b). Applying KVL to the loop,
-40 + (8 + 12 + 20)i + 200 = 0 leads to i = -4
vx 12i = -48 V
12 Ω
+ vx − 10Ω 20Ω40Ω5A 8A 2A
(a) 8 Ω 12 Ω 20 Ω
+ −
+ −
+ vx − 40V 200V i
(b)
Chapter 4, Solution 28. Transforming only the current sources leads to Fig. (a). Continuing with source transformations finally produces the circuit in Fig. (d).
3 Ωio 10 V
+ −
12 V
+ −
+ −
4 Ω 5 Ω2 Ω
10Ω
12 V (a) 4 Ω
5 Ω io
+ − 12V
4 Ω
+ − 11V io
io
+ − 12V 10Ω
4 Ω
2.2A10Ω
io
+ − 22 V
(b)
+ − 12V 10Ω
10 Ω
(c) (d) Applying KVL to the loop in fig. (d),
-12 + 9io + 11 = 0, produces io = 1/9 = 111.11 mA
Chapter 4, Solution 29. Transform the dependent voltage source to a current source as shown in Fig. (a). 2||4 = (4/3) k ohms
4 kΩ
It is clear that i = 3 mA which leads to vo = 1000i = 3 V If the use of source transformations was not required for this problem, the actual answer could have been determined by inspection right away since the only current that could have flowed through the 1 k ohm resistor is 3 mA. Chapter 4, Solution 30 Transform the dependent current source as shown below.
ix 24Ω 60Ω 10Ω + + 12V 30Ω 7ix - -
+
vo
−
1 kΩ +
vo
−
3 mA
2vo (4/3) kΩ
(b)
i
− +
3 mA
1.5vo
1 kΩ
2 kΩ
(a)
Combine the 60-ohm with the 10-ohm and transform the dependent source as shown below.
ix 24Ω + 12V 30Ω 70Ω 0.1ix - Combining 30-ohm and 70-ohm gives 30//70 = 70x30/100 = 21-ohm. Transform the dependent current source as shown below.
ix 24Ω 21Ω + + 12V 2.1ix - - Applying KVL to the loop gives
mA 8.2541.47
1201.21245 ==→=+− xxx iii
Chapter 4, Solution 31. Transform the dependent source so that we have the circuit in Fig. (a). 6||8 = (24/7) ohms. Transform the dependent source again to get the circuit in Fig. (b). 3 Ω
6 Ω
+ − vx + −
8 Ω vx/3 12V (a)
(24/7) Ω3 Ω
i +
+ − vx + −
(8/7)vx 12V
(b)
From Fig. (b),
vx = 3i, or i = vx/3. Applying KVL,
-12 + (3 + 24/7)i + (24/21)vx = 0
12 = [(21 + 24)/7]vx/3 + (8/7)vx, leads to vx = 84/23 = 3.625 V Chapter 4, Solution 32. As shown in Fig. (a), we transform the dependent current source to a voltage source,
15 Ω 10 Ω− +
50 Ω
+ −
5ix
40 Ω 60V
(a)
15 Ω
ix
50 Ω 50 Ω
(b)
+ −
ix 25 Ω
+ − 60V 2.5ix
15 Ω
−
60V 0.1ix
(c)
In Fig. (b), 50||50 = 25 ohms. Applying KVL in Fig. (c), -60 + 40ix – 2.5ix = 0, or ix = 1.6 A Chapter 4, Solution 33.
(a) RTh = 10||40 = 400/50 = 8 ohms
VTh = (40/(40 + 10))20 = 16 V (b) RTh = 30||60 = 1800/90 = 20 ohms
2 + (30 – v1)/60 = v1/30, and v1 = VTh
120 + 30 – v1 = 2v1, or v1 = 50 V
VTh = 50 V
Chapter 4, Solution 34. To find RTh, consider the circuit in Fig. (a).
3 A
RTh = 20 + 10||40 = 20 + 400/50 = 28 ohms (b)
v2
+
VTh
20 Ω
40 Ω
10 Ω v1
+ − 40VRTh
20 Ω
40 Ω
10 Ω
(a)
To find VTh, consider the circuit in Fig. (b). At node 1, (40 – v1)/10 = 3 + [(v1 – v2)/20] + v1/40, 40 = 7v1 – 2v2 (1) At node 2, 3 + (v1- v2)/20 = 0, or v1 = v2 – 60 (2) Solving (1) and (2), v1 = 32 V, v2 = 92 V, and VTh = v2 = 92 V
Chapter 4, Solution 35. To find RTh, consider the circuit in Fig. (a).
RTh = Rab = 6||3 + 12||4 = 2 + 3 =5 ohms To find VTh, consider the circuit shown in Fig. (b).
RTh At node 1, 2 + (12 – v1)/6 = v1/3, or v1 = 8 At node 2, (19 – v2)/4 = 2 + v2/12, or v2 = 33/4 But, -v1 + VTh + v2 = 0, or VTh = v1 – v2 = 8 – 33/4 = -0.25
b
RTh = 5 Ω
10 Ω
+ − vo a
+ −
4 Ω
+ −
+
v2
−
+
v1
−
12Ω
v2+ VTh
3 Ω
2 A
v1
+ − 12V
6 Ω
(b)
a b
4 Ω12 Ω6 Ω 3 Ω
(a)
19V
VTh = (-1/4)V
vo = VTh/2 = -0.25/2 = -125 mV
Chapter 4, Solution 36. Remove the 30-V voltage source and the 20-ohm resistor.
a
b
10Ω +
VTh
+ − 50V
a
b
40Ω
RTh
10Ω
40Ω
(a) (b) From Fig. (a), RTh = 10||40 = 8 ohms From Fig. (b), VTh = (40/(10 + 40))50 = 40V
+ −
a
b
12 Ω
(c)
i
+ − 40V
8 Ω
30V The equivalent circuit of the original circuit is shown in Fig. (c). Applying KVL, 30 – 40 + (8 + 12)i = 0, which leads to i = 500mA
Chapter 4, Solution 37 RN is found from the circuit below.
20 Ω a 40Ω 12Ω b
Ω=+= 10)4020//(12NR IN is found from the circuit below.
2A
20 Ω a + 40Ω 120V 12Ω - IN b Applying source transformation to the current source yields the circuit below.
20Ω 40Ω + 80 V - + 120V IN - Applying KVL to the loop yields
A 6667.060/4006080120 ==→=++− NN II
Chapter 4, Solution 38 We find Thevenin equivalent at the terminals of the 10-ohm resistor. For RTh, consider the circuit below. 1Ω
4Ω 5Ω RTh 16 Ω
Ω=+=++= 541)164//(51ThR For VTh, consider the circuit below. 1Ω
V1 4Ω V2 5Ω + 3A 16Ω VTh
+ -
12 V - At node 1,
21211 4548
4163 VV
VVV−=→
−+= (1)
At node 2,
21221 95480
512
4VVVVV
+−=→=−
+− (2)
Solving (1) and (2) leads to 2.192 ==VVTh
Thus, the given circuit can be replaced as shown below. 5 Ω + + 19.2V Vo 10Ω - - Using voltage division,
8.12)2.19(510
10=
+=oV V
Chapter 4, Solution 39. To find RTh, consider the circuit in Fig. (a).
- 1 – 3 + 10io = 0, or io = 0.4
RTh = 1/io = 2.5 ohms To find VTh, consider the circuit shown in Fig. (b).
[(4 – v)/10] + 2 = 0, or v = 24 But, v = VTh + 3vab = 4VTh = 24, which leads to VTh = 6 V Chapter 4, Solution 40. To find RTh, consider the circuit in Fig. (a).
40V
a 10 Ω
(a)
+ −
1V50V
+
v1
−
+ −
3vab 10 Ω io
b
a
(b)
+ −
440 ΩV
+ −
3vab 20 Ω
2A
+ −
v
40Ω
RTh
a b 10 Ω 20 Ω
+
v2
−
+ VTh
8 A
+ −
(b)
10 Ω
+
vab = VTh
−
b
(a)
RTh = 10||40 + 20 = 28 ohms To get VTh, consider the circuit in Fig. (b). The two loops are independent. From loop 1,
v1 = (40/50)50 = 40 V For loop 2, -v2 + 20x8 + 40 = 0, or v2 = 200 But, VTh + v2 – v1 = 0, VTh = v1 = v2 = 40 – 200 = -160 volts This results in the following equivalent circuit. 28 Ω
vx = [12/(12 + 28)](-160) = -48 V Chapter 4, Solution 41 To find RTh, consider the circuit below 14Ω a 6Ω 5 Ω b
+ −
+
vx
−
12 Ω-160V
NTh RR =Ω=+= 4)614//(5 Applying source transformation to the 1-A current source, we obtain the circuit below.
6Ω - 14V + 14Ω VTh a + 6V 3A 5Ω - b At node a,
V 85
3146
614−=→+=
+−+
ThThTh VVV
A 24/)8( −=−==Th
ThN R
VI
Thus, A 2 V,8,4 −=−=Ω== NThNTh IVRR
Chapter 4, Solution 42. To find RTh, consider the circuit in Fig. (a). 20 Ω 30 Ω
10 Ω
20 Ω 10 Ω
10Ω 10 Ω
a
10 Ω10 Ω
30 Ω 30 Ωb
ba (a) (b) 20||20 = 10 ohms. Transform the wye sub-network to a delta as shown in Fig. (b). 10||30 = 7.5 ohms. RTh = Rab = 30||(7.5 + 7.5) = 10 ohms. To find VTh, we transform the 20-V and the 5-V sources. We obtain the circuit shown in Fig. (c).
+ − 50V
10 Ω
10 Ω10 Ω
10 Ω − +
10 V
a 10 Ω+ b
+ − 30V
i2i1
(c) For loop 1, -30 + 50 + 30i1 – 10i2 = 0, or -2 = 3i1 – i2 (1) For loop 2, -50 – 10 + 30i2 – 10i1 = 0, or 6 = -i1 + 3i2 (2) Solving (1) and (2), i1 = 0, i2 = 2 A Applying KVL to the output loop, -vab – 10i1 + 30 – 10i2 = 0, vab = 10 V
VTh = vab = 10 volts Chapter 4, Solution 43. To find RTh, consider the circuit in Fig. (a).
RTh
a b
5 Ω+
vb
−
+
va
−
+ VTh
10 Ω
+ − 50V
10 Ω
5 Ω10Ω
a b
10Ω
(a)
2 A
(b)
RTh = 10||10 + 5 = 10 ohms
To find VTh, consider the circuit in Fig. (b).
vb = 2x5 = 10 V, va = 20/2 = 10 V But, -va + VTh + vb = 0, or VTh = va – vb = 0 volts Chapter 4, Solution 44. (a) For RTh, consider the circuit in Fig. (a).
RTh = 1 + 4||(3 + 2 + 5) = 3.857 ohms For VTh, consider the circuit in Fig. (b). Applying KVL gives,
10 – 24 + i(3 + 4 + 5 + 2), or i = 1
VTh = 4i = 4 V 3Ω 1Ω
+ a
+ −
VTh3Ω 1Ω a 4 Ω 24V
+ −
b RTh4 Ω 10V 2 Ω b 2 Ω
i 5 Ω 5 Ω
(b) (a) (b) For RTh, consider the circuit in Fig. (c).
3Ω 1Ω 3Ω 1Ω
+ − 24V
2 Ω
2A
c
b
5 Ω
+
VTh
vo4 Ω
2 Ω
5 Ω
4 Ω
RTh
c
b
(c) (d)
RTh = 5||(2 + 3 + 4) = 3.214 ohms To get VTh, consider the circuit in Fig. (d). At the node, KCL gives,
[(24 – vo)/9] + 2 = vo/5, or vo = 15
VTh = vo = 15 V Chapter 4, Solution 45. For RN, consider the circuit in Fig. (a).
6 Ω 6 Ω
RN6 Ω 4 Ω 4A 6 Ω 4 Ω
IN
(a) (b)
RN = (6 + 6)||4 = 3 ohms For IN, consider the circuit in Fig. (b). The 4-ohm resistor is shorted so that 4-A current is equally divided between the two 6-ohm resistors. Hence, IN = 4/2 = 2 A Chapter 4, Solution 46. (a) RN = RTh = 8 ohms. To find IN, consider the circuit in Fig. (a). 10 Ω 60 Ω
4 Ω
+ − 30V 2A IN
30 Ω
+ −
Isc
20V
(a) (b) IN = Isc = 20/10 = 2 A
(b) To get IN, consider the circuit in Fig. (b). IN = Isc = 2 + 30/60 = 2.5 A
Chapter 4, Solution 47 Since VTh = Vab = Vx, we apply KCL at the node a and obtain
V 19.1126/15026012
30==→+=
−ThTh
ThTh VVVV
To find RTh, consider the circuit below.
12Ω Vx a 2Vx 60Ω
1A
At node a, KCL gives
4762.0126/601260
21 ==→++= xxx
x VVV
V
5.24762.0/19.1,4762.01
===Ω==Th
ThN
xTh R
VI
VR
Thus, A 5.2,4762.0,19.1 =Ω=== NNThTh IRRVV
Chapter 4, Solution 48. To get RTh, consider the circuit in Fig. (a).
+
VTh
−
10Io
+ −
Io
2A4 Ω
2 Ω
Io
2 Ω
4 Ω
+ − +
V
−
10Io
1A
(a) (b) From Fig. (a), Io = 1, 6 – 10 – V = 0, or V = -4
RN = RTh = V/1 = -4 ohms
To get VTh, consider the circuit in Fig. (b),
Io = 2, VTh = -10Io + 4Io = -12 V
IN = VTh/RTh = 3A Chapter 4, Solution 49. RN = RTh = 28 ohms To find IN, consider the circuit below,
3A
vo
io
20 Ω
+ −
40 Ω
10 Ω
Isc = IN 40V At the node, (40 – vo)/10 = 3 + (vo/40) + (vo/20), or vo = 40/7 io = vo/20 = 2/7, but IN = Isc = io + 3 = 3.286 A Chapter 4, Solution 50. From Fig. (a), RN = 6 + 4 = 10 ohms
6 Ω 6 Ω
4 Ω 12V
+ −
4 Ω2A
Isc = IN
(b) (a) From Fig. (b), 2 + (12 – v)/6 = v/4, or v = 9.6 V
-IN = (12 – v)/6 = 0.4, which leads to IN = -0.4 A Combining the Norton equivalent with the right-hand side of the original circuit produces the circuit in Fig. (c).
i
4A5 Ω
0.4A 10 Ω
(c) i = [10/(10 + 5)] (4 – 0.4) = 2.4 A Chapter 4, Solution 51. (a) From the circuit in Fig. (a),
RN = 4||(2 + 6||3) = 4||4 = 2 ohms
2 Ω
+ − 120V
+
6A
VTh
4 Ω
3 Ω
6 Ω
RTh
4 Ω
3 Ω
6 Ω
2 Ω (a) (b) For IN or VTh, consider the circuit in Fig. (b). After some source transformations, the circuit becomes that shown in Fig. (c).
i
2 Ω
+ − 12V
+ −
+ VTh
4 Ω2 Ω
40V (c) Applying KVL to the circuit in Fig. (c),
-40 + 8i + 12 = 0 which gives i = 7/2
VTh = 4i = 14 therefore IN = VTh/RN = 14/2 = 7 A
(b) To get RN, consider the circuit in Fig. (d).
RN = 2||(4 + 6||3) = 2||6 = 1.5 ohms
i
2 Ω
+ − 12V
+VTh
RN
4 Ω
3 Ω 2 Ω
6 Ω (d) (e) To get IN, the circuit in Fig. (c) applies except that it needs slight modification as in Fig. (e).
i = 7/2, VTh = 12 + 2i = 19, IN = VTh/RN = 19/1.5 = 12.667 A Chapter 4, Solution 52. For RTh, consider the circuit in Fig. (a). a
Io
b
2 kΩ3 kΩ 20Io
RTh (a)
+ + − Io
a
b
2 kΩ VTh 20Io
3 kΩ 6V (b) For Fig. (a), Io = 0, hence the current source is inactive and
RTh = 2 k ohms
For VTh, consider the circuit in Fig. (b).
Io = 6/3k = 2 mA
VTh = (-20Io)(2k) = -20x2x10-3x2x103 = -80 V Chapter 4, Solution 53. To get RTh, consider the circuit in Fig. (a). 0.25vo0.25vo
1/2 a 2 Ω
1A
+
vo
−
2 Ω
(b) b
+
vab
−
1/2
a
b
2 Ω
1A
+
vo
−
3 Ω
(a)
6 Ω From Fig. (b),
vo = 2x1 = 2V, -vab + 2x(1/2) +vo = 0
vab = 3V
RN = vab/1 = 3 ohms To get IN, consider the circuit in Fig. (c). 0.25vo
6 Ω a 2 Ω
+
vo
−
3 Ω
+ −
18V Isc = IN b (c)
[(18 – vo)/6] + 0.25vo = (vo/2) + (vo/3) or vo = 4V
But, (vo/2) = 0.25vo + IN, which leads to IN = 1 A
Chapter 4, Solution 54
To find VTh =Vx, consider the left loop.
xoxo ViVi 2100030210003 +=→=++− (1) For the right loop, (2) oox iixV 20004050 −=−=Combining (1) and (2), mA13000400010003 −=→−=−= oooo iiii 222000 =→=−= Thox ViV To find RTh, insert a 1-V source at terminals a-b and remove the 3-V independent source, as shown below.
1 k iΩ x .
io + + + 2Vx 40io Vx 50 Ω 1V
- - -
mA210002
,1 −=−== xox
ViV
-60mAA501mA80
5040 =+−=+= x
oxV
ii
Ω−=−== 67.16060.0/11
xTh iR
Chapter 4, Solution 55. To get RN, apply a 1 mA source at the terminals a and b as shown in Fig. (a).
80I+ −
vab/1000 +
vab
−
8 kΩ
a I
50 kΩ
1mA
b (a)
We assume all resistances are in k ohms, all currents in mA, and all voltages in volts. At node a,
(vab/50) + 80I = 1 (1) Also,
-8I = (vab/1000), or I = -vab/8000 (2) From (1) and (2), (vab/50) – (80vab/8000) = 1, or vab = 100
RN = vab/1 = 100 k ohms To get IN, consider the circuit in Fig. (b).
vab/1000
I
80I
a
50 kΩ
+
vab
−
+ −
+ −
8 kΩ
IN 2V b (b) Since the 50-k ohm resistor is shorted,
IN = -80I, vab = 0 Hence, 8i = 2 which leads to I = (1/4) mA
IN = -20 mA Chapter 4, Solution 56. We first need RN and IN.
16V
2A 4 Ω
2 Ω IN
1 Ω
− +
+ − 20V
4 Ω2 Ω RN
a
b
1 Ω
(a) (b)
To find RN, consider the circuit in Fig. (a).
RN = 1 + 2||4 = (7/3) ohms To get IN, short-circuit ab and find Isc from the circuit in Fig. (b). The current source can be transformed to a voltage source as shown in Fig. (c).
vo
i
RN
a
IN
1 Ω
− + 16V 2V
4 Ω
2 Ω IN − +
+ −
20V
3 Ω b (c) (d)
(20 – vo)/2 = [(vo + 2)/1] + [(vo + 16)/4], or vo = 16/7
IN = (vo + 2)/1 = 30/7 From the Norton equivalent circuit in Fig. (d),
i = RN/(RN + 3), IN = [(7/3)/((7/3) + 3)](30/7) = 30/16 = 1.875 A Chapter 4, Solution 57. To find RTh, remove the 50V source and insert a 1-V source at a – b, as shown in Fig. (a).
B A i
10 Ω6 Ω +
vx
− 0.5vx
a
b(a)
2 Ω
+ −
1V 3 Ω We apply nodal analysis. At node A,
i + 0.5vx = (1/10) + (1 – vx)/2, or i + vx = 0.6 (1) At node B,
(1 – vo)/2 = (vx/3) + (vx/6), and vx = 0.5 (2)
From (1) and (2), i = 0.1 and
RTh = 1/i = 10 ohms To get VTh, consider the circuit in Fig. (b).
10 Ω6 Ω+
vx
− 0.5vx
a
b(b)
2 Ω
+ −
v1 3 Ω v2 +
VTh
−
50V At node 1, (50 – v1)/3 = (v1/6) + (v1 – v2)/2, or 100 = 6v1 – 3v2 (3) At node 2, 0.5vx + (v1 – v2)/2 = v2/10, vx = v1, and v1 = 0.6v2 (4) From (3) and (4),
v2 = VTh = 166.67 V
IN = VTh/RTh = 16.667 A
RN = RTh = 10 ohms Chapter 4, Solution 58. This problem does not have a solution as it was originally stated. The reason for this is that the load resistor is in series with a current source which means that the only equivalent circuit that will work will be a Norton circuit where the value of RN = infinity. IN can be found by solving for Isc.
voib
+ −
β ib
R2
R1 Isc VS Writing the node equation at node vo,
ib + βib = vo/R2 = (1 + β)ib
But ib = (Vs – vo)/R1
vo = Vs – ibR1
Vs – ibR1 = (1 + β)R2ib, or ib = Vs/(R1 + (1 + β)R2)
Isc = IN = -βib = -βVs/(R1 + (1 + β)R2) Chapter 4, Solution 59. RTh = (10 + 20)||(50 + 40) 30||90 = 22.5 ohms To find VTh, consider the circuit below.
10 Ω
50 Ω
+ VTh
8A
i1 i2 20 Ω
40 Ω
i1 = i2 = 8/2 = 4, 10i1 + VTh – 20i2 = 0, or VTh = 20i2 –10i1 = 10i1 = 10x4 VTh = 40V, and IN = VTh/RTh = 40/22.5 = 1.7778 A Chapter 4, Solution 60. The circuit can be reduced by source transformations. 2A
5 Ω
10 Ω18 V
+ −
12 V
+ −
10 V
+ −
2A
b a
3A
2A
10 Ω
5 Ω
3A
10 V
+ − 3.333Ωa b 3.333Ω a b
Norton Equivalent Circuit Thevenin Equivalent Circuit Chapter 4, Solution 61. To find RTh, consider the circuit in Fig. (a). Let R = 2||18 = 1.8 ohms, RTh = 2R||R = (2/3)R = 1.2 ohms. To get VTh, we apply mesh analysis to the circuit in Fig. (d).
2 Ω a
2 Ω
6 Ω
6 Ω
2 Ω
6 Ω
b
(a)
1.8 Ω
1.8 Ω
a
b
1.8 Ω RTh
18 Ω
18 Ω 18 Ω2 Ω
a
b
(b)
2 Ω
2 Ω
(c)
+
+ − 12V
i3
i2i12 Ω
6 Ω
6 Ω
a
b
2 Ω
6 Ω
12V
− +
+ − 12V
2 Ω
VTh
(d) -12 – 12 + 14i1 – 6i2 – 6i3 = 0, and 7 i1 – 3 i2 – 3i3 = 12 (1) 12 + 12 + 14 i2 – 6 i1 – 6 i3 = 0, and -3 i1 + 7 i2 – 3 i3 = -12 (2) 14 i3 – 6 i1 – 6 i2 = 0, and -3 i1 – 3 i2 + 7 i3 = 0 (3) This leads to the following matrix form for (1), (2) and (3),
−=
−−−−−−
012
12
iii
733373337
3
2
1
100733373337=
−−−−−−
=∆ , 12070331233127
2 −=−
−−−−
=∆
i2 = ∆/∆2 = -120/100 = -1.2 A
VTh = 12 + 2i2 = 9.6 V, and IN = VTh/RTh = 8 A Chapter 4, Solution 62. Since there are no independent sources, VTh = 0 V To obtain RTh, consider the circuit below.
2
2vo
0.1io ix
v1io
40 Ω
10 Ω
+ −
+ − VS
+vo −1
20 Ω
At node 2,
ix + 0.1io = (1 – v1)/10, or 10ix + io = 1 – v1 (1) At node 1, (v1/20) + 0.1io = [(2vo – v1)/40] + [(1 – v1)/10] (2) But io = (v1/20) and vo = 1 – v1, then (2) becomes,
1.1v1/20 = [(2 – 3v1)/40] + [(1 – v1)/10]
2.2v1 = 2 – 3v1 + 4 – 4v1 = 6 – 7v1 or v1 = 6/9.2 (3) From (1) and (3),
10ix + v1/20 = 1 – v1
10ix = 1 – v1 – v1/20 = 1 – (21/20)v1 = 1 – (21/20)(6/9.2)
ix = 31.52 mA, RTh = 1/ix = 31.73 ohms.
Chapter 4, Solution 63. Because there are no independent sources, IN = Isc = 0 A RN can be found using the circuit below. 3 Ω
io10 Ω
+ −
+
vo
−
v1
0.5vo20 Ω
1V Applying KCL at node 1, 0.5vo + (1 – v1)/3 = v1/30, but vo = (20/30)v1 Hence, 0.5(2/3)(30)v1 + 10 – 10v1 =v1, or v1 = 10 and io = (1 – v1)/3 = -3 RN = 1/io = -1/3 = -333.3 m ohms Chapter 4, Solution 64.
With no independent sources, VTh = 0 V. To obtain RTh, consider the circuit shown below.
1 Ω
ix
io4 Ω
+ −
+ –
10ix
vo
2 Ω
1V
ix = [(1 – vo)/1] + [(10ix – vo)/4], or 2vo = 1 + 3ix (1)
But ix = vo/2. Hence,
2vo = 1 + 1.5vo, or vo = 2, io = (1 – vo)/1 = -1
Thus, RTh = 1/io = -1 ohm
Chapter 4, Solution 65 At the terminals of the unknown resistance, we replace the circuit by its Thevenin equivalent.
V 24)32(412
12,53212//42 =+
=Ω=+=+= ThTh VR
Thus, the circuit can be replaced by that shown below.
5Ω Io
+ + 24 V Vo
- - Applying KVL to the loop,
oooo I524V0VI524 −=→=++− Chapter 4, Solution 66. We first find the Thevenin equivalent at terminals a and b. We find RTh using the circuit in Fig. (a).
3 Ω
5 Ω
b a
RTh
2 Ω
2 Ω 10V − +
i 20V
+ −
30V
3 Ω
+VTh
a b
− +
5 Ω
(a) (b)
RTh = 2||(3 + 5) = 2||8 = 1.6 ohms By performing source transformation on the given circuit, we obatin the circuit in (b).
We now use this to find VTh.
10i + 30 + 20 + 10 = 0, or i = -5
VTh + 10 + 2i = 0, or VTh = 2 V
p = VTh2/(4RTh) = (2)2/[4(1.6)] = 625 m watts
Chapter 4, Solution 67. We need to find the Thevenin equivalent at terminals a and b. From Fig. (a),
RTh = 4||6 + 8||12 = 2.4 + 4.8 = 7.2 ohms From Fig. (b),
10i1 – 30 = 0, or i1 = 3
+
−
4 Ω
12 Ω 8 Ω
6 Ω
i2
i1
+ −
+
−
30V
RTh
4 Ω
12 Ω 8 Ω
6 Ω
VTh
+
−
(b)(a)
20i2 + 30 = 0, or i2 = 1.5, VTh = 6i1 + 8i2 = 6x3 – 8x1.5 = 6 V For maximum power transfer,
p = VTh2/(4RTh) = (6)2/[4(7.2)] = 1.25 watts
Chapter 4, Solution 68. This is a challenging problem in that the load is already specified. This now becomes a "minimize losses" style problem. When a load is specified and internal losses can be adjusted, then the objective becomes, reduce RThev as much as possible, which will result in maximum power transfer to the load.
-+
-+
Removing the 10 ohm resistor and solving for the Thevenin Circuit results in:
RTh = (Rx20/(R+20)) and a Voc = VTh = 12x(20/(R +20)) + (-8) As R goes to zero, RTh goes to zero and VTh goes to 4 volts, which produces the maximum power delivered to the 10-ohm resistor.
P = vi = v2/R = 4x4/10 = 1.6 watts Notice that if R = 20 ohms which gives an RTh = 10 ohms, then VTh becomes -2 volts and the power delivered to the load becomes 0.1 watts, much less that the 1.6 watts. It is also interesting to note that the internal losses for the first case are 122/20 = 7.2 watts and for the second case are = to 12 watts. This is a significant difference. Chapter 4, Solution 69. We need the Thevenin equivalent across the resistor R. To find RTh, consider the circuit below. 22 kΩ v1 Assume that all resistances are in k ohms and all currents are in mA.
1mA 3vo30 kΩ40 kΩ
+
vo
−
10 kΩ
10||40 = 8, and 8 + 22 = 30
1 + 3vo = (v1/30) + (v1/30) = (v1/15)
15 + 45vo = v1
But vo = (8/30)v1, hence,
15 + 45x(8v1/30) v1, which leads to v1 = 1.3636
RTh = v1/1 = -1.3636 k ohms To find VTh, consider the circuit below. 10 kΩ 22 kΩvo v1
(100 – vo)/10 = (vo/40) + (vo – v1)/22 (1)
3vo30 kΩ40 kΩ
+
vo
−
+ −
+
VTh
−
100V
[(vo – v1)/22] + 3vo = (v1/30) (2)
Solving (1) and (2),
v1 = VTh = -243.6 volts
p = VTh2/(4RTh) = (243.6)2/[4(-1363.6)] = -10.882 watts
Chapter 4, Solution 70 We find the Thevenin equivalent across the 10-ohm resistor. To find VTh, consider the circuit below.
3Vx
5Ω 5Ω + + 4V 15Ω VTh - 6Ω - + Vx -
From the figure,
V3)4(515
15,0 =+
== Thx VV
To find RTh, consider the circuit below:
3Vx
5Ω 5Ω V1 V2 + 4V 15Ω 1A - 6Ω + Vx - At node 1,
122111 73258616,
5153
54 VVxVVVVVV
xx −=→==−
++=− (1)
At node 2,
9505
31 2121 −=→=
−++ VV
VVVx (2)
Solving (1) and (2) leads to V2 = 101.75 V
mW 11.2275.1014
94
,75.1011
2
max2 ===Ω==
xRV
pVRTh
ThTh
Chapter 4, Solution 71. We need RTh and VTh at terminals a and b. To find RTh, we insert a 1-mA source at the terminals a and b as shown below. Assume that all resistances are in k ohms, all currents are in mA, and all voltages are in volts. At node a,
a
b
− +
1mA 120vo40 kΩ1 kΩ
+
vo
−
10 kΩ
3 kΩ
1 = (va/40) + [(va + 120vo)/10], or 40 = 5va + 480vo (1) The loop on the left side has no voltage source. Hence, vo = 0. From (1), va = 8 V.
RTh = va/1 mA = 8 kohms To get VTh, consider the original circuit. For the left loop,
vo = (1/4)8 = 2 V For the right loop, vR = VTh = (40/50)(-120vo) = -192 The resistance at the required resistor is
R = RTh = 8 kohms
p = VTh2/(4RTh) = (-192)2/(4x8x103) = 1.152 watts
Chapter 4, Solution 72. (a) RTh and VTh are calculated using the circuits shown in Fig. (a) and (b) respectively. From Fig. (a), RTh = 2 + 4 + 6 = 12 ohms From Fig. (b), -VTh + 12 + 8 + 20 = 0, or VTh = 40 V
12V 4 Ω4 Ω 6 Ω 2 Ω 6 Ω − + +
VTh
−
+ −
RTh 2 Ω 8V 20V
+ − (b)(a)
(b) i = VTh/(RTh + R) = 40/(12 + 8) = 2A (c) For maximum power transfer, RL = RTh = 12 ohms (d) p = VTh
2/(4RTh) = (40)2/(4x12) = 33.33 watts. Chapter 4, Solution 73 Find the Thevenin’s equivalent circuit across the terminals of R.
10 Ω 25Ω RTh 20Ω 5Ω
Ω==+= 833.1030/3255//2520//10ThR
10 Ω 25Ω + + VTh - 60 V + + - Va Vb 20Ω 5Ω - -
10)60(305,40)60(
3020
==== ba VV
V 3010400 =−=−=→=++− baThbTha VVVVVV
W77.20833.104
304
22
max ===xR
Vp
Th
Th
Chapter 4, Solution 74. When RL is removed and Vs is short-circuited,
RTh = R1||R2 + R3||R4 = [R1 R2/( R1 + R2)] + [R3 R4/( R3 + R4)]
RL = RTh = (R1 R2 R3 + R1 R2 R4 + R1 R3 R4 + R2 R3 R4)/[( R1 + R2)( R3 + R4)] When RL is removed and we apply the voltage division principle,
Voc = VTh = vR2 – vR4
= ([R2/(R1 + R2)] – [R4/(R3 + R4)])Vs = [(R2R3) – (R1R4)]/[(R1 + R2)(R3 + R4)]Vs
pmax = VTh2/(4RTh)
= [(R2R3) – (R1R4)]2/[(R1 + R2)(R3 + R4)]2Vs
2[( R1 + R2)( R3 + R4)]/[4(a)]
where a = (R1 R2 R3 + R1 R2 R4 + R1 R3 R4 + R2 R3 R4)
pmax =
[(R2R3) – (R1R4)]2Vs2/[4(R1 + R2)(R3 + R4) (R1 R2 R3 + R1 R2 R4 + R1 R3 R4 + R2 R3 R4)]
Chapter 4, Solution 75.
We need to first find RTh and VTh. R
+ −
+ − 2V
R
R
+ − 3V
+
VTh
−
vo
R R R RTh
1V (a) (b) Consider the circuit in Fig. (a).
(1/RTh) = (1/R) + (1/R) + (1/R) = 3/R
RTh = R/3 From the circuit in Fig. (b),
((1 – vo)/R) + ((2 – vo)/R) + ((3 – vo)/R) = 0
vo = 2 = VTh For maximum power transfer,
RL = RTh = R/3
Pmax = [(VTh)2/(4RTh)] = 3 mW
RTh = [(VTh)2/(4Pmax)] = 4/(4xPmax) = 1/Pmax = R/3
R = 3/(3x10-3) = 1 k ohms Chapter 4, Solution 76. Follow the steps in Example 4.14. The schematic and the output plots are shown below. From the plot, we obtain,
V = 92 V [i = 0, voltage axis intercept]
R = Slope = (120 – 92)/1 = 28 ohms
Chapter 4, Solution 77. (a) The schematic is shown below. We perform a dc sweep on a current source, I1, connected between terminals a and b. We label the top and bottom of source I1 as 2 and 1 respectively. We plot V(2) – V(1) as shown.
VTh = 4 V [zero intercept]
RTh = (7.8 – 4)/1 = 3.8 ohms
(b) Everything remains the same as in part (a) except that the current source, I1, is connected between terminals b and c as shown below. We perform a dc sweep on I1 and obtain the plot shown below. From the plot, we obtain,
V = 15 V [zero intercept]
R = (18.2 – 15)/1 = 3.2 ohms
Chapter 4, Solution 78.
he schematic is shown below. We perform a dc sweep on the current source, I1,
VTh = -80 V
Tconnected between terminals a and b. The plot is shown. From the plot we obtain,
[zero intercept]
RTh = (1920 – (-80))/1 = 2 k ohms
Chapter 4, Solution 79. After drawing and saving the schematic as shown below, we perform a dc sweep on I1 connected across a and b. The plot is shown. From the plot, we get,
V = 167 V [zero intercept]
R = (177 – 167)/1 = 10 ohms
Chapter 4, Solution 80. The schematic in shown below. We label nodes a and b as 1 and 2 respectively. We perform dc sweep on I1. In the Trace/Add menu, type v(1) – v(2) which will result in the plot below. From the plot,
VTh = 40 V [zero intercept]
RTh = (40 – 17.5)/1 = 22.5 ohms [slope]
Chapter 4, Solution 81. The schematic is shown below. We perform a dc sweep on the current source, I2, connected between terminals a and b. The plot of the voltage across I2 is shown below. From the plot,
VTh = 10 V [zero intercept]
RTh = (10 – 6.4)/1 = 3.4 ohms.
Chapter 4, Solution 82.
VTh = Voc = 12 V, Isc = 20 A
RTh = Voc/Isc = 12/20 = 0.6 ohm. 0.6 Ω
i = 12/2.6 , p = i2R = (12/2.6)2(2) = 42.6 watts
i + − 2 Ω12V
Chapter 4, Solution 83.
VTh = Voc = 12 V, Isc = IN = 1.5 A
RTh = VTh/IN = 8 ohms, VTh = 12 V, RTh = 8 ohms
Chapter 4, Solution 84 Let the equivalent circuit of the battery terminated by a load be as shown below. RTh IL + + VTh - VL RL
-
For open circuit,
V 8.10, ===→∞= LocThL VVVR When RL = 4 ohm, VL=10.5,
7.24/8.10 ===L
LL R
VI
But
Ω=−
=−
=→+= 4444.07.2
8.1012
L
LThThThLLTh I
VVRRIVV
Chapter 4, Solution 85 (a) Consider the equivalent circuit terminated with R as shown below. RTh a + + VTh Vab R - - b
ThTh
ThTh
ab VR
VRRRV
+=→
+=
10106
or ThTh VR 10660 =+ (1)
where RTh is in k-ohm.
Similarly,
ThThThTh
VRVR
301236030
3012 =+→+
= (2)
Solving (1) and (2) leads to
Ω== kRV ThTh 30 V, 24
(b) V 6.9)24(3020
20=
+=abV
Chapter 4, Solution 86. We replace the box with the Thevenin equivalent. RTh
+
v
−
i
R
+ −
VTh
VTh = v + iRTh When i = 1.5, v = 3, which implies that VTh = 3 + 1.5RTh (1) When i = 1, v = 8, which implies that VTh = 8 + 1xRTh (2) From (1) and (2), RTh = 10 ohms and VTh = 18 V. (a) When R = 4, i = VTh/(R + RTh) = 18/(4 + 10) = 1.2857 A (b) For maximum power, R = RTH
Pmax = (VTh)2/4RTh = 182/(4x10) = 8.1 watts Chapter 4, Solution 87. (a) im = 9.975 mA im = 9.876 mA
+
vm
−
Rs Rm Rs Rs Rm
Is Is
(a) (b)
From Fig. (a),
vm = Rmim = 9.975 mA x 20 = 0.1995 V
Is = 9.975 mA + (0.1995/Rs) (1) From Fig. (b),
vm = Rmim = 20x9.876 = 0.19752 V
Is = 9.876 mA + (0.19752/2k) + (0.19752/Rs)
= 9.975 mA + (0.19752/Rs) (2) Solving (1) and (2) gives,
Rs = 8 k ohms, Is = 10 mA (b)
im’ = 9.876 mA
8k||4k = 2.667 k ohms
im’ = [2667/(2667 + 20)](10 mA) = 9.926 mA Chapter 4, Solution 88
To find RTh, consider the circuit below. RTh 5kΩ A B
30k 20kΩ Ω
Rs
(b)
RsIs Rm
10kΩ Ω=++= kRTh 445//201030
To find VTh , consider the circuit below.
5kΩ A B io +
30k 20kΩ Ω 4mA 60 V - 10kΩ
V 72,48)60(2520,120430 =−===== BAThBA VVVVxV
Chapter 4, Solution 89 It is easy to solve this problem using Pspice. (a) The schematic is shown below. We insert IPROBE to measure the desired ammeter reading. We insert a very small resistance in series IPROBE to avoid problem. After the circuit is saved and simulated, the current is displaced on IPROBE as A99.99 µ .
(b) By interchanging the ammeter and the 12-V voltage source, the schematic is shown below. We obtain exactly the same result as in part (a).
Chapter 4, Solution 90.
Rx = (R3/R1)R2 = (4/2)R2 = 42.6, R2 = 21.3
which is (21.3ohms/100ohms)% = 21.3% Chapter 4, Solution 91.
Rx = (R3/R1)R2 (a) Since 0 < R2 < 50 ohms, to make 0 < Rx < 10 ohms requires that when R2
= 50 ohms, Rx = 10 ohms.
10 = (R3/R1)50 or R3 = R1/5
so we select R1 = 100 ohms and R3 = 20 ohms (b) For 0 < Rx < 100 ohms
100 = (R3/R1)50, or R3 = 2R1
So we can select R1 = 100 ohms and R3 = 200 ohms
Chapter 4, Solution 92. For a balanced bridge, vab = 0. We can use mesh analysis to find vab. Consider the circuit in Fig. (a), where i1 and i2 are assumed to be in mA.
2 kΩ
5 kΩ
i2
i1
+ −
+ vab −
a b
3 kΩ 6 kΩ 220V 10 kΩ 0 (a)
220 = 2i1 + 8(i1 – i2) or 220 = 10i1 – 8i2 (1) 0 = 24i2 – 8i1 or i2 = (1/3)i1 (2) From (1) and (2),
i1 = 30 mA and i2 = 10 mA Applying KVL to loop 0ab0 gives
5(i2 – i1) + vab + 10i2 = 0 V Since vab = 0, the bridge is balanced. When the 10 k ohm resistor is replaced by the 18 k ohm resistor, the gridge becomes unbalanced. (1) remains the same but (2) becomes
0 = 32i2 – 8i1, or i2 = (1/4)i1 (3) Solving (1) and (3),
i1 = 27.5 mA, i2 = 6.875 mA
vab = 5(i1 – i2) – 18i2 = -20.625 V
VTh = vab = -20.625 V To obtain RTh, we convert the delta connection in Fig. (b) to a wye connection shown in Fig. (c).
2 kΩ
a RTh b
18 kΩ
6 kΩR2
R1a RTh b
5 kΩ 18 kΩ R3
6 kΩ3 kΩ
(b) (c)
R1 = 3x5/(2 + 3 + 5) = 1.5 k ohms, R2 = 2x3/10 = 600 ohms,
R3 = 2x5/10 = 1 k ohm.
RTh = R1 + (R2 + 6)||(R3 + 18) = 1.5 + 6.6||9 = 6.398 k ohms
RL = RTh = 6.398 k ohms
Pmax = (VTh)2/(4RTh) = (20.625)2/(4x6.398) = 16.622 mWatts Chapter 4, Solution 93.
Rs Ro ix
βRoix+ −
ix
+ −
VS
-Vs + (Rs + Ro)ix + βRoix = 0
ix = Vs/(Rs + (1 + β)Ro)
Chapter 4, Solution 94.
(a) Vo/Vg = Rp/(Rg + Rs + Rp) (1)
Req = Rp||(Rg + Rs) = Rg
Rg = Rp(Rg + Rs)/(Rp + Rg + Rs)
RgRp + Rg2 + RgRs = RpRg + RpRs
RpRs = Rg(Rg + Rs) (2)
From (1), Rp/α = Rg + Rs + Rp Rg + Rs = Rp((1/α) – 1) = Rp(1 - α)/α (1a)
Combining (2) and (1a) gives, Rs = [(1 - α)/α]Req (3) = (1 – 0.125)(100)/0.125 = 700 ohms
From (3) and (1a),
Rp(1 - α)/α = Rg + [(1 - α)/α]Rg = Rg/α
Rp = Rg/(1 - α) = 100/(1 – 0.125) = 114.29 ohms
(b)
RTh
+ −
I RLVTh
VTh = Vs = 0.125Vg = 1.5 V
RTh = Rg = 100 ohms
I = VTh/(RTh + RL) = 1.5/150 = 10 mA
Chapter 4, Solution 95.
Let 1/sensitivity = 1/(20 k ohms/volt) = 50 µA For the 0 – 10 V scale,
Rm = Vfs/Ifs = 10/50 µA = 200 k ohms For the 0 – 50 V scale,
Rm = 50(20 k ohms/V) = 1 M ohm
+ − VTh
RTh
Rm
VTh = I(RTh + Rm)
(a) A 4V reading corresponds to
I = (4/10)Ifs = 0.4x50 µA = 20 µA
VTh = 20 µA RTh + 20 µA 250 k ohms
= 4 + 20 µA RTh (1)
(b) A 5V reading corresponds to
I = (5/50)Ifs = 0.1 x 50 µA = 5 µA
VTh = 5 µA x RTh + 5 µA x 1 M ohm
VTh = 5 + 5 µA RTh (2) From (1) and (2)
0 = -1 + 15 µA RTh which leads to RTh = 66.67 k ohms
From (1),
VTh = 4 + 20x10-6x(1/(15x10-6)) = 5.333 V
Chapter 4, Solution 96. (a) The resistance network can be redrawn as shown in Fig. (a),
+ − VTh
RTh
+
Vo
−
+
VTh
−
R
10 Ω 8 Ω 10 Ω
40Ω 10 Ω
i2i1
+ − 60 Ω
8 Ω
9V R
(a) (b)
RTh = 10 + 10 + 60||(8 + 8 + 10||40) = 20 + 60||24 = 37.14 ohms Using mesh analysis, -9 + 50i1 - 40i2 = 0 (1) 116i2 – 40i1 = 0 or i1 = 2.9i2 (2) From (1) and (2), i2 = 9/105
VTh = 60i2 = 5.143 V From Fig. (b),
Vo = [R/(R + RTh)]VTh = 1.8
R/(R + 37.14) = 1.8/5.143 which leads to R = 20 ohms (b) R = RTh = 37.14 ohms Imax = VTh/(2RTh) = 5.143/(2x37.14) = 69.23 mA Chapter 4, Solution 97.
4 kΩ
4 kΩ
+ −
+B
VTh
− E
12V
RTh = R1||R2 = 6||4 = 2.4 k ohms
VTh = [R2/(R1 + R2)]vs = [4/(6 + 4)](12) = 4.8 V Chapter 4, Solution 98. The 20-ohm, 60-ohm, and 14-ohm resistors form a delta connection which needs to be connected to the wye connection as shown in Fig. (b),
b
a
RTh
R2
R3
30 Ω
R1
b
a RTh
20 Ω
60 Ω
30 Ω
14 Ω
(a) (b)
R1 = 20x60/(20 + 60 + 14) = 1200/94 = 12.97 ohms
R2 = 20x14/94 = 2.98 ohms
R3 = 60x14/94 = 8.94 ohms
RTh = R3 + R1||(R2 + 30) = 8.94 + 12.77||32.98 = 18.15 ohms
To find VTh, consider the circuit in Fig. (c).
16 V
I1
IT
IT
+ −
b
a +
VTh
20 Ω
60 Ω
30 Ω
14 Ω
(c)
IT = 16/(30 + 15.74) = 350 mA
I1 = [20/(20 + 60 + 14)]IT = 94.5 mA
VTh = 14I1 + 30IT = 11.824 V
I40 = VTh/(RTh + 40) = 11.824/(18.15 + 40) = 203.3 mA
P40 = I402R = 1.654 watts
Chapter 5, Solution 1. (a) Rin = 1.5 MΩ (b) Rout = 60 Ω (c) A = 8x104
Therefore AdB = 20 log 8x104 = 98.0 dB Chapter 5, Solution 2. v0 = Avd = A(v2 - v1) = 105 (20-10) x 10-6 = 0.1V Chapter 5, Solution 3. v0 = Avd = A(v2 - v1) = 2 x 105 (30 + 20) x 10-6 = 10V Chapter 5, Solution 4.
v0 = Avd = A(v2 - v1)
v2 - v1 = V2010x24
Av
50 µ−=
−=
If v1 and v2 are in mV, then
v2 - v1 = -20 mV = 0.02 1 - v1 = -0.02
v1 = 1.02 mV Chapter 5, Solution 5.
+ v0
-
-
vd
+
R0
Rin
I
vi -+
Avd +-
-vi + Avd + (Ri - R0) I = 0 (1) But vd = RiI, -vi + (Ri + R0 + RiA) I = 0
vd = i0
ii
R)A1(RRv++
(2)
-Avd - R0I + v0 = 0
v0 = Avd + R0I = (R0 + RiA)I = i0
ii0
R)A1(Rv)ARR(
+++
45
54
i0
i0
i
0 10)101(100
10x10100R)A1(R
ARRvv
⋅++
+=
+++
=
≅ ( ) =⋅+
45
9
10101
10=
001,100000,100 0.9999990
Chapter 5, Solution 6.
- vd
+ + vo
-
R0
Rin
I
vi
+ -
Avd +-
(R0 + Ri)R + vi + Avd = 0 But vd = RiI, vi + (R0 + Ri + RiA)I = 0
I = i0
i
R)A1(Rv++
− (1)
-Avd - R0I + vo = 0 vo = Avd + R0I = (R0 + RiA)I Substituting for I in (1),
v0 =
++
+
i0
i0
R)A1(RARR
− vi
= ( )( ) 65
356
10x2x10x21501010x2x10x250
++⋅+
−−
≅ mV10x2x001,20010x2x000,200
6
6−
v0 = -0.999995 mV Chapter 5, Solution 7. 100 kΩ
1 210 kΩ
-+
+ Vd -
+ Vout
-
Rout = 100 Ω
Rin AVd +-
VS
At node 1, (VS – V1)/10 k = [V1/100 k] + [(V1 – V0)/100 k] 10 VS – 10 V1 = V1 + V1 – V0
which leads to V1 = (10VS + V0)/12 At node 2, (V1 – V0)/100 k = (V0 – AVd)/100
But Vd = V1 and A = 100,000, V1 – V0 = 1000 (V0 – 100,000V1)
0= 1001V0 – 100,000,001[(10VS + V0)/12]
0 = -83,333,334.17 VS - 8,332,333.42 V0
which gives us (V0/ VS) = -10 (for all practical purposes) If VS = 1 mV, then V0 = -10 mV Since V0 = A Vd = 100,000 Vd, then Vd = (V0/105) V = -100 nV
Chapter 5, Solution 8. (a) If va and vb are the voltages at the inverting and noninverting terminals of the op
amp.
va = vb = 0
1mA = k2v0 0−
v0 = -2V
(b)
10 kΩ
2V
+ -+ va -
10 kΩ
ia
+ vo -
+ vo -
va
vb
ia
2 kΩ
2V -+
1V -+
- +
(b) (a)
Since va = vb = 1V and ia = 0, no current flows through the 10 kΩ resistor. From Fig. (b), -va + 2 + v0 = 0 va = va - 2 = 1 - 2 = -1V Chapter 5, Solution 9. (a) Let va and vb be respectively the voltages at the inverting and noninverting terminals of the op amp va = vb = 4V At the inverting terminal,
1mA = k2
0v4 − v0 = 2V
Since va = vb = 3V, -vb + 1 + vo = 0 vo = vb - 1 = 2V
+ vb -
+ vo -
+ -
(b) 1V
Chapter 5, Solution 10.
Since no current enters the op amp, the voltage at the input of the op amp is vs. Hence
vs = vo 2v
101010 o=
+
s
o
vv
= 2
Chapter 5, Solution 11.
8 kΩ
vb = V2)3(510
10=
+
io
b
a
+ −
5 kΩ
2 kΩ
4 kΩ
+
vo
−
10 kΩ
−+
3 V
At node a,
8
vv2v3 oaa −
=−
12 = 5va – vo
But va = vb = 2V,
12 = 10 – vo vo = -2V
–io = mA142
822
4v0
8vv ooa =+
+=
−+
−
i o = -1mA
Chapter 5, Solution 12. 4 kΩ
b
a
+ −
2 kΩ
1 kΩ
+
vo
− 4 kΩ
−+
1.2V
At node b, vb = ooo v32v
32v
244
==+
At node a, 4
vv1
v2.1 oaa −=
−, but va = vb = ov
32
4.8 - 4 x ooo vv32v
32
−= vo = V0570.27
8.4x3=
va = vb = 76.9v
32
o =
is = 7
2.11
v2. a −=
1 −
p = vsis = 1.2 =
−7
2.1 -205.7 mW
Chapter 5, Solution 13. By voltage division,
i1 i2
90 kΩ
10 kΩ
b
a
+ −
100 kΩ
4 kΩ
50 kΩ
+−
io
+
vo
− 1 V
va = V9.0)1(100
=90
vb = 3
vv
150o
o =50
But va = vb 9.03
v0 = vo = 2.7V
io = i1 + i2 = =+k150
vk10
v oo 0.27mA + 0.018mA = 288 µA
Chapter 5, Solution 14. Transform the current source as shown below. At node 1,
10
vv20
vv5
v10 o1211 −+
−=
−
But v2 = 0. Hence 40 - 4v1 = v1 + 2v1 - 2vo 40 = 7v1 - 2vo (1)
20 kΩ
vo
10 kΩ
+ −
v1 −+
v2
5 kΩ
10 kΩ
+
vo
−
10V
At node 2, 0v,10
vv20
vv2
o221 =−
=− or v1 = -2vo (2)
From (1) and (2), 40 = -14vo - 2vo vo = -2.5V Chapter 5, Solution 15
(a) Let v1 be the voltage at the node where the three resistors meet. Applying KCL at this node gives
3321
3
1
2
1 11Rv
RRv
Rvv
Rvi oo
s −
+=
−+= (1)
At the inverting terminal,
111
10 RivRvi ss −=→
−= (2)
Combining (1) and (2) leads to
++−=→−=
++
2
3131
33
1
2
11RRRRR
iv
Rv
RR
RRi
s
oos
(b) For this case,
Ω=Ω
++−= k 92- k
2540204020 x
iv
s
o
Chapter 5, Solution 16 10kΩ ix 5kΩ va iy - vb + vo + 2kΩ 0.5V - 8kΩ
Let currents be in mA and resistances be in kΩ . At node a,
oaoaa vvvvv
−=→−
=−
31105
5.0 (1)
But
aooba vvvvv8
1028
8=→
+== (2)
Substituting (2) into (1) gives
148
81031 =→−= aaa vvv
Thus,
A 28.14mA 70/15
5.0µ−=−=
−= a
xv
i
A 85.71mA 148
46.0)
810(6.0)(6.0
102µ==−=−=
−+
−= xvvvv
vvvvi aaao
aoboy
Chapter 5, Solution 17.
(a) G = =−=−=5
12RR
vv
1
2
i
o -2.4
(b) 5
80vv
i
o −= = -16
(c) =−=5
2000vv
i
o -400
Chapter 5, Solution 18. Converting the voltage source to current source and back to a voltage source, we have the circuit shown below:
3202010 = kΩ
1 MΩ
3v2
32050
1000v io ⋅
+−= =−=
17200
v1
ov -11.764
Chapter 5, Solution 19. We convert the current source and back to a voltage source.
3442 =
5 kΩ
vo
0V
(4/3) kΩ 10 kΩ
−+
4 kΩ
+ − (2/3)V
(20/3) kΩ
+
vo
−
−+
50 kΩ
+ − 2vi/3
=
−=32
k34x4
k10vo -1.25V
=−
+=k10
0vk5
vi oo
o -0.375mA
Chapter 5, Solution 20. 8 kΩ
+ − vs
+ −
4 kΩ
+
vo
−
−+
2 kΩ4 kΩ
a b
9 V At node a,
4
vv8
vv4v9 baoaa −
+−
=−
18 = 5va – vo - 2vb (1)
At node b,
2
vv4
vv obba −=
− va = 3vb - 2vo (2)
But vb = vs = 0; (2) becomes va = –2vo and (1) becomes
-18 = -10vo – vo vo = -18/(11) = -1.6364V
Chapter 5, Solution 21.
Eqs. (1) and (2) remain the same. When vb = vs = 3V, eq. (2) becomes
va = 3 x 3 - 2v0 = 9 - 2vo
Substituting this into (1), 18 = 5 (9-2vo) – vo – 6 leads to
vo = 21/(11) = 1.909V Chapter 5, Solution 22.
Av = -Rf/Ri = -15.
If Ri = 10kΩ, then Rf = 150 kΩ. Chapter 5, Solution 23
At the inverting terminal, v=0 so that KCL gives
121
000RR
vv
Rv
RRv f
s
o
f
os −= →−
+=−
Chapter 5, Solution 24
v1 Rf
R1 R2 - vs + - + + R4 R3 vo v2 - We notice that v1 = v2. Applying KCL at node 1 gives
f
os
ff
os
Rv
Rvv
RRRRvv
Rvv
Rv
=−
++→=
−+
−+
21
21
1
2
1
1
1 1110)(
(1)
Applying KCL at node 2 gives
ss v
RRR
vRvv
Rv
43
31
4
1
3
1 0+
=→=−
+ (2)
Substituting (2) into (1) yields
sf
fo vRRR
RRR
RR
RRRv
−
+
−+=
243
3
2
43
1
3 1
i.e.
−
+
−+=
243
3
2
43
1
3 1RRR
RRR
RR
RRRk
ff
Chapter 5, Solution 25.
vo = 2 V
−
+
va
-va + 3 + vo = 0 Chapter 5, Solution 26
+
0.4V -
8.028
84.0 =+
== oob vvv
Hence,
05
5.05
===kk
vooi
+
which lea
+
vb 8kΩ
→
mA 1.
+
vo
ds to va = vo + 3 = 5 V.
- io + 5k Ω 2kΩ vo
-
V 5.08.0/4.0 ==ov
Chapter 5, Solution 27. (a) Let va be the voltage at the noninverting terminal. va = 2/(8+2) vi = 0.2vi
ia0 v2.10v20
1 =
+
1000v =
G = v0/(vi) = 10.2
(b) vi = v0/(G) = 15/(10.2) cos 120πt = 1.471 cos 120πt V Chapter 5, Solution 28.
−+
+ −
At node 1, k50vv
k10v0 o11 −
=−
But v1 = 0.4V, -5v1 = v1 – vo, leads to vo = 6v1 = 2.4V Alternatively, viewed as a noninverting amplifier, vo = (1 + (50/10)) (0.4V) = 2.4V io = vo/(20k) = 2.4/(20k) = 120 µA
Chapter 5, Solution 29 R1 va + vb - + + vi R2 R2 vo - R1 -
obia vRR
RvvRR
Rv21
1
21
2 ,+
=+
=
But oiba vRR
RvRR
Rv21
1
21
2
+=
+→=v
Or
1
2
RR
vv
i
o =
Chapter 5, Solution 30. The output of the voltage becomes vo = vi = 12 Ω= k122030 By voltage division,
V2.0)2.1(6012
12vx =+
=
===k202.0
k20v
i xx 10µA
===k20
04.0Rvp
2x 2µW
Chapter 5, Solution 31. After converting the current source to a voltage source, the circuit is as shown below: 12 kΩ
2
vo6 kΩ
+−
+ − 6 kΩ
3 kΩ 1 v1
vo 12 V At node 1,
12
vv6
vv3
v o1o1112 −+
−=
− 48 = 7v1 - 3vo (1)
At node 2,
xoo1 i6
0v6
vv=
−=
− v1 = 2vo (2)
From (1) and (2),
1148vo =
==k6
vi o
x 0.7272mA
Chapter 5, Solution 32. Let vx = the voltage at the output of the op amp. The given circuit is a non-inverting amplifier.
=xv
+
10501 (4 mV) = 24 mV
Ω= k203060
By voltage division,
vo = mV122
vv
2020o
o ==+20
ix = ( ) ==+ k40
mV24k2020
vx 600nA
p = =−
3
62o
10x6010x
=144
Rv
204nW
Chapter 5, Solution 33. After transforming the current source, the current is as shown below: 1 kΩ This is a noninverting amplifier.
3 kΩ
vi
va+−
+ − 2 kΩ
4 kΩ vo
4 V
iio v23v
211v =
+=
Since the current entering the op amp is 0, the source resistor has a OV potential drop. Hence vi = 4V.
V6)4(23vo ==
Power dissipated by the 3kΩ resistor is
==k3
36Rv2
o 12mW
=−
=−
=k1
64R
vvi oa
x -2mA
Chapter 5, Solution 34
0R
vvR
vv
2
in1
1
in1 =−
+− (1)
but
o43
3a v
RRRv+
= (2)
Combining (1) and (2),
0vRRv
RRvv a
2
12
2
1a1 =−+−
22
11
2
1a v
RRv
RR1v +=
+
22
11
2
1
43
o3 vRRv
RR1
RRvR
+=
+
+
+
+
+= 2
2
11
2
13
43o v
RRv
RR1R
RRv
vO = )vRv()RR(R
RR221
213
43 ++
+
Chapter 5, Solution 35.
10RR1
vv
Ai
f
i
ov =+== Rf = 9Ri
If Ri = 10kΩ, Rf = 90kΩ
Chapter 5, Solution 36 V abTh V=
But abs VRR
R
21
1
+=v . Thus,
ssabTh vRRv
RRRVV )1(
1
2
1
21 +=+
==
To get RTh, apply a current source Io at terminals a-b as shown below.
v1 + v2 - a + R2
vo io R1 - b Since the noninverting terminal is connected to ground, v1 = v2 =0, i.e. no current passes through R1 and consequently R2 . Thus, vo=0 and
0==o
oTh i
vR
Chapter 5, Solution 37.
++−= 3
3
f2
2
f1
1
fo v
RR
vRR
vRR
v
−++−= )3(
3030)2(
2030)1(
1030
vo = -3V
Chapter 5, Solution 38.
+++−= 4
4
f3
3
f2
2
f1
1
fo v
RR
vRR
vRR
vRR
v
−++−+−= )100(
5050)50(
1050)20(
2050)10(
2550
= -120mV Chapter 5, Solution 39
This is a summing amplifier.
2233
22
11
5.29)1(5050
2050)2(
1050 vvv
RR
vRR
vRR
v fffo −−=
−++−=
++−=
Thus, V 35.295.16 22 =→−−=−= vvvo
Chapter 5, Solution 40
R1 R2 va + R3 vb - + + v1 + - v2 Rf vo - + v3 R - - Applying KCL at node a,
)111(03213
3
2
2
1
1
3
3
2
2
1
1
RRRv
Rv
Rv
Rv
Rvv
Rvv
Rvv
aaaa ++=++→=
−+
−+
− (1)
But
of
ba vRRRv+
==v (2)
Substituting (2) into (1)gives
)111(3213
3
2
2
1
1
RRRRRRv
Rv
Rv
Rv
f
o +++
=++
or
)111/()(3213
3
2
2
1
1
RRRRv
Rv
Rv
RRR
v fo ++++
+=
Chapter 5, Solution 41. Rf/Ri = 1/(4) Ri = 4Rf = 40kΩ The averaging amplifier is as shown below: Chapter 5, Solution 42
v1 R2 = 40 kΩ
v2 R3 = 40 kΩ
v3 R4 = 40 kΩ
v4
10 kΩ
−+
R1 = 40 kΩ
vo
Ω== k 10R31R 1f
Chapter 5, Solution 43. In order for
+++= 4
4
f3
3
f2
2
f1
1
fo v
RR
vRR
vRR
vRR
v
to become
( )4321o vvvv41v +++−=
41
RR
i
f = ===4
124
RR if 3kΩ
Chapter 5, Solution 44. R4
At node b, 0R
vvR
vv
2
2b
1
1b =−
+−
21
2
2
1
1
b
R1
R1
Rv
Rv
v+
+= (1)
b
a
R1 v1
R2 v2
R3
vo
−+
At node a, 4
oa
3
a
Rvv
Rv0 −
=−
34
oa R/R1
vv
+= (2)
But va = vb. We set (1) and (2) equal.
21
1112
34
o
RRvRvR
R/R1v
++
=+
or
vo = ( )( ) ( )1112
213
43 vRvRRRR
RR+
++
Chapter 5, Solution 45. This can be achieved as follows:
( )
+−−= 21o v
2/RRv
3/RRv
( )
+−−= 2
2
f1
1
f vRR
vRR
i.e. Rf = R, R1 = R/3, and R2 = R/2 Thus we need an inverter to invert v1, and a summer, as shown below (R<100kΩ).
R/3
R/2 v2
R
−+
R v1
-v1
R
−+
vo Chapter 5, Solution 46.
33
f2
2
x1
1
f32
1o v
RR
)v(RR
vRR
v21)v(
31
3v
v +−+=+−+=−
i.e. R3 = 2Rf, R1 = R2 = 3Rf. To get -v2, we need an inverter with Rf = Ri. If Rf = 10kΩ, a solution is given below.
v1
30 kΩ
20 kΩv3
10 kΩ
−+
10 kΩ v2
-v2
10 kΩ
−+
30 kΩ
vo
Chapter 5, Solution 47.
If a is the inverting terminal at the op amp and b is the noninverting terminal, then,
V6vv,V6)8(13
3v bab ===+
= and at node a,4
vv2
v10 oaa −=
−
which leads to vo = –2 V and io = k4
)vv(k5
v oao −− = –0.4 – 2 mA = –2.4 mA
Chapter 5, Solution 48. Since the op amp draws no current from the bridge, the bridge may be treated separately as follows: v1
v2
i2
i1
+ −
For loop 1, (10 + 30) i1 = 5 i1 = 5/(40) = 0.125µA For loop 2, (40 + 60) i2 = -5 i2 = -0.05µA But, 10i + v1 - 5 = 0 v1 = 5 - 10i = 3.75mV 60i + v2 + 5 = 0 v2 = -5 - 60i = -2mV As a difference amplifier,
( ) [ ]mV)2(75.32080vv
RR
v 121
2o −−=−=
= 23mV
Chapter 5, Solution 49. R1 = R3 = 10kΩ, R2/(R1) = 2 i.e. R2 = 2R1 = 20kΩ = R4
Verify: 11
22
43
21
1
2o v
RR
vR/R1R/R1
RR
v −++
=
( )1212 vv2v2v5.01)5.01(2 −=−
++
=
Thus, R1 = R3 = 10kΩ, R2 = R4 = 20kΩ
Chapter 5, Solution 50. (a) We use a difference amplifier, as shown below:
( ) ( ,vv2vvRR
v 12121
2o −=−= ) i.e. R2/R1 = 2
R1 v2
R2
v1
R2
−+
R1
vo
If R1 = 10 kΩ then R2 = 20kΩ (b) We may apply the idea in Prob. 5.35.
210 v2v2v −=
( )
+−−= 21 v
2/RRv
2/RR
( )
+−−= 2
2
f1
1
f vRR
vRR
i.e. Rf = R, R1 = R/2 = R2
We need an inverter to invert v1 and a summer, as shown below. We may let R = 10kΩ.
R/2
R/2 v2
R
−+
R v1
-v1
R
−+
vo
Chapter 5, Solution 51.
We achieve this by cascading an inverting amplifier and two-input inverting summer as shown below: Verify: vo = -va - v1 But va = -v2. Hence
R
R v1
R
−+
R v2
va
R
−+
vo
vo = v2 - v1.
io2
112 vv
RRvv ==−
=i
o
vv
1
2
RR
(b)
At node A, 2/Rvv
Rvv
2/Rvv
1
aA
g
AB
1
A1 −=
−+
−
v1
R2
vb
va
vB
vA
R1/2 v2
R1/2 Rg
R2
+
vo
−
−+
R1/2 R1/2
−
vi
+
or ( ) aAABg
1A1 vvvv
R2Rvv −=−+− (1)
At node B, g
bB
1
AB
1
B2
Rvv
2/Rvv
2/Rvv −
+−
=−
or bBABg
1B2 vv)vv(
R2Rvv −=−−− (2)
Subtracting (1) from (2),
( ) abABABg
1AB12 vvvvvv
R2R2vvv +−−=−−+−−v
Since, va = vb,
( )2v
vvR2R
12
vv iAB
g
112 =−
+=
−
or
g
1
iAB
R2R1
12vvv
+⋅=− (3)
But for the difference amplifier,
( )AB1
2o vv
2/RR
−=v
or o2
1AB v
R2Rvv =− (4)
Equating (3) and (4),
g
1
io
2
1
R2R1
12vv
R2R
+⋅=
g
11
2
i
o
R2R1
1RR
vv
+⋅=
(c) At node a, 2/R
vvR
vv
2
Aa
1
a1 −=
−
A2
1a
2
1a1 v
RR2v
RR2v −=−v (1)
At node b, B2
1b
2
1b2 v
RR2v
RR2vv −=− (2)
Since va = vb, we subtract (1) from (2),
2v)vv(
RR2v i
AB2
112 =−
−=−v
or i1
2AB v
R2Rvv −
=− (3)
At node A,
2/Rvv
Rvv
2/Rvv oA
g
AB
2
Aa −=
−+
−
( ) oAABg
2Aa vvvv
R2Rvv −=−+− (4)
At node B, 2/R0v
Rvv
2/Rvv B
g
ABBb −=
−−
−
( ) BABg
2Bb vvv
R2Rv =−−−v (5)
Subtracting (5) from (4),
( ) oBAABg
2AB vvvvv
RRv −−=−+−v
( ) og
2AB v
R2R
1vv2 −=
+− (6)
Combining (3) and (6),
og
2i
1
2 vR2
R1vRR
−=
+
−
+=
g
2
1
2
i
o
R2R
1RR
vv
Chapter 5, Solution 54. (a) A0 = A1A2A3 = (-30)(-12.5)(0.8) = 300 (b) A = A1A2A3A4 = A0A4 = 300A4
But dB60ALog20 10 = 3ALog10 =
A = 103 = 1000 A4 = A/(300) = 3.333 Chapter 5, Solution 55. Let A1 = k, A2 = k, and A3 = k/(4) A = A1A2A3 = k3/(4) 42ALog20 10 = A = 101.2ALog10 = 2 ⋅1 = 125.89 k3 = 4A = 503.57 k = 956.57. 75033 = Thus A1 = A2 = 7.956, A3 = 1.989
Chapter 5, Solution 56. There is a cascading system of two inverting amplifiers.
sso v6v612
412v =
−−
=
mAv3k2
vi s
so ==
(a) When vs = 12V, io = 36mA (b) When vs = 10 cos 377t V, io = 30 cos 377t mA
Chapter 5, Solution 57
The first stage is a difference amplifier. Since R1/R2 = R3/R4,
mA10)41(50
100)vv(RRv 12
1
2o =+=−=′
The second stage is a non-inverter.
)given(mV40mA1040R1v
40R1v oo =
+=′
+=
Which leads to, R = 120 kΩ
Chapter 5, Solution 58. By voltage division, the input to the voltage follower is:
V45.0)6.0(13
3v1 =+
=
Thus
15.3v7v5
10v210v 111o −=−=−
−=
=−
=k4v0
i oo 0.7875mA
Chapter 5, Solution 59. Let a be the node between the two op amps. va = vo
The first stage is a summer
oosa vv2010v
510v =−
−=
1.5vs = -2vs or
=−
=2v
5.1vs
o -1.333
Chapter 5, Solution 60. Transform the current source as shown below: 4 kΩ Assume all currents are in mA. The first stage is a summer
v1
3 Ω
5 kΩ
3 kΩ
+−
10 kΩ
−+
+ −
io
5is
2 kΩ
( ) osos1 v5.2i10v4
10i5510
−−=−−
=v (1)
By voltage division,
oo1 v21v
333
=+
=v (2)
Alternatively, we notice that the second stage is a non-inverter.
11o v2v33
1=
+=v
From (1) and (2), 0 3voso v5.2i10v5. −−= o = 10is
3i10
i2 soo −=−=v ==
35
ii
s
o 1.667
Chapter 5, Solution 61.
Let v01 be the voltage at the left end of R5. The first stage is an inverter, while the second stage is a summer.
11
201 v
RRv −=
015
40 v
RR
v −= 23
4 vRR
−
v1 = 151RR42 v
RR2
3
4 vRR
−
Chapter 5, Solution 62. Let v1 = output of the first op amp v2 = output of the second op amp The first stage is a summer
i1
21 v
RR
v −= – of
2 vRR (1)
The second stage is a follower. By voltage division
143
42o v
RRR
vv+
== o4
431 v
RRR
v+
= (2)
From (1) and (2),
i1
2o
4
3 vRR
vRR
1 −=
+ o
f
2 vRR
−
i1
2o
f
2
4
3 vRR
vRR
RR
1 −=
++
4
2
4
31
2
i
o
RR
RR1
1RR
vv
++⋅−=
( )4321
42
RRRRRR++
−=
Chapter 5, Solution 63. The two op amps are summer. Let v1 be the output of the first op amp. For the first stage,
o3
2i
1
21 v
RRv
RRv −−= (1)
For the second stage,
io
41
5
4o v
RRv
RRv −−= (2)
Combining (1) and (2),
i6
4o
3
2
5
4i
1
2
5
4o v
RR
vRR
RR
vRR
RR
v −
+
=
i6
4
51
42
53
42o v
RR
RRRR
RRRR
1v
−=
−
53
42
6
4
31
42
i
o
RRRR
1
RR
RRRR
vv
−
−=
Chapter 5, Solution 64
G4 G G3 G1 1 G 2 - - + 0V + v 0V + + vs G2 vo
- -
At node 1, v1=0 so that KCL gives
GvvGvG os −=+ 41 (1) At node 2,
GvvGvG os −=+ 32 (2) From (1) and (2),
ososos vGGvGGvGvGvGvG )()( 43213241 −=−→+=+ or
43
21
GGGG
vv
s
o
−−
=
Chapter 5, Solution 65
The output of the first op amp (to the left) is 6 mV. The second op amp is an inverter so that its output is
mV -18mV)6(1030' =−=ov
The third op amp is a noninverter so that
mV 6.21'4048
84040' −==→+
= oooo vvvv
Chapter 5, Solution 66.
)2(10100)4(
2040
20100)6(
25110vo −
−−
−=
=−+−= 204024 -4V Chapter 5, Solution 67.
vo = )2.0(2080)5.0(
2040−
−
8080
−
=−= 8.02.3 2.4V Chapter 5, Solution 68.
If Rq = ∞, the first stage is an inverter.
mV30)10(5
15Va −=−=
when Va is the output of the first op amp. The second stage is a noninverting amplifier.
=−+=
+= )30)(31(v
261v ao -120mV
Chapter 5, Solution 69.
In this case, the first stage is a summer
ooa v5.130v1015)10(
515v −−=−−=
For the second stage,
( )oaao v5.1304v4v261v −−==
+=
120v7 o −= =−=7
120ov -17.143mV
Chapter 5, Solution 70.
The output of amplifier A is
9)2(1030)10(
1030vA −=−−=
The output of amplifier B is
14)4(1020)3(
1020vB −=−−=
40 kΩ
V2)14(1060
60vb −=−+
=
vA
vB
a
b
60 kΩ
20 kΩ
−+
10 kΩ
vo
At node a, 40
vv20
vv oaaA −=
−
But va = vb = -2V, 2(-9+2) = -2-vo
Therefore, vo = 12V
Chapter 5, Solution 71 20kΩ 5k 100kΩ Ω
- 40kΩ +
+ v2 2V 80kΩ - - 10kΩ + + vo 20kΩ - - 10kΩ + v1 + - v3 + 3V 50kΩ - 30kΩ
8)30501(,8)2(
520,3 1321 =+=−=−== vvvv
V 10)1020(80
10040
10032 =+−−=
+−= vvvo
Chapter 5, Solution 72.
Since no current flows into the input terminals of ideal op amp, there is no voltage drop across the 20 kΩ resistor. As a voltage summer, the output of the first op amp is v01 = 0.4 The second stage is an inverter
012 v100150v −=
= =− )4.0(5.2 -1V Chapter 5, Solution 73.
The first stage is an inverter. The output is
V9)8.1(1050v01 −=−−=
The second stage is == 012 vv -9V
Chapter 5, Solution 74. Let v1 = output of the first op amp v2 = input of the second op amp.
The two sub-circuits are inverting amplifiers
V6)6.0(10100
1 −=−=v
V8)4.0(6.1
322 −=−=v
=+−
−=−
=k20
86k20vv 21
oi 100 µA
Chapter 5, Solution 75. The schematic is shown below. Pseudo-components VIEWPOINT and IPROBE are involved as shown to measure vo and i respectively. Once the circuit is saved, we click Analysis | Simulate. The values of v and i are displayed on the pseudo-components as:
i = 200 µA
(vo/vs) = -4/2 = -2 The results are slightly different than those obtained in Example 5.11.
Chapter 5, Solution 76. The schematic is shown below. IPROBE is inserted to measure io. Upon simulation, the value of io is displayed on IPROBE as
io = -374.78 µA
Chapter 5, Solution 77. The schematic is shown below. IPROBE is inserted to measure io. Upon simulation, the value of io is displayed on IPROBE as
io = -374.78 µA
Chapter 5, Solution 78. The circuit is constructed as shown below. We insert a VIEWPOINT to display vo. Upon simulating the circuit, we obtain,
vo = 667.75 mV
Chapter 5, Solution 79. The schematic is shown below. A pseudo-component VIEWPOINT is inserted to display vo. After saving and simulating the circuit, we obtain,
vo = -14.61 V
Chapter 5, Solution 80. The schematic is shown below. VIEWPOINT is inserted to display vo. After simulation, we obtain,
vo = 12 V
Chapter 5, Solution 81. The schematic is shown below. We insert one VIEWPOINT and one IPROBE to measure vo and io respectively. Upon saving and simulating the circuit, we obtain,
vo = 343.37 mV
io = 24.51 µA
Chapter 5, Solution 82.
The maximum voltage level corresponds to
11111 = 25 – 1 = 31 Hence, each bit is worth (7.75/31) = 250 mV
Chapter 5, Solution 83. The result depends on your design. Hence, let RG = 10 k ohms, R1 = 10 k ohms, R2 = 20 k ohms, R3 = 40 k ohms, R4 = 80 k ohms, R5 = 160 k ohms, R6 = 320 k ohms, then,
-vo = (Rf/R1)v1 + --------- + (Rf/R6)v6
= v1 + 0.5v2 + 0.25v3 + 0.125v4 + 0.0625v5 + 0.03125v6
(a) |vo| = 1.1875 = 1 + 0.125 + 0.0625 = 1 + (1/8) + (1/16) which implies, [v1 v2 v3 v4 v5 v6] = [100110]
(b) |vo| = 0 + (1/2) + (1/4) + 0 + (1/16) + (1/32) = (27/32) = 843.75 mV (c) This corresponds to [1 1 1 1 1 1]. |vo| = 1 + (1/2) + (1/4) + (1/8) + (1/16) + (1/32) = 63/32 = 1.96875 V
Chapter 5, Solution 84. For (a), the process of the proof is time consuming and the results are only approximate, but close enough for the applications where this device is used. (a) The easiest way to solve this problem is to use superposition and to solve
for each term letting all of the corresponding voltages be equal to zero. Also, starting with each current contribution (ik) equal to one amp and working backwards is easiest.
2R R R R
+ − ik
2R
v2 v4
+ − v3
+ −
+ −
2R 2R
v1
R
For the first case, let v2 = v3 = v4 = 0, and i1 = 1A. Therefore, v1 = 2R volts or i1 = v1/(2R). Second case, let v1 = v3 = v4 = 0, and i2 = 1A. Therefore, v2 = 85R/21 volts or i2 = 21v2/(85R). Clearly this is not (1/4th), so where is the difference? (21/85) = 0.247 which is a really good approximation for 0.25. Since this is a practical electronic circuit, the result is good enough for all practical purposes. Now for the third case, let v1 = v2 = v4 = 0, and i3 = 1A. Therefore, v3 = 8.5R volts or i3 = v3/(8.5R). Clearly this is not (1/8th), so where is the difference? (1/8.5) = 0.11765 which is a really good approximation for 0.125. Since this is a practical electronic circuit, the result is good enough for all practical purposes. Finally, for the fourth case, let v1 = v2 = v4 = 0, and i3 = 1A.
Therefore, v4 = 16.25R volts or i4 = v4/(16.25R). Clearly this is not (1/16th), so where is the difference? (1/16.25) = 0.06154 which is a really good approximation for 0.0625. Since this is a practical electronic circuit, the result is good enough for all practical purposes. Please note that a goal of a lot of electronic design is to come up with practical circuits that are economical to design and build yet give the desired results.
(b) If Rf = 12 k ohms and R = 10 k ohms,
-vo = (12/20)[v1 + (v2/2) + (v3/4) + (v4/8)]
= 0.6[v1 + 0.5v2 + 0.25v3 + 0.125v4] For [v1 v2 v3 v4] = [1 0 11],
|vo| = 0.6[1 + 0.25 + 0.125] = 825 mV
For [v1 v2 v3 v4] = [0 1 0 1],
|vo| = 0.6[0.5 + 0.125] = 375 mV
Chapter 5, Solution 85.
Av = 1 + (2R/Rg) = 1 + 20,000/100 = 201 Chapter 5, Solution 86.
vo = A(v2 – v1) = 200(v2 – v1)
(a) vo = 200(0.386 – 0.402) = -3.2 V (b) vo = 200(1.011 – 1.002) = 1.8 V
Chapter 5, Solution 87.
The output, va, of the first op amp is,
va = (1 + (R2/R1))v1 (1) Also, vo = (-R4/R3)va + (1 + (R4/R3))v2 (2)
Substituting (1) into (2), vo = (-R4/R3) (1 + (R2/R1))v1 + (1 + (R4/R3))v2 Or, vo = (1 + (R4/R3))v2 – (R4/R3 + (R2R4/R1R3))v1 If R4 = R1 and R3 = R2, then, vo = (1 + (R4/R3))(v2 – v1) which is a subtractor with a gain of (1 + (R4/R3)). Chapter 5, Solution 88. We need to find VTh at terminals a – b, from this,
vo = (R2/R1)(1 + 2(R3/R4))VTh = (500/25)(1 + 2(10/2))VTh
= 220VTh Now we use Fig. (b) to find VTh in terms of vi.
30 kΩ
80 kΩ
20 kΩ
40 kΩ
30 kΩ
b
a
vi + −
b
a
vi
20 kΩ
40 kΩ 80 kΩ
(a) (b)
va = (3/5)vi, vb = (2/3)vi
VTh = vb – va (1/15)vi
(vo/vi) = Av = -220/15 = -14.667
Chapter 5, Solution 89. If we use an inverter, R = 2 k ohms,
(vo/vi) = -R2/R1 = -6
R = 6R = 12 k ohms Hence the op amp circuit is as shown below. 12 kΩ
2 kΩ−+
+ −
+
vo
−
vi Chapter 5, Solution 90. Transforming the current source to a voltage source produces the circuit below, At node b, vb = (2/(2 + 4))vo = vo/3
20 kΩ
io
b
a
2 kΩ
5 kΩ −+
+
vo
−
+ −
4 kΩ
5is At node a, (5is – va)/5 = (va – vo)/20 But va = vb = vo/3. 20is – (4/3)vo = (1/3)vo – vo, or is = vo/30 io = [(2/(2 + 4))/2]vo = vo/6 io/is = (vo/6)/(vo/30) = 5
Chapter 5, Solution 91.
io
i2
i1
is
−+
vo
R1
R2
io = i1 + i2 (1) But i1 = is (2) R1 and R2 have the same voltage, vo, across them. R1i1 = R2i2, which leads to i2 = (R1/R2)i1 (3) Substituting (2) and (3) into (1) gives, io = is(1 + R1/R2) io/is = 1 + (R1/R2) = 1 + 8/1 = 9 Chapter 5, Solution 92
The top op amp circuit is a non-inverter, while the lower one is an inverter. The output at the top op amp is
v1 = (1 + 60/30)vi = 3vi while the output of the lower op amp is v2 = -(50/20)vi = -2.5vi Hence, vo = v1 – v2 = 3vi + 2.5vi = 5.5vi
vo/vi = 5.5
Chapter 5, Solution 93. R3
+
vi
−
+
vL
−
vb
R4
RL
io
iL
−+
va R1
R2
+
vo
−
At node a, (vi – va)/R1 = (va – vo)/R3 vi – va = (R1/R2)(va – vo) vi + (R1/R3)vo = (1 + R1/R3)va (1) But va = vb = vL. Hence, (1) becomes vi = (1 + R1/R3)vL – (R1/R3)vo (2)
io = vo/(R4 + R2||RL), iL = (RL/(R2 + RL))io = (R2/(R2 + RL))(vo/( R4 + R2||RL)) Or, vo = iL[(R2 + RL)( R4 + R2||RL)/R2 (3) But, vL = iLRL (4) Substituting (3) and (4) into (2),
vi = (1 + R1/R3) iLRL – R1[(R2 + RL)/(R2R3)]( R4 + R2||RL)iL
= [((R3 + R1)/R3)RL – R1((R2 + RL)/(R2R3)(R4 + (R2RL/(R2 + RL))]iL = (1/A)iL
Thus,
A =
+
+
+−
+
L2
L24
32
L21L
3
1
RRRRR
RRRRRR
RR1
1
Chapter 6, Solution 1.
( =+−== −− t3t3 e6e25dtdvCi ) 10(1 - 3t)e-3t A
p = vi = 10(1-3t)e-3t ⋅ 2t e-3t = 20t(1 - 3t)e-6t W
Chapter 6, Solution 2.
2211 )120)(40(
21Cv
21w ==
w2 = 221 )80)(40(
21
2=Cv1
( ) =−=−=∆ 22
21 8012020www 160 kW Chapter 6, Solution 3.
i = C =−
= −
516028010x40
dtdv 3 480 mA
Chapter 6, Solution 4.
)0(vidtC1v
t
o+= ∫
∫ +1tdt4sin621
=
= 1 - 0.75 cos 4t Chapter 6, Solution 5.
v = ∫ +t
o)0(vidt
C1
For 0 < t < 1, i = 4t,
∫−=t
o6 t410x201v dt + 0 = 100t2 kV
v(1) = 100 kV
For 1 < t < 2, i = 8 - 4t,
∫ +−= −
t
16 )1(vdt)t48(10x201v
= 100 (4t - t2 - 3) + 100 kV
Thus v (t) =
<<−−
<<
2t1,kV)2tt4(1001t0,kVt100
2
2
Chapter 6, Solution 6.
610x30dtdvCi −== x slope of the waveform.
For example, for 0 < t < 2,
310x210
dtdv
−=
i = mA15010x210x10x30
dtdv
36 == −
−C
Thus the current i is sketched below.
t (msec)
150
12 10 2
8
6
4
-150
i(t) (mA) Chapter 6, Solution 7.
∫∫ +=+= −−
t
o
33o 10dt10tx4
10x501)t(vidt
C1v
= =+1050t2 2
0.04k2 + 10 V
Chapter 6, Solution 8.
(a) tt BCeACedtdvC 600100 600100 −− −−==i (1)
BABCACi 656001002)0( −−=→−−== (2)
BAvv +=→= −+ 50)0()0( (3) Solving (2) and (3) leads to A=61, B=-11
(b) J 5250010421)0(
21 32 === − xxxCvEnergy
(c ) From (1),
A 4.264.241041160010461100 60010060031003 tttt eeexxxexxxi −−−−−− −−=−−= Chapter 6, Solution 9.
v(t) = ( ) ( )∫ −− +=+−t
o
tt Vet120dte1621
1
v(2) = 12(2 + e-2) = 25.62 V
p = iv = 12 (t + e-t) 6 (1-e-t) = 72(t-e-2t)
p(2) = 72(2-e-4) = 142.68 W Chapter 6, Solution 10
dtdvx
dtdvCi 3102 −==
<<<<<<
=s4t316t,-64
s 3t116,s10,16
µµµtt
v
<<<<
<<=
s4t3,16x10-s 3t10,
s10,1016
6
6
µµµtx
dtdv
<<<<
<<=
s4t3kA, 32-s 3t10,
s10,kA 32)(
µµµt
ti
Chapter 6, Solution 11.
v = ∫ +t
o)0(vidt
C1
For 0 < t < 1,
∫ == −−
t
o
36 t10dt10x40
10x41v kV
v(1) = 10 kV For 1 < t < 2,
kV10)1(vvdtC1v
t
1=+= ∫
For 2 < t < 3,
∫ +−= −−
t
2
36 )2(vdt)10x40(
10x41v
= -10t + 30kV Thus
v(t) =
<<+−<<<<⋅
3t2,kV30t102t1,kV101t0,kVt10
Chapter 6, Solution 12.
π−π== − 4sin)(4(60x10x3dtdvCi 3 t)
= - 0.7e π sin 4πt A P = vi = 60(-0.72)π cos 4π t sin 4π t = -21.6π sin 8π t W
W = ∫ ∫ t dt ππ−=t
o81
o8sin6.21pdt
= πππ 8
86. cos21 8/1
o = -5.4J
Chapter 6, Solution 13.
Under dc conditions, the circuit becomes that shown below:
i250 Ω
20 Ω
+ − 60V
+
v1
−
i1
30 Ω
10 Ω
+
v2
−
i2 = 0, i1 = 60/(30+10+20) = 1A v1 = 30i2 = 30V, v2 = 60-20i1 = 40V
Thus, v1 = 30V, v2 = 40V Chapter 6, Solution 14. (a) Ceq = 4C = 120 mF
(b) 304
C4
C1
eq
== Ceq = 7.5 mF
Chapter 6, Solution 15. In parallel, as in Fig. (a), v1 = v2 = 100
C2+
v2
−
C1 + − 100V
+
v2
−
C2
+ − v1
C1 +
v1
−
+ −
100V (b)(a)
w20 = == − 262 100x10x20x21Cv
21 0.1J
w30 = =− 26 100x10x30x21 0.15J
(b) When they are connected in series as in Fig. (b):
,60100x5030V
CCC
v21
21 ==
+= v2 = 40
w20 = =− 26 60x10x30x21 36 mJ
w30 = =− 26 4010x302
xx1 24 mJ
Chapter 6, Solution 16
F 203080
8014 µ=→=+
+= CCCxCeq
Chapter 6, Solution 17. (a) 4F in series with 12F = 4 x 12/(16) = 3F 3F in parallel with 6F and 3F = 3+6+3 = 12F 4F in series with 12F = 3F i.e. Ceq = 3F
(b) Ceq = 5 + [6 || (4 + 2)] = 5 + (6 || 6) = 5 + 3 = 8F (c) 3F in series with 6F = (3 x 6)/9 = 6F
131
61
21
C1
eq
=++=
Ceq = 1F
Chapter 6, Solution 18.
For the capacitors in parallel = 15 + 5 + 40 = 60 µF 1
eqC
Hence 101
601
301
201
C1
eq
=++=
Ceq = 10 µF Chapter 6, Solution 19. We combine 10-, 20-, and 30- µ F capacitors in parallel to get 60µ F. The 60 -µ F capacitor in series with another 60- µ F capacitor gives 30 µ F. 30 + 50 = 80µ F, 80 + 40 = 120 µ F The circuit is reduced to that shown below.
12 120
12 80
120-µ F capacitor in series with 80µ F gives (80x120)/200 = 48 48 + 12 = 60 60-µ F capacitor in series with 12µ F gives (60x12)/72 = 10µ F Chapter 6, Solution 20.
3 in series with 6 = 6x3/(9) = 2 2 in parallel with 2 = 4 4 in series with 4 = (4x4)/8 = 2 The circuit is reduced to that shown below:
20
1 6
2
8
6 in parallel with 2 = 8 8 in series with 8 = 4 4 in parallel with 1 = 5 5 in series with 20 = (5x20)/25 = 4 Thus Ceq = 4 mF
Chapter 6, Solution 21. 4µF in series with 12µF = (4x12)/16 = 3µF 3µF in parallel with 3µF = 6µF 6µF in series with 6µF = 3µF 3µF in parallel with 2µF = 5µF 5µF in series with 5µF = 2.5µF
Hence Ceq = 2.5µF Chapter 6, Solution 22. Combining the capacitors in parallel, we obtain the equivalent circuit shown below:
a b
40 µF 60 µF 30 µF
20 µF Combining the capacitors in series gives C , where 1
eq
101
301
201
601
C11eq
=++= C = 10µF 1eq
Thus
Ceq = 10 + 40 = 50 µF
Chapter 6, Solution 23.
(a) 3µF is in series with 6µF 3x6/(9) = 2µF v4µF = 1/2 x 120 = 60V v2µF = 60V
v6µF = =(3+
)6036
20V
v3µF = 60 - 20 = 40V
(b) Hence w = 1/2 Cv2 w4µF = 1/2 x 4 x 10-6 x 3600 = 7.2mJ w2µF = 1/2 x 2 x 10-6 x 3600 = 3.6mJ w6µF = 1/2 x 6 x 10-6 x 400 = 1.2mJ w3µF = 1/2 x 3 x 10-6 x 1600 = 2.4mJ
Chapter 6, Solution 24.
20µF is series with 80µF = 20x80/(100) = 16µF
14µF is parallel with 16µF = 30µF (a) v30µF = 90V
v60µF = 30V v14µF = 60V
v20µF = =+
60x8020
80 48V
v80µF = 60 - 48 = 12V
(b) Since w = 2Cv21
w30µF = 1/2 x 30 x 10-6 x 8100 = 121.5mJ w60µF = 1/2 x 60 x 10-6 x 900 = 27mJ w14µF = 1/2 x 14 x 10-6 x 3600 = 25.2mJ w20µF = 1/2 x 20 x 10-6 x (48)2 = 23.04mJ w80µF = 1/2 x 80 x 10-6 x 144 = 5.76mJ
Chapter 6, Solution 25.
(a) For the capacitors in series,
Q1 = Q2 C1v1 = C2v2 1
2
2
1
CC
vv
=
vs = v1 + v2 = 21
2122
1
2 vC
CCvvCC +
=+ s21
12 v
CCC+
=v
Similarly, s21
21 v
CCCv+
=
(b) For capacitors in parallel
v1 = v2 = 2
2
1
1
CQ
CQ
=
Qs = Q1 + Q2 = 22
2122
2
1 QC
CCQQCC +
=+
or
Q2 = 21
2
CCC+
s21
11 Q
CCC
Q+
=
i = dtdQ s
21
11 i
CCC+
=i , s21
22 i
CCC+
=i
Chapter 6, Solution 26.
(a) Ceq = C1 + C2 + C3 = 35µF (b) Q1 = C1v = 5 x 150µC = 0.75mC
Q2 = C2v = 10 x 150µC = 1.5mC Q3 = C3v = 20 x 150 = 3mC
(c) w = J150x35x21
222
eq µ=vC1 = 393.8mJ
Chapter 6, Solution 27.
(a) 207
201
101
51
C1
C1
C1
C1
321eq
=++=++=
Ceq = =µF720 2.857µF
(b) Since the capacitors are in series,
Q1 = Q2 = Q3 = Q = Ceqv = =µV200x720 0.5714mV
(c) w = =µ= J200x720x
21
222
eq vC1 57.143mJ
Chapter 6, Solution 28. We may treat this like a resistive circuit and apply delta-wye transformation, except that R is replaced by 1/C.
Ca
Cb
Cc
50 µF 20 µF
301
401
301
301
101
401
101
C1
a
+
+
=
= 102
401
101
40=++
3
Ca = 5µF
302
101
12001
3001
4001
C1
6
=++
=
Cb = 15µF
154
401
12001
3001
4001
C1
c
=++
=
Cc = 3.75µF Cb in parallel with 50µF = 50 + 15 = 65µF Cc in series with 20µF = 23.75µF
65µF in series with 23.75µF = F39.1775.88
75.23x65µ=
17.39µF in parallel with Ca = 17.39 + 5 = 22.39µF Hence Ceq = 22.39µF Chapter 6, Solution 29. (a) C in series with C = C/(2) C/2 in parallel with C = 3C/2
2C3 in series with C =
5C3
2C5
2C3Cx=
5C3 in parallel with C = C + =
5C3 1.6 C
(b) 2C
Ceq 2C
C1
C21
C21
C1
eq
=+=
Ceq = C
Chapter 6, Solution 30.
vo = ∫ +t
o)0(iidt
C1
For 0 < t < 1, i = 60t mA,
kVt100tdt6010x3
10v 2t
o6
3
o =+= ∫−
−
vo(1) = 10kV For 1< t < 2, i = 120 - 60t mA,
vo = ∫ +−−
− t
1 o6
3
)1(vdt)t60120(10x3
10 t + = [40t – 10t2 kV10] 1
= 40t – 10t2 - 20
<<−−
<<=
2t1,kV20t10t401t0,kVt10
)t(v2
2
o
Chapter 6, Solution 31.
<<+−<<<<
=5t3,t10503t1,mA201t0,tmA20
)t(is
Ceq = 4 + 6 = 10µF
)0(vidtC1v
t
oeq
+= ∫
For 0 < t < 1,
∫−
−
=t
o6
3
t2010x10
10v dt + 0 = t2 kV
For 1 < t < 3,
∫ +−=+=t
1
3
kV1)1t(2)1(vdt201010v
kV1t2 −= For 3 < t < 5,
∫ +−=t
3
3
)3(vdt)5t(101010v
kV11t5tkV55t 2t3
2 +−=++−=
<<+−
<<−<<
=
5t3,kV11t5t3t1,kV1t21t0,kVt
)t(v2
2
dtdv10x6
dtdvCi 6
11−==
=
<<−<<<<
5t3,mA30123t1,mA121t0,tmA12
dtdv10x4
dtdvCi 6
21−==
<<−<<<<
=5t3,mA20t83t1,mA81t0,tmA8
Chapter 6, Solution 32.
(a) Ceq = (12x60)/72 = 10 µ F
13001250501250)0(301012
10 2
00
21
26
3
1 +−=+−=+= −−−−
−
∫ tt
ttt eevdtex
v
23025020250)0(301060
10 2
00
22
26
3
2 −=+=+= −−−−
−
∫ tt
ttt eevdtex
v
(b) At t=0.5s,
03.138230250,15.84013001250 12
11 −=−==+−= −− evev
J 235.4)15.840(101221 26
12 == − xxxw Fµ
J 1905.0)03.138(102021 26
20 =−= − xxxw Fµ
J 381.0)03.138(104021 26
40 =−= − xxxFµw
Chapter 6, Solution 33
Because this is a totally capacitive circuit, we can combine all the capacitors using the property that capacitors in parallel can be combined by just adding their values and we combine capacitors in series by adding their reciprocals. 3F + 2F = 5F 1/5 +1/5 = 2/5 or 2.5F The voltage will divide equally across the two 5F capacitors. Therefore, we get: VTh = 7.5 V, CTh = 2.5 F
Chapter 6, Solution 34.
i = 6e-t/2
2/t3 e21)6(10x10
dtdiL −−
==v
= -30e-t/2 mV v(3) = -300e-3/2 mV = -0.9487 mV p = vi = -180e-t mW
p(3) = -180e-3 mW = -0.8 mW
Chapter 6, Solution 35.
dtdiLv = ==
∆∆=
−
)2/(6.010x60
t/iVL
3
200 mH
Chapter 6, Solution 36.
V)t2sin)(2)(12(10x41
dtdiLv 3 −== −
= - 6 sin 2t mV p = vi = -72 sin 2t cos 2t mW But 2 sin A cos A = sin 2A
p = -36 sin 4t mW Chapter 6, Solution 37.
t100cos)100(4x10x12dtdiLv 3−==
= 4.8 cos 100t V p = vi = 4.8 x 4 sin 100t cos 100t = 9.6 sin 200t
w = t200sin6.9pdtt
o
200/11
o∫ ∫=
Jt200cos200
6.9 200/11o−=
= =−π− mJ)1(cos48 96 mJ Chapter 6, Solution 38.
( )dtte2e10x40dtdiLv t2t23 −−− −==
= 240 0t,mVe)t1( t2 >− −
Chapter 6, Solution 39
)0(iidtL1i
dtdiLv t
0 +∫=→=
1dt)4t2t3(10x200
1i t0
23 +∫ ++= −
1)t4tt(5t
023 +++=
i(t) = 5t3 + 5t2 + 20t + 1 A Chapter 6, Solution 40
dtdix
dtdiLv 31020 −==
<<+<<
<<=
ms 4t310t,40-ms 3t110t,-20
ms 10,10 tti
<<<<<<
=ms 4t3,10x10ms 3t1,10x10-
ms 10,1010
3
3
3 tx
dtdi
<<<<<<
=ms 4t3V, 200ms 3t1V, 200-
ms 10,V 200 tv
which is sketched below. v(t) V 200 0 1 2 3 4 t(ms) -200
Chapter 6, Solution 41.
( )∫∫ +−
=+= −t
o
t2t
03.0dt2120
21)0(ivdt
L1i
= Aetet tto
t 745103021 22 .. −+=+
+ −−10
At t = ls, i = 10 - 4.7 + 5e-2 = 5.977 A
L21w = i2 = 35.72J
Chapter 6, Solution 42.
∫ ∫ −=+=t
o
t
o1dt)t(v
51)0(ivdt
L1i
For 0 < t < 1, ∫ −=−=t
01t21dt
510i A
For 1 < t < 2, i = 0 + i(1) = 1A
For 2 < t < 3, i = ∫ +=+ 1t2)2(idt1051 2
t
= 2t - 3 A For 3 < t < 4, i = 0 + i(3) = 3 A
For 4 < t < 5, i = ∫ +=+t
4
t4 3t2)4(idt10
51
= 2t - 5 A
Thus,
2 1 , 0 11 , 1 2
( ) 2 3 , 2 33 , 3 42 5, 4
t A tA t
i t t A tA tt t
− <
5
<< <
= − < < < < − < <
Chapter 6, Solution 43.
w = L )(Li21)t(Li
21idt
2t
−∞−=∫ ∞−
( ) 010x60x10x80x21 33 −= −−
= 144 µJ Chapter 6, Solution 44.
( )∫ ∫ −+=+=t
ot
t
oo 1dt)t2cos104(51tivdt
L1i
= 0.8t + sin 2t -1
Chapter 6, Solution 45.
i(t) = ∫ +t
o)0(i)t(v
L1
For 0 < t < 1, v = 5t
∫−=t
o3 t510x101i dt + 0
= 0.25t2 kA
For 1 < t < 2, v = -10 + 5t
∫ ++−= −
t
13 )1(idt)t510(10x101i
∫= ( +−t
1kA25.0dt)1t5.0
= 1 - t + 0.25t2 kA
<<+−
<<=
2t1,kAt25.0t11t0,kAt25.0
)t(i2
2
Chapter 6, Solution 46. Under dc conditions, the circuit is as shown below: 2 Ω
+
vC
−3 A
iL
4 Ω
By current division,
=+
= )3(24
4iL 2A, vc = 0V
L21w L = =
= 22
L )2(21
21i 1J
C21w c = == )v)(2(
21v2
c 0J
Chapter 6, Solution 47. Under dc conditions, the circuit is equivalent to that shown below: R
+ vC −
5 A
iL
2 Ω
,2R
10)5(2R
2iL +=
+=
2RR10Riv Lc +
==
2
262
cc )2R(R100x10x80Cv
21w
+== −
232
1L )2R(100x10x2Li
21w
+== −
If wc = wL,
2
3
2
26
)2R(100x10x2
)2Rx(R100x10x80
+=
−− 80 x 10-3R2 = 2
R = 5Ω
Chapter 6, Solution 48. Under dc conditions, the circuit is as shown below:
+
vC2
−
+ −
+
vC1
−
iL1
6 Ω
4 Ω
iL2 30V
=+
==64
30ii2L1L 3A
==
1L1C i6v 18V
=2Cv 0V
Chapter 6, Solution 49.
(a) ( ) =+=++= 36544165Leq 7H (b) ( ) ==+= 41266112Leq 3H (c) ( ) ==+= 446324Leq 2H
Chapter 6, Solution 50.
( )63124510Leq ++=
= 10 + 5||(3 + 2) = 10 + 2.5 = 12.5 mH
Chapter 6, Solution 51.
101
301
201
601
L1
=++= L = 10 mH
( )45
35x10102510Leq =+=
= 7.778 mH
Chapter 6, Solution 52. 3//2//6 = 1H, 4//12 = 3H After the parallel combinations, the circuit becomes that shown below. 3H a 1H 1 H b Lab = (3+1)//1 = (4x1)/5 = 0.8 H
Chapter 6, Solution 53.
[ ])48(6)128(58106Leq +++++= 416)44(816 +=++= Leq = 20 mH
Chapter 6, Solution 54.
( )126010)39(4Leq +++= 34)40(124 +=++= Leq = 7H Chapter 6, Solution 55.
(a) L//L = 0.5L, L + L = 2L
LLLLLxLLLLLeq 4.1
5.025.025.0//2 =
++=+=
(b) L//L = 0.5L, L//L + L//L = L Leq = L//L = 0.5L
Chapter 6, Solution 56.
3L
L31LLL ==
Hence the given circuit is equivalent to that shown below: L
L/3 L
L/3
=+
=
+=
L35L
L35Lx
L32LLLeq L
85
Chapter 6, Solution 57.
Let dtdiLeqv = (1)
221 vdtdi4vvv +=+= (2)
i = i1 + i2 i2 = i – i1 (3)
3
vdtdi
ordtdi
3v 2112 == (4)
and
0dtdi5
dtdi2v 2
2 =++−
dtdi5
dtdi2v 2
2 += (5)
Incorporating (3) and (4) into (5),
3
v5dtdi7
dtdi5
dtdi5
dtdi2v 21
2 −=−+=
dtdi7
351v2 =
+
dtdi
835v2 =
Substituting this into (2) gives
dtdi
835
dtdi4v +=
dtdi
867
=
Comparing this with (1),
==867Leq 8.375H
Chapter 6, Solution 58.
===dtdi3
dtdiLv 3 x slope of i(t).
Thus v is sketched below:
6
t (s)
7 5 1 4 3 2
-6
v(t) (V) 6 Chapter 6, Solution 59.
(a) ( )dtdiLLv 21s +=
21
s
LLv
dtdi
+=
,dtdiL11 =v
dtdiL22 =v
,vLL
Lv s
21
11 += s
21
2L v
LLL
v+
=
(b) dtdi
Ldtdi
Lvv 22
112i ===
21s iii +=
( )
21
21
21
21s
LLLLv
Lv
Lv
dtdi
dtdi
dtdi +
=+=+=
∫∫ =+
== dtdtdi
LLLL
L1vdt
L1i s
21
21
111 s
21
2 iLL
L+
=+
== ∫∫ dtdtdi
LLLL
L1vdt
L1i s
21
21
222 s
21
1 iLL
L+
Chapter 6, Solution 60
8
155//3 ==eqL
( ) tteqo ee
dtd
dtdiLv 22 154
815 −− −===
∫∫ −−− +=+=−+=+=t
tt
ttt
ooo eeeidttvLIi
0
2
0
22
0
A 5.15.05.12)15(512)0()(
Chapter 6, Solution 61.
(a) is = i1 + i2 i )0(i)0(i)0( 21s += 6 i)0(i4 2+= 2(0) = 2mA (b) Using current division:
( ) ==+
= − t2s1 e64.0i
203020i 2.4e-2t mA
=−= 1s2 iii 3.6e-2t mA
(c) mH1250
20x302030 ==
( ) === −−− 3t231 10xe6
dtd10x10
dtdiLv -120e-2t µV
( ) === −−− 3t232 10xe6
dtd10x12
dtdiLv -144e-2t µV
(d) ( )6t43mH10 10xe3610x30x
21w −−−=
Je8.021t
t4 µ=
−=
= 24.36nJ
( ) 2/1t6t43
mH30 10xe76.510x30x21w =
−−−=
= 11.693nJ
( ) 2/1t6t43
mH20 10xe96.1210x20x21w =
−−−=
= 17.54 nJ
Chapter 6, Solution 62.
(a) mH 4080
60202560//2025 =+=+=xLeq
∫∫ +−−=+=+=→= −−−
− ttt
eqeq ieidte
xidttv
Li
dtdiLv
0
333
3
)0()1(1.0)0(121040
10)0()(1
Using current division,
iiiii41,
43
8060
21 ===
01333.0)0(01.0)0(75.0)0(43)0(1 −=→−=→= iiii
mA 67.2125e- A )08667.01.0(41 3t-3
2 +=+−= − tei
mA 33.367.2125)0(2 −=+−=i
(b) mA 6575e- A )08667.01.0(43 3t-3
1 +=+−= − tei
mA 67.2125e- -3t2 +=i
Chapter 6, Solution 63. We apply superposition principle and let
21 vvvo += where v1 and v2 are due to i1 and i2 respectively.
<<−<<
===63,2
30,22 11
1 tt
dtdi
dtdiLv
<<−<<<<
===64,4
42,020,4
2 222
ttt
dtdi
dtdiLv
v1 v2 2 4 0 3 6 t 0 2 4 6 t -2 -4 Adding v1 and v2 gives vo, which is shown below. vo(t) V 6 2 0 2 3 4 6 t (s) -2 -6 Chapter 6, Solution 64.
(a) When the switch is in position A, i=-6 =i(0) When the switch is in position B,
8/1/,34/12)( ====∞ RLi τ
A 93)]()0([)()( 8/ tt eeiiiti −− −=∞−+∞= ι (b) -12 + 4i(0) + v=0, i.e. v=12 – 4i(0) = 36 V (c) At steady state, the inductor becomes a short circuit so that v= 0 V
Chapter 6, Solution 65.
(a) === 22115 )4(x5x
21iL
21w 40 W
=−= 220 )2)(20(
21w 40 W
(b) w = w5 + w20 = 80 W
(c) ( )( )3t20011 10xe502005
dtdvLi −−−==
= -50e-200tA
( )3t20022 10xe50)200(20
dtdvLi −−−==
= -200e-200tA
( )3t20022 10xe50)200(20
dtdvLi −−−==
= -200e-200t A (d) i = i1 + i2 = -250e-200t A
Chapter 6, Solution 66. mH60243640601620Leq =+=++=
dtdiLv =
∫ +=t
o)0(ivdt
L1i
∫= 121 dt + 0 mA −
t
o3 t4sin10x60
=−= tot4cos50i 50(1 - cos 4t) mA
mH244060 =
mV)t4cos1)(50(dtd10x24
dtdiLv 3 −== −
= 4.8 sin 4t mV
Chapter 6, Solution 67.
∫−= viRC1vo dt, RC = 50 x 103 x 0.04 x 10-6 = 2 x 10-3
∫−
= t50sin10210v
3
o dt
vo = 100 cos 50t mV Chapter 6, Solution 68.
∫−= viRC1vo dt + v(0), RC = 50 x 103 x 100 x 10-6 = 5
vo = ∫ −=+−t
ot20dt10
51
The op amp will saturate at vo = 12± -12 = -2t t = 6s
Chapter 6, Solution 69. RC = 4 x 106 x 1 x 10-6 = 4
∫ ∫−=−= dtv41dtv
RC1v iio
For 0 < t < 1, vi = 20, ∫ =−=t
oo dt2041v -5t mV
For 1 < t < 2, vi = 10, ∫ −−−=+−=t
1o 5)1t(5.2)1(vdt1041v
= -2.5t - 2.5mV
For 2 < t < 4, vi = - 20, ∫ −−=++=t
2o 5.7)2t(5)2(vdt2041v
= 5t - 17.5 mV
For 4 < t < 5m, vi = -10, 5.2)4t(5.2)4(vdt1041v
t
4o +−=+= ∫
= 2.5t - 7.5 mV
For 5 < t < 6, vi = 20, ∫ +−−=+−=t
5o 5)5t(5)5(vdt2041v
= - 5t + 30 mV
Thus vo(t) is as shown below:
2 5
6
5
751 432
5
0 Chapter 6, Solution 70.
One possibility is as follows:
50RC
=1
Let R = 100 kΩ, F2.010x100x50
13 µ==C
Chapter 6, Solution 71. By combining a summer with an integrator, we have the circuit below:
∫ ∫ ∫−−−= dtvCR
1dtvCR
1dtvCR
1v 22
22
11
o
−+
For the given problem, C = 2µF, R1C = 1 R1 = 1/(C) = 1006/(2) = 500 kΩ R2C = 1/(4) R2 = 1/(4C) = 500kΩ/(4) = 125 kΩ R3C = 1/(10) R3 = 1/(10C) = 50 kΩ
Chapter 6, Solution 72.
The output of the first op amp is
∫−= i1 vRC1v dt = ∫ −=−
t
o63 2t100idt
10x2x10x101
−
= - 50t
∫−= io vRC1v dt = ∫ −−
t
o63 dt)t50(10x5.0x10x20
1−
= 2500t2 At t = 1.5ms, == −62
o 10x)5.1(2500v 5.625 mV Chapter 6, Solution 73.
Consider the op amp as shown below: Let va = vb = v
At node a, R
vvR
v0 o−=
− 2v - vo = 0 (1)
v +
vo
−
b + − vi C
R
v R
a
R
−+
R
At node b, dtdvC
Rvv
Rvv oi +
−=
−
dtdvRCvv2v oi +−= (2)
Combining (1) and (2),
dt
dv2
RCvvv oooi +−=
or
∫= io vRC2v dt
showing that the circuit is a noninverting integrator. Chapter 6, Solution 74.
RC = 0.01 x 20 x 10-3 sec
secmdtdv2.0
dtdvRCv i
o −=−=
<<−<<<<−
=4t3,V23t1,V21t0,V2
vo
Thus vo(t) is as sketched below:
3
t (ms)
2 1
-2
vo(t) (V) 2
Chapter 6, Solution 75.
,dt
dvRCv i0 −= 5.210x10x10x250RC 63 == −
=−= )t12(dtd5.2vo -30 mV
Chapter 6, Solution 76.
,dt
dvRCv io −= RC = 50 x 103 x 10 x 10-6 = 0.5
<<<<−
==5t5,55t0,10
dtdv5.0v i
o
The input is sketched in Fig. (a), while the output is sketched in Fig. (b).
t (ms)
5
5
0 10
(b)
15
-10
t (ms)
vi(t) (V)
5
5
0 10
(a)
15
vo(t) (V) Chapter 6, Solution 77. i = iR + iC
( )oF
0i v0dtdC
Rv0
R0v
−+−
=−
110x10CR 66F == −
Hence
+−=
dtdv
vv ooi
Thus vi is obtained from vo as shown below: –dvo(t)/dt – vo(t) (V)
4
-4
t (ms)
1
4
0 2 3
vi(t) (V)
3
8
2 1
t (ms)
-8
4 -4
4
-4
t (ms)
1
4
0 2 3
Chapter 6, Solution 78.
ooo
2
vdtdv2
t2sin10dtvd
−−=
Thus, by combining integrators with a summer, we obtain the appropriate analog computer as shown below:
Chapter 6, Solution 79.
We can write the equation as
)(4)( tytfdtdy
−=
which is implemented by the circuit below. 1 V t=0 C R R R R/4 R dy/dt - - -
+ -y + + R dy/dt
f(t)
R
t = 0
-dvo/dt
dvo/dt
d2vo/dt2
d2vo/dt2
-sin2t
2vo
C C
R
R/10
R
R
R
R/2
R
R
− +
−+
− +
−+
− +
+ − sin2t
−+
vo
R
Chapter 6, Solution 80.
From the given circuit,
dt
dvk200k1000v
k5000k1000)t(f
dtvd o
o2o
2
ΩΩ
−ΩΩ
−=
or
)t(fv2dt
dv5
dtvd
oo
2o
2
=++
Chapter 6, Solution 81
We can write the equation as
)(252
2
tfvdtvd
−−=
which is implemented by the circuit below. C C R R - R R/5 d2v/dt2 + - -dv/dt + v - + d2v/dt2 R/2 f(t)
Chapter 6, Solution 82 The circuit consists of a summer, an inverter, and an integrator. Such circuit is shown below. 10R R R R - + - vo + R C=1/(2R) R - + + vs - Chapter 6, Solution 83.
Since two 10µF capacitors in series gives 5µF, rated at 600V, it requires 8 groups in parallel with each group consisting of two capacitors in series, as shown below:
+
600
−
Answer: 8 groups in parallel with each group made up of 2 capacitors in series.
Chapter 6, Solution 84.
tqI∆∆
=∆ ∆I x ∆t = ∆q
∆q = 0.6 x 4 x 10-6
= 2.4µC
62.4 10 150
(36 20)q xC nv
−∆= = =∆ −
F
Chapter 6, Solution 85. It is evident that differentiating i will give a waveform similar to v. Hence,
dtdiLv =
<<−
<<=
2t1,t481t0,t4
i
<<−
<<==
2t1,L41t0,L4
dtdiLv
But,
<<−
<<=
2t1,mV51t0,mV5
v
Thus, 4L = 5 x 10-3 L = 1.25 mH in a 1.25 mH inductor Chapter 6, Solution 86. (a) For the series-connected capacitor
Cs = 8C
C1....
CC
=+++
111
For the parallel-connected strings,
=µ=== F3
1000x108C10
C10C sseq 1250µF
(b) vT = 8 x 100V = 800V
( ) 262Teq )800(10x1250
21vC
21w −==
= 400J
Chapter 7, Solution 1.
Applying KVL to Fig. 7.1.
0RidtiC1 t
-=+∫
∞
Taking the derivative of each term,
0dtdi
RCi
=+
or RCdt
idi
−=
Integrating,
RCt-
I)t(i
ln0
=
RCt-0eI)t(i =
RCt-0eRI)t(Ri)t(v ==
or RCt-0eV)t(v =
Chapter 7, Solution 2.
CR th=τ where is the Thevenin equivalent at the capacitor terminals. thR
Ω=+= 601280||120R th =××=τ -3105.060 ms30
Chapter 7, Solution 3.
(a) ms 10102105,510//10 63 ===Ω== −xxxCRkR ThTh τ (b) 6s3.020,208)255//(20 ===Ω=++= xCRR ThTh τ
Chapter 7, Solution 4.
eqeqCR=τ
where 21
21eq CC
CC+
=C , 21
21eq RR
RRR
+=
=τ)CC)(RR(
CCRR
2121
2121
++
Chapter 7, Solution 5.
τ= 4)-(t-e)4(v)t(v where 24)4(v = , 2)1.0)(20(RC ===τ
24)-(t-e24)t(v = == 26-e24)10(v V195.1
Chapter 7, Solution 6.
Ve4)t(v25210x2x10x40RC,ev)t(v
V4)24(210
2)0(vv
t5.12
36/to
o
−
−τ−
=
===τ=
=+
==
Chapter 7, Solution 7.
τ= t-e)0(v)t(v , CR th=τwhere is the Thevenin resistance across the capacitor. To determine , we insert a 1-V voltage source in place of the capacitor as shown below.
thR thR
8 Ω i2 i
i1
10 Ω0.5 V +
v = 1
−
+ −
+ − 1 V
1.0101
i1 == , 161
85.01
i2 =−
=
8013
161
1.0iii 21 =+=+=
1380
i1
R th ==
138
1.01380
CR th =×==τ
=)t(v V20 813t-e
Chapter 7, Solution 8.
(a) 41
RC ==τ
dtdv
Ci- =
=→= Ce-4))(10(Ce0.2- -4t-4t mF5
==C41
R Ω50
(b) ===τ41
RC s25.0
(c) =×== )100)(105(21
CV21
)0(w 3-20C mJ250
(d) ( )τ−=×= 02t-20
20R e1CV
21
CV21
21
w
21
ee15.0 00 8t-8t- =→−=
or 2e 08t =
== )2(ln81
t 0 ms6.86
Chapter 7, Solution 9.
τ= t-e)0(v)t(v , CR eq=τ
Ω=++=++= 82423||68||82R eq
2)8)(25.0(CR eq ===τ
=)t(v Ve20 2t-
Chapter 7, Solution 10.
10 Ω
10 mF +
v
−
i
15 Ωio
iT
4 Ω
A215
)3)(10(ii10i15 oo ==→=
i.e. if i , then A3)0( = A2)0(io = A5)0(i)0(i)0(i oT =+=
V502030)0(i4)0(i10)0(v T =+=+= across the capacitor terminals.
Ω=+=+= 106415||104R th 1.0)1010)(10(CR -3
th =×==τ -10tt- e50e)0(v)t(v == τ
)e500-)(1010(dtdv
Ci 10t-3-C ×==
=Ci Ae5- -10t By applying the current division principle,
==+
= CC i-0.6)i-(1510
15)t(i Ae3 -10t
Chapter 7, Solution 11.
Applying KCL to the RL circuit,
0Rv
dtvL1
=+∫
Differentiating both sides,
0vLR
dtdv
0dtdv
R1
Lv
=+→=+ LRt-eAv =
If the initial current is , then 0I
ARI)0(v 0 ==
τ= t-0 eRIv ,
RL
=τ
∫∞
=t
-dt)t(v
L1
i
t-
t-0 eL
RI-i ∞
ττ
= τ= t-
0 eRI-i τ= t-
0 eI)t(i Chapter 7, Solution 12. When t < 0, the switch is closed and the inductor acts like a short circuit to dc. The 4 Ω resistor is short-circuited so that the resulting circuit is as shown in Fig. (a).
3 Ω
i(0-)
+ −
12 V 2 H 4 Ω
(a) (b)
A43
12)0(i ==−
Since the current through an inductor cannot change abruptly, A4)0(i)0(i)0(i === +−
When t > 0, the voltage source is cut off and we have the RL circuit in Fig. (b).
5.042
RL
===τ
Hence, == τt-e)0(i)t(i Ae4 -2t
Chapter 7, Solution 13.
thRL
=τ
where is the Thevenin resistance at the terminals of the inductor. thR
Ω=+=+= 37162120||8030||70R th
=×
=τ37102 -3
s08.81 µ
Chapter 7, Solution 14
Converting the wye-subnetwork to delta gives 16Ω R2 80mH R1 R3 30Ω
Ω==Ω==Ω==++
= 17010
1700,3450
1700,8520/170020
105050202010321 RRxxxR
30//170 = (30x170)/200 = 25.5 Ω , 34//16=(34x16)/50 =10.88Ω
sxRLxRTh
Th m 14.3476.251080,476.25
38.12138.3685)88.105.25//(85
3
===Ω==+=−
τ
Chapter 7, Solution 15
(a) sRLRTh
Th 25.020/5,2040//1012 ===Ω=+= τ
(b) ms 5.040/)1020(,408160//40 3 ===Ω=+= −xRLRTh
Th τ
Chapter 7, Solution 16.
eq
eq
RL
=τ
(a) LLeq = and 31
31312
31
312eq RR
RR)RR(RRR
RRRR
+++
=+
+=
=τ31312
31
RR)RR(R)RR(L
+++
(b) where 21
21eq LL
LL+
=L and 21
21213
21
213eq RR
RR)RR(RRR
RRRR
+++
=+
+=
=τ)RR)RR(R()LL(
)RR(LL
2121321
2121
++++
Chapter 7, Solution 17.
τ= t-e)0(i)t(i , 161
441
RL
eq===τ
-16te2)t(i =
16t-16t-
o e2)16-)(41(e6dtdi
Li3)t(v +=+=
=)t(vo Ve2- -16t
Chapter 7, Solution 18. If , the circuit can be redrawn as shown below. 0)t(v =
+
vo(t)
− i(t)
Req0.4 H
56
3||2R eq == , 31
65
52
RL
=×==τ -3tt- ee)0(i)t(i == τ
=== 3t-o e-3)(
52-
dtdi
-L)t(v Ve2.1 -3t
Chapter 7, Solution 19.
i1 i2
i2 i1
i/2 10 Ω 40 Ω
− +
1 V i
To find we replace the inductor by a 1-V voltage source as shown above. thR
0i401i10 21 =+− But 2iii 2 += and 1ii =i.e. i2i2i 21 ==
301
i0i201i10 =→=+−
Ω== 30i1
R th
s2.0306
RL
th===τ
=)t(i Ae2 -5t
Chapter 7, Solution 20.
(a). L50R501
RL
=→==τ
dtdi
Lv- =
=→= Le-50))(30(Le150- -50t-50t H1.0 == L50R Ω5
(b). ===501
RL
τ ms20
(c). === 22 )30)(1.0(21
)0(iL21
w J45
(d). Let p be the fraction
( )τ−=⋅ 02t-00 e1IL
21
pIL21
3296.0e1e1p -0.450(2)(10)- =−=−= i.e. =p %33
Chapter 7, Solution 21.
The circuit can be replaced by its Thevenin equivalent shown below.
Rth
+ −
Vth 2 H
V40)60(4080
80Vth =
+=
R3
80R80||40R th +=+=
R38040
RV
)(i)0(iIth
th
+==∞==
1380R
40)2(
21
IL21
w2
2 =
+==
340
R1380R
40=→=
+
=R Ω33.13
Chapter 7, Solution 22.
τ= t-e)0(i)t(i , eqR
L=τ
Ω=+= 5120||5R eq , 52
=τ
=)t(i Ae10 -2.5t Using current division, the current through the 20 ohm resistor is
2.5t-o e-2
5i-
-i)(205
5i ==
+=
== oi20)t(v Ve04- -2.5t
Chapter 7, Solution 23.
Since the 2 Ω resistor, 1/3 H inductor, and the (3+1) Ω resistor are in parallel, they always have the same voltage.
-1.5)0(i5.113
222
i- =→=+
+=
The Thevenin resistance at the inductor’s terminals is thR
34
)13(||2R th =+= , 41
3431
RL
th===τ
0t,e-1.5e)0(i)t(i -4tt- >== τ
4t-oL e/3)-1.5(-4)(1
dtdi
Lvv ===
=ov 0t,Ve2 -4t >
=+
= Lx v13
1v 0t,Ve5.0 -4t >
Chapter 7, Solution 24.
(a) =)t(v u(t)5-
(b) [ ] [ ])5t(u)3t(u10)3t(u)t(u-10)t(i −−−+−−=
= )5t(u10)3t(u20)t(u10- −−−+
(c) [ ] [ ])3t(u)2t(u)2t(u)1t(u)1t()t(x −−−+−−−−= [ ])4t(u)3t(u)t4( −−−−+
= )4t(u)4t()3t(u)3t()2t(u)2t()1t(u)1t( −−+−−−−−−−− = )4t(r)3t(r)2t(r)1t(r −+−−−−−
(d) [ ])1t(u)t(u5)t-(u2)t(y −−−= = )1t(u5)t(u5)t-(u2 −+−
Chapter 7, Solution 25.
v(t) = [u(t) + r(t – 1) – r(t – 2) – 2u(t – 2)] V Chapter 7, Solution 26.
(a) [ ])t(u)1t(u)t(u)1t(u)t(v1 −−+−+=
=)t(v1 )1t(u)t(u2)1t(u −+−+
(b) [ ])4t(u)2t(u)t4()t(v2 −−−−= )4t(u)4t()2t(u)4t(-)t(v2 −−+−−=
=)t(v2 4)r(t2)r(t2)u(t2 −+−−−
(c) [ ] [ ]6)u(t4)u(t44)u(t2)u(t2)t(v3 −−−+−−−= =)t(v3 6)u(t44)u(t22)u(t2 −−−+−
(d) [ ] )2t(ut1)u(t-t)2t(u)1t(u-t)t(v4 −+−=−−−=
)2t(u)22t()1t(u)11t-()t(v4 −+−+−−+= =)t(v4 2)u(t22)r(t1)u(t1)r(t- −+−+−−−
Chapter 7, Solution 27.
v(t) is sketched below.
1
2
4 3 2 0 1 t -1
v(t)
Chapter 7, Solution 28. i(t) is sketched below.
1
4 3 2 0 1 t
-1
i(t)
Chapter 7, Solution 29
x(t) (a) 3.679 0 1 t (b) y(t) 27.18
0 t
(c) )1(6536.0)1(4cos)1(4cos)( −−=−=−= tttttz δδδ , which is sketched below. z(t) 0 1 t -0.653 ( )tδ Chapter 7, Solution 30.
(a) ==−δ =∫ 1t210
02 t4dt)1t(t4 4
(b) =π=π=−δπ =
∞
∞∫ cos)t2cos(dt)5.0t()t2cos( 5.0t-1-
Chapter 7, Solution 31.
(a) [ ] ===−δ =
∞
∞∫ 16-
2t4t-
-4t- eedt)2t(e 22 -910112×
(b) [ ] ( ) =++=π++=δπ+δ+δ =
∞
∞∫ 115)t2cos(e5dt)t(t2cos)t(e)t(5 0t
t--
t- 7 Chapter 7, Solution 32.
(a) 11)(111
−=== ∫∫ tdduttt
λλλλ
(b) 5.42
)1(0)1( 41
4
1
21
0
4
0
=−=−+=− ∫∫ ttdttdtdttr∫
(c ) 16)6()2()6( 22
5
1
2 =−=−− =∫ ttdttt δ
Chapter 7, Solution 33.
)0(idt)t(vL1
)t(it
0+= ∫
0dt)2t(201010
10)t(i
t
03-
-3
+−δ×
= ∫
=)t(i A)2t(u2 −
Chapter 7, Solution 34.
(a) [ ]
)1t()1t(01)1t()1t()1t(u
)1t(u)1t()1t(u )1t(udtd
−δ=+δ•+•−δ=+δ−
++−δ=+−
(b) [ ]
)6t(u)2t(01)6t(u)2t()6t(r
)2t(u)6t(u)2t(u )6t(rdtd
−=−δ•+•−=−δ−
+−−=−−
(c)
[ ]
)3t(5366.0)3t(u t4cos4)3t(3x4sin)3t(u t4cos4
)3t(t4sin)3t(u t4cos4)3t(u t4sindtd
−δ−−=−δ+−=
−δ+−=−
Chapter 7, Solution 35.
(a) 35t-eA)t(v = , -2A)0(v == =)t(v Ve2- 35t-
(b) 32teA)t(v = , 5A)0(v == =)t(v Ve5 32t
Chapter 7, Solution 36.
(a) 0t,eBA)t(v -t >+=
1A = , B10)0(v +== or B -1= =)t(v 0t,Ve1 -t >−
(b) 0t,eBA)t(v 2t >+=
-3A = , B-3-6)0(v +== or -3B = =)t(v ( ) 0t,Ve13- 2t >+
Chapter 7, Solution 37. Let v = vh + vp, vp =10.
4/041 t
hh Aevvhv −•
=→=+
tAev 25.010 −+= 8102)0( −=→+== AAv tev 25.0810 −−=
(a) s4=τ (b) 10)( =∞v V (c ) te 25.0810 −−=v
Chapter 7, Solution 38
Let i = ip +ih
)(03 3 tuAeiii thhh
−•
=→=+
Let 32)(2)(3,0),( =→===
•
ktutkuitku ppi
)(32 tui p =
)()32( 3 tuAei t += −
If i(0) =0, then A + 2/3 = 0, i.e. A=-2/3. Thus
)()1(32 3 tuei t−−=
Chapter 7, Solution 39.
(a) Before t = 0,
=+
= )20(14
1)t(v V4
After t = 0, [ ] τ∞−+∞= t-e)(v)0(v)(v)t(v
8)2)(4(RC ===τ , 4)0(v = , 20)(v =∞ 8t-e)208(20)t(v −+=
=)t(v Ve1220 8t-− (b) Before t = 0, 21 vvv += , where is due to the 12-V source and is
due to the 2-A source. 1v 2v
V12v1 = To get v , transform the current source as shown in Fig. (a). 2
V-8v2 = Thus,
=−= 812v V4 After t = 0, the circuit becomes that shown in Fig. (b).
2 F 2 F 4 Ω
12 V
+ −
+ − v2 8 V
+ −
3 Ω 3 Ω
(a) (b)
[ ] τ∞−+∞= t-e)(v)0(v)(v)t(v 12)(v =∞ , 4)0(v = , 6)3)(2(RC ===τ
6t-e)124(12)t(v −+= =)t(v Ve812 6t-−
Chapter 7, Solution 40.
(a) Before t = 0, =v V12 .
After t = 0, [ ] τ∞−+∞= t-e)(v)0(v)(v)t(v 4)(v =∞ , 12)0(v = , 6)3)(2(RC ===τ
6t-e)412(4)t(v −+= =)t(v Ve84 6t-+
(b) Before t = 0, =v V12 .
After t = 0, [ ] τ∞−+∞= t-e)(v)0(v)(v)t(v After transforming the current source, the circuit is shown below.
t = 0
4 Ω
2 Ω
+ −
12 V 5 F
12)0(v = , 12)(v =∞ , 10)5)(2(RC ===τ
=v V12 Chapter 7, Solution 41.
0)0(v = , 10)12(1630
)(v ==∞
536
)30)(6()1)(30||6(CR eq ===
[ ] τ∞−+∞= t-e)(v)0(v)(v)t(v
5t-e)100(10)t(v −+=
=)t(v V)e1(10 -0.2t−
Chapter 7, Solution 42.
(a) [ ] τ∞−+∞= t-
oooo e)(v)0(v)(v)t(v
0)0(vo = , 8)12(24
4)(vo =
+=∞
eqeqCR=τ , 34
4||2R eq ==
4)3(34
==τ 4t-
o e88)t(v −= =)t(vo V)e1(8 -0.25t−
(b) For this case, 0)(vo =∞ so that
τ= t-oo e)0(v)t(v
8)12(24
4)0(vo =
+= , 12)3)(4(RC ===τ
=)t(vo Ve8 12t- Chapter 7, Solution 43. Before t = 0, the circuit has reached steady state so that the capacitor acts like an open circuit. The circuit is equivalent to that shown in Fig. (a) after transforming the voltage source.
40v
2i5.0 o−= , 80v
i o=
Hence, 645
320v
40v
280v
21
ooo ==→−=
==80v
i o A8.0
After t = 0, the circuit is as shown in Fig. (b).
τ= t-CC e)0(v)t(v , CR th=τ
To find , we replace the capacitor with a 1-V voltage source as shown in Fig. (c). thR
0.5i vC
+ − 0.5i
i
1 V 80 Ω
(c)
801
80v
i C == , 80
5.0i5.0io ==
Ω=== 1605.0
80i1
Ro
th , 480CR th ==τ
V64)0(vC = 480t-
C e64)t(v =
480t-CC e64
4801
-3dt
dv-Ci-i5.0
===
=)t(i Ae8.0 480t- Chapter 7, Solution 44.
Ω== 23||6R eq , 4RC ==τ [ ] τ∞−+∞= t-e)(v)0(v)(v)t(v
Using voltage division,
V10)30(63
3)0(v =
+= , V4)12(
633
)(v =+
=∞
Thus, 4t-4t- e64e)410(4)t(v +=−+=
=
== 4t-e41-
)6)(2(dtdv
C)t(i Ae3- -0.25t
Chapter 7, Solution 45.
For t < 0, 0)0(v0)t(u5vs =→==
For t > 0, , 5vs = 45
)5(124
4)(v =
+=∞
1012||47R eq =+= , 5)21)(10(CR eq ===τ
[ ] τ∞−+∞= t-e)(v)0(v)(v)t(v
=)t(v V)e1(25.1 5t-−
5t-e
51-
45-
21
dtdv
C)t(i
==
=)t(i Ae125.0 5t-
Chapter 7, Solution 46.
30566)(,0)0(,225.0)62( ===∞==+== xivvsxCR sThτ
)1(30)]()0([)()( 2// tt eevvvtv −− −=∞−+∞= τ V Chapter 7, Solution 47.
For t < 0, , 0)t(u = 0)1t(u =− , 0)0(v = For 0 < t < 1, 1)1.0)(82(RC =+==τ
0)0(v = , 24)3)(8()(v ==∞ [ ] τ∞−+∞= t-e)(v)0(v)(v)t(v
( )t-e124)t(v −= For t > 1, ( ) 17.15e124)1(v 1- =−=
30)(v024-)v(6- =∞→=∞+ 1)--(te)3017.15(30)t(v −+=
1)--(te83.1430)t(v −= Thus,
=)t(v( )
>−<<−1t,Ve83.1430
1t0,Ve124-1)(t-
t-
Chapter 7, Solution 48.
For t < 0, , 1-t)(u = V10)0(v = For t > 0, , 0-t)(u = 0)(v =∞
301020R th =+= , 3)1.0)(30(CR th ===τ [ ] τ∞−+∞= t-e)(v)0(v)(v)t(v
=)t(v Ve10 3t-
3t-e1031-
)1.0(dtdv
C)t(i
==
=)t(i Ae31- 3t-
Chapter 7, Solution 49.
For 0 < t < 1, , 0)0(v = 8)4)(2()(v ==∞
1064R eq =+= , 5)5.0)(10(CR eq ===τ [ ] τ∞−+∞= t-e)(v)0(v)(v)t(v
( ) Ve18)t(v 5t-−= For t > 1, ( ) 45.1e18)1(v -0.2 =−= , 0)(v =∞
[ ] τ−∞−+∞= )1t(-e)(v)1(v)(v)t(v Ve45.1)t(v 5)1t(- −=
Thus,
=)t(v ( )
><<−
− 1t,Ve45.11t0,Ve18
5)1t(-
5t-
Chapter 7, Solution 50.
For the capacitor voltage, [ ] τ∞−+∞= t-e)(v)0(v)(v)t(v
0)0(v = For t < 0, we transform the current source to a voltage source as shown in Fig. (a).
1 kΩ1 kΩ
+
v
−
+ −
30 V 2 kΩ
(a)
V15)30(112
2)(v =
++=∞
Ω=+= k12||)11(R th
41
1041
10CR 3-3th =××==τ
( ) 0t,e115)t(v -4t >−=
We now obtain i from v(t). Consider Fig. (b). x
ix
1/4 mF
v 1 kΩ
1 kΩ
iT
30 mA 2 kΩ
(b)
Tx imA30i −=
But dtdv
CRv
i3
T +=
( ) Ae-15)(-4)(1041
mAe15.7)t(i 4t-3-4t-T ×+−=
( ) mAe15.7)t(i -4tT +=
Thus,
mAe5.75.730)t(i -4tx −−=
=)t(i x ( ) 0t,mAe35.7 -4t >− Chapter 7, Solution 51.
Consider the circuit below.
fter the switch is closed, applying KVL gives
R-di=
L
R
i
t = 0
+ −
+
v
−
VS
A
dtLRiVS +=
di
−=
RV
iR-dtdi
L S or
dtLRVi S−
tegrating both sides, In
tLR-
Riln I0
=
−
V )t(iS
τ=
−− t-
RVIRVi
lnS0
S
τ=−− t-
S0
S eRVIRVi
or
τ
−+= t-S0
S eRV
IRV
)t(i
which is the same as Eq. (7.60).
hapter 7, Solution 52.
C
A21020
)0(i == , A2)(i =∞
[ ] τ∞−+∞= t-e)(i)0(i)(i)t(i
=)t A2 (i
hapter 7, Solution 53.
(a) Before t = 0,
C
=+
=23
25i A5
After t = 0, τ= t-e)0(i)t(i L
224
R===τ , 5)0(i =
=)t(i Ae5 2t-
(b) Before t = 0, the inductor acts as a short circuit so that the 2 Ω and 4 Ω resistors are short-circuited.
=)t(i A6 After t = 0, we have an RL circuit.
= e)0(i)t(i , 23
R==τ τt- L
=)t(i Ae6 3t2-
Chapter 7, Solution 54. (a) Before t = 0, i is obtained by current division or
=+
= )2(44
4)t(i A1
After t = 0, [ ] τ∞−+∞= t-e)(i)0(i)(i)t(i
eqRL
=τ , Ω=+= 712||44R eq
21
75.3==τ
1)0(i = , 76
)2(34
3)2(
12||4412||4
)(i =+
=+
=∞
t2-e76
176
)t(i
−+=
=)t(i ( ) Ae671 2t-−
(b) Before t = 0, =+
=32
10)t(i A2
After t = 0, 5.42||63R eq =+=
94
5.42
RL
eq===τ
2)0(i = To find , consider the circuit below, at t = when the inductor becomes a short circuit,
)(i ∞
v
2 Ω
24 V
6 Ω
+ −
+ −
i
2 H 10 V
3 Ω
9v3v
6v24
2v10
=→=−
+−
A33v
)(i ==∞ 4t9-e)32(3)t(i −+=
=)t(i Ae3 4t9-−
Chapter 7, Solution 55. For t < 0, consider the circuit shown in Fig. (a).
+
v
−
4io io
+ −
io 0.5 H
+ −
0.5 H
+
v
−
+ −
i 3 Ω 8 Ω
2 Ω 2 Ω24 V 20 V
(a) (b)
24i0i424i3 ooo =→=−+
== oi4)t(v V96 A482v
i ==
For t > 0, consider the circuit in Fig. (b).
[ ] τ∞−+∞= t-e)(i)0(i)(i)t(i
48)0(i = , A228
20)(i =
+=∞
Ω=+= 1082R th , 201
105.0
RL
th===τ
-20t-20t e462e)248(2)t(i +=−+= == )t(i2)t(v Ve924 -20t+
Chapter 7, Solution 56.
Ω=+= 105||206R eq , 05.0RL==τ
[ ] τ∞−+∞= t-e)(i)0(i)(i)t(i i(0) is found by applying nodal analysis to the following circuit.
0.5 H
+ −
12 Ω
5 Ω
20 Ω
6 Ω
+
v
−
vx i
20 V
2 A
12v6
v20v
12v
5v20
2 xxxxx =→++=
−+
A26
v)0(i x ==
45||20 =
6.1)4(64
4)(i =
+=∞
-20t0.05t- e4.06.1e)6.12(6.1)t(i +=−+=
Since ,
20t-e-20)()4.0(21
dtdi
L)t(v ==
=)t(v Ve4- -20t Chapter 7, Solution 57.
At , the circuit has reached steady state so that the inductors act like short circuits.
−= 0t
+ −
6 Ω i
5 Ω
i1 i2
20 Ω30 V
31030
20||5630
i ==+
= , 4.2)3(2520
i1 == , 6.0i2 =
A4.2)0(i1 = , A6.0)0(i2 =
For t > 0, the switch is closed so that the energies in L and flow through the closed switch and become dissipated in the 5 Ω and 20 Ω resistors.
1 2L
1t-11 e)0(i)t(i τ= ,
21
55.2
RL
1
11 ===τ
=)t(i1 Ae4.2 -2t
2t-22 e)0(i)t(i τ= ,
51
204
RL
2
22 ===τ
=)t(i2 Ae6.0 -5t
Chapter 7, Solution 58.
For t < 0, 0)t(vo =
For t > 0, , 10)0(i = 531
20)(i =
+=∞
Ω=+= 431R th , 161
441
RL
th===τ
[ ] τ∞−+∞= t-e)(i)0(i)(i)t(i =)t(i ( ) Ae15 -16t+
( ) 16t-16t-o e-16)(5)(
41
e115dtdi
Li3)t(v ++=+=
=)t(vo Ve515 -16t− Chapter 7, Solution 59.
Let I be the current through the inductor. For t < 0, , 0vs = 0)0(i =
For t > 0, 63||64Req =+= , 25.065.1
RL
eq===τ
1)3(42
2)(i =
+=∞
[ ] τ∞−+∞= t-e)(i)0(i)(i)t(i -4te1)t(i −=
)-4)(-e)(5.1(dtdi
L)t(v 4t-o ==
=)t(vo Ve6 -4t
Chapter 7, Solution 60.
Let I be the inductor current. For t < 0, 0)0(i0)t(u =→=
For t > 0, Ω== 420||5Req , 248
RL
eq===τ
4)(i =∞ [ ] τ∞−+∞= t-e)(i)0(i)(i)t(i
( )2t-e14)t(i −=
2t-e21-
)4-)(8(dtdi
L)t(v
==
=)t(v Ve16 -0.5t Chapter 7, Solution 61.
The current source is transformed as shown below.
4 Ω
20u(-t) + 40u(t)
+ −
0.5 H
81
421
RL
===τ , 5)0(i = , 10)(i =∞
[ ] τ∞−+∞= t-e)(i)0(i)(i)t(i =)t(i Ae510 -8t−
8t-e)8-)(5-(
21
dtdi
L)t(v
==
=)t(v Ve20 -8t Chapter 7, Solution 62.
16||3
2RL
eq===τ
For 0 < t < 1, so that 0)1t(u =−
0)0(i = , 61
)(i =∞
( )t-e161
)t(i −=
For t > 1, ( ) 1054.0e161
)1(i 1- =−=
21
61
31
)(i =+=∞ 1)--(te)5.01054.0(5.0)t(i −+=
1)--(te3946.05.0)t(i −= Thus,
=)t(i ( )
>−
<<−
1tAe3946.05.0
1t0Ae161
-1)(t-
t-
Chapter 7, Solution 63.
For t < 0, , 1)t-(u = 25
10)0(i ==
For t > 0, , 0-t)(u = 0)(i =∞
Ω== 420||5R th , 81
45.0
RL
th===τ
[ ] τ∞−+∞= t-e)(i)0(i)(i)t(i =)t(i Ae2 -8t
8t-e)2)(8-(
21
dtdi
L)t(v
==
=)t(v Ve8- -8t Chapter 7, Solution 64.
Let i be the inductor current. For t < 0, the inductor acts like a short circuit and the 3 Ω resistor is short-circuited so that the equivalent circuit is shown in Fig. (a).
v 6 Ω
(b)
i
3 Ω
2 Ω
+ − 10 Ω
(a)
i
3 Ω
6 Ω
+ −
io
10 Ω
A667.16
10)0(ii ===
For t > 0, Ω=+= 46||32R th , 144
RL
th===τ
To find , consider the circuit in Fig. (b). )(i ∞
610
v2v
3v
6v10
=→+=−
65
2v
)(ii ==∞=
[ ] τ∞−+∞= t-e)(i)0(i)(i)t(i
( ) Ae165
e65
610
65
)t(i t-t- −=
−+=
ov is the voltage across the 4 H inductor and the 2 Ω resistor
t-t-t-o e
610
610
e-1)(65
)4(e6
106
10dtdi
Li2)t(v −=
++=+=
=)t(vo ( ) Ve1667.1 -t− Chapter 7, Solution 65.
Since [ ])1t(u)t(u10vs −−= , this is the same as saying that a 10 V source is turned on at t = 0 and a -10 V source is turned on later at t = 1. This is shown in the figure below.
For 0 < t < 1, , 0)0(i = 25
10)(i ==∞
vs
-10
10
1
t
420||5R th == , 21
42
RL
th===τ
[ ] τ∞−+∞= t-e)(i)0(i)(i)t(i
( ) Ae12)t(i -2t−= ( ) 729.1e12)1(i -2 =−=
For t > 1, since 0v0)(i =∞ s =
τ−= )1t(-e)1(i)t(i Ae729.1)t(i )1t(-2 −=
Thus,
=)t(i( )
><<−
− 1tAe729.11t0Ae12
)1t(2-
2t-
Chapter 7, Solution 66.
Following Practice Problem 7.14, τ= t-
T eV)t(v
-4)0(vVT == , 501
)102)(1010(CR 6-3f =××==τ
-50te-4)t(v = 0t,e4)t(v-)t(v -50t
o >==
=×
== 50t-3
o
oo e
10104
R)t(v
)t(i 0t,mAe4.0 -50t >
Chapter 7, Solution 67.
The op amp is a voltage follower so that vvo = as shown below.
R
vo
vo −+
+
vo
−
C
v1 R
R
At node 1,
o1o111o v
32
vR
vvR
0vR
vv=→
−+
−=
−
At the noninverting terminal,
0R
vvdt
dvC 1oo =
−+
ooo1oo v
31v
32vvv
dtdv
RC =−=−=−
RC3v
dtdv oo −=
3RCt-To eV)t(v =
V5)0(vV oT == , 100
3)101)(1010)(3(RC3 6-3 =××==τ
=)t(vo Ve5 3100t-
Chapter 7, Solution 68. This is a very interesting problem and has both an important ideal solution as well as an important practical solution. Let us look at the ideal solution first. Just before the switch closes, the value of the voltage across the capacitor is zero which means that the voltage at both terminals input of the op amp are each zero. As soon as the switch closes, the output tries to go to a voltage such that the input to the op amp both go to 4 volts. The ideal op amp puts out whatever current is necessary to reach this condition. An infinite (impulse) current is necessary if the voltage across the capacitor is to go to 8 volts in zero time (8 volts across the capacitor will result in 4 volts appearing at the negative terminal of the op amp). So vo will be equal to 8 volts for all t > 0. What happens in a real circuit? Essentially, the output of the amplifier portion of the op amp goes to whatever its maximum value can be. Then this maximum voltage appears across the output resistance of the op amp and the capacitor that is in series with it. This results in an exponential rise in the capacitor voltage to the steady-state value of 8 volts.
vC(t) = Vop amp max(1 – e-t/(RoutC)) volts, for all values of vC less than 8 V,
= 8 V when t is large enough so that the 8 V is reached. Chapter 7, Solution 69.
Let v be the capacitor voltage. x
For t < 0, 0)0(vx =
For t > 0, the 20 kΩ and 100 kΩ resistors are in series since no current enters the op amp terminals. As ∞→t , the capacitor acts like an open circuit so that
1348
)4(1010020
10020)(vx =
+++
=∞
Ω=+= k12010020R th , 3000)1025)(10120(CR 3-3th =××==τ
[ ] τ∞−+∞= t-xxxx e)(v)0(v)(v)t(v
( )3000t-x e1
1348
)t(v −=
== )t(v120100
)t(v xo ( ) Ve11340 3000t-−
Chapter 7, Solution 70.
Let v = capacitor voltage. For t < 0, the switch is open and 0)0(v = . For t > 0, the switch is closed and the circuit becomes as shown below.
2
1
C R
+ −v + −
vo
+−
vS
s21 vvv == (1)
dtdv
CR
v0 s =−
(2)
where (3) vvvvvv soos −=→−= From (1),
0RCv
dtdv s ==
∫ =+=RC
vt-)0(vdtv
RC1-
v ss
Since v is constant,
1.0)105)(1020(RC -63 =××=
mVt-200mV0.1
t20-v ==
From (3),
t20020vvv so +=−= =ov mV)t101(20 +
Chapter 7, Solution 71.
Let v = voltage across the capacitor. Let v = voltage across the 8 kΩ resistor. o
For t < 2, so that 0v = 0)2(v = . For t > 2, we have the circuit shown below.
+
vo
−
10 kΩ
+ −
20 kΩ
+
v
−
100 mF
10 kΩ−+ io
4 V 8 kΩ
Since no current enters the op amp, the input circuit forms an RC circuit.
1000)10100)(1010(RC 3-3 =××==τ [ ] τ−∞−+∞= )2t(-e)(v)2(v)(v)t(v
( )1000)2t(-e14)t(v −−= As an inverter,
( )1e2vk20k10-
v 1000)2t(-o −== −
==8v
i oo ( ) A1e25.0 1000)2t(- −−
Chapter 7, Solution 72.
The op amp acts as an emitter follower so that the Thevenin equivalent circuit is shown below.
C
+ −v + −
io
3u(t) R
Hence,
[ ] τ∞−+∞= t-e)(v)0(v)(v)t(v V2-)0(v = , V3)(v =∞ , 1.0)1010)(1010(RC -63 =××==τ
-10t-10t e53e3)--2(3)t(v −=+=
10t-6-o e)10-)(5-)(1010(
dtdv
Ci ×==
=oi 0t,mAe5.0 -10t > Chapter 7, Solution 73.
Consider the circuit below.
Rf
+
vo
−
+ −v
+ −
R1 −+
C v2v1 v3
v1
At node 2,
dtdv
CR
vv
1
21 =−
(1)
At node 3,
f
o3
Rvv
dtdv
C−
= (2)
But and 0v3 = 232 vvvv =−= . Hence, (1) becomes
dtdv
CR
vv
1
1 =−
dtdv
CRvv 11 =−
or CR
vCR
vdtdv
1
1
1=+
which is similar to Eq. (7.42). Hence,
( )
>−+<
= τ 0tevvv0tv
)t(v t-1T1
T
where and 1)0(vvT == 4v1 = 2.0)1020)(1010(CR 6-3
1 =××==τ
>−<
=0te340t1
)t(v 5t-
From (2),
)e15)(1020)(1020(dtdv
CR-v 5t-6-3fo ××==
0t,e-6v -5to >= =ov V)t(ue6- -5t
Chapter 7, Solution 74. Let v = capacitor voltage.
Rf
+
vo
−
+ −v
+ −
R1 −+
C v2v1 v3
v1
For t < 0, 0)0(v =For t > 0, . Consider the circuit below. A10is µ=
Rv
dtdv
Cis += (1)
[ ] τ∞−+∞= t-e)(v)0(v)(v)t(v (2) It is evident from the circuit that
1.0)1050)(102(RC 36 =××==τ −
is
+
vo
−
C Rf
−+
R
is
At steady state, the capacitor acts like an open circuit so that i passes through R. Hence,
s
V5.0)1050)(1010(Ri)(v 36s =××==∞ −
Then,
( ) Ve15.0)t(v -10t−= (3)
But fsof
os Ri-v
Rv0
i =→−
= (4)
Combining (1), (3), and (4), we obtain
dtdv
CRvRR-
v ff
o −=
dtdv
)102)(1010(v51-
v 6-3o ××−=
( )-10t-2-10to e10-)5.0)(102(e0.1-0.1v ×−+=
1.0e2.0v -10to −= =ov ( ) V1e21.0 -10t −
Chapter 7, Solution 75. Let v = voltage at the noninverting terminal. 1
Let 2v = voltage at the inverting terminal.
For t > 0, 4vvv s21 ===
o1
s iR
v0=
−, Ω= k20R1
Ri-v oo = (1)
Also, dtdv
CRv
i2
o += , Ω= k10R 2 , F2C µ=
i.e. dtdv
CRv
Rv-
21
s += (2)
This is a step response.
[ ] τ∞−+∞= t-e)(v)0(v)(v)t(v , 1)0(v =
where 501
)102)(1010(CR 6-32 =××==τ
At steady state, the capacitor acts like an open circuit so that i passes through
. Hence, as o
2R ∞→t
2o
1
s
R)(v
iRv- ∞
==
i.e. -2)4(2010-
vRR-
)(v s1
2 ===∞
-50te2)(1-2)t(v ++=
-50te3-2)t(v += But os vvv −=
or t50-so e324vvv −+=−=
=ov Ve36 t50-−
mA-0.220k
4-Rv-
i1
so ===
or =+=dtdv
CRv
i2
o mA0.2-
Chapter 7, Solution 76. The schematic is shown below. For the pulse, we use IPWL and enter the corresponding values as attributes as shown. By selecting Analysis/Setup/Transient, we let Print Step = 25 ms and Final Step = 2 s since the width of the input pulse is 1 s. After saving and simulating the circuit, we select Trace/Add and display –V(C1:2). The plot of V(t) is shown below.
Chapter 7, Solution 77. The schematic is shown below. We click Marker and insert Mark Voltage Differential at the terminals of the capacitor to display V after simulation. The plot of V is shown below. Note from the plot that V(0) = 12 V and V(∞) = -24 V which are correct.
Chapter 7, Solution 78. (a) When the switch is in position (a), the schematic is shown below. We insert
IPROBE to display i. After simulation, we obtain,
i(0) = 7.714 A from the display of IPROBE.
(b) When the switch is in position (b), the schematic is as shown below. For inductor I1, we let IC = 7.714. By clicking Analysis/Setup/Transient, we let Print Step = 25 ms and Final Step = 2 s. After Simulation, we click Trace/Add in the probe menu and display I(L1) as shown below. Note that i(∞) = 12A, which is correct.
Chapter 7, Solution 79. When the switch is in position 1, io(0) = 12/3 = 4A. When the switch is in position 2,
3/80,3/84//)53(A, 5.035
4)( ===+=−=+
−=∞LR
Ri ThTho τ
A 5.45.0)]()0([)()( 80/3/ tt
oooo eeiiiti −− +−=∞−+∞= τ Chapter 7, Solution 80.
(a) When the switch is in position A, the 5-ohm and 6-ohm resistors are short-
circuited so that 0)0()0()0( 21 === ovii but the current through the 4-H inductor is iL(0) =30/10 = 3A. (b) When the switch is in position B,
5.04/2,26//3 ===Ω==LR
R ThTh τ
A 330)]()0([)()( 25.0// ttt
LLLL eeeiiiti −−− =+=∞−+∞= τ
(c) A 0)(93)(,A 2
51030)( 21 =∞−=∞=+
=∞ Liii
V 0)()( =∞→= oL
o vdtdiLtv
Chapter 7, Solution 81. The schematic is shown below. We use VPWL for the pulse and specify the attributes as shown. In the Analysis/Setup/Transient menu, we select Print Step = 25 ms and final Step = 3 S. By inserting a current marker at one termial of LI, we automatically obtain the plot of i after simulation as shown below.
hapter 7, Solution 82.
C
=××
=τ
=→=τ 6-
-3
10100103
CRRC Ω30
Chapter 7, Solution 83.
sxxxRCvv 51010151034,0)0(,120)( 66 =====∞ −τ
)1(1206.85)]()0([)()( 510// tt eevvvtv −− −=→∞−+∞= τ Solving for t gives st 16.637488.3ln510 == speed = 4000m/637.16s = 6.278m/s
Chapter 7, Solution 84.
Let Io be the final value of the current. Then 50/18/16.0/),1()( / ===−= − LReIti t
o ττ
. ms 33.184.0
1ln501)1(6.0 50 ==→−= − teII t
oo
Chapter 7, Solution 85.
(a) s24)106)(104(RC -66 =××==τ Since [ ] τ∞−+∞= t-e)(v)0(v)(v)t(v
[ ] τ∞−=∞− 1t-1 e)(v)0(v)(v)t(v (1)
[ ] τ∞−=∞− 2t-2 e)(v)0(v)(v)t(v (2)
Dividing (1) by (2),
τ−=∞−∞− )tt(
2
1 12e)(v)t(v)(v)t(v
∞−∞−
τ=−=)(v)t(v)(v)t(v
lnttt2
1120
==
−−
= )2(ln241203012075
ln24t0 s63.16
(b) Since , the light flashes repeatedly every tt0 <==τ RC s24
Chapter 7, Solution 86.
[ ] τ∞−+∞= t-e)(v)0(v)(v)t(v 12)(v =∞ , 0)0(v = ( )τ−= t-e112)t(v
( )τ−== 0t-0 e1128)t(v
31
ee1128
00 t-t- =→−= ττ
)3(lnt0 τ= For , Ω= k100R
s2.0)102)(10100(RC -63 =××==τ s2197.0)3(ln2.0t0 ==
For , Ω= M1R
s2)102)(101(RC -66 =××==τ s197.2)3(ln2t0 ==
Thus,
s197.2ts2197.0 0 <<
Chapter 7, Solution 87.
Let i be the inductor current.
For t < 0, A2.1100120
)0(i ==−
For t > 0, we have an RL circuit
1.0400100
50RL
=+
==τ , 0)(i =∞
[ ] τ∞−+∞= t-e)(i)0(i)(i)t(i -10te2.1)t(i =
At t = 100 ms = 0.1 s,
== -1e2.1)1.0(i A441.0 which is the same as the current through the resistor.
Chapter 7, Solution 88.
(a) s60)10200)(10300(RC -123 µ=××==τAs a differentiator,
ms6.0s60010T =µ=τ> i.e. =minT ms6.0
(b) s60RC µ==τAs an integrator,
s61.0T µ=τ< i.e. =maxT s6 µ
Chapter 7, Solution 89.
Since s1T1.0 µ=<τ
s1RL
µ<
)101)(10200(10RL -63-6 ××=×<
mH200L <
Chapter 7, Solution 90.
We determine the Thevenin equivalent circuit for the capacitor C . s
ips
sth v
RRR
v+
= , psth R||RR =
Rth
+ −
Vth Cs
The Thevenin equivalent is an RC circuit. Since
ps
sith RR
R101
v101
v+
=→=
===96
R91
R ps ΩM32
Also,
s15CR sth µ==τ
where Ω=+
== M6.0326)32(6
R||RR spth
=××
=τ
= 6
6-
ths 100.6
1015R
C pF25
Chapter 7, Solution 91.
mA2405012
)0(io == , 0)(i =∞
[ ] τ∞−+∞= t-e)(i)0(i)(i)t(i τ= t-e240)t(i
R2
RL==τ
τ== 0t-0 e24010)t(i
)24(lnt24e 0t0 τ=→=τ
R2
573.1)24(ln
5)24(ln
t 0 ====τ
==573.12
R Ω271.1
Chapter 7, Solution 92.
<<×
<<×⋅×==
DR6-
R3-9-
ttt10510-
tt0102
10
104dtdv
Ci
µ+<<<<µ
=s5ms2tms2mA8-
ms2t0A20)t(i
which is sketched below.
20 µA
2 ms
5 µs
-8 mA
t
i(t)
(not to scale)
Chapter 8, Solution 1. (a) At t = 0-, the circuit has reached steady state so that the equivalent circuit is shown in Figure (a).
+
vL
−
6 Ω
10 H
+
v
−(a)
6 Ω
+ −
6 Ω
VS
10 µF
(b)
i(0-) = 12/6 = 2A, v(0-) = 12V
At t = 0+, i(0+) = i(0-) = 2A, v(0+) = v(0-) = 12V (b) For t > 0, we have the equivalent circuit shown in Figure (b).
vL = Ldi/dt or di/dt = vL/L Applying KVL at t = 0+, we obtain,
vL(0+) – v(0+) + 10i(0+) = 0
vL(0+) – 12 + 20 = 0, or vL(0+) = -8 Hence, di(0+)/dt = -8/2 = -4 A/s Similarly, iC = Cdv/dt, or dv/dt = iC/C
iC(0+) = -i(0+) = -2
dv(0+)/dt = -2/0.4 = -5 V/s (c) As t approaches infinity, the circuit reaches steady state.
i(∞) = 0 A, v(∞) = 0 V
Chapter 8, Solution 2. (a) At t = 0-, the equivalent circuit is shown in Figure (a). 25 kΩ 20 kΩ
iR + −
+
v
−
iL
60 kΩ
80V
(a) 25 kΩ 20 kΩ
iR + −
iL 80V
(b)
60||20 = 15 kohms, iR(0-) = 80/(25 + 15) = 2mA. By the current division principle,
iL(0-) = 60(2mA)/(60 + 20) = 1.5 mA
vC(0-) = 0 At t = 0+,
vC(0+) = vC(0-) = 0
iL(0+) = iL(0-) = 1.5 mA
80 = iR(0+)(25 + 20) + vC(0-)
iR(0+) = 80/45k = 1.778 mA But, iR = iC + iL
1.778 = iC(0+) + 1.5 or iC(0+) = 0.278 mA
(b) vL(0+) = vC(0+) = 0
But, vL = LdiL/dt and diL(0+)/dt = vL(0+)/L = 0
diL(0+)/dt = 0
Again, 80 = 45iR + vC
0 = 45diR/dt + dvC/dt
But, dvC(0+)/dt = iC(0+)/C = 0.278 mohms/1 µF = 278 V/s
Hence, diR(0+)/dt = (-1/45)dvC(0+)/dt = -278/45
diR(0+)/dt = -6.1778 A/s
Also, iR = iC + iL
diR(0+)/dt = diC(0+)/dt + diL(0+)/dt
-6.1788 = diC(0+)/dt + 0, or diC(0+)/dt = -6.1788 A/s (c) As t approaches infinity, we have the equivalent circuit in Figure
(b).
iR(∞) = iL(∞) = 80/45k = 1.778 mA
iC(∞) = Cdv(∞)/dt = 0. Chapter 8, Solution 3. At t = 0-, u(t) = 0. Consider the circuit shown in Figure (a). iL(0-) = 0, and vR(0-) = 0. But, -vR(0-) + vC(0-) + 10 = 0, or vC(0-) = -10V. (a) At t = 0+, since the inductor current and capacitor voltage cannot change abruptly,
the inductor current must still be equal to 0A, the capacitor has a voltage equal to –10V. Since it is in series with the +10V source, together they represent a direct short at t = 0+. This means that the entire 2A from the current source flows through the capacitor and not the resistor. Therefore, vR(0+) = 0 V.
(b) At t = 0+, vL(0+) = 0, therefore LdiL(0+)/dt = vL(0+) = 0, thus, diL/dt = 0A/s,
iC(0+) = 2 A, this means that dvC(0+)/dt = 2/C = 8 V/s. Now for the value of dvR(0+)/dt. Since vR = vC + 10, then dvR(0+)/dt = dvC(0+)/dt + 0 = 8 V/s.
40 Ω 40 Ω
+ − 10V
+
vC
−
10 Ω
2A
iL +
vR
−
+
vR
−
+ − 10V
+
vC
− 10 Ω
(b) (a) (c) As t approaches infinity, we end up with the equivalent circuit shown in Figure (b).
iL(∞) = 10(2)/(40 + 10) = 400 mA
vC(∞) = 2[10||40] –10 = 16 – 10 = 6V
vR(∞) = 2[10||40] = 16 V Chapter 8, Solution 4. (a) At t = 0-, u(-t) = 1 and u(t) = 0 so that the equivalent circuit is shown in Figure (a).
i(0-) = 40/(3 + 5) = 5A, and v(0-) = 5i(0-) = 25V. Hence, i(0+) = i(0-) = 5A v(0+) = v(0-) = 25V
3 Ω
5 Ω
i +
v
−
+ −
40V
(a)
0.25 H3 Ω
iRiC
+ −
+ vL −i
5 Ω0.1F
4 A
40V (b) (b) iC = Cdv/dt or dv(0+)/dt = iC(0+)/C For t = 0+, 4u(t) = 4 and 4u(-t) = 0. The equivalent circuit is shown in Figure (b). Since i and v cannot change abruptly,
iR = v/5 = 25/5 = 5A, i(0+) + 4 =iC(0+) + iR(0+)
5 + 4 = iC(0+) + 5 which leads to iC(0+) = 4
dv(0+)/dt = 4/0.1 = 40 V/s Chapter 8, Solution 5. (a) For t < 0, 4u(t) = 0 so that the circuit is not active (all initial conditions = 0).
iL(0-) = 0 and vC(0-) = 0.
For t = 0+, 4u(t) = 4. Consider the circuit below.
iL
iC + vL −
1 H
+
v
−
4 Ω 0.25F +
vC
−
Ai
6 Ω
4A
Since the 4-ohm resistor is in parallel with the capacitor,
i(0+) = vC(0+)/4 = 0/4 = 0 A Also, since the 6-ohm resistor is in series with the inductor, v(0+) = 6iL(0+) = 0V.
(b) di(0+)/dt = d(vR(0+)/R)/dt = (1/R)dvR(0+)/dt = (1/R)dvC(0+)/dt
= (1/4)4/0.25 A/s = 4 A/s
v = 6iL or dv/dt = 6diL/dt and dv(0+)/dt = 6diL(0+)/dt = 6vL(0+)/L = 0
Therefore dv(0+)/dt = 0 V/s
(c) As t approaches infinity, the circuit is in steady-state.
i(∞) = 6(4)/10 = 2.4 A
v(∞) = 6(4 – 2.4) = 9.6 V Chapter 8, Solution 6. (a) Let i = the inductor current. For t < 0, u(t) = 0 so that
i(0) = 0 and v(0) = 0.
For t > 0, u(t) = 1. Since, v(0+) = v(0-) = 0, and i(0+) = i(0-) = 0. vR(0+) = Ri(0+) = 0 V
Also, since v(0+) = vR(0+) + vL(0+) = 0 = 0 + vL(0+) or vL(0+) = 0 V. (1)
(b) Since i(0+) = 0, iC(0+) = VS/RS
But, iC = Cdv/dt which leads to dv(0+)/dt = VS/(CRS) (2)
From (1), dv(0+)/dt = dvR(0+)/dt + dvL(0+)/dt (3) vR = iR or dvR/dt = Rdi/dt (4)
But, vL = Ldi/dt, vL(0+) = 0 = Ldi(0+)/dt and di(0+)/dt = 0 (5)
From (4) and (5), dvR(0+)/dt = 0 V/s
From (2) and (3), dvL(0+)/dt = dv(0+)/dt = Vs/(CRs) (c) As t approaches infinity, the capacitor acts like an open circuit, while the inductor acts like a short circuit.
vR(∞) = [R/(R + Rs)]Vs
vL(∞) = 0 V
Chapter 8, Solution 7.
s2 + 4s + 4 = 0, thus s1,2 = 2
4x444 2 −±− = -2, repeated roots.
v(t) = [(A + Bt)e-2t], v(0) = 1 = A
dv/dt = [Be-2t] + [-2(A + Bt)e-2t]
dv(0)/dt = -1 = B – 2A = B – 2 or B = 1.
Therefore, v(t) = [(1 + t)e-2t] V
Chapter 8, Solution 8.
s2 + 6s + 9 = 0, thus s1,2 = 2
3666 2 −±− = -3, repeated roots.
i(t) = [(A + Bt)e-3t], i(0) = 0 = A
di/dt = [Be-3t] + [-3(Bt)e-3t]
di(0)/dt = 4 = B.
Therefore, i(t) = [4te-3t] A
Chapter 8, Solution 9.
s2 + 10s + 25 = 0, thus s1,2 = 2
101010 −±− = -5, repeated roots.
i(t) = [(A + Bt)e-5t], i(0) = 10 = A
di/dt = [Be-5t] + [-5(A + Bt)e-5t]
di(0)/dt = 0 = B – 5A = B – 50 or B = 50.
Therefore, i(t) = [(10 + 50t)e-5t] A
Chapter 8, Solution 10.
s2 + 5s + 4 = 0, thus s1,2 = 2
16255 −±− = -4, -1.
v(t) = (Ae-4t + Be-t), v(0) = 0 = A + B, or B = -A
dv/dt = (-4Ae-4t - Be-t)
dv(0)/dt = 10 = – 4A – B = –3A or A = –10/3 and B = 10/3.
Therefore, v(t) = (–(10/3)e-4t + (10/3)e-t) V Chapter 8, Solution 11.
s2 + 2s + 1 = 0, thus s1,2 = 2
442 −±− = -1, repeated roots.
v(t) = [(A + Bt)e-t], v(0) = 10 = A
dv/dt = [Be-t] + [-(A + Bt)e-t]
dv(0)/dt = 0 = B – A = B – 10 or B = 10.
Therefore, v(t) = [(10 + 10t)e-t] V
Chapter 8, Solution 12. (a) Overdamped when C > 4L/(R2) = 4x0.6/400 = 6x10-3, or C > 6 mF (b) Critically damped when C = 6 mF (c) Underdamped when C < 6mF
Chapter 8, Solution 13. Let R||60 = Ro. For a series RLC circuit,
ωo = LC1 =
4x01.01 = 5
For critical damping, ωo = α = Ro/(2L) = 5
or Ro = 10L = 40 = 60R/(60 + R)
which leads to R = 120 ohms
Chapter 8, Solution 14. This is a series, source-free circuit. 60||30 = 20 ohms
α = R/(2L) = 20/(2x2) = 5 and ωo = LC1 =
04.01 = 5
ωo = α leads to critical damping
i(t) = [(A + Bt)e-5t], i(0) = 2 = A
v = Ldi/dt = 2[Be-5t] + [-5(A + Bt)e-5t]
v(0) = 6 = 2B – 10A = 2B – 20 or B = 13.
Therefore, i(t) = [(2 + 13t)e-5t] A
Chapter 8, Solution 15. This is a series, source-free circuit. 60||30 = 20 ohms
α = R/(2L) = 20/(2x2) = 5 and ωo = LC1 =
04.01 = 5
ωo = α leads to critical damping
i(t) = [(A + Bt)e-5t], i(0) = 2 = A
v = Ldi/dt = 2[Be-5t] + [-5(A + Bt)e-5t]
v(0) = 6 = 2B – 10A = 2B – 20 or B = 13.
Therefore, i(t) = [(2 + 13t)e-5t] A
Chapter 8, Solution 16. At t = 0, i(0) = 0, vC(0) = 40x30/50 = 24V For t > 0, we have a source-free RLC circuit.
α = R/(2L) = (40 + 60)/5 = 20 and ωo = LC1 =
5.2x1013−
= 20
ωo = α leads to critical damping
i(t) = [(A + Bt)e-20t], i(0) = 0 = A
di/dt = [Be-20t] + [-20(Bt)e-20t],
but di(0)/dt = -(1/L)[Ri(0) + vC(0)] = -(1/2.5)[0 + 24]
Hence, B = -9.6 or i(t) = [-9.6te-20t] A
Chapter 8, Solution 17.
.iswhich,20
412
10L2
R
10
251
411
LC1
240)600(4)VRI(L1
dt)0(di
6015x4V)0(v,0I)0(i
o
o
00
00
ω>===α
===ω
−=+−=+−=
=====
( )t268t32.3721
2121
t32.372
t68.21
2o
2
ee928.6)t(i
A928.6AtoleadsThis
240A32.37A68.2dt
)0(di,AA0)0(i
eAeA)t(i
32.37,68.23102030020s
−−
−−
−=
−=−=
−=−−=+==
+=
−−=±−=±−=ω−α±α−=
getwe,60dt)t(iC1)t(v,Since t
0 +∫=
v(t) = (60 + 64.53e-2.68t – 4.6412e-37.32t) V
Chapter 8, Solution 18. When the switch is off, we have a source-free parallel RLC circuit.
5.02
1,2125.0
11=====
RCxLCo αω
936.125.04case dunderdampe 22d =−=−=→< αωωωα oo
Io(0) = i(0) = initial inductor current = 20/5 = 4A Vo(0) = v(0) = initial capacitor voltage = 0 V
)936.1sin936.1cos()sincos()( 215.0
21 tAtAetAtAetv tdd
t +=+= −− αα ωω v(0) =0 = A1
)936.1cos936.1936.1sin936.1()936.1sin936.1cos)(5.0( 215.0
215.0 tAtAetAtAe
dtdv tt +−++−= −− αα
066.2936.15.041
)40()()0(221 −=→+−=−=
+−=
+−= AAA
RCRIV
dtdv oo
Thus,
tetv t 936.1sin066.2)( 5.0−−= Chapter 8, Solution 19. For t < 0, the equivalent circuit is shown in Figure (a). 10 Ω i
+ −
+
v
−
i L C
120V (a) (b)
i(0) = 120/10 = 12, v(0) = 0
+
v
−
For t > 0, we have a series RLC circuit as shown in Figure (b) with R = 0 = α.
ωo = LC1 =
41 = 0.5 = ωd
i(t) = [Acos0.5t + Bsin0.5t], i(0) = 12 = A
v = -Ldi/dt, and -v/L = di/dt = 0.5[-12sin0.5t + Bcos0.5t],
which leads to -v(0)/L = 0 = B
Hence, i(t) = 12cos0.5t A and v = 0.5
However, v = -Ldi/dt = -4(0.5)[-12sin0.5t] = 24sin0.5t V
Chapter 8, Solution 20. For t < 0, the equivalent circuit is as shown below.
2 Ω
+ − 12
− +vC
i
v(0) = -12V and i(0) = 12/2 = 6A For t > 0, we have a series RLC circuit.
α = R/(2L) = 2/(2x0.5) = 2
ωo = 1/ 2241x5.0/1LC ==
Since α is less than ωo, we have an under-damped response.
24822od =−=α−ω=ω
i(t) = (Acos2t + Bsin2t)e-2t
i(0) = 6 = A
di/dt = -2(6cos2t + Bsin2t)e-2t + (-2x6sin2t + 2Bcos2t)e-αt
di(0)/dt = -12 + 2B = -(1/L)[Ri(0) + vC(0)] = -2[12 – 12] = 0
Thus, B = 6 and i(t) = (6cos2t + 6sin2t)e-2t A Chapter 8, Solution 21. By combining some resistors, the circuit is equivalent to that shown below. 60||(15 + 25) = 24 ohms.
12 Ω
+ − +
v
−
t = 0 i
24 Ω
6 Ω 3 H
24V (1/27)F At t = 0-, i(0) = 0, v(0) = 24x24/36 = 16V For t > 0, we have a series RLC circuit. R = 30 ohms, L = 3 H, C = (1/27) F
α = R/(2L) = 30/6 = 5
27/1x3/1LC/1o ==ω = 3, clearly α > ωo (overdamped response)
s1,2 = 222o
2 355 −±−=ω−α±α− = -9, -1
v(t) = [Ae-t + Be-9t], v(0) = 16 = A + B (1)
i = Cdv/dt = C[-Ae-t - 9Be-9t]
i(0) = 0 = C[-A – 9B] or A = -9B (2) From (1) and (2), B = -2 and A = 18.
Hence, v(t) = (18e-t – 2e-9t) V
Chapter 8, Solution 22. α = 20 = 1/(2RC) or RC = 1/40 (1)
22od 50 α−ω==ω which leads to 2500 + 400 = ωo
2 = 1/(LC) Thus, LC 1/2900 (2) In a parallel circuit, vC = vL = vR But, iC = CdvC/dt or iC/C = dvC/dt
= -80e-20tcos50t – 200e-20tsin50t + 200e-20tsin50t – 500e-20tcos50t = -580e-20tcos50t
iC(0)/C = -580 which leads to C = -6.5x10-3/(-580) = 11.21 µF
R = 1/(40C) = 106/(2900x11.21) = 2.23 kohms
L = 1/(2900x11.21) = 30.76 H
Chapter 8, Solution 23. Let Co = C + 0.01. For a parallel RLC circuit,
α = 1/(2RCo), ωo = 1/ oLC
α = 1 = 1/(2RCo), we then have Co = 1/(2R) = 1/20 = 50 mF
ωo = 1/ 5.0x5.0 = 6.32 > α (underdamped)
Co = C + 10 mF = 50 mF or 40 mF Chapter 8, Solution 24. For t < 0, u(-t) 1, namely, the switch is on.
v(0) = 0, i(0) = 25/5 = 5A For t > 0, the voltage source is off and we have a source-free parallel RLC circuit.
α = 1/(2RC) = 1/(2x5x10-3) = 100
ωo = 1/ 310x1.0/1LC −= = 100
ωo = α (critically damped)
v(t) = [(A1 + A2t)e-100t]
v(0) = 0 = A1
dv(0)/dt = -[v(0) + Ri(0)]/(RC) = -[0 + 5x5]/(5x10-3) = -5000
But, dv/dt = [(A2 + (-100)A2t)e-100t]
Therefore, dv(0)/dt = -5000 = A2 – 0 v(t) = -5000te-100t V
Chapter 8, Solution 25. In the circuit in Fig. 8.76, calculate io(t) and vo(t) for t>0.
(1/4)F
+ −
8 Ω
2 Ω
t=0, note this is a make before break
switch so the inductor current is
not interrupted.
1 H io(t) 30V
Figure 8.78 For Problem 8.25. At t = 0-, vo(0) = (8/(2 + 8)(30) = 24 For t > 0, we have a source-free parallel RLC circuit.
α = 1/(2RC) = ¼
ωo = 1/ 241x1/1LC ==
Since α is less than ωo, we have an under-damped response.
9843.1)16/1(422od =−=α−ω=ω
vo(t) = (A1cosωdt + A2sinωdt)e-αt
+
vo(t)
−
vo(0) = 24 = A1 and io(t) = C(dvo/dt) = 0 when t = 0.
dvo/dt = -α(A1cosωdt + A2sinωdt)e-αt + (-ωdA1sinωdt + ωdA2cosωdt)e-αt
at t = 0, we get dvo(0)/dt = 0 = -αA1 + ωdA2
Thus, A2 = (α/ωd)A1 = (1/4)(24)/1.9843 = 3.024
vo(t) = (24cosωdt + 3.024sinωdt)e-t/4 volts
Chapter 8, Solution 26.
s2 + 2s + 5 = 0, which leads to s1,2 = 2
2042 −±− = -1±j4
i(t) = Is + [(A1cos4t + A2sin4t)e-t], Is = 10/5 = 2
i(0) = 2 = = 2 + A1, or A1 = 0
di/dt = [(A2cos4t)e-t] + [(-A2sin4t)e-t] = 4 = 4A2, or A2 = 1
i(t) = 2 + sin4te-t A
Chapter 8, Solution 27.
s2 + 4s + 8 = 0 leads to s = 2j22
32164±−=
−±−
v(t) = Vs + (A1cos2t + A2sin2t)e-2t
8Vs = 24 means that Vs = 3
v(0) = 0 = 3 + A1 leads to A1 = -3
dv/dt = -2(A1cos2t + A2sin2t)e-2t + (-2A1sin2t + 2A2cos2t)e-2t
0 = dv(0)/dt = -2A1 +2A2 or A2 = A1 = -3
v(t) = [3 – 3(cos2t + sin2t)e-2t] volts
Chapter 8, Solution 28.
The characteristic equation is s2 + 6s + 8 with roots
2,42
323662,1 −−=
−±−=s
Hence,
tts BeAeIti 42)( −− ++=
5.1128 =→= ss II
BAi ++=→= 5.100)0( (1)
tt BeAe
dtdi 42 42 −− −−=
BABAdtdi 210422)0(
++=→−−== (2)
Solving (1) and (2) leads to A=-2 and B=0.5.
tt eeti 42 5.025.1)( −− +−= A Chapter 8, Solution 29.
(a) s2 + 4 = 0 which leads to s1,2 = ±j2 (an undamped circuit)
v(t) = Vs + Acos2t + Bsin2t
4Vs = 12 or Vs = 3
v(0) = 0 = 3 + A or A = -3
dv/dt = -2Asin2t + 2Bcos2t
dv(0)/dt = 2 = 2B or B = 1, therefore v(t) = (3 – 3cos2t + sin2t) V
(b) s2 + 5s + 4 = 0 which leads to s1,2 = -1, -4
i(t) = (Is + Ae-t + Be-4t)
4Is = 8 or Is = 2
i(0) = -1 = 2 + A + B, or A + B = -3 (1)
di/dt = -Ae-t - 4Be-4t
di(0)/dt = 0 = -A – 4B, or B = -A/4 (2)
From (1) and (2) we get A = -4 and B = 1
i(t) = (2 – 4e-t + e-4t) A
(c) s2 + 2s + 1 = 0, s1,2 = -1, -1
v(t) = [Vs + (A + Bt)e-t], Vs = 3.
v(0) = 5 = 3 + A or A = 2
dv/dt = [-(A + Bt)e-t] + [Be-t]
dv(0)/dt = -A + B = 1 or B = 2 + 1 = 3
v(t) = [3 + (2 + 3t)e-t] V Chapter 8, Solution 30.
22
222
1 800,500 oo ss ωααωαα −−−=−=−+−=−=
LRss
26502130021 ==→−=−=+ αα
Hence,
mH 8.1536502
2002
===x
RLα
LCss oo
145.6232300 2221 ==→−==− ωωα
F25.16)45.632(
12 µ==L
C
Chapter 8, Solution 31. For t = 0-, we have the equivalent circuit in Figure (a). For t = 0+, the equivalent circuit is shown in Figure (b). By KVL,
v(0+) = v(0-) = 40, i(0+) = i(0-) = 1 By KCL, 2 = i(0+) + i1 = 1 + i1 which leads to i1 = 1. By KVL, -vL + 40i1 + v(0+) = 0 which leads to vL(0+) = 40x1 + 40 = 80
vL(0+) = 80 V, vC(0+) = 40 V 40 Ω 10 Ω i1
0.5H
+
v
−50V
+ −
+
vL
−
40 Ω 10 Ω
i +
v
−
50V
+ −
(a) (b) Chapter 8, Solution 32. For t = 0-, the equivalent circuit is shown below. 2 A
i
6 Ω
+ −v
i(0-) = 0, v(0-) = -2x6 = -12V For t > 0, we have a series RLC circuit with a step input.
α = R/(2L) = 6/2 = 3, ωo = 1/ 04.0/1LC =
s = 4j32593 ±−=−±−
Thus, v(t) = Vf + [(Acos4t + Bsin4t)e-3t]
where Vf = final capacitor voltage = 50 V
v(t) = 50 + [(Acos4t + Bsin4t)e-3t]
v(0) = -12 = 50 + A which gives A = -62
i(0) = 0 = Cdv(0)/dt
dv/dt = [-3(Acos4t + Bsin4t)e-3t] + [4(-Asin4t + Bcos4t)e-3t]
0 = dv(0)/dt = -3A + 4B or B = (3/4)A = -46.5
v(t) = 50 + [(-62cos4t – 46.5sin4t)e-3t] V Chapter 8, Solution 33. We may transform the current sources to voltage sources. For t = 0-, the equivalent circuit is shown in Figure (a).
1 H
i
+ − 30V
+
v
−
4F
i
+ − 5 Ω
10 Ω
+
v
−
10 Ω 30V (a) (b)
i(0) = 30/15 = 2 A, v(0) = 5x30/15 = 10 V
For t > 0, we have a series RLC circuit.
α = R/(2L) = 5/2 = 2.5
4/1LC/1o ==ω = 0.25, clearly α > ωo (overdamped response)
s1,2 = 25.025.65.22o
2 −±−=ω−α±α− = -4.95, -0.05
v(t) = Vs + [A1e-4.95t + A2e-0.05t], v = 20.
v(0) = 10 = 20 + A1 + A2 (1)
i(0) = Cdv(0)/dt or dv(0)/dt = 2/4 = 1/2
Hence, ½ = -4.95A1 – 0.05A2 (2) From (1) and (2), A1 = 0, A2 = -10.
v(t) = 20 – 10e-0.05t V Chapter 8, Solution 34. Before t = 0, the capacitor acts like an open circuit while the inductor behaves like a short circuit.
i(0) = 0, v(0) = 20 V For t > 0, the LC circuit is disconnected from the voltage source as shown below.
+ − Vx
(1/16)F
(¼) H
i This is a lossless, source-free, series RLC circuit.
α = R/(2L) = 0, ωo = 1/ LC = 1/41
161
+ = 8, s = ±j8
Since α is less than ωo, we have an underdamped response. Therefore,
i(t) = A1cos8t + A2sin8t where i(0) = 0 = A1
di(0)/dt = (1/L)vL(0) = -(1/L)v(0) = -4x20 = -80 However, di/dt = 8A2cos8t, thus, di(0)/dt = -80 = 8A2 which leads to A2 = -10 Now we have i(t) = -10sin8t A
Chapter 8, Solution 35. At t = 0-, iL(0) = 0, v(0) = vC(0) = 8 V For t > 0, we have a series RLC circuit with a step input.
α = R/(2L) = 2/2 = 1, ωo = 1/ LC = 1/ 5/1 = 5
s1,2 = 2j12
o2 ±−=ω−α±α−
v(t) = Vs + [(Acos2t + Bsin2t)e-t], Vs = 12.
v(0) = 8 = 12 + A or A = -4, i(0) = Cdv(0)/dt = 0.
But dv/dt = [-(Acos2t + Bsin2t)e-t] + [2(-Asin2t + Bcos2t)e-t]
0 = dv(0)/dt = -A + 2B or 2B = A = -4 and B = -2
v(t) = 12 – (4cos2t + 2sin2t)e-t V. Chapter 8, Solution 36. For t = 0-, 3u(t) = 0. Thus, i(0) = 0, and v(0) = 20 V. For t > 0, we have the series RLC circuit shown below.
20 V 2 Ω0.2 F
i 10 Ω
+ −
+ −
5 H 10 Ω +
v
−
15V
α = R/(2L) = (2 + 5 + 1)/(2x5) = 0.8
ωo = 1/ LC = 1/ 2.0x5 = 1
s1,2 = 6.0j8.02o
2 ±−=ω−α±α−
v(t) = Vs + [(Acos0.6t + Bsin0.6t)e-0.8t]
Vs = 15 + 20 = 35V and v(0) = 20 = 35 + A or A = -15
i(0) = Cdv(0)/dt = 0
But dv/dt = [-0.8(Acos0.6t + Bsin0.6t)e-0.8t] + [0.6(-Asin0.6t + Bcos0.6t)e-0.8t]
0 = dv(0)/dt = -0.8A + 0.6B which leads to B = 0.8x(-15)/0.6 = -20
v(t) = 35 – [(15cos0.6t + 20sin0.6t)e-0.8t] V i = Cdv/dt = 0.2[0.8(15cos0.6t + 20sin0.6t)e-0.8t] + [0.6(15sin0.6t – 20cos0.6t)e-0.8t]
i(t) = [(5sin0.6t)e-0.8t] A
Chapter 8, Solution 37. For t = 0-, the equivalent circuit is shown below.
6 Ω
+ − 10V
+ −
i2
i1
+
v(0)
−
6 Ω6 Ω
30V
18i2 – 6i1 = 0 or i1 = 3i2 (1) -30 + 6(i1 – i2) + 10 = 0 or i1 – i2 = 10/3 (2)
From (1) and (2). i1 = 5, i2 = 5/3
i(0) = i1 = 5A
-10 – 6i2 + v(0) = 0
v(0) = 10 + 6x5/3 = 20
For t > 0, we have a series RLC circuit.
R = 6||12 = 4
ωo = 1/ LC = 1/ )8/1)(2/1( = 4
α = R/(2L) = (4)/(2x(1/2)) = 4
α = ωo, therefore the circuit is critically damped
v(t) = Vs +[(A + Bt)e-4t], and Vs = 10
v(0) = 20 = 10 + A, or A = 10
i = Cdv/dt = -4C[(A + Bt)e-4t] + C[(B)e-4t]
i(0) = 5 = C(-4A + B) which leads to 40 = -40 + B or B = 80
i(t) = [-(1/2)(10 + 80t)e-4t] + [(10)e-4t]
i(t) = [(5 – 40t)e-4t] A
Chapter 8, Solution 38. At t = 0-, the equivalent circuit is as shown.
2 A
10 Ω
i
i1
5 Ω
+
v
− 10 Ω
i(0) = 2A, i1(0) = 10(2)/(10 + 15) = 0.8 A
v(0) = 5i1(0) = 4V
For t > 0, we have a source-free series RLC circuit.
R = 5||(10 + 10) = 4 ohms
ωo = 1/ LC = 1/ )4/3)(3/1( = 2
α = R/(2L) = (4)/(2x(3/4)) = 8/3
s1,2 = =ω−α±α− 2o
2 -4.431, -0.903
i(t) = [Ae-4.431t + Be-0.903t]
i(0) = A + B = 2 (1)
di(0)/dt = (1/L)[-Ri(0) + v(0)] = (4/3)(-4x2 + 4) = -16/3 = -5.333
Hence, -5.333 = -4.431A – 0.903B (2)
From (1) and (2), A = 1 and B = 1.
i(t) = [e-4.431t + e-0.903t] A
Chapter 8, Solution 39. For t = 0-, the equivalent circuit is shown in Figure (a). Where 60u(-t) = 60 and 30u(t) = 0.
+ − 30V 20 Ω
+ −
+ v −
20 Ω
30 Ω 0.5F 0.25H 30 Ω 60V (a) (b)
v(0) = (20/50)(60) = 24 and i(0) = 0
For t > 0, the circuit is shown in Figure (b).
R = 20||30 = 12 ohms
ωo = 1/ LC = 1/ )4/1)(2/1( = 8
α = R/(2L) = (12)/(0.5) = 24
Since α > ωo, we have an overdamped response.
s1,2 = =ω−α±α− 2o
2 -47.833, -0.167 Thus, v(t) = Vs + [Ae-47.833t + Be-0.167t], Vs = 30 v(0) = 24 = 30 + A + B or -6 = A + B (1)
i(0) = Cdv(0)/dt = 0
But, dv(0)/dt = -47.833A – 0.167B = 0 B = -286.43A (2)
From (1) and (2), A = 0.021 and B = -6.021
v(t) = 30 + [0.021e-47.833t – 6.021e-0.167t] V Chapter 8, Solution 40. At t = 0-, vC(0) = 0 and iL(0) = i(0) = (6/(6 + 2))4 = 3A For t > 0, we have a series RLC circuit with a step input as shown below.
+ −v
6 Ω+ −
+ − 12V
24V
i 14 Ω0.02 F 2 H
ωo = 1/ LC = 1/ 02.0x2 = 5
α = R/(2L) = (6 + 14)/(2x2) = 5
Since α = ωo, we have a critically damped response.
v(t) = Vs + [(A + Bt)e-5t], Vs = 24 – 12 = 12V
v(0) = 0 = 12 + A or A = -12
i = Cdv/dt = C[Be-5t] + [-5(A + Bt)e-5t]
i(0) = 3 = C[-5A + B] = 0.02[60 + B] or B = 90
Thus, i(t) = 0.02[90e-5t] + [-5(-12 + 90t)e-5t]
i(t) = (3 – 9t)e-5t A
Chapter 8, Solution 41. At t = 0-, the switch is open. i(0) = 0, and
v(0) = 5x100/(20 + 5 + 5) = 50/3 For t > 0, we have a series RLC circuit shown in Figure (a). After source transformation, it becomes that shown in Figure (b).
1 H
+ − 20V
+
v
−
i 4 Ω
20 Ω 10 µF
10 H
5A
(a)
5 Ω
0.04F (b)
ωo = 1/ LC = 1/ 25/1x1 = 5
α = R/(2L) = (4)/(2x1) = 2
s1,2 = =ω−α±α− 2o
2 -2 ± j4.583 Thus, v(t) = Vs + [(Acosωdt + Bsinωdt)e-2t],
where ωd = 4.583 and Vs = 20
v(0) = 50/3 = 20 + A or A = -10/3
i(t) = Cdv/dt = C(-2) [(Acosωdt + Bsinωdt)e-2t] + Cωd[(-Asinωdt + Bcosωdt)e-2t]
i(0) = 0 = -2A + ωdB
B = 2A/ωd = -20/(3x4.583) = -1.455
i(t) = C[(0cosωdt + (-2B - ωdA)sinωdt)]e-2t
= (1/25)[(2.91 + 15.2767) sinωdt)]e-2t
i(t) = 0.7275sin(4.583t)e-2t A
Chapter 8, Solution 42. For t = 0-, we have the equivalent circuit as shown in Figure (a).
i(0) = i(0) = 0, and v(0) = 4 – 12 = -8V
+
v
−
6 Ω
− + 12V
1 H i 4V + − − +
+
v(0)
−
12V
1 Ω
5 Ω
0.04F
(a) (b) For t > 0, the circuit becomes that shown in Figure (b) after source transformation.
ωo = 1/ LC = 1/ 25/1x1 = 5
α = R/(2L) = (6)/(2) = 3
s1,2 = =ω−α±α− 2o
2 -3 ± j4 Thus, v(t) = Vs + [(Acos4t + Bsin4t)e-3t], Vs = -12
v(0) = -8 = -12 + A or A = 4
i = Cdv/dt, or i/C = dv/dt = [-3(Acos4t + Bsin4t)e-3t] + [4(-Asin4t + Bcos4t)e-3t]
i(0) = -3A + 4B or B = 3
v(t) = -12 + [(4cos4t + 3sin4t)e-3t] A
Chapter 8, Solution 43.
For t>0, we have a source-free series RLC circuit.
Ω===→= 85.08222
xxLRLR αα
836649003022 =−=→=−= ood ωαωω
mF 392.25.0836
1112 ===→=
xLC
LC oo ω
ω
Chapter 8, Solution 44.
4
910
1010011,500
121000
2======
−xLCxLR
oωα
→α>ωo underdamped. Chapter 8, Solution 45.
ωo = 1/ LC = 1/ 5.0x1 = 2
α = R/(2L) = (1)/(2x2x0.5) = 0.5
Since α < ωo, we have an underdamped response.
s1,2 = =ω−α±α− 2o
2 -0.5 ± j1.323 Thus, i(t) = Is + [(Acos1.323t + Bsin1.323t)e-0.5t], Is = 4
i(0) = 1 = 4 + A or A = -3
v = vC = vL = Ldi(0)/dt = 0
di/dt = [1.323(-Asin1.323t + Bcos1.323t)e-0.5t] + [-0.5(Acos1.323t + Bsin1.323t)e-0.5t]
di(0)/dt = 0 = 1.323B – 0.5A or B = 0.5(-3)/1.323 = -1.134 Thus, i(t) = 4 – [(3cos1.323t + 1.134sin1.323t)e-0.5t] A
Chapter 8, Solution 46. For t = 0-, u(t) = 0, so that v(0) = 0 and i(0) = 0. For t > 0, we have a parallel RLC circuit with a step input, as shown below.
5µF8mH
+
v
−
i 2 kΩ
6mA
α = 1/(2RC) = (1)/(2x2x103 x5x10-6) = 50
ωo = 1/ LC = 1/ 63 10x5x10x8 − = 5,000
Since α < ωo, we have an underdamped response.
s1,2 = ≅ω−α±α− 2o
2 -50 ± j5,000 Thus, i(t) = Is + [(Acos5,000t + Bsin5,000t)e-50t], Is = 6mA
i(0) = 0 = 6 + A or A = -6mA
v(0) = 0 = Ldi(0)/dt
di/dt = [5,000(-Asin5,000t + Bcos5,000t)e-50t] + [-50(Acos5,000t + Bsin5,000t)e-50t]
di(0)/dt = 0 = 5,000B – 50A or B = 0.01(-6) = -0.06mA Thus, i(t) = 6 – [(6cos5,000t + 0.06sin5,000t)e-50t] mA Chapter 8, Solution 47. At t = 0-, we obtain, iL(0) = 3x5/(10 + 5) = 1A
and vo(0) = 0.
For t > 0, the 20-ohm resistor is short-circuited and we have a parallel RLC circuit with a step input.
α = 1/(2RC) = (1)/(2x5x0.01) = 10
ωo = 1/ LC = 1/ 01.0x1 = 10
Since α = ωo, we have a critically damped response.
s1,2 = -10
Thus, i(t) = Is + [(A + Bt)e-10t], Is = 3
i(0) = 1 = 3 + A or A = -2
vo = Ldi/dt = [Be-10t] + [-10(A + Bt)e-10t]
vo(0) = 0 = B – 10A or B = -20
Thus, vo(t) = (200te-10t) V
Chapter 8, Solution 48. For t = 0-, we obtain i(0) = -6/(1 + 2) = -2 and v(0) = 2x1 = 2.
For t > 0, the voltage is short-circuited and we have a source-free parallel RLC circuit.
α = 1/(2RC) = (1)/(2x1x0.25) = 2
ωo = 1/ LC = 1/ 25.0x1 = 2
Since α = ωo, we have a critically damped response.
s1,2 = -2
Thus, i(t) = [(A + Bt)e-2t], i(0) = -2 = A
v = Ldi/dt = [Be-2t] + [-2(-2 + Bt)e-2t]
vo(0) = 2 = B + 4 or B = -2
Thus, i(t) = [(-2 - 2t)e-2t] A
and v(t) = [(2 + 4t)e-2t] V
Chapter 8, Solution 49. For t = 0-, i(0) = 3 + 12/4 = 6 and v(0) = 0.
For t > 0, we have a parallel RLC circuit with a step input.
α = 1/(2RC) = (1)/(2x5x0.05) = 2
ωo = 1/ LC = 1/ 05.0x5 = 2
Since α = ωo, we have a critically damped response.
s1,2 = -2
Thus, i(t) = Is + [(A + Bt)e-2t], Is = 3
i(0) = 6 = 3 + A or A = 3
v = Ldi/dt or v/L = di/dt = [Be-2t] + [-2(A + Bt)e-2t]
v(0)/L = 0 = di(0)/dt = B – 2x3 or B = 6
Thus, i(t) = 3 + [(3 + 6t)e-2t] A Chapter 8, Solution 50. For t = 0-, 4u(t) = 0, v(0) = 0, and i(0) = 30/10 = 3A. For t > 0, we have a parallel RLC circuit.
i
40 Ω 6A 3A
10 mF+
v
−
10 Ω
10 H
Is = 3 + 6 = 9A and R = 10||40 = 8 ohms
α = 1/(2RC) = (1)/(2x8x0.01) = 25/4 = 6.25
ωo = 1/ LC = 1/ 01.0x4 = 5
Since α > ωo, we have a overdamped response.
s1,2 = =ω−α±α− 2o
2 -10, -2.5
Thus, i(t) = Is + [Ae-10t] + [Be-2.5t], Is = 9
i(0) = 3 = 9 + A + B or A + B = -6
di/dt = [-10Ae-10t] + [-2.5Be-2.5t],
v(0) = 0 = Ldi(0)/dt or di(0)/dt = 0 = -10A – 2.5B or B = -4A
Thus, A = 2 and B = -8
Clearly, i(t) = 9 + [2e-10t] + [-8e-2.5t] A Chapter 8, Solution 51. Let i = inductor current and v = capacitor voltage.
At t = 0, v(0) = 0 and i(0) = io.
For t > 0, we have a parallel, source-free LC circuit (R = ∞).
α = 1/(2RC) = 0 and ωo = 1/ LC which leads to s1,2 = ± jωo
v = Acosωot + Bsinωot, v(0) = 0 A
iC = Cdv/dt = -i
dv/dt = ωoBsinωot = -i/C
dv(0)/dt = ωoB = -io/C therefore B = io/(ωoC)
v(t) = -(io/(ωoC))sinωot V where ωo = LC Chapter 8, Solution 52.
RC21300 ==α (1)
LCood1575.264300400400 2222 ==−=→=−= ωαωω (2)
From (2),
F71.2851050)575.264(
132 µ== −xx
C
From (1),
Ω=== 833.5)3500(30021
21
xCR
α
Chapter 8, Solution 53.
C1 R2
+ −v1
i2i1
+ −
C2
+
vo
−
R1
vS
i2 = C2dvo/dt (1) i1 = C1dv1/dt (2) 0 = R2i2 + R1(i2 – i1) +vo (3) Substituting (1) and (2) into (3) we get, 0 = R2C2dvo/dt + R1(C2dvo/dt – C1dv1/dt) (4) Applying KVL to the outer loop produces,
vs = v1 + i2R2 + vo = v1 + R2C2dvo/dt + vo, which leads to v1 = vs – vo – R2C2dvo/dt (5) Substituting (5) into (4) leads to,
0 = R1C2dvo/dt + R1C2dvo/dt – R1C1(dvs/dt – dvo/dt – R2C2d2vo/dt2)
Hence, (R1C1R2C2)(d2vo/dt2) + (R1C1 + R2C2 +R1C2)(dvo/dt) = R1C1(dvs/dt) Chapter 8, Solution 54. Let i be the inductor current.
dtdvvi 5.0
4+=− (1)
dtdii += 2v (2)
Substituting (1) into (2) gives
035.221
41
2 2
2
2
2
=++→+++=− vdtdv
dtvd
dtvd
dtdv
dtdvvv
199.125.1035.22 jsss ±−=→=++
tBetAev tt 199.1sin199.1cos 25.125.1 −− +=
v(0) = 2=A. Let w=1.199
)cossin()sincos(25.1 25.125.125.125.1 wtBewtAewwtBewtAedtdv tttt −−−− +−++−=
085.2199.1
225.125.10)0(==→+−==
XBBwAdtdv
V 199.1sin085.2199.1cos2 25.125.1 tetev tt −− +=
Chapter 8, Solution 55. At the top node, writing a KCL equation produces,
i/4 +i = C1dv/dt, C1 = 0.1
5i/4 = C1dv/dt = 0.1dv/dt i = 0.08dv/dt (1)
But, v = )idt)C/1(i2( 2 ∫+− , C2 = 0.5 or -dv/dt = 2di/dt + 2i (2)
Substituting (1) into (2) gives,
-dv/dt = 0.16d2v/dt2 + 0.16dv/dt
0.16d2v/dt2 + 0.16dv/dt + dv/dt = 0, or d2v/dt2 + 7.25dv/dt = 0
Which leads to s2 + 7.25s = 0 = s(s + 7.25) or s1,2 = 0, -7.25
v(t) = A + Be-7.25t (3) v(0) = 4 = A + B (4)
From (1), i(0) = 2 = 0.08dv(0+)/dt or dv(0+)/dt = 25
But, dv/dt = -7.25Be-7.25t, which leads to,
dv(0)/dt = -7.25B = 25 or B = -3.448 and A = 4 – B = 4 + 3.448 = 7.448
Thus, v(t) = 7.45 – 3.45e-7.25t V Chapter 8, Solution 56. For t < 0, i(0) = 0 and v(0) = 0. For t > 0, the circuit is as shown below. 4 Ω
io
i 6 Ω
+ −
0.04F
i 20 0.25H Applying KVL to the larger loop,
-20 +6io +0.25dio/dt + 25 ∫ + dt)ii( o = 0 Taking the derivative,
6dio/dt + 0.25d2io/dt2 + 25(io + i) = 0 (1) For the smaller loop, 4 + 25 ∫ + dt)ii( o = 0 Taking the derivative, 25(i + io) = 0 or i = -io (2) From (1) and (2) 6dio/dt + 0.25d2io/dt2 = 0
This leads to, 0.25s2 + 6s = 0 or s1,2 = 0, -24
io(t) = (A + Be-24t) and io(0) = 0 = A + B or B = -A
As t approaches infinity, io(∞) = 20/10 = 2 = A, therefore B = -2
Thus, io(t) = (2 - 2e-24t) = -i(t) or i(t) = (-2 + 2e-24t) A Chapter 8, Solution 57. (a) Let v = capacitor voltage and i = inductor current. At t = 0-, the switch is closed and the circuit has reached steady-state.
v(0-) = 16V and i(0-) = 16/8 = 2A At t = 0+, the switch is open but, v(0+) = 16 and i(0+) = 2. We now have a source-free RLC circuit.
R 8 + 12 = 20 ohms, L = 1H, C = 4mF.
α = R/(2L) = (20)/(2x1) = 10
ωo = 1/ LC = 1/ )36/1(x1 = 6
Since α > ωo, we have a overdamped response.
s1,2 = =ω−α±α− 2o
2 -18, -2
Thus, the characteristic equation is (s + 2)(s + 18) = 0 or s2 + 20s +36 = 0. (b) i(t) = [Ae-2t + Be-18t] and i(0) = 2 = A + B (1)
To get di(0)/dt, consider the circuit below at t = 0+.
+
vL
−
12 Ω
(1/36)F+
v
−
i 8 Ω
1 H
-v(0) + 20i(0) + vL(0) = 0, which leads to,
-16 + 20x2 + vL(0) = 0 or vL(0) = -24
Ldi(0)/dt = vL(0) which gives di(0)/dt = vL(0)/L = -24/1 = -24 A/s
Hence -24 = -2A – 18B or 12 = A + 9B (2) From (1) and (2), B = 1.25 and A = 0.75
i(t) = [0.75e-2t + 1.25e-18t] = -ix(t) or ix(t) = [-0.75e-2t - 1.25e-18t] A
v(t) = 8i(t) = [6e-2t + 10e-18t] A Chapter 8, Solution 58.
(a) Let i =inductor current, v = capacitor voltage i(0) =0, v(0) = 4
V/s 85.0
)04()]0()0([)0(−=
+−=
+−=
RCRiv
dtdv
(b) For , the circuit is a source-free RLC parallel circuit. 0≥t
2125.0
11,115.02
12
1======
xLCxxRC oωα
732.11422 =−=−= αωω od
Thus,
)732.1sin732.1cos()( 21 tAtAetv t += − v(0) = 4 = A1
tAetAetAetAedtdv tttt 732.1cos732.1732.1sin732.1sin732.1732.1cos 2211
−−−− +−−−=
309.2732.18)0(221 −=→+−=−= AAA
dtdv
)732.1sin309.2732.1cos4()( ttetv t −= − V
Chapter 8, Solution 59.
Let i = inductor current and v = capacitor voltage v(0) = 0, i(0) = 40/(4+16) = 2A For t>0, the circuit becomes a source-free series RLC with
2,216/14
11,242
162
==→====== oo xLCxLR ωαωα
tt BteAeti 22)( −− += i(0) = 2 = A
ttt BteBeAedtdi 222 22 −−− −+−=
4),032(412)]0()0([12)0(
−=+−=+−→+−=+−= BBAvRiL
BAdtdi
tt teeti 22 42)( −− −= t
ttt
ttt
tt
teedttedtevidtC
v0
2
00
22
0
2
0
)12(4
64166432)0(1−−−−=−=+= −−−− ∫∫∫
ttev 232 −= V
Chapter 8, Solution 60.
At t = 0-, 4u(t) = 0 so that i1(0) = 0 = i2(0) (1)
Applying nodal analysis,
4 = 0.5di1/dt + i1 + i2 (2) Also, i2 = [1di1/dt – 1di2/dt]/3 or 3i2 = di1/dt – di2/dt (3) Taking the derivative of (2), 0 = d2i1/dt2 + 2di1/dt + 2di2/dt (4) From (2) and (3), di2/dt = di1/dt – 3i2 = di1/dt – 3(4 – i1 – 0.5di1/dt)
= di1/dt – 12 + 3i1 + 1.5di1/dt Substituting this into (4),
d2i1/dt2 + 7di1/dt + 6i1 = 24 which gives s2 + 7s + 6 = 0 = (s + 1)(s + 6)
Thus, i1(t) = Is + [Ae-t + Be-6t], 6Is = 24 or Is = 4
i1(t) = 4 + [Ae-t + Be-6t] and i1(0) = 4 + [A + B] (5)
i2 = 4 – i1 – 0.5di1/dt = i1(t) = 4 + -4 - [Ae-t + Be-6t] – [-Ae-t - 6Be-6t]
= [-0.5Ae-t + 2Be-6t] and i2(0) = 0 = -0.5A + 2B (6) From (5) and (6), A = -3.2 and B = -0.8
i1(t) = 4 + [-3.2e-t – 0.8e-6t] A
i2(t) = [1.6e-t – 1.6e-6t] A Chapter 8, Solution 61. For t > 0, we obtain the natural response by considering the circuit below.
0.25F+
vC
−
a iL
4 Ω
1 H 6 Ω At node a, vC/4 + 0.25dvC/dt + iL = 0 (1) But, vC = 1diL/dt + 6iL (2) Combining (1) and (2),
(1/4)diL/dt + (6/4)iL + 0.25d2iL/dt2 + (6/4)diL/dt + iL = 0
d2iL/dt2 + 7diL/dt + 10iL = 0
s2 + 7s + 10 = 0 = (s + 2)(s + 5) or s1,2 = -2, -5
Thus, iL(t) = iL(∞) + [Ae-2t + Be-5t],
where iL(∞) represents the final inductor current = 4(4)/(4 + 6) = 1.6
iL(t) = 1.6 + [Ae-2t + Be-5t] and iL(0) = 1.6 + [A+B] or -1.6 = A+B (3)
diL/dt = [-2Ae-2t - 5Be-5t]
and diL(0)/dt = 0 = -2A – 5B or A = -2.5B (4)
From (3) and (4), A = -8/3 and B = 16/15
iL(t) = 1.6 + [-(8/3)e-2t + (16/15)e-5t]
v(t) = 6iL(t) = 9.6 + [-16e-2t + 6.4e-5t] V
vC = 1diL/dt + 6iL = [ (16/3)e-2t - (16/3)e-5t] + 9.6 + [-16e-2t + 6.4e-5t]
vC = 9.6 + [-(32/3)e-2t + 1.0667e-5t]
i(t) = vC/4 = 2.4 + [-2.667e-2t + 0.2667e-5t] A Chapter 8, Solution 62. This is a parallel RLC circuit as evident when the voltage source is turned off.
α = 1/(2RC) = (1)/(2x3x(1/18)) = 3
ωo = 1/ LC = 1/ 18/1x2 = 3
Since α = ωo, we have a critically damped response.
s1,2 = -3
Let v(t) = capacitor voltage
Thus, v(t) = Vs + [(A + Bt)e-3t] where Vs = 0
But -10 + vR + v = 0 or vR = 10 – v
Therefore vR = 10 – [(A + Bt)e-3t] where A and B are determined from initial conditions.
Chapter 8, Solution 63. - R R v1 + vo vs v2 C C At node 1,
dtdvC
Rvvs 11 =
− (1)
At node 2,
dtdv
CRvv oo =
−2 (2)
As a voltage follower, vvv == 21 . Hence (2) becomes
dtdv
RCvv oo += (3)
and (1) becomes
dtdvRCvvs += (4)
Substituting (3) into (4) gives
2
222
dtvdCR
dtdvRC
dtdvRCvv ooo
os +++=
or
sooo vvdtdvRC
dtvdCR =++ 22
222
Chapter 8, Solution 64. C2
vs R1 2 1
v1
R2
−+
C1
vo
At node 1, (vs – v1)/R1 = C1 d(v1 – 0)/dt or vs = v1 + R1C1dv1/dt (1) At node 2, C1dv1/dt = (0 – vo)/R2 + C2d(0 – vo)/dt or –R2C1dv1/dt = vo + C2dvo/dt (2) From (1) and (2), (vs – v1)/R1 = C1 dv1/dt = -(1/R2)(vo + C2dvo/dt) or v1 = vs + (R1/R2)(vo + C2dvo/dt) (3) Substituting (3) into (1) produces,
vs = vs + (R1/R2)(vo + C2dvo/dt) + R1C1dvs + (R1/R2)(vo + C2dvo/dt)/dt
= vs + (R1/R2)(vo)+ (R1C2/R2) dvo/dt) + R1C1dvs/dt + (R1R1C1/R2)dvo/dt + (R1
2 C1C2/R2)[d2vo/dt2]
Simplifying we get,
d2vo/dt2 + [(1/ R1C1) + (1/ C2)]dvo/dt + [1/(R1C1C2)](vo) = - [R2/(R1C2)]dvs/dt
Chapter 8, Solution 65. At the input of the first op amp,
(vo – 0)/R = Cd(v1 – 0) (1) At the input of the second op amp, (-v1 – 0)/R = Cdv2/dt (2) Let us now examine our constraints. Since the input terminals are essentially at ground, then we have the following,
vo = -v2 or v2 = -vo (3) Combining (1), (2), and (3), eliminating v1 and v2 we get,
0v100dt
vdv
CR1
dtvd
o2o
2
o222o
2
=−=
−
Which leads to s2 – 100 = 0
Clearly this produces roots of –10 and +10. And, we obtain,
vo(t) = (Ae+10t + Be-10t)V
At t = 0, vo(0+) = – v2(0+) = 0 = A + B, thus B = –A
This leads to vo(t) = (Ae+10t – Ae-10t)V. Now we can use v1(0+) = 2V.
From (2), v1 = –RCdv2/dt = 0.1dvo/dt = 0.1(10Ae+10t + 10Ae-10t)
v1(0+) = 2 = 0.1(20A) = 2A or A = 1
Thus, vo(t) = (e+10t – e-10t)V It should be noted that this circuit is unstable (clearly one of the poles lies in the right-half-plane). Chapter 8, Solution 66. C2
R4
R2
+–
C1
vS 1
2
R3
R1
vo Note that the voltage across C1 is v2 = [R3/(R3 + R4)]vo This is the only difference between this problem and Example 8.11, i.e. v = kv, where k = [R3/(R3 + R4)].
At node 1,
(vs – v1)/R1 = C2[d(v1 – vo)/dt] + (v1 – v2)/R2
vs/R1 = (v1/R1) + C2[d(v1)/dt] – C2[d(vo)/dt] + (v1 – kvo)/R2 (1) At node 2,
(v1 – kvo)/R2 = C1[d(kvo)/dt] or v1 = kvo + kR2C1[d(vo)/dt] (2) Substituting (2) into (1), vs/R1 = (kvo/R1) + (kR2C1/R1)[d(vo)/dt] + kC2[d(vo)/dt] + kR2C1C2[d2(vo)/dt2] – (kvo/R2) + kC1[d(vo)/dt] – (kvo/R2) + C2[d(vo)/dt] We now rearrange the terms. [d2(vo)/dt2] + [(1/C2R1) + (1/ R2C2) + (1/R2C1) – (1/ kR2C1)][d(vo)/dt] + [vo/(R1R2C1C2)] = vs/(kR1R2C1C2) If R1 = R2 10 kohms, C1 = C2 = 100 µF, R3 = 20 kohms, and R4 = 60 kohms,
k = [R3/(R3 + R4)] = 1/3
R1R2C1C2 = 104 x104 x10-4 x10-4 = 1
(1/C2R1) + (1/ R2C2) + (1/R2C1) – (1/ kR2C1) = 1 + 1 + 1 – 3 = 3 – 3 = 0 Hence, [d2(vo)/dt2] + vo = 3vs = 6, t > 0, and s2 + 1 = 0, or s1,2 = ±j
vo(t) = Vs + [Acost + B sint], Vs = 6
vo(0) = 0 = 6 + A or A = –6
dvo/dt = –Asint + Bcost, but dvo(0)/dt = 0 = B
Hence, vo(t) = 6(1 – cost)u(t) volts.
Chapter 8, Solution 67. At node 1,
dt
)0v(dCdt
)vv(dC
Rvv 1
2o1
11
1in −+
−=
− (1)
At node 2, 2
o12 R
v0dt
)0v(dC−
=− , or
22
o1
RCv
dtdv −
= (2)
From (1) and (2),
2
o1
o11
o
22
111in R
vR
dtdv
CRdt
dvRCCRvv −−−=−
2
o1
o11
o
22
11in1 R
vR
dtdv
CRdt
dvRCCRvv +++= (3)
C1
From (2) and (3),
0Vvin
C2
R2
−+v1
1 2 R1
vo
dtdv
RR
dtvd
CRdt
dvRCCR
dtdv
dtdv
RCv o
2
12
o2
11o
22
11in1
22
o +++==−
dtdv
CR1
RRCCv
dtdv
C1
C1
R1
dtvd in
111221
oo
2122
o2
−=+
++
But C1C2R1R2 = 10-4 x10-4 x104 x104 = 1
210x10
2CR
2C1
C1
R1
4412212
===
+ −
dtdvv
dtdv
2dt
vd ino
o2
o2
−=++
Which leads to s2 + 2s + 1 = 0 or (s + 1)2 = 0 and s = –1, –1
Therefore, vo(t) = [(A + Bt)e-t] + Vf
As t approaches infinity, the capacitor acts like an open circuit so that
Vf = vo(∞) = 0
vin = 10u(t) mV and the fact that the initial voltages across each capacitor is 0
means that vo(0) = 0 which leads to A = 0.
vo(t) = [Bte-t]
dtdvo = [(B – Bt)e-t] (4)
From (2), 0RC
)0(vdt
)0(dv
22
oo =+
−=+
From (1) at t = 0+,
dt)0(dv
CR
01 o1
1
+−=
− which leads to 1RC1
dt)0(dv
11
o −=−=+
Substituting this into (4) gives B = –1
Thus, v(t) = –te-tu(t) V
Chapter 8, Solution 68. The schematic is as shown below. The unit step is modeled by VPWL as shown. We insert a voltage marker to display V after simulation. We set Print Step = 25 ms and final step = 6s in the transient box. The output plot is shown below.
Chapter 8, Solution 69. The schematic is shown below. The initial values are set as attributes of L1 and C1. We set Print Step to 25 ms and the Final Time to 20s in the transient box. A current marker is inserted at the terminal of L1 to automatically display i(t) after simulation. The result is shown below.
Chapter 8, Solution 70. The schematic is shown below.
After the circuit is saved and simulated, we obtain the capacitor voltage v(t) as shown below.
Chapter 8, Solution 71. The schematic is shown below. We use VPWL and IPWL to model the 39 u(t) V and 13 u(t) A respectively. We set Print Step to 25 ms and Final Step to 4s in the Transient box. A voltage marker is inserted at the terminal of R2 to automatically produce the plot of v(t) after simulation. The result is shown below.
Chapter 8, Solution 72. When the switch is in position 1, we obtain IC=10 for the capacitor and IC=0 for the inductor. When the switch is in position 2, the schematic of the circuit is shown below.
When the circuit is simulated, we obtain i(t) as shown below.
Chapter 8, Solution 73. (a) For t < 0, we have the schematic below. When this is saved and simulated, we
obtain the initial inductor current and capacitor voltage as
iL(0) = 3 A and vc(0) = 24 V.
(b) For t > 0, we have the schematic shown below. To display i(t) and v(t), we insert current and voltage markers as shown. The initial inductor current and capacitor voltage are also incorporated. In the Transient box, we set Print Step = 25 ms and the Final Time to 4s. After simulation, we automatically have io(t) and vo(t) displayed as shown below.
Chapter 8, Solution 74.
10Ω 5Ω + 20 V 2F 4H - Hence the dual circuit is shown below. 2H 4F 0.2Ω 20A 0.1 Ω
Chapter 8, Solution 75. The dual circuit is connected as shown in Figure (a). It is redrawn in Figure (b).
0.5 H 2 F
12A
24A
0.25 Ω
0.1 Ω
10 µF
12A + − 24V
10 H
10 Ω
0.25 Ω
24A
+ − 12V
0.5 F
10 H
(a)
4 Ω
0.1 Ω
(b)
Chapter 8, Solution 76. The dual is obtained from the original circuit as shown in Figure (a). It is redrawn in Figure (b).
120 A
2 V
+ −
2 A
30 Ω
1/3 Ω
10 Ω
0.1 Ω
120 V
– +
60 V
+ − 60 A
4F
1 F 4 H 1 H
20 Ω
0.05 Ω (a) 0.05 Ω
60 A 1 H
120 A
1/4 F
0.1 Ω
+ − 2V
1/30 Ω (b)
Chapter 8, Solution 77. The dual is constructed in Figure (a) and redrawn in Figure (b).
+ − 5 V
1/4 F
1 H
2 Ω
1 Ω
1/3 Ω12 A
1 Ω
12 A
1/3 Ω
3 Ω1/2 Ω
5 V
– + 5 A
+ − 12V1/4 F
1 F 1/4 H
1 H
(a)
1 Ω
2 Ω
(b)
Chapter 8, Solution 78. The voltage across the igniter is vR = vC since the circuit is a parallel RLC type.
vC(0) = 12, and iL(0) = 0.
α = 1/(2RC) = 1/(2x3x1/30) = 5
ωo 30/1x10x60/1LC/1 3−== = 22.36
α < ωo produces an underdamped response.
2o
22,1s ω−α±α−= = –5 ± j21.794
vC(t) = e-5t(Acos21.794t + Bsin21.794t) (1)
vC(0) = 12 = A dvC/dt = –5[(Acos21.794t + Bsin21.794t)e-5t] + 21.794[(–Asin21.794t + Bcos21.794t)e-5t] (2)
dvC(0)/dt = –5A + 21.794B
But, dvC(0)/dt = –[vC(0) + RiL(0)]/(RC) = –(12 + 0)/(1/10) = –120
Hence, –120 = –5A + 21.794B, leads to B (5x12 – 120)/21.794 = –2.753
At the peak value, dvC(to)/dt = 0, i.e.,
0 = A + Btan21.794to + (A21.794/5)tan21.794to – 21.794B/5
(B + A21.794/5)tan21.794to = (21.794B/5) – A
tan21.794to = [(21.794B/5) – A]/(B + A21.794/5) = –24/49.55 = –0.484
Therefore, 21.7945to = |–0.451|
to = |–0.451|/21.794 = 20.68 ms
Chapter 8, Solution 79.
For critical damping of a parallel RLC circuit,
LCRCo1
21
=→=ωα
Hence,
F434144425.0
4 2 µ===xR
LC
Chapter 8, Solution 80.
t1 = 1/|s1| = 0.1x10-3 leads to s1 = –1000/0.1 = –10,000
t2 = 1/|s2| = 0.5x10-3 leads to s1 = –2,000
2o
21s ω−α−α−=
2o
22s ω−α+α−=
s1 + s2 = –2α = –12,000, therefore α = 6,000 = R/(2L)
L = R/12,000 = 60,000/12,000 = 5H
2o
22s ω−α+α−= = –2,000
2o
2 ω−α−α = 2,000
2o
2000,6 ω−α− = 2,000
2o
2 ω−α = 4,000
α2 – = 16x102oω
6
2oω = α2 – 16x106 = 36x106 – 16x106
ωo = 103 LC/120 =
C = 1/(20x106x5) = 10 nF
Chapter 8, Solution 81.
t = 1/α = 0.25 leads to α = 4
But, α 1/(2RC) or, C = 1/(2αR) = 1/(2x4x200) = 625 µF
22od α−ω=ω
232322
d2o 010x42(16)10x42( π≅+π=α+ω=ω = 1/(LC)
This results in L = 1/(64π2x106x625x10-6) = 2.533 µH
Chapter 8, Solution 82. For t = 0-, v(0) = 0. For t > 0, the circuit is as shown below.
+
vo
−
C1
+
v
−
a
R2
R1 C2 At node a,
(vo – v/R1 = (v/R2) + C2dv/dt
vo = v(1 + R1/R2) + R1C2 dv/dt
60 = (1 + 5/2.5) + (5x106 x5x10-6)dv/dt
60 = 3v + 25dv/dt
v(t) = Vs + [Ae-3t/25]
where 3Vs = 60 yields Vs = 20
v(0) = 0 = 20 + A or A = –20
v(t) = 20(1 – e-3t/25)V
Chapter 8, Solution 83. i = iD + Cdv/dt (1) –vs + iR + Ldi/dt + v = 0 (2) Substituting (1) into (2),
vs = RiD + RCdv/dt + Ldi/dt + LCd2v/dt2 + v = 0
LCd2v/dt2 + RCdv/dt + RiD + Ldi/dt = vs
d2v/dt2 + (R/L)dv/dt + (R/LC)iD + (1/C)di/dt = vs/LC
Chapter 9, Solution 1.
(a) angular frequency ω = 103 rad/s
(b) frequency f = πω2
= 159.2 Hz
(c) period T = f
= 1
6.283 ms
(d) Since sin(A) = cos(A – 90°),
vs = 12 sin(103t + 24°) = 12 cos(103t + 24° – 90°) vs in cosine form is vs = 12 cos(103t – 66°) V
(e) vs(2.5 ms) = 12 )24)105.2)(10sin(( 3-3 °+×
= 12 sin(2.5 + 24°) = 12 sin(143.24° + 24°) = 2.65 V
Chapter 9, Solution 2.
(a) amplitude = 8 A (b) ω = 500π = 1570.8 rad/s
(c) f = πω2
= 250 Hz
(d) Is = 8∠-25° A
Is(2 ms) = )25)102)(500cos((8 3- °−×π= 8 cos(π − 25°) = 8 cos(155°) = -7.25 A
Chapter 9, Solution 3.
(a) 4 sin(ωt – 30°) = 4 cos(ωt – 30° – 90°) = 4 cos(ωt – 120°) (b) -2 sin(6t) = 2 cos(6t + 90°) (c) -10 sin(ωt + 20°) = 10 cos(ωt + 20° + 90°) = 10 cos(ωt + 110°)
Chapter 9, Solution 4.
(a) v = 8 cos(7t + 15°) = 8 sin(7t + 15° + 90°) = 8 sin(7t + 105°) (b) i = -10 sin(3t – 85°) = 10 cos(3t – 85° + 90°) = 10 cos(3t + 5°)
Chapter 9, Solution 5. v1 = 20 sin(ωt + 60°) = 20 cos(ωt + 60° − 90°) = 20 cos(ωt − 30°) v2 = 60 cos(ωt − 10°)
This indicates that the phase angle between the two signals is 20° and that v1 lags v2.
Chapter 9, Solution 6.
(a) v(t) = 10 cos(4t – 60°) i(t) = 4 sin(4t + 50°) = 4 cos(4t + 50° – 90°) = 4 cos(4t – 40°) Thus, i(t) leads v(t) by 20°.
(b) v1(t) = 4 cos(377t + 10°)
v2(t) = -20 cos(377t) = 20 cos(377t + 180°) Thus, v2(t) leads v1(t) by 170°.
(c) x(t) = 13 cos(2t) + 5 sin(2t) = 13 cos(2t) + 5 cos(2t – 90°)
X = 13∠0° + 5∠-90° = 13 – j5 = 13.928∠-21.04° x(t) = 13.928 cos(2t – 21.04°) y(t) = 15 cos(2t – 11.8°) phase difference = -11.8° + 21.04° = 9.24° Thus, y(t) leads x(t) by 9.24°.
Chapter 9, Solution 7. If f(φ) = cosφ + j sinφ,
)(fj)sinj(cosjcosj-sinddf
φ=φ+φ=φ+φ=φ
φ= djfdf
Integrating both sides ln f = jφ + ln A f = Aejφ = cosφ + j sinφ f(0) = A = 1 i.e. f(φ) = ejφ = cosφ + j sinφ
Chapter 9, Solution 8.
(a) 4j3
4515−
°∠+ j2 =
°∠°∠
53.13-54515
+ j2
= 3∠98.13° + j2 = -0.4245 + j2.97 + j2 = -0.4243 + j4.97
(b) (2 + j)(3 – j4) = 6 – j8 + j3 + 4 = 10 – j5 = 11.18∠-26.57°
j4)-j)(3(220-8
+°∠
+j125-
10+
= °∠
°∠26.57-11.1820-8
+14425
)10)(12j5-(+
−
= 0.7156∠6.57° − 0.2958 − j0.71
= 0.7109 + j0.08188 − 0.2958 − j0.71
= 0.4151 − j0.6281 (c) 10 + (8∠50°)(13∠-68.38°) = 10+104∠-17.38°
= 109.25 – j31.07 Chapter 9, Solution 9.
(a) 2 +8j54j3
−+
= 2 +6425
)8j5)(4j3(+
++
= 2 +89
3220j24j15 −++
= 1.809 + j0.4944
(b) 4∠-10° +°∠
−632j1
= 4∠-10° +°∠
°∠6363.43-236.2
= 4∠-10° + 0.7453∠-69.43° = 3.939 – j0.6946 + 0.2619 – j0.6978 = 4.201 – j1.392
(c) °∠−°∠°∠+°∠
50480920-6108
= 064.3j571.2863.8j5628.1052.2j638.53892.1j879.7
−−+−++
= 799.5j0083.1
6629.0j517.13+−
− =
°∠°∠
86.99886.581.2-533.13
= 2.299∠-102.67° = -0.5043 – j2.243
Chapter 9, Solution 10.
(a) z 9282.64z and ,566.8z ,86 321 jjj −−=−=−=
93.1966.10321 jzzz −=++
(b) 499.7999.93
21 jzzz
+=
Chapter 9, Solution 11.
(a) = (-3 + j4)(12 + j5) 21zz = -36 – j15 + j48 – 20 = -56 + j33
(b) ∗2
1
zz
= 5j124j3-
−+
= 25144
)5j12)(4j3(-+
++ = -0.3314 + j0.1953
(c) = (-3 + j4) + (12 + j5) = 9 + j9 21 zz +
21 zz − = (-3 + j4) – (12 + j5) = -15 – j
21
21
zzzz
−+
= )j15(-)j1(9
++
= 22 115j)-15)(j1(9-
−+
= 226
)14j16(9- +
= -0.6372 – j0.5575
Chapter 9, Solution 12.
(a) = (-3 + j4)(12 + j5) 21zz = -36 – j15 + j48 – 20 = -56 + j33
(b) ∗2
1
zz
= 5j124j3-
−+
= 25144
)5j12)(4j3(-+
++ = -0.3314 + j0.1953
(c) = (-3 + j4) + (12 + j5) = 9 + j9 21 zz +
21 zz − = (-3 + j4) – (12 + j5) = -15 – j
21
21
zzzz
−+
= )j15(-)j1(9
++
= 22 115j)-15)(j1(9-
−+
= 226
)14j16(9- +
= -0.6372 – j0.5575 Chapter 9, Solution 13.
(a) 1520.02749.1)2534.08425.0()4054.04324.0 jjj( +−=−−++−
(b) 0833.215024
3050−=
∠−∠
o
o
(c) (2+j3)(8-j5) –(-4) = 35 +j14
Chapter 9, Solution 14.
(a) 5116.05751.01115
143 jjj
+−=+−
−
(b) 55.11922.17.213406.246
9.694424186)5983.1096.16)(8467(
)8060)(8056.13882.231116.62( jjjj
jjj−−=
+−
=++
−−++
(c) ( ) 89.2004.256)120260(42 2 jjj −−=−+−
Chapter 9, Solution 15.
(a) j1-5-3j26j10
+−+
= -10 – j6 + j10 – 6 + 10 – j15
= -6 – j11
(b) °∠°∠°∠°−∠
45301610-4-3020
= 60∠15° + 64∠-10°
= 57.96 + j15.529 + 63.03 – j11.114 = 120.99 – j4.415
(c)
j1j0jj1
j1j1j1j
0jj1
−−−
+−
−−
= 1 )j1(j)j1(j0101 22 ++−+−−++
= 1 )j1j1(1 ++−− = 1 – 2 = -1
Chapter 9, Solution 16.
(a) -10 cos(4t + 75°) = 10 cos(4t + 75° − 180°) = 10 cos(4t − 105°)
The phasor form is 10∠-105° (b) 5 sin(20t – 10°) = 5 cos(20t – 10° – 90°)
= 5 cos(20t – 100°) The phasor form is 5∠-100°
(c) 4 cos(2t) + 3 sin(2t) = 4 cos(2t) + 3 cos(2t – 90°)
The phasor form is 4∠0° + 3∠-90° = 4 – j3 = 5∠-36.87° Chapter 9, Solution 17.
(a) Let A = 8∠-30° + 6∠0° = 12.928 – j4 = 13.533∠-17.19°
a(t) = 13.533 cos(5t + 342.81°)
(b) We know that -sinα = cos(α + 90°). Let B = 20∠45° + 30∠(20° + 90°)
= 14.142 + j14.142 – 10.261 + j28.19 = 3.881 + j42.33 = 42.51∠84.76°
b(t) = 42.51 cos(120πt + 84.76°) (c) Let C = 4∠-90° + 3∠(-10° – 90°)
= -j4 – 0.5209 – j2.954 = 6.974∠265.72°
c(t) = 6.974 cos(8t + 265.72°) Chapter 9, Solution 18.
(a) = )t(v1 60 cos(t + 15°) (b) = 6 + j8 = 10∠53.13° 2V
)t(v2 = 10 cos(40t + 53.13°) (c) = )t(i1 2.8 cos(377t – π/3) (d) = -0.5 – j1.2 = 1.3∠247.4° 2I
)t(i 2 = 1.3 cos(103t + 247.4°) Chapter 9, Solution 19.
(a) 3∠10° − 5∠-30° = 2.954 + j0.5209 – 4.33 + j2.5 = -1.376 + j3.021 = 3.32∠114.49°
Therefore, 3 cos(20t + 10°) – 5 cos(20t – 30°) = 3.32 cos(20t + 114.49°)
(b) 4∠-90° + 3∠-45° = -j40 + 21.21 – j21.21
= 21.21 – j61.21 = 64.78∠-70.89°
Therefore, 40 sin(50t) + 30 cos(50t – 45°) = 64.78 cos(50t – 70.89°) (c) Using sinα = cos(α − 90°),
20∠-90° + 10∠60° − 5∠-110° = -j20 + 5 + j8.66 + 1.7101 + j4.699 = 6.7101 – j6.641 = 9.44∠-44.7°
Therefore, 20 sin(400t) + 10 cos(400t + 60°) – 5 sin(400t – 20°) = 9.44 cos(400t – 44.7°)
Chapter 9, Solution 20. (a) oooo jj 399.4966.82139.383.32464.340590604 −∠=−−−−=∠−−−∠=V Hence,
)399.4377cos(966.8 otv −= (b) 5,90208010 =−∠+∠= ωω ooo jI , i.e. ooI 04.1651.49204010 ∠=∠+=
)04.165cos(51.49 oti +=
Chapter 9, Solution 21.
(a) oooo jF 86.343236.8758.48296.690304155 ∠=+=−−∠−∠=
)86.3430cos(324.8)( ottf +=
(b) G ooo j 49.62565.59358.4571.2504908 −∠=−=∠+−∠=
)49.62cos(565.5)( ottg −=
(c) ( ) 40,9050101=−∠+∠= ω
ωoo
jH
i.e. ooo jH 6.1162795.0125.025.0180125.09025.0 −∠=−−=−∠+−∠=
)6.11640cos(2795.0)( otth −= Chapter 9, Solution 22.
Let f(t) = ∫∞−
−+t
dttvdtdvtv )(24)(10
oVjVVjVF 3020,5,2410 −∠==−+= ωω
ω
ojjVjVjVF 97.921.440)1032.17)(6.1910(4.02010 −∠=−−=−+=
)97.925cos(1.440)( ottf −=
Chapter 9, Solution 23.
(a) v(t) = 40 cos(ωt – 60°) (b) V = -30∠10° + 50∠60°
= -4.54 + j38.09 = 38.36∠96.8° v(t) = 38.36 cos(ωt + 96.8°)
(c) I = j6∠-10° = 6∠(90° − 10°) = 6∠80°
i(t) = 6 cos(ωt + 80°)
(d) I = j2
+ 10∠-45° = -j2 + 7.071 – j7.071
= 11.5∠-52.06° i(t) = 11.5 cos(ωt – 52.06°)
Chapter 9, Solution 24.
(a)
1,010j
=ω°∠=ω
+V
V
10)j1( =−V
°∠=+=−
= 45071.75j5j1
10V
Therefore, v(t) = 7.071 cos(t + 45°) (b)
4),9010(20j4
5j =ω°−°∠=ω
++ωV
VV
°∠=
++ 80-20
4j4
54jV
°∠=+
°∠= 96.110-43.3
3j580-20
V
Therefore, v(t) = 3.43 cos(4t – 110.96°)
Chapter 9, Solution 25.
(a) 2,45-432j =ω°∠=+ω II
°∠=+ 45-4)4j3(I
°∠=°∠°∠
=+
°∠= 98.13-8.0
13.53545-4
j4345-4
I
Therefore, i(t) = 0.8 cos(2t – 98.13°) (b)
5,2256jj
10 =ω°∠=+ω+ω
III
°∠=++ 225)65j2j-( I
°∠=°∠
°∠=
+°∠
= 56.4-745.056.26708.6
2253j6
225I
Therefore, i(t) = 0.745 cos(5t – 4.56°) Chapter 9, Solution 26.
2,01j
2j =ω°∠=ω
++ωI
II
12j1
22j =
++I
°∠=+
= 87.36-4.05.1j2
1I
Therefore, i(t) = 0.4 cos(2t – 36.87°) Chapter 9, Solution 27.
377,10-110j
10050j =ω°∠=ω
++ωV
VV
°∠=
−+ 10-110377100j
50377jV
°∠=°∠ 10-110)45.826.380(V °∠= 45.92-289.0V
Therefore, v(t) = 0.289 cos(377t – 92.45°).
Chapter 9, Solution 28.
===8
)t377cos(110R
)t(v)t(i s 13.75 cos(377t) A.
Chapter 9, Solution 29.
5.0j-)102)(10(j
1Cj
16-6 =
×=
ω=Z
°∠=°∠°∠== 65-2)90-5.0)(254(IZV
Therefore v(t) = 2 sin(106t – 65°) V.
Chapter 9, Solution 30.
Z 2j)104)(500(jLj 3- =×=ω=
°∠=°∠°∠
== 155-30902
65-60ZV
I
Therefore, i(t) = 30 cos(500t – 155°) A. Chapter 9, Solution 31.
i(t) = 10 sin(ωt + 30°) = 10 cos(ωt + 30° − 90°) = 10 cos(ωt − 60°) Thus, I = 10∠-60°
v(t) = -65 cos(ωt + 120°) = 65 cos(ωt + 120° − 180°) = 65 cos(ωt − 60°)
Thus, V = 65∠-60°
Ω=°∠°∠
== 5.660-1060-65
IV
Z
Since V and I are in phase, the element is a resistor with R = 6.5 Ω.
Chapter 9, Solution 32. V = 180∠10°, I = 12∠-30°, ω = 2
Ω+=°∠=°∠°∠
== 642.9j49.11041530-1201180
IV
Z
One element is a resistor with R = 11.49 Ω. The other element is an inductor with ωL = 9.642 or L = 4.821 H.
Chapter 9, Solution 33. 2
L2R vv110 +=
2R
2L v110v −=
=−= 22L 85110v 69.82 V
Chapter 9, Solution 34.
if 0vo = LC1
C1
L =ω→ω
=ω
=××
=ω−− )102)(105(
133
100 rad/s
Chapter 9, Solution 35.
°∠= 05sV2j)1)(2(jLj ==ω
2j-)25.0)(2(j
1Cj
1==
ω
=°∠°∠=°∠=+−
= )05)(901(0522j
2j2j22j
so VV 5∠90°
Thus, 5 cos(2t + 90°) = =)t(vo -5 sin(2t) V
Chapter 9, Solution 36.
Let Z be the input impedance at the source.
2010100200mH 100 3 jxxjLj ==→ −ω
5002001010
11F10 6 jxxjCj
−==→ −ωµ
1000//-j500 = 200 –j400 1000//(j20 + 200 –j400) = 242.62 –j239.84
ojZ 104.6225584.23962.2242 −∠=−=
mA 896.361.26104.62255
1060 oo
o
I −∠=−∠−∠
=
)896.3200cos(1.266 oti −=
Chapter 9, Solution 37.
5j)1)(5(jLj ==ω
j-)2.0)(5(j
1Cj
1==
ω
Let Z , j-1 = 5j210j
5j2)5j)(2(
5j||22 +=
+==Z
Then, s21
2x I
ZZZ
I+
= , where °∠= 0s 2I
°∠=+
=
++
+= 3212.2
8j520j
)2(
5j210j
j-
5j210j
xI
Therefore, =)t(i x 2.12 sin(5t + 32°) A
Chapter 9, Solution 38.
(a) 2j-)6/1)(3(j
1Cj
1F
61
==ω
→
°∠=°∠−
= 43.18-472.4)4510(2j4
2j-I
Hence, i(t) = 4.472 cos(3t – 18.43°) A
°∠=°∠== 43.18-89.17)43.18-472.4)(4(4IV Hence, v(t) = 17.89 cos(3t – 18.43°) V
(b) 3j-)12/1)(4(j
1Cj
1F
121
==ω
→
12j)3)(4(jLjH3 ==ω→
°∠=−
°∠== 87.3610
j34050
ZV
I
Hence, i(t) = 10 cos(4t + 36.87°) A
°∠=°∠+
= 69.336.41)050(j128
12jV
Hence, v(t) = 41.6 cos(4t + 33.69°) V Chapter 9, Solution 39.
10j810j5j
)10j-)(5j(8)10j-(||5j8 +=
−+=+=Z
°∠=°∠
=+
°∠== 34.51-124.3
34.51403.620
j108040
ZV
I
°∠==−
= 34.51-248.6210j5j
10j-1 III
°∠=== 66.1283.124-5j-5j
2 III
Therefore, =)t(i1 6.248 cos(120πt – 51.34°) A
=)t(i2 3.124 cos(120πt + 128.66°) A
Chapter 9, Solution 40.
(a) For ω , 1=j)1)(1(jLjH1 ==ω→
20j-)05.0)(1(j
1Cj
1F05.0 ==
ω→
802.0j98.120j2
40j-j)20j-(||2j +=
−+=+=Z
°∠=°∠
°∠=
+°∠
== 05.22-872.105.22136.2
04j0.8021.9804
o ZV
I
Hence, i =)t(o 1.872 cos(t – 22.05°) A
(b) For ω , 5=5j)1)(5(jLjH1 ==ω→
4j-)05.0)(5(j
1Cj
1F05.0 ==
ω→
2.4j6.12j1
4j-5j)4j-(||25j +=
−+=+=Z
°∠=°∠
°∠=
+°∠
== 14.69-89.014.69494.4
04j41.6
04o Z
VI
Hence, i =)t(o 0.89 cos(5t – 69.14°) A
(c) For ω , 10=10j)1)(10(jLjH1 ==ω→
2j-)05.0)(10(j
1Cj
1F05.0 ==
ω→
9j12j2
4j-10j)2j-(||210j +=
−+=+=Z
°∠=°∠
°∠=
+°∠
== 66.38-4417.066.839.055
049j1
04o Z
VI
Hence, i =)t(o 0.4417 cos(10t – 83.66°) A
Chapter 9, Solution 41.
, 1=ωj)1)(1(jLjH1 ==ω→
j-)1)(1(j
1Cj
1F1 ==
ω→
j21
1j-1)j-(||)j1(1 −=
++=++=Z
j210s
−==
ZV
I , II )j1(c +=
°∠=−
−=−=+= 18.43-325.6
j2)10)(j1(
)j1()j1)(j-( IIV
Thus, v(t) = 6.325 cos(t – 18.43°) V
Chapter 9, Solution 42.
ω 200=
100j-)1050)(200(j
1Cj
1F50 6- =
×=
ω→µ
20j)1.0)(200(jLjH1.0 ==ω→
20j40j2-1
j100-j10050
)(50)(-j100-j100||50 −==
−=
°∠=°∠=°∠−++
= 9014.17)060(7020j
)060(20j403020j
20joV
Thus, =)t(vo 17.14 sin(200t + 90°) V or =)t(vo 17.14 cos(200t) V
Chapter 9, Solution 43.
ω 2=2j)1)(2(jLjH1 ==ω→
5.0j-)1)(2(j
1Cj
1F1 ==
ω→
°∠=°∠+
=+−
−= 69.33328.304
5.1j15.1j
15.0j2j5.0j2j
o II
Thus, =)t(io 3.328 cos(2t + 33.69°) A
Chapter 9, Solution 44.
ω 200=2j)1010)(200(jLjmH10 -3 =×=ω→
j-)105)(200(j
1Cj
1mF5 3- =
×=
ω→
4.0j55.010
j35.0j25.0
j31
2j1
41
−=+
+−=−
++=Y
865.0j1892.14.0j55.0
11+=
−==
YZ
°∠=+°∠
=+
°∠= 7.956-96.0
865.0j1892.606
506Z
I
Thus, i(t) = 0.96 cos(200t – 7.956°) A
Chapter 9, Solution 45. We obtain I by applying the principle of current division twice. o
I I2 I2 Io
Z1 -j2 ΩZ2 2 Ω
(a) (b)
2j-1 =Z , 3j1j2-2
j4-j42||-j2)(4j2 +=+=+=Z
j1j10-
)05(3j12j-
2j-
21
12 +
=°∠++
=+
= IZZ
ZI
=+
=
+
==
1110-
j1j10-
j-1j-
j2-2j2-
2o II -5 A
Chapter 9, Solution 46.
°∠=→°+= 405)40t10cos(5i ss I
j-)1.0)(10(j
1Cj
1F1.0 ==
ω→
2j)2.0)(10(jLjH2.0 ==ω→
Let 6.1j8.02j4
8j2j||41 +=
+==Z , j32 −=Z
)405(6.0j8.3
j1.60.8s
21
1o °∠
++
=+
= IZZ
ZI
°∠=°∠
°∠°∠= 46.94325.2
97.8847.3)405)(43.63789.1(
oI
Thus, =)t(io 2.325 cos(10t + 94.46°) A
Chapter 9, Solution 47.
First, we convert the circuit into the frequency domain.
-j10
Ix
+ −
2 Ω j4 5∠0˚ 20 Ω
°∠=°−∠
=−+
=
++−+−
+= 63.524607.0
63.52854.105
626.8j588.425
4j2010j)4j20(10j2
5Ix
is(t) = 0.4607cos(2000t +52.63˚) A
Chapter 9, Solution 48.
Converting the circuit to the frequency domain, we get: 10 Ω 30 ΩV1
+ − j20
Ix -j20 20∠-40˚
We can solve this using nodal analysis.
A)4.9t100sin(4338.0i
4.94338.020j30
29.24643.15I
29.24643.1503462.0j12307.0
402V
402)01538.0j02307.005.0j1.0(V
020j300V
20j0V
104020V
x
x
1
1
111
°+=
°∠=−
°−∠=
°−∠=−
°∠=
°−∠=++−
=−−
+−
+°−∠−
Chapter 9, Solution 49.
4j1
)j1)(2j(2)j1(||2j2T =
+−
+=−+=Z
1 ΩI Ix
j2 Ω -j Ω
IIIj1
2jj12j
2jx +
=−+
= , where 21
05.x =°∠0=I
4jj1
2jj1
x
+=
+= II
°∠=−=+
=+
== 45-414.1j1j
j1)4(
4jj1
Ts ZIV
=)t(vs 1.414 sin(200t – 45°) V Chapter 9, Solution 50. Since ω = 100, the inductor = j100x0.1 = j10 Ω and the capacitor = 1/(j100x10-3) = -j10Ω.
j10 Ix
-j10 +
vx
−
20 Ω
5∠40˚
Using the current dividing rule:
V)50t100cos(50v5050I20V
505.2405.2j40510j2010j
10jI
x
xx
x
°−=°−∠==
°−∠=°∠−=°∠++−
−=
Chapter 9, Solution 51.
5j-)1.0)(2(j
1Cj
1F1.0 ==
ω→
j)5.0)(2(jLjH5.0 ==ω→ The current I through the 2-Ω resistor is
4j32j5j11 s
s −=
++−=
III , where °∠= 010I
°∠=−= 13.53-50)4j3)(10(sI Therefore,
=)t(is 50 cos(2t – 53.13°) A Chapter 9, Solution 52.
5.2j5.2j1
5j5j5
25j5j||5 +=
+=
+=
101 =Z , 5.2j5.25.2j5.25j-2 −=++=Z
I2
Z1 Z2IS
sss21
12 j5
45.2j5.12
10III
ZZZ
I−
=−
=+
=
)5.2j5.2( += 2o IV
ss j5)j1(10
)j1)(5.2(j5
4308 II
−+
=+
−
=°∠
=+
−°∠=
)j1(10)j5)(308(
sI 2.884∠-26.31° A
Chapter 9, Solution 53.
Convert the delta to wye subnetwork as shown below. Z1 Z2 Io 2 Ω Z3 + 10Ω
60 V 8o30−∠ Ω - Z
,3077.24615.0210
46,7692.01532.021042
21 jjxjZj
jxjZ +−=
−=−=
−−
=
2308.01538.1210
123 j
jZ +=
−=
6062.0726.4)3077.25385.9//()2308.01538.9()10//()8( 23 jjjZZ +=++=++
163.0878.66062.0726.42 1 jjZZ −=+++=
A 64.28721.83575.188.63060
Z3060I o
o
ooo −∠=
−∠
−∠=
−∠=
Chapter 9, Solution 54.
Since the left portion of the circuit is twice as large as the right portion, the equivalent circuit is shown below. Vs
−
V1
+
+
V2
−
+ −
Z 2 Z
)j1(2)j1(o1 −=−= IV )j1(42 12 −== VV
)j1(621s −=+= VVV =sV 8.485∠-45° V
Chapter 9, Solution 55.
-j4 Ω
I I1
+
Vo
−
I2+ −
Z 12 Ω
-j20 V j8 Ω
-j0.58j4
8jo
1 ===V
I
j8j4-
)8j((-j0.5)j4-
)8j(12 +=
+=
+=
ZZZII
5.0j8
j8
-j0.521 +=++=+=ZZ
III
)8j(12j20- 1 ++= ZII
)8j(2j-
2j
812j20- ++
+= ZZ
−=21
j23
j26-4- Z
°∠=°∠°∠
=−
= 279.6864.1618.43-5811.1
25.26131.26
21
j23
j26-4-Z
Z = 2.798 – j16.403 Ω
Chapter 9, Solution 56.
30H3 jLj =→ ω
30/13F jCj
−=→ω
15/11.5F jCj
−=→ω
06681.0
1530
1530
)15///(30 jjj
jxjjj −=
−
−
=−
Ω−=−+−−−
=−−
= m 333606681.02033.0
)06681.02(033.0)06681.02//(30
jjjjjjjZ
Chapter 9, Solution 57.
2H2 jLj =→ ω
jCj
−=→ω11F
2.1j6.2j22j)j2(2j1)j2//(2j1Z +=
−+−
+=−+=
S 1463.0j3171.0Z
1Y −==
Chapter 9, Solution 58.
(a) 2j-)1010)(50(j
1Cj
1mF10 3- =
×=
ω→
5.0j)1010)(50(jLjmH10 -3 =×=ω→
)2j1(||15.0jin −+=Z
2j22j1
5.0jin −−
+=Z
)j3(25.05.0jin −+=Z =inZ 0.75 + j0.25 Ω
(b) 20j)4.0)(50(jLjH4.0 ==ω→
10j)2.0)(50(jLjH2.0 ==ω→
20j-)101)(50(j
1Cj
1mF1 3- =
×=
ω→
For the parallel elements,
20j-1
10j1
2011
p++=
Z
10j10p +=Z Then,
=inZ 10 + j20 + = pZ 20 + j30 Ω Chapter 9, Solution 59.
)4j2(||)2j1(6eq +−+=Z
)4j2()2j1()4j2)(2j1(
6eq ++−+−
+=Z
5385.1j308.26eq −+=Z
=eqZ 8.308 – j1.5385 Ω
Chapter 9, Solution 60.
Ω+=−++=+−++= 878.91.51122.5097.261525)1030//()5020()1525( jjjjjjZ
Chapter 9, Solution 61.
All of the impedances are in parallel.
3j11
5j1
2j11
j111
eq +++
++
−=
Z
4.0j8.0)3.0j1.0()2.0j-()4.0j2.0()5.0j5.0(1
eq
−=−++−++=Z
=−
=4.0j8.0
1eqZ 1 + j0.5 Ω
Chapter 9, Solution 62.
2 20j)102)(1010(jLjmH -33 =××=ω→
100j-)101)(1010(j
1Cj
1F1 6-3 =
××=
ω→µ
50 Ω j20 Ω
+
+ −V +
Vin
−
1∠0° A 2V
-j100 Ω
50)50)(01( =°∠=V
)50)(2()100j20j50)(01(in +−+°∠=V 80j15010080j50in −=+−=V
=°∠
=01in
in
VZ 150 – j80 Ω
Chapter 9, Solution 63. First, replace the wye composed of the 20-ohm, 10-ohm, and j15-ohm impedances with the corresponding delta.
5.22j1020
450j200z,333.13j3015j
450j200z
45j2010
300j150j200z
32
1
+=+
=−=+
=
+=++
=
–j12 Ω –j16 Ω8 Ω
z2
z3
–j16 Ω
z1
10 Ω
ZT 10 Ω Now all we need to do is to combine impedances.
Ω−=−+−+−=
−=−
−=−
−−=−
93.6j69.34)821.3j7.21938.8j721.8(z12j8Z
821.3j70.21)16j10(z
938.8j721.833.29j40
)16j10)(333.13j30()16j10(z
1T
3
2
Chapter 9, Solution 64.
A7.104527.14767.1j3866.0Z
9030I
5j192j6
)8j6(10j4Z
T
T
°∠=+−=°∠
=
Ω−=−+−
+=
Chapter 9, Solution 65.
)4j3(||)6j4(2T +−+=Z
2j7)4j3)(6j4(
2T −+−
+=Z
=TZ 6.83 + j1.094 Ω = 6.917∠9.1° Ω
=°∠
°∠==
1.9917.610120
TZV
I 17.35∠0.9° A
Chapter 9, Solution 66.
)j12(145170
5j60)10j40)(5j20(
)10j40(||)5j20(T −=+
+−=+−=Z
=TZ 14.069 – j1.172 Ω = 14.118∠-4.76°
°∠=°∠
°∠== 76.9425.4
76.4-118.149060
TZV
I
I
I1
20 Ω
+ −Vab
I2
j10 Ω
IIIj122j8
5j6010j40
1 ++
=++
=
IIIj12j4
5j605j20
2 +−
=+−
=
21ab 10j20- IIV +=
IIVj12
40j10j12
)40j(160-ab +
++
++
=
IIV145
j)(150)-12(j12
150-ab
+=
+=
)76.9725.4)(24.175457.12(ab °∠°∠=V
=abV 52.94∠273° V
Chapter 9, Solution 67.
(a) 20j)1020)(10(jLjmH20 -33 =×=ω→
80j-)105.12)(10(j
1Cj
1F5.12 6-3 =
×=
ω→µ
)80j60(||20j60in −+=Z
60j60)80j60)(20j(
60in −−
+=Z
°∠=+= 22.20494.6733.23j33.63inZ
==in
in
1Z
Y 0.0148∠-20.22° S
(b) 10j)1010)(10(jLjmH10 -33 =×=ω→
50j-)1020)(10(j
1Cj
1F20 6-3 =
×=
ω→µ
2060||30 =
)10j40(||2050j-in ++=Z
10j60)10j40)(20(
50j-in ++
+=Z
°∠=−= 56.74-75.5092.48j5.13inZ
==in
in
1Z
Y 0.0197∠74.56° S = 5.24 + j18.99 mS
Chapter 9, Solution 68.
4j-
1j3
12j5
1eq +
++
−=Y
)25.0j()1.0j3.0()069.0j1724.0(eq +−++=Y
=eqY 0.4724 + j0.219 S
Chapter 9, Solution 69.
)2j1(41
2j-1
411
o+=+=
Y
6.1j8.05
)2j1)(4(2j1
4o −=
−=
+=Y
6.0j8.0jo −=+Y
)6.0j8.0()333.0j()1(6.0j8.0
13j-
1111
o
+++=−
++=′Y
°∠=+=′ 41.27028.2933.0j8.11
oY
2271.0j4378.041.27-4932.0o −=°∠=′Y
773.4j4378.05jo +=+′Y
97.22773.4j4378.0
5.0773.4j4378.0
1211
eq
−+=
++=
Y
2078.0j5191.01
eq−=
Y
=−
=3126.0
2078.0j5191.0eqY 1.661 + j0.6647 S
Chapter 9, Solution 70.
Make a delta-to-wye transformation as shown in the figure below.
cb
n
a
Zan
Zcn Zbn
2 Ω
Zeq
8 Ω
-j5 Ω
9j75j15
)10j15)(10(15j1010j5)15j10)(10j-(
an −=+−
=++−
+=Z
5.3j5.45j15
)15j10)(5(bn +=
++
=Z
3j1-5j15
)10j-)(5(cn −=
+=Z
)5j8(||)2( cnbnaneq −+++= ZZZZ
)8j7(||)5.3j5.6(9j7eq −++−=Z
5.4j5.13)8j7)(5.3j5.6(
9j7eq −−+
+−=Z
2.0j511.59j7eq −+−=Z
=−= 2.9j51.12eqZ 15.53∠-36.33° Ω
Chapter 9, Solution 71.
We apply a wye-to-delta transformation.
j4 Ω
Zeq
-j2 ΩZbc
Zab
Zac
a b
c
1 Ω
j12j
2j22j
4j2j2ab −=
+=
+−=Z
j12
2j2ac +=
+=Z
j2-2j-2j2
bc +=+
=Z
8.0j6.13j1
)j1)(4j()j1(||4j||4j ab −=
+−
=−=Z
2.0j6.0j2
)j1)(1()j1(||1||1 ac +=
++
=+=Z
6.0j2.2||1||4j acab −=+ ZZ
6.0j2.21
2j2-1
2j-11
eq −+
++=
Z
1154.0j4231.025.0j25.05.0j ++−−= °∠=+= 66.644043.03654.0j173.0
=eqZ 2.473∠-64.66° Ω = 1.058 – j2.235 Ω
Chapter 9, Solution 72.
Transform the delta connections to wye connections as shown below. a
R3
R2R1
-j18 Ω -j9 Ω
j2 Ω
j2 Ωj2 Ω
b
6j-18j-||9j- = ,
Ω=++
= 8102020)20)(20(
R1 , Ω== 450
)10)(20(R 2 , Ω== 4
50)10)(20(
3R
44)j6(j2||)82j(j2ab ++−++=Z
j4)(4||)2j8(j24ab −+++=Z
j2-12)4jj2)(4(8
j24ab
−+++=Z
4054.1j567.3j24ab −++=Z
=abZ 7.567 + j0.5946 Ω
Chapter 9, Solution 73. Transform the delta connection to a wye connection as in Fig. (a) and then transform the wye connection to a delta connection as in Fig. (b).
a
R3
R2R1
-j18 Ω -j9 Ω
j2 Ω
j2 Ωj2 Ω
b
8.4j-10j48
6j8j8j)6j-)(8j(
1 ==−+
=Z
-j4.812 == ZZ
4.6jj1064-
10j)8j)(8j(
3 ===Z
=++++++ ))(2())(4()4)(2( 313221 ZZZZZZ
6.9j4.46)4.6j)(8.4j2()4.6j)(8.4j4()8.4j4)(8.4j2( +=−+−+−−
25.7j5.14.6j
6.9j4.46a −=
+=Z
688.6j574.38.4j4
6.9j4.46b +=
−+
=Z
945.8j727.18.4j2
6.9j4.46c +=
−+
=Z
3716.3j07407688.12j574.3
)88.61583.7)(906(||6j b +=
+°∠°∠
=Z
602.2j186.025.11j5.1
j7.25)-j4)(1.5(||4j- a −=
−−
=Z
1693.5j5634.0945.20j727.1
)07.7911.9)(9012(||12j c +=
+°∠°∠
=Z
)||12j||4j-(||)||6j( cabeq ZZZZ +=
)5673.2j7494.0(||)3716.3j7407.0(eq ++=Z =eqZ 1.508∠75.42° Ω = 0.3796 + j1.46 Ω
Chapter 9, Solution 74.
One such RL circuit is shown below.
Z
+
Vi = 1∠0° j20 Ω
20 Ω 20 ΩV
j20 Ω+
Vo
−
We now want to show that this circuit will produce a 90° phase shift.
)3j1(42j120j20-
40j20)20j20)(20j(
)20j20(||20j +=++
=++
=+=Z
)j1(31
3j63j1
)01(12j2412j4
20 i +=++
=°∠++
=+
= VZ
ZV
°∠==
+
+
=+
= 903333.03j
)j1(31
j1j
20j2020j
o VV
This shows that the output leads the input by 90°.
Chapter 9, Solution 75.
Since , we need a phase shift circuit that will cause the output to lead the input by 90°.
)90tsin()tcos( °+ω=ωThis is achieved by the RL circuit shown
below, as explained in the previous problem.
10 Ω 10 Ω
+
Vi
−
j10 Ω j10 Ω+
Vo
−
This can also be obtained by an RC circuit.
Chapter 9, Solution 76.
Let Z = R – jX, where fC2
1C
1Xπ
=ω
=
394.9566116R|Z|XXR|Z| 222222 ===−=→+=
F81.27394.95x60x2
1fX21C µ=
π=
π=
Chapter 9, Solution 77.
(a) ic
co jXR
jX-VV
−=
where 979.3)1020)(102)(2(
1C1
X 9-6c =××π
=ω
=
))53.979(tan-90(979.35
979.3j3.979-5
j3.979- 1-22
i
o +°∠+
==VV
)51.38-90(83.1525
979.3
i
o °−°∠+
=VV
°∠= 51.49-6227.0i
o
VV
Therefore, the phase shift is 51.49° lagging
(b) )RX(tan-90-45 c-1+°=°=θ
C1
XR)RX(tan45 cc1-
ω==→=°
RC1
f2 =π=ω
=×π
=π
=)1020)(5)(2(
1RC21
f 9- 1.5915 MHz
Chapter 9, Solution 78. 8+j6 R Z -jX
[ ] 5)6(8
)]6(8[)6(8// =−++−+
=−+=XjRXjRXjRZ
i.e 8R + j6R – jXR = 5R + 40 + j30 –j5X Equating real and imaginary parts:
8R = 5R + 40 which leads to R=13.33Ω 6R-XR =30-5 which leads to X=4.125Ω.
Chapter 9, Solution 79.
(a) Consider the circuit as shown.
40 Ω
Z1
V1V2
j30 Ω+
Vi
−
j60 Ω+
Vo
−
j10 Ω
30 Ω20 Ω
Z2
21j390j30
)60j30)(30j()60j30(||30j1 +=
++
=+=Z
°∠=+=++
=+= 21.80028.9896.8j535.131j43
)21j43)(10j()40(||10j 12 ZZ
Let V . °∠= 01i
896.8j535.21)01)(21.80028.9(
20 i2
22 +
°∠°∠=
+= V
ZZ
V
°∠= 77.573875.02V
°∠°∠°∠
=++
=+
=03.2685.47
)77.573875.0)(87.81213.21(21j43
21j340 22
1
11 VV
ZZ
V
°∠= 61.1131718.01V
111o )j2(52
2j12j
60j3060j
VVVV +=+
=+
=
)6.1131718.0)(56.268944.0(o °∠°∠=V °∠= 2.1401536.0oV
Therefore, the phase shift is 140.2°
(b) The phase shift is leading. (c) If , then V120i =V
°∠=°∠= 2.14043.18)2.1401536.0)(120(oV V and the magnitude is 18.43 V.
Chapter 9, Solution 80. Ω=×π=ω→ 4.75j)10200)(60)(2(jLjmH200 3-
)0120(4.75j50R
4.75j4.75j50R
4.75jio °∠
++=
++= VV
(a) When Ω= 100R ,
°∠°∠°∠
=°∠+
=69.2688.167
)0120)(904.75()0120(
4.75j1504.75j
oV
=oV 53.89∠63.31° V
(b) When Ω= 0R ,
°∠°∠°∠
=°∠+
=45.5647.90
)0120)(904.75()0120(
4.75j504.75j
oV
=oV 100∠33.55° V
(c) To produce a phase shift of 45°, the phase of = 90° + 0° − α = 45°. oVHence, α = phase of (R + 50 + j75.4) = 45°. For α to be 45°, R + 50 = 75.4 Therefore, R = 25.4 Ω
Chapter 9, Solution 81.
Let Z , 11 R=2
22 Cj1
Rω
+=Z , 33 R=Z , and x
xx Cj1
Rω
+=Z .
21
3x Z
ZZ
Z =
ω+=
ω+
22
1
3
xx Cj
1R
RR
Cj1
R
=== )600(400
1200R
RR
R 21
3x 1.8 kΩ
=×
==→
= )103.0(
1200400
CRR
CC1
RR
C1 6-
23
1x
21
3
x 0.1 µF
Chapter 9, Solution 82.
=×
== )1040(2000100
CRR
C 6-s
2
1x 2 µF
Chapter 9, Solution 83.
=×
== )10250(1200500
LRR
L 3-s
1
2x 104.17 mH
Chapter 9, Solution 84.
Let s
11 Cj1
||Rω
=Z , 22 R=Z , 33 R=Z , and . xxx LjR ω+=Z
1CRjR
Cj1
R
CjR
s1
1
s1
s
1
1 +ω=
ω+
ω=Z
Since 21
3x Z
ZZ
Z = ,
)CRj1(R
RRR
1CRjRRLjR s1
1
32
1
s132xx ω+=
+ω=ω+
Equating the real and imaginary components,
1
32x R
RRR =
)CR(R
RRL s1
1
32x ω=ω implies that
s32x CRRL =
Given that , Ω= k40R1 Ω= k6.1R 2 , Ω= k4R 3 , and F45.0Cs µ=
=Ω=Ω== k16.0k40
)4)(6.1(R
RRR
1
32x 160 Ω
=== )45.0)(4)(6.1(CRRL s32x 2.88 H Chapter 9, Solution 85.
Let , 11 R=Z2
22 Cj1
Rω
+=Z , 33 R=Z , and 4
44 Cj1
||Rω
=Z .
jCRRj-
1CRjR
44
4
44
44 −ω
=+ω
=Z
Since 324121
34 ZZZZZ
ZZ
Z =→= ,
ω−=
−ω 223
44
14
Cj
RRjCR
RRj-
2
3232
424
24414
CjR
RR1CR
)jCR(RRj-ω
−=+ω
+ω
Equating the real and imaginary components,
3224
24
241 RR
1CRRR
=+ω
(1)
2
324
24
24
241
CR
1CRCRR
ω=
+ωω
(2) Dividing (1) by (2),
2244
CRCR
1ω=
ω
4422
2
CRCR1
=ω
4422 CRCR1
f2 =π=ω
4242 CCRR21
fπ
=
Chapter 9, Solution 86.
84j-1
95j1
2401
++=Y
0119.0j01053.0j101667.4 3- +−×=Y
°∠=
+==
2.183861.41000
37.1j1667.410001
YZ
Z = 228∠-18.2° Ω
Chapter 9, Solution 87.
)102)(102)(2(j-
50Cj
150 6-31 ××π
+=ω
+=Z
79.39j501 −=Z
)1010)(102)(2(j80Lj80 -33
2 ××π+=ω+=Z
66.125j802 +=Z
1003 =Z
321
1111ZZZZ
++=
66.125j801
79.39j501
10011
++
−+=
Z
)663.5j605.3745.9j24.1210(101 3- −+++=Z
3-10)082.4j85.25( ×+= °∠×= 97.81017.26 -3
Z = 38.21∠-8.97° Ω
Chapter 9, Solution 88.
(a) 20j12030j20j- −++=Z
Z = 120 – j10 Ω
(b) If the frequency were halved, Cf2
1C1
π=
ωLf2L
would cause the capacitive
impedance to double, while π=ω would cause the inductive impedance to halve. Thus,
40j12015j40j- −++=Z Z = 120 – j65 Ω
Chapter 9, Solution 89.
ω
+ω=Cj
1R||LjinZ
ω−ω+
ω+=
ω+ω+
ω
+ω=
C1
LjR
RLjCL
Cj1
LjR
Cj1
RLj
inZ
22
in
C1
LR
C1
LjRRLjCL
ω−ω+
ω−ω−
ω+=Z
To have a resistive impedance, 0)Im( in =Z . Hence,
0C1
LCL
RL 2 =
ω−ω
−ω
C1
LCR 2
ω−ω=ω
1LCCR 2222 −ω=ω
C1CR
L 2
222
ω+ω
=
(1) Ignoring the +1 in the numerator in (1),
=×== )1050()200(CRL 9-22 2 mH Chapter 9, Solution 90.
Let , °∠= 0145sV L377jL)60)(2(jLjX =π=ω=
jXR800145
jXR80s
++°∠
=++
=V
I
jXR80)145)(80(
801 ++== IV
jXR80)145)(80(
50++
= (1)
jXR80)0145)(jXR(
)jXR(o ++°∠+
=+= IV
jXR80)145)(jXR(
110++
+= (2)
From (1) and (2),
jXR80
11050
+=
=+5
11)80(jXR
30976XR 22 =+ (3)
From (1),
23250
)145)(80(jXR80 ==++
53824XRR1606400 22 =+++
47424XRR160 22 =++ (4)
Subtracting (3) from (4),
=→= R16448R160 102.8 Ω From (3),
204081056830976X2 =−=
=→== LL37786.142X 0.3789 H
Chapter 9, Solution 91.
Lj||RCj
1in ω+
ω=Z
LjRLRj
Cj-
in ω+ω
+ω
=Z
222
222
LRLRjRL
Cj-
ω+ω+ω
+ω
=
To have a resistive impedance, 0)Im( in =Z . Hence,
0LR
LRC1-
222
2
=ω+
ω+
ω
222
2
LRLR
C1
ω+ω
=ω
22
222
LRLR
Cω
ω+=
where 7102f2 ×π=π=ω
)109)(1020)(104()10400)(104(109
C 46142
121424
×××π××π+×
= −
−
nF72
169C 2
2
ππ+
=
C = 235 pF
Chapter 9, Solution 92.
(a) Ω∠=∠
∠== −
oo
o
o xYZZ 5.134.471
104845075100
6
(b) ooo xxZY 5.612121.0104845075100 6 ∠=∠∠== −γ
Chapter 9, Solution 93.
Ls 2 ZZZZ ++= )9.186.05.0(j)2.238.01( +++++=Z
20j25+=Z
°∠°∠
==66.3802.32
0115SL Z
VI
=LI 3.592∠-38.66° A
Chapter 10, Solution 1. ω
1=
°∠→°− 45-10)45tcos(10 °∠→°+ 60-5)30tsin(5
jLjH1 =ω→
j-Cj
1F1 =
ω→
The circuit becomes as shown below.
Vo3 Ω
+ − 10∠-45° V +
− 2 Io
j Ω
5∠-60° V
Applying nodal analysis,
j-j)60-5(
3)45-10( ooo VVV
=−°∠
+−°∠
oj60-1545-10j V=°∠+°∠ °∠=°∠+°∠= 9.24715.73150-1545-10oV
Therefore, =)t(vo 15.73 cos(t + 247.9°) V
Chapter 10, Solution 2.
ω 10=°∠→π− 45-4)4t10cos(4
°∠→π+ 150-20)3t10sin(20 10jLjH1 =ω→
5j-2.0j
1Cj
1F02.0 ==
ω→
The circuit becomes that shown below.
Io
Vo10 Ω
j10 Ω -j5 Ω20∠-150° V + − 4∠-45° A
Applying nodal analysis,
5j-10j45-4
10)150-20( ooo VVV
+=°∠+−°∠
o)j1(1.045-4150-20 V+=°∠+°∠
°∠=+
°∠+°∠== 98.150816.2
)j1(j45-4150-2
10jo
o
VI
Therefore, =)t(io 2.816 cos(10t + 150.98°) A Chapter 10, Solution 3.
ω 4=°∠→ 02)t4cos(2
-j1690-16)t4sin(16 =°∠→ 8jLjH2 =ω→
3j-)121)(4(j
1Cj
1F121 ==
ω→
The circuit is shown below.
Vo
+ − 2∠0° A-j16 V
4 Ω -j3 Ω 6 Ω
1 Ω
j8 Ω
Applying nodal analysis,
8j612
3j416j- ooo
++=+
−− VVV
o8j61
3j41
123j4
16j-V
+
+−
+=+−
°∠=°∠°∠
=+−
= 02.35-835.388.12207.115.33-682.4
04.0j22.156.2j92.3
oV
Therefore, =)t(vo 3.835 cos(4t – 35.02°) V
Chapter 10, Solution 4.
16 4,10-16)10t4sin( =ω°∠→°−
4jLjH1 =ω→
j-)41)(4(j
1Cj
1F25.0 ==
ω→
j4 Ω
1 Ω
Ix -j Ω
16∠-10° V + −
V1
0.5 Ix
+
Vo
−
j121
4j)10-16( 1
x1
−=+
−°∠ VI
V
But
4j)10-16( 1
x
VI
−°∠=
So, j18j
))10-16((3 11
−=
−°∠ VV
4j1-10-48
1 +°∠
=V
Using voltage division,
°∠=+°∠
=−
= 04.69-232.8)4j1j)(--(1
10-48j1
11o VV
Therefore, =)t(vo 8.232 sin(4t – 69.04°) V
Chapter 10, Solution 5. Let the voltage across the capacitor and the inductor be Vx and we get:
03j
V2j
V4
3010I5.0V xxxx =+−
+°∠−−
xx
xxx V5.0j2j
VIbut3030I5.1V)4j6j3( =−
=°∠=−−+
Combining these equations we get:
A38.97615.425.1j3
30305.0jI
25.1j33030Vor3030V)75.0j2j3(
x
xx
°∠=+
°∠=
+°∠
=°∠=−+
Chapter 10, Solution 6. Let Vo be the voltage across the current source. Using nodal analysis we get:
010j20
V3
20V4V oxo =
++−
− where ox V
10j2020V+
=
Combining these we get:
30j60V)35.0j1(010j20
V3
10j20V4
20V
oooo +=−+→=+
+−+
−
=+−
=+−+
=5.0j2
)3(20Vor5.0j2
30j60V xo 29.11∠–166˚ V.
Chapter 10, Solution 7. At the main node,
++
+
=−−+−
→+−
+∠=+
−−∠
501
30j
20j401V
3j196.520j40
058.31j91.11550V
30jV306
20j40V15120 o
o
V 15408.1240233.0j04.0
7805.4j1885.3V o−∠=+−−
=
Chapter 10, Solution 8.
,200=ω
20j1.0x200jLjmH100 ==ω→
100j10x50x200j
1Cj
1F50 6 −==ω
→µ−
The frequency-domain version of the circuit is shown below.
0.1 Vo 40Ω V1 Io V2 + -j100Ω
6 o15∠ 20Ω Vo j20Ω -
At node 1,
40VV
100jV
20VV1.0156 2111
1o −
+−
+=+∠
or 21 025.0)01.0025.0(5529.17955.5 VVjj −+−=+ (1)
At node 2,
212
121 V)2j1(V30
20jV
V1.040
VV−+=→+=
− (2)
From (1) and (2),
BAVor0
)5529.1j7955.5(VV
)2j1(3025.0)01.0j025.0(
2
1 =
+=
−
−+−
Using MATLAB,
V = inv(A)*B leads to V 09.1613.110,23.12763.70 21 jVj +−=−−=
o21
o 17.82276.740
VVI −∠=
−=
Thus, A )17.82t200cos(276.7)t(i o
o −= Chapter 10, Solution 9.
10 33 10,010)t10cos( =ω°∠→
10jLjmH10 =ω→
20j-)1050)(10(j
1Cj
1F50 6-3 =
×=
ω→µ
Consider the circuit shown below.
V120 Ω V2
-j20 Ω
20 Ω
Io
30 Ω
+
Vo
−
4 Io+ − 10∠0° V
j10 Ω At node 1,
20j-202010 2111 VVVV −
+=−
21 j)j2(10 VV −+= (1)
At node 2,
10j3020)4(
20j-2121
++=
− VVVV, where
201
o
VI = has been substituted.
21 )8.0j6.0()j4-( VV +=+
21 j4-8.0j6.0
VV++
= (2)
Substituting (2) into (1)
22 jj4-
)8.0j6.0)(j2(10 VV −
+++
=
or 2.26j6.0
1702 −=V
°∠=−
⋅+
=+
= 26.70154.62.26j6.0
170j3
310j30
302o VV
Therefore, =)t(vo 6.154 cos(103 t + 70.26°) V Chapter 10, Solution 10.
2000,100j10x50x2000jLj mH 50 3 =ω==ω→ −
250j10x2x2000j
1Cj
1F2 6 −==ω
→µ−
Consider the frequency-domain equivalent circuit below. V1 -j250 V2 36<0o 2kΩ j100 0.1V1 4kΩ
At node 1,
212111 V004.0jV)006.0j0005.0(36
250jVV
100jV
2000V
36 −−=→−−
++= (1)
At node 2,
212
121 V)004.0j00025.0(V)004.0j1.0(0
4000V
V1.0250j
VV++−=→+=
−−
(2)
Solving (1) and (2) gives
o2o 43.931.89515.893j6.535VV ∠=+−==
vo (t) = 8.951 sin(2000t +93.43o) kV
Chapter 10, Solution 11.
cos( 2,01)t2 =ω°∠→
°∠→°+ 60-8)30t2sin(8
2jLjH1 =ω→ j-)21)(2(j
1Cj
1F2 ==
ω→1
4jLjH2 =ω→ 2j-)41)(2(j
1Cj
1F4 ==
ω→1
Consider the circuit below.
-j Ω
-j Ω
2 Io
2 Io
2 Io 2 Io 2 I
2 2
2 Io
2
At node 1,
2jj-2)60-8( 2111 VVVV −
+=−°∠
21 j)j1(60-8 VV ++=°∠ (1)
At node 2,
02j4j)60-8(
2j1 221 =
−−°∠
+−
+VVV
12 5.0j60-4 VV ++°∠= (2)
Substituting (2) into (1), 1)5.1j1(30460-81 V+=°∠−°∠+
5.1j130460-81
1 +°∠−°∠+
=V
°∠=−
°∠−°∠+== 46.55-024.5
j5.130460-81
j-1
o
VI
Therefore, =)t(io 5.024 cos(2t – 46.55°) Chapter 10, Solution 12.
20 1000,020)t1000sin( =ω°∠→
10jLjmH10 =ω→
20j-)1050)(10(j
1Cj
1F50 6-3 =
×=
ω→µ
The frequency-domain equivalent circuit is shown below.
2 Io
-j20 Ω20 Ω
V2V1
20∠0° A
10 Ω
Io
j10 Ω
At node 1,
1020220 211
o
VVVI
−++= ,
where
10j2
o
VI =
102010j2
20 2112 VVVV −++=
21 )4j2(3400 VV +−= (1)
At node 2,
10j20j-1010j2 22212 VVVVV
+=−
+
21 )2j3-(2j VV +=
or (2) 21 )5.1j1( VV +=Substituting (2) into (1),
222 )5.0j1()4j2()5.4j3(400 VVV +=+−+=
5.0j1400
2 +=V
°∠=+
== 6.116-74.35)5.0j1(j
4010j
2o
VI
Therefore, =)t(io 35.74 sin(1000t – 116.6°) A Chapter 10, Solution 13. Nodal analysis is the best approach to use on this problem. We can make our work easier by doing a source transformation on the right hand side of the circuit. –j2 Ω 18 Ω j6 Ω
+ − 50∠0º V
+ −
+
Vx
−
3 Ω
40∠30º V
06j1850V
3V
2j3040V xxx =
+−
++−
°∠−
which leads to Vx = 29.36∠62.88˚ A.
Chapter 10, Solution 14.
At node 1,
°∠=−
+−
+−
30204j10
02j-
0 1211 VVVV
100j2.1735.2j)5.2j1(- 21 +=−+ VV (1)
At node 2,
°∠=−
++ 30204j5j-2j
1222 VVVV
100j2.1735.2j5.5j- 12 +=+ VV (2)
Equations (1) and (2) can be cast into matrix form as
°∠°∠
=
+3020030200-
5.5j-5.2j5.2j5.2j1
2
1
VV
°∠=−=+
=∆ 38.15-74.205.5j205.5j-5.2j5.2j5.2j1
°∠=°∠=°∠°∠
=∆ 120600)30200(3j5.5j-302005.2j30200-
1
°∠=+°∠=°∠°∠+
=∆ 7.1081020)5j1)(30200(302005.2j30200-5.2j1
2
°∠=∆∆
= 38.13593.2811V
°∠=∆∆
= 08.12418.4922V
)5.31381.15)(905.0(2j-1 °∠°∠==
VI
=I 7.906∠43.49° A
Chapter 10, Solution 16.
At node 1,
5j-10202j 21211 VVVVV −
+−
+=
21 )4j2()4j3(40j VV +−+= At node 2,
10jj1
5j-1022121 VVVVV
=++−
+−
21 )j1()2j1(-)j1(10 VV +++=+ Thus,
++++
=
+ 2
1
j1j2)(1-j2)(12-4j3
)j1(1040j
VV
°∠=−=++++
=∆ 11.31-099.5j5j1j2)(1-j2)(12-4j3
°∠=+−=+++
=∆ 96.12062.116100j60j1j)(110j2)(12-40j
1
°∠=+=++
+=∆ 29.129142.13j110-90
j)(110j2)(1-40j4j3
2
=∆∆
= 11V 22.87∠132.27° V
=∆∆
= 22V 27.87∠140.6° V
Chapter 10, Solution 17.
Consider the circuit below.
At node 1,
IoV2V1
j4 Ω
+ − 100∠20° V
2 Ω
-j2 Ω
1 Ω
3 Ω
234j20100 2111 VVVV −
+=−°∠
21 2j)10j3(
320100 V
V−+=°∠ (1)
At node 2,
2j-2120100 2212 VVVV
=−
+−°∠
21 )5.0j5.1(5.0-20100 VV ++=°∠ (2)
From (1) and (2),
+
+=
°∠°∠
2
1
2j-310j1)j3(5.05.0-
2010020100
VV
5.4j1667.02j-310j1
5.0j5.15.0-−=
++
=∆
2.286j45.55-2j-20100
5.0j5.1201001 −=
°∠+°∠
=∆
5.364j95.26-20100310j1201005.0-
2 −=°∠+°∠
=∆
°∠=∆∆
= 08.13-74.6411V
°∠=∆∆
= 35.6-17.8122V
9j3333.031.78j5.28-
222121
o −+
=∆∆−∆
=−
=VV
I
=oI 9.25∠-162.12°
Chapter 10, Solution 18.
Consider the circuit shown below.
j6 Ω
2 Ω +
Vo
−
-j2 Ω
4 Ω j5 Ω
4∠45° A -j Ω+
Vx
−
2 Vx
8 ΩV1 V2
At node 1,
6j82454 211
+−
+=°∠VVV
21 )3j4()3j29(45200 VV −−−=°∠ (1)
At node 2,
2j5j4j-2
6j822
x21
−++=+
+− VV
VVV
, where VV =
1x
21 )41j12()3j104( VV +=−
21 3j10441j12
VV−+
= (2)
Substituting (2) into (1),
22 )3j4(3j104)41j12(
)3j29(45200 VV −−−+
−=°∠
2)17.8921.14(45200 V°∠=°∠
°∠°∠
=17.8921.14
452002V
222o 258j6-
3j42j-
2j5j42j-
VVVV−
=+
=−+
=
°∠°∠
⋅°∠
=17.8921.14
4520025
13.23310oV
=oV 5.63∠189° V
Chapter 10, Solution 19.
We have a supernode as shown in the circuit below.
j2 Ω
-j4 Ω+
Vo
−
V2V3
V1 4 Ω
2 Ω 0.2 Vo
Notice that . 1o VV =At the supernode,
2j24j-4311223 VVVVVV −
++=−
321 )2j1-()j1()2j2(0 VVV ++++−= (1)
At node 3,
42j2.0 2331
1
VVVVV
−=
−+
0)2j1-()2j8.0( 321 =+++− VVV (2)
Subtracting (2) from (1),
21 j2.10 VV += (3) But at the supernode,
21 012 VV +°∠= or (4) 1212 −= VVSubstituting (4) into (3),
)12(j2.10 11 −+= VV
o1 j2.112j
VV =+
=
°∠°∠
=81.39562.1
9012oV
=oV 7.682∠50.19° V
Chapter 10, Solution 20.
The circuit is converted to its frequency-domain equivalent circuit as shown below.
R
Cj1ω
+
Vo
−
jωL+ − Vm∠0°
Let LC1
Lj
Cj1
Lj
CL
Cj1
||Lj 2ω−ω
=
ω+ω
=ω
ω=Z
m2m
2
2
mo VLj)LC1(R
LjV
LC1Lj
R
LC1Lj
VR ω+ω−
ω=
ω−ω
+
ω−ω
=+
=Z
ZV
ω−ω
−°∠ω+ω−
ω=
)LC1(RL
tan90L)LC1(R
VL2
1-22222
moV
If , then
φ∠= AoV
=A22222
m
L)LC1(RVL
ω+ω−
ω
and =φ)LC1(R
Ltan90 2
1-
ω−ω
−°
Chapter 10, Solution 21.
(a) RCjLC1
1
Cj1
LjR
Cj1
2i
o
ω+ω−=
ω+ω+
ω=
VV
At , 0=ω ==11
i
o
VV
1
As ω , ∞→ =i
o
VV
0
At LC1
=ω , =⋅
=
LC1
jRC
1
i
o
VV
CL
Rj-
(b) RCjLC1
LC
Cj1
LjR
Lj2
2
i
o
ω+ω−ω−
=
ω+ω+
ω=
VV
At , 0=ω =i
o
VV
0
As ω , ∞→ ==11
i
o
VV
1
At LC1
=ω , =⋅
−=
LC1
jRC
1
i
o
VV
CL
Rj
Chapter 10, Solution 22.
Consider the circuit in the frequency domain as shown below.
R1
Cj1ω
R2 +
Vo
−jωL
+ −
Vs
Let Cj
1||)LjR( 2 ω
ω+=Z
LCRj1LjR
Cj1
LjR
)LjR(Cj
1
22
2
2
2
ω−ω+ω+
=
ω+ω+
ω+ω
=Z
CRjLC1LjR
R
CRjLC1LjR
R2
22
1
22
2
1s
o
ω+ω−ω+
+
ω+ω−ω+
=+
=Z
ZVV
=s
o
VV
)CRRL(jLCRRRLjR
2112
21
2
+ω+ω−+ω+
Chapter 10, Solution 23.
0CVj
Cj1Lj
VR
VV s =ω+
ω+ω
+−
s2 VRCVj1LC
RCVjV =ω++ω−
ω+
s2
232VV
LC1RLCjRCjRCjLC1
=
ω−
ω−ω+ω+ω−
)LC2(RCjLC1V)LC1(V 22
s2
ω−ω+ω−
ω−=
Chapter 10, Solution 24.
For mesh 1,
22
121
s Cj1
Cj1
Cj1
IIVω
−
ω+
ω= (1)
For mesh 2,
22
12 Cj
1LjR
Cj1
0 II
ω+ω++
ω−
= (2)
Putting (1) and (2) into matrix form,
ω+ω+
ω−
ω−
ω+
ω=
2
1
22
221s
Cj1LjR
Cj1
Cj1
Cj1
Cj1
0 IIV
212
221 CC1
Cj1
LjRCj1
Cj1
ω+
ω+ω+
ω+
ω=∆
ω+ω+=∆
2s1 Cj
1LjRV and
2
s2 Cjω=∆
V
=∆∆
= 11I
212
221
2s
CC1
Cj1
LjRCj1
Cj1
Cj1
LjRV
ω+
ω+ω+
ω+
ω
ω+ω+
=∆∆
= 22I
212
221
2
s
CC1
Cj1
LjRCj1
Cj1
CjV
ω+
ω+ω+
ω+
ω
ω
Chapter 10, Solution 25.
2=ω
°∠→ 010)t2cos(10
-j690-6)t2sin(6 =°∠→ 4jLjH2 =ω→
2j-)41)(2(j
1Cj
1F25.0 ==
ω→
The circuit is shown below.
4 Ω j4 Ω
For loop 1,
I2+ − 10∠0° V +
− -j2 ΩI1
Io
6∠-90° V
02j)2j4(10- 21 =+−+ II
21 j)j2(5 II +−= (1) For loop 2,
0)6j-()2j4j(2j 21 =+−+ II 321 =+ II (2)
In matrix form (1) and (2) become
=
−35
11jj2
2
1
II
)j1(2 −=∆ , 3j51 −=∆ , 3j12 −=∆
°∠=+=−
=∆∆−∆
=−= 45414.1j1)j1(2
42121o III
Therefore, =)t(io 1.414 cos(2t + 45°) A Chapter 10, Solution 26.
We apply mesh analysis to the circuit shown below. For mesh 1,
0204010- 21 =−+ II
21 241 II −= (1) For the supermesh,
0)10j30(20)20j20( 312 =++−− III 0)j3()2j2(2- 321 =++−+ III (2)
At node A, 21o III −= (3)
At node B, o32 4III += (4)
Substituting (3) into (4) 2132 44 IIII −+=
123 45 III −= (5) Substituting (5) into (2) gives
21 )3j17()4j14(-0 II +++= (6) From (1) and (6),
++
=
2
1
3j17)4j14(-2-4
01
II
4j40 +=∆
3j17j3170
2-11 +=
+=∆ , 4j14
0j4)(14-14
2 +=+
=∆
4j408j245
45 12123 +
+=
∆∆−∆
=−= III
°∠=++
== 25.70154.6j10
)4j1(1530 3o IV
Therefore, =)t(vo 6.154 cos(103 t + 70.25°) V Chapter 10, Solution 27.
For mesh 1,
020j)20j10j(3040- 21 =+−+°∠ II
21 2jj-304 II +=°∠ (1) For mesh 2,
020j)20j40(050 12 =+−+°∠ II
21 )2j4(2j-5 II −−= (2) From (1) and (2),
−
=
°∠
2
1
)2j4(-2j-2jj-
5304
II
°∠=+=∆ 56.116472.4j42-
°∠=−−°∠=∆ 8.21101.2110j)2j4)(304(-1
°∠=°∠+=∆ 27.15444.412085j-2
=∆∆
= 11I 4.698∠95.24° A
=∆∆
= 22I 0.9928∠37.71° A
Chapter 10, Solution 28.
25.0j4x1j
1Cj
1F1,4jLjH1 −==ω
→=ω→
The frequency-domain version of the circuit is shown below, where
o2
o1 3020V,010V −∠=∠= .
1 j4 j4 1 -j0.25 + + V1 I1 1 I2 V2 - -
o2
o1 3020V,010V −∠=∠=
Applying mesh analysis,
21 I)25.0j1(I)75.3j2(10 −−+= (1)
(2) 21
o I)75.3j2(I)025.0j1(3020 ++−−=−∠− From (1) and (2), we obtain
++−+−+
=
+− 2
1II
75.3j225.0j125.0j175.3j2
10j32.1710
Solving this leads to
o2
o1 1536505.4111.2j1438.4I,69.356747.19769.0j3602.1I ∠=+−=−∠=−=
Hence,
A )15346cos(651.4i A, )69.35t4cos(675.1i o2
o1 +=−=
Chapter 10, Solution 29.
For mesh 1,
02030)j2()5j5( 21 =°∠−+−+ II
21 )j2()5j5(2030 II +−+=°∠ (1) For mesh 2,
0)j2()6j3j5( 12 =+−−+ II
21 )3j5()j2(-0 II −++= (2) From (1) and (2),
+
++=
°∠
2
1
j3-5j)(2-j)(2-5j5
02030
II
°∠=+=∆ 21.948.376j37
°∠=°∠°∠=∆ 96.10-175)96.30-831.5)(2030(1 °∠=°∠°∠=∆ 56.4608.67)56.26356.2)(2030(2
=∆∆
= 11I 4.67∠-20.17° A
=∆∆
= 22I 1.79∠37.35° A
Chapter 10, Solution 30. Consider the circuit shown below.
For mesh 1,
I3
I1
I2
Io
j4 Ω
+ − 10∠0° V
2 Ω
3 Ω
1 Ω
-j2 Ω
321 34j)4j3(20100 III −−+=°∠ (1) For mesh 2,
321 2j)4j3(4j-0 III −++= (2) For mesh 3,
321 )2j5(23-0 III −+−= (3) Put (1), (2), and (3) into matrix form.
°∠=
++
00
20100
j2-52-3-j2-j43j4-3-j4-4j3
3
2
1
III
30j106j2-52-3-
j2-j43j4-3-j4-4j3
+=++
=∆
)26j8)(20100(j2-503-
j2-0j4-3-201004j3
2 +°∠=°∠+
=∆
)20j9)(20100(02-3-0j43j4-
20100j4-4j3
3 +°∠=+°∠+
=∆
30j106)6j1)(20100(23
23o +−°∠
=∆∆−∆
=−= III
=oI 5.521∠-76.34° A
Chapter 10, Solution 31.
Consider the network shown below.
80 Ω j60 Ω 20 ΩIo
-j40 ΩI1 I3-j40 Ω+ − 100∠120° V +
− I2 60∠-30° V
For loop 1,
040j)40j80(20100- 21 =+−+°∠ II
21 4j)j2(42010 II +−=°∠ (1) For loop 2,
040j)80j60j(40j 321 =+−+ III
321 220 III +−= (2)
For loop 3, 040j)40j20(30-60 23 =+−+°∠ II
32 )2j1(24j30-6- II −+=°∠ (3) From (2),
123 22 III −= Substituting this equation into (3),
21 )2j1()2j1(2-30-6- II ++−=°∠ (4) From (1) and (4),
+−
−=
°∠°∠
2
1
2j1)2j1(2-4j)j2(4
30-6-12010
II
°∠=+=++
−=∆ 3274.3720j32
2j14j2-4j-4j8
°∠=+=°∠+°∠−
=∆ 44.9325.8211.82j928.4-30-6-4j2-
120104j82
=∆∆
== 22o II 2.179∠61.44° A
Chapter 10, Solution 32.
Consider the circuit below.
j4 Ω
3 Vo -j2 Ω
Io
I1 I2 + −
+
Vo
−
2 Ω4∠-30° V
For mesh 1,
03)30-4(2)4j2( o1 =+°∠−+ VI where )30-4(2 1o IV −°∠= Hence,
0)30-4(630-8)4j2( 11 =−°∠+°∠−+ II
1)j1(30-4 I−=°∠ or °∠= 15221I
)30-4)(2(2j-
32j-
31
oo I
VI −°∠==
)152230-4(3jo °∠−°∠=I =oI 8.485∠15° A
==32j- o
o
IV 5.657∠-75° V
Chapter 10, Solution 33.
Consider the circuit shown below.
5 A
I1 + −
2 Ω
I
-j2 Ω 2 I I2
I4
I3
j Ω
-j20 V 4 Ω
For mesh 1,
02j)2j2(20j 21 =+−+ II 10j-j)j1( 21 =+− II (1)
For the supermesh,
0j42j)2jj( 4312 =−++− IIII (2) Also,
)(22 2123 IIIII −==−
213 2 III −= (3) For mesh 4,
54 =I (4) Substituting (3) and (4) into (2),
5j)j4-()2j8( 21 =+−+ II (5) Putting (1) and (5) in matrix form,
=
−+
−5j10j-
j42j8jj1
2
1
II
5j3- −=∆ , 40j5-1 +=∆ , 85j15-2 +=∆
=−−
=∆∆−∆
=−=5j3-
45j102121 III 7.906∠43.49° A
Chapter 10, Solution 34.
The circuit is shown below.
-j2 Ω
j4 Ω
8 Ω
10 Ω
3 A I2
I3
I1+ − 40∠90° V
5 Ω
Io
20 Ω
j15 Ω
For mesh 1,
0)4j10()2j8()2j18(40j- 321 =+−−−++ III (1) For the supermesh,
0)2j18()19j30()2j13( 132 =+−++− III (2) Also,
332 −= II (3) Adding (1) and (2) and incorporating (3),
0)15j20()3(540j- 33 =++−+ II
°∠=++
= 48.38465.13j58j3
3I
== 3o II 1.465∠38.48° A
Chapter 10, Solution 35.
Consider the circuit shown below.
4 Ω j2 Ω
1 Ω
I1
I3
I2+ −
-j3 Ω8 Ω
-j4 A
10 Ω
20 V
-j5 Ω
For the supermesh,
0)3j9()8j11(820- 321 =−−−++ III (1) Also,
4j21 += II (2) For mesh 3,
0)3j1(8)j13( 213 =−−−− III (3) Substituting (2) into (1),
32j20)3j9()8j19( 32 −=−−− II (4) Substituting (2) into (3),
32j)j13()3j9(- 32 =−+− II (5) From (4) and (5),
−=
−−−−
32j32j20
j13)3j9(-)3j9(-8j19
3
2
II
69j167 −=∆ , 148j3242 −=∆
°∠°∠
=−−
=∆∆
=45.22-69.18055.24-2.356
69j167148j3242
2I
=2I 1.971∠-2.1° A
Chapter 10, Solution 36.
Consider the circuit below.
j4 Ω -j3 Ω
I1 I2
I3
+
Vo
−
2 Ω2 Ω
+ − 4∠90° A 2 Ω 12∠0° V
2∠0° A
Clearly, 4j9041 =°∠=I and 2-3 =I
For mesh 2, 01222)3j4( 312 =+−−− III
01248j)3j4( 2 =++−− I
64.0j52.3-3j48j16-
2 −=−+
=I
Thus, 28.9j04.7)64.4j52.3)(2()(2 21o +=+=−= IIV
=oV 11.648∠52.82° V Chapter 10, Solution 37. I1 + Ix 120 V Z o90−∠ - I2 Z=80-j35Ω Iz - Iy
o30120 −∠ V Z + I3 For mesh x,
120jZIZI zx −=− (1) For mesh y,
60j92.10330120ZIZI ozy +−=∠−=− (2)
For mesh z,
0ZI3ZIZI zyx =+−− (3) Putting (1) to (3) together leads to the following matrix equation:
BAI0
60j92.103120j
III
)105j240()35j80()35j80()35j80()35j80(0)35j80(0)35j80(
z
y
x=→
+−
−=
−+−+−+−−+−−
Using MATLAB, we obtain
+
+==
j0.15251.3657-j0.954- 2.1806-j1.4115 1.9165-
B*inv(A)I
A 6.1433802.24115.1j9165.1II ox1 ∠=+−==
A 37.963802.23655.2j2641.0III oxy2 −∠=−−=−=
A 63.233802.2954.0j1806.2II oy3 ∠=+=−=
Chapter 10, Solution 38.
Consider the circuit below.
Io
I1
j2 Ω
1 Ω
2 Ω-j4 Ω
10∠90° V
1 Ω
I2
I3 I44∠0° A
+ − 2∠0° A
A
Clearly, 21 =I (1)
For mesh 2, 090104j2)4j2( 412 =°∠++−− III (2)
Substitute (1) into (2) to get 5j22j)2j1( 42 −=+− II
For the supermesh, 04j)4j1(2j)2j1( 2413 =+−+−+ IIII
4j)4j1()2j1(4j 432 =−+++ III (3) At node A,
443 −= II (4) Substituting (4) into (3) gives
)3j1(2)j1(2j 42 +=−+ II (5) From (2) and (5),
+−
=
−
−6j25j2
j12j2j2j1
4
2
II
3j3−=∆ , 11j91 −=∆
)j10-(31
3j3)11j9(--
- 12o +=
−−
=∆∆
== II
=oI 3.35∠174.3° A
Chapter 10, Solution 39.
For mesh 1,
o321 6412I15jI8I)15j28( ∠=+−− (1)
For mesh 2, 0I16jI)9j8(I8 321 =−−+− (2)
For mesh 3, 0I)j10(I16jI15j 321 =++− (3) In matrix form, (1) to (3) can be cast as
BAIor006412
III
)j10(16j15j16j)9j8(8
15j8)15j28( o
3
2
1=
∠=
+−−−−
−−
Using MATLAB, I = inv(A)*B
A 6.1093814.03593.0j128.0I o1 ∠=+−=
A 4.1243443.02841.0j1946.0I o2 ∠=+−=
A 42.601455.01265.0j0718.0I o3 −∠=−=
A 5.481005.00752.0j0666.0III o21x ∠=+=−=
Chapter 10, Solution 40.
Let i , where i is due to the dc source and is due to the ac source. For
, consider the circuit in Fig. (a). 2O1OO ii += 1O 2Oi
1Oi
4 Ω 2 Ω
iO1 + −
8 V
(a)Clearly,
A428i 1O == For , consider the circuit in Fig. (b). 2Oi
4 Ω 2 Ω
10∠0° V j4 Ω
IO2
+ −
(b)If we transform the voltage source, we have the circuit in Fig. (c), where Ω= 342||4 .
2 Ω2.5∠0° A 4 Ω
IO2
j4 Ω
(c) By the current division principle,
)05.2(4j34
342O °∠
+=I
°∠=−= 56.71-79.075.0j25.02OI
Thus, A)56.71t4cos(79.0i 2O °−= Therefore,
=+= 2O1OO iii 4 + 0.79 cos(4t – 71.56°) A
Chapter 10, Solution 41. Let vx = v1 + v2. For v1 we let the DC source equal zero. 5 Ω 1 Ω
j100V)j55j1(tosimplifieswhich01
Vj
V5
20V1
111 =+−=+−
+−
V1 = 2.56∠–39.8˚ or v1 = 2.56sin(500t – 39.8˚) V
Setting the AC signal to zero produces:
+ – 20∠0˚ –j
+
V1
−
1 Ω
+ – 6 V
+
V2
−
5 Ω
The 1-ohm resistor in series with the 5-ohm resistor creating a simple voltage divider yielding: v2 = (5/6)6 = 5 V.
vx = 2.56sin(500t – 39.8˚) + 5 V.
Chapter 10, Solution 42. Let ix = i1 + i2, where i1 and i2 which are generated by is and vs respectively. For i1 we let is = 6sin2t A becomes Is = 6∠0˚, where ω =2.
°−∠=−=−−
=−++
−= 63.41983.431.3j724.3
2j52j1126
4j22j34j2I1
i1= 4.983sin(2t – 41.63˚) A –j4 2 Ω
j2
3 Ω
i1 is For i2, we transform vs = 12cos(4t – 30˚) into the frequency domain and get Vs = 12∠–30˚.
Thus, °∠=++−°−∠
= 2.8385.54j32j2
30122I or i2 = 5.385cos(4t + 8.2˚) A
–j2 2 Ω
+ − Vs
j4
3 Ω
i2
ix = [5.385cos(4t + 8.2˚) + 4.983sin(2t – 41.63˚)] A.
Chapter 10, Solution 43.
Let i , where i is due to the dc source and is due to the ac source. For , consider the circuit in Fig. (a).
2O1OO ii += 1O 2Oi
1Oi
4 Ω 2 Ω
iO1 + −
8 V
(a)Clearly,
A428i 1O == For , consider the circuit in Fig. (b). 2Oi
4 Ω 2 Ω
10∠0° V j4 Ω
IO2
+ −
(b)If we transform the voltage source, we have the circuit in Fig. (c), where Ω= 342||4 .
2 Ω2.5∠0° A 4 Ω
IO2
j4 Ω
(c) By the current division principle,
)05.2(4j34
342O °∠
+=I
°∠=−= 56.71-79.075.0j25.02OI Thus, A)56.71t4cos(79.0i 2O °−= Therefore,
=+= 2O1OO iii 4 + 0.79 cos (89)(4t – 71.56°) A
Chapter 10, Solution 44. Let v , where v21x vv += 1 and v2 are due to the current source and voltage source respectively.
For v1 , , 6=ω 30jLjH 5 =ω→ The frequency-domain circuit is shown below. 20Ω j30 + 16Ω V1 Is -
Let o5.1631.12497.3j8.1130j36
)30j20(16)30j20//(16 ∠=+=+
Z +=+=
V )5.26t6cos(7.147v5.267.147)5.1631.12)(1012(ZIV o1
ooos1 +=→∠=∠∠==
For v2 , , 2=ω 10jLjH 5 =ω→ The frequency-domain circuit is shown below. 20Ω j10 + 16Ω V2 + Vs
- - -
Using voltage division,
V )52.15t2sin(41.21v52.1541.2110j36
)050(16V10j2016
16V o2
oo
s2 −=→−∠=+∠
=++
=
Thus,
V )52.15t2sin(41.21)5.26t6cos(7.147v oox −++=
Chapter 10, Solution 45.
Let I , where I is due to the voltage source and is due to the current source. For I , consider the circuit in Fig. (a).
21o II +=
1
1 2I
10 Ω IT
+ − 20∠-150° V -j5 Ωj10 Ω
I1
(a)
10j-5j-||10j =
j1150-2
10j10150-20
T −°∠
=−
°∠=I
Using current division,
°∠+=−
°∠⋅=
−= 150-)j1(-
j1150-2
5j5j-
5j10j5j-
T1 II
For , consider the circuit in Fig. (b). 2I
-j5 Ωj10 Ω
I2
4∠-45° A 10 Ω
(b)
j210j-
5j-||10−
=
Using current division,
°∠+=°∠+−−
= 45-)j1(2-)45-4(10j)j2(10j-
)j2(10j-2I
°∠−°∠=+= 022105-2-21o III °∠=+= 98.150816.2366.1j462.2-oI
Therefore, =oi 2.816 cos(10t + 150.98°) A Chapter 10, Solution 46.
Let v , where , , and are respectively due to the 10-V dc source, the ac current source, and the ac voltage source. For consider the circuit in Fig. (a).
321o vvv ++= 1v 2v 3v
1v
2 H 6 Ω
1/12 F + −
+
v1
−
10 V
(a) The capacitor is open to dc, while the inductor is a short circuit. Hence,
V10v1 = For , consider the circuit in Fig. (b). 2v
2=ω 4jLjH2 =ω→
6j-)12/1)(2(j
1Cj
1F
121
==ω
→
4∠0° A+
V2
−
-j6 Ω 6 Ω j4 Ω
(b)
Applying nodal analysis,
2222
4j
6j
61
4j6j-64 V
VVV
−+=++=
°∠=−
= 56.2645.215.0j1
242V
Hence, V)56.26t2sin(45.21v2 °+= For , consider the circuit in Fig. (c). 3v
3=ω 6jLjH2 =ω→
4j-)12/1)(3(j
1Cj
1F
121
==ω
→
6 Ω j6 Ω
+
V3
−
-j4 Ω+ − 12∠0° V
(c)
At the non-reference node,
6j4j-612 333 VVV
+=−
°∠=+
= 56.26-73.105.0j1
123V
Hence, V)56.26t3cos(73.10v3 °−= Therefore, =ov 10 + 21.45 sin(2t + 26.56°) + 10.73 cos(3t – 26.56°) V Chapter 10, Solution 47.
Let i , where i , i , and are respectively due to the 24-V dc source, the ac voltage source, and the ac current source. For , consider the circuit in Fig. (a).
321o iii ++= 1 2 3i
1i
− +
2 Ω
1 Ω 1/6 F 2 H 24 V
i1
4 Ω
(a) Since the capacitor is an open circuit to dc,
A424
24i1 =
+=
For , consider the circuit in Fig. (b). 2i
1=ω 2jLjH2 =ω→
6j-Cj
1F
61
=ω
→
1 Ω j2 Ω-j6 Ω
2 Ω I2I1+ − 10∠-30° V
I2
4 Ω
(b)For mesh 1,
02)6j3(30-10- 21 =−−+°∠ II
21 2)j21(330-10 II −−=°∠ (1)
For mesh 2,
21 )2j6(2-0 II ++=
21 )j3( II += (2)
Substituting (2) into (1)
215j1330-10 I−=°∠ °∠= 1.19504.02I
Hence, A)1.19tsin(504.0i2 °+= For , consider the circuit in Fig. (c). 3i
3=ω 6jLjH2 =ω→
2j-)6/1)(3(j
1Cj
1F
61
==ω
→
1 Ω j6 Ω-j2 Ω
2∠0° A2 Ω
I3
4 Ω
(c)
2j3)2j1(2
)2j1(||2−−
=−
Using current division,
3j13)2j1(2
2j3)2j1(2
6j4
)02(2j3
)2j1(2
3 +−
=
−−
++
°∠⋅−−
=I
°∠= 43.76-3352.03I Hence A)43.76t3cos(3352.0i3 °−= Therefore, =oi 4 + 0.504 sin(t + 19.1°) + 0.3352 cos(3t – 76.43°) A Chapter 10, Solution 48.
Let i , where i is due to the ac voltage source, is due to the dc voltage source, and is due to the ac current source. For , consider the circuit in Fig. (a).
3O2O1OO iii ++=
3Oi1O 2Oi
1Oi
2000=ω °∠→ 050)t2000cos(50
80j)1040)(2000(jLjmH40 3- =×=ω→
25j-)1020)(2000(j
1Cj
1F20 6- =
×=
ω→µ
I IO1
50∠0° V 80 Ω+ −
-j25 Ω
100 Ω
j80 Ω 60 Ω(a)
3160)10060(||80 =+
33j3230
25j80j316050
+=
−+=I
Using current division,
°∠°∠
==+
=9.4546
1801031-
16080I80-
1O II
°∠= 1.134217.01OI Hence, A)1.134t2000cos(217.0i 1O °+= For , consider the circuit in Fig. (b). 2Oi
+ −
100 Ω
24 V
iO2
80 Ω
60 Ω
(b)
A1.01006080
24i 2O =
++=
For , consider the circuit in Fig. (c). 3Oi
4000=ω °∠→ 02)t4000cos(2
160j)1040)(4000(jLjmH40 3- =×=ω→
5.12j-)1020)(4000(j
1Cj
1F20 6- =
×=
ω→µ
-j12.5 Ω
80 Ω
60 Ω
I2
I1
I3
2∠0° A
j160 Ω
IO3
100 Ω
(c) For mesh 1,
21 =I (1) For mesh 2,
080160j)5.12j160j80( 312 =−−−+ III Simplifying and substituting (1) into this equation yields
32j8)75.14j8( 32 =−+ II (2) For mesh 3,
08060240 213 =−− III Simplifying and substituting (1) into this equation yields
5.13 32 −= II (3) Substituting (3) into (2) yields
125.54j12)25.44j16( 3 +=+ I
°∠=++
= 38.71782.125.44j16
125.54j123I
°∠== 38.71782.1-- 33O II
Hence, A)38.7t4000sin(1782.1-i 3O °+= Therefore, =Oi 0.1 + 0.217 cos(2000t + 134.1°) – 1.1782 sin(4000t + 7.38°) A Chapter 10, Solution 49.
200,308)30t200sin(8 =ω°∠→°+ j)105)(200(jLjmH5 3- =×=ω→
5j-)101)(200(j
1Cj
1mF1 3- =
×=
ω→
After transforming the current source, the circuit becomes that shown in the figure below.
5 Ω 3 Ω I
+ − 40∠30° V
j Ω
-j5 Ω
°∠=−
°∠=
−++°∠
= 56.56472.44j8
30405jj35
3040I
=i 4.472 sin(200t + 56.56°) A
Chapter 10, Solution 50.
55 10,050)t10cos(50 =ω°∠→ 40j)104.0)(10(jLjmH4.0 3-5 =×=ω→
50j-)102.0)(10(j
1Cj
1F2.0 6-5 =
×=
ω→µ
After transforming the voltage source, we get the circuit in Fig. (a).
j40 Ω
20 Ω+
Vo
−
-j50 Ω2.5∠0° A 80 Ω
(a)
Let 5j2
100j-50j-||20
−==Z
and 5j2
250j-)05.2(s −
=°∠= ZV
With these, the current source is transformed to obtain the circuit in Fig.(b).
Z j40 Ω
+
Vo
−
+ − 80 ΩVs
(b)
By voltage division,
5j2250j-
40j805j2
100j-80
40j8080
so −⋅
++−
=++
= VZ
V
°∠=−
= 6.40-15.3642j36
)250j-(8oV
Therefore, =ov 36.15 cos(105 t – 40.6°) V
Chapter 10, Solution 51.
The original circuit with mesh currents and a node voltage labeled is shown below.
Io
40 Ωj10 Ω -j20 Ω4∠-60° V 1.25∠0° A
The following circuit is obtained by transforming the voltage sources.
Io
4∠-60° V -j20 Ωj10 Ω 40 Ω 1.25∠0° A
Use nodal analysis to find . xV
x401
20j-1
10j1
025.160-4 V
++=°∠+°∠
x)05.0j025.0(464.3j25.3 V−=−
°∠=+=−−
= 61.1697.8429.24j42.8105.0j025.0
464.3j25.3xV
Thus, from the original circuit,
10j)29.24j42.81()20j64.34(
10j3040 x
1
+−+=
−°∠=
VI
=+=−
= 678.4j429.0-10j
29.4j78.46-1I 4.698∠95.24° A
4029.24j42.31
40050x
2
+=
°∠−=
VI
=°∠=+= 7.379928.06072.0j7855.02I 0.9928∠37.7° A Chapter 10, Solution 52.
We transform the voltage source to a current source.
12j64j2060
s −=+
°∠=I
The new circuit is shown in Fig. (a).
-j2 Ω
6 Ω
2 ΩIs = 6 – j12 A
-j3 Ωj4 Ω
4 Ω
Ix
5∠90° A
(a)
Let 8.1j4.24j8
)4j2(6)4j2(||6s +=
++
=+=Z
)j2(1818j36)8.1j4.2)(12j6(sss −=−=+−== ZIV With these, we transform the current source on the left hand side of the circuit to a voltage source. We obtain the circuit in Fig. (b).
Zs -j2 Ω
4 Ω
-j3 Ω
+ −
Vs
Ix
j5 A
(b) Let )j12(2.02.0j4.22jso −=−=−= ZZ
207.6j517.15)j12(2.0
)j2(18
o
so −=
−−
==ZV
I
With these, we transform the voltage source in Fig. (b) to a current source. We obtain the circuit in Fig. (c).
Zo
-j3 Ω
4 Ω
Ix
Io j5 A
(c)
Using current division,
)207.1j517.15(2.3j4.62.0j4.2
)5j(3j4 o
o
ox −
−−
=+−+
= IZ
ZI
=+= 5625.1j5xI 5.238∠17.35° A Chapter 10, Solution 53.
We transform the voltage source to a current source to obtain the circuit in Fig. (a).
4 Ω j2 Ω+
Vo
−
2 Ω
j4 Ω-j3 Ω
5∠0° A -j2 Ω
(a)
Let 6.1j8.02j4
8j2j||4s +=
+==Z
j4)6.1j8.0)(5()05( ss +=+=°∠= ZV 8 With these, the current source is transformed so that the circuit becomes that shown in Fig. (b).
Zs -j3 Ω j4 Ω
+
Vo
−
+ −
Vs 2 Ω -j2 Ω
(b)Let 4.1j8.03jsx −=−= ZZ
6154.4j0769.34.1j8.0
8j4
s
sx +−=
−+
==ZVI
With these, we transform the voltage source in Fig. (b) to obtain the circuit in Fig. (c).
j4 Ω
+
Vo
−
Zx -j2 Ω2 ΩIx
(c)
Let 5714.0j8571.04.1j8.28.2j6.1
||2 xy −=−−
== ZZ
7143.5j)5714.0j8571.0()6154.4j0769.3(yxy =−⋅+−== ZIV With these, we transform the current source to obtain the circuit in Fig. (d).
Zy j4 Ω
+ − -j2 Ω
+
Vo
−
Vy
(d)
Using current division,
=−+−
=−+
=2j4j5714.0j8571.0
)7143.5j(2j-2j4j
2j-y
yo V
ZV (3.529 – j5.883) V
Chapter 10, Solution 54.
059.2224.133050
)30(50)30//(50 jjjxj −=
−−
=−
We convert the current source to voltage source and obtain the circuit below. 13.24 – j22.059Ω
40Ω j20Ω
+ + - 115.91 –j31.06V I V - + -
134.95-j74.912 V
Applying KVL gives -115.91 + j31.058 + (53.24-j2.059)I -134.95 + j74.912 = 0
or 8055.17817.4059.224.53
97.10586.250 jjjI +−=
−+−
=
But I)20j40(VV0VI)20j40(V s +−=→=+++−
V 15406.124)8055.1j7817.4)(20j40(05.31j91.115V o−∠=+−+−−=
which agrees with the result in Prob. 10.7. Chapter 10, Solution 55.
(a) To find , consider the circuit in Fig. (a). thZ
j20 Ω 10 Ω
Zth-j10 Ω
(a)
10j20j)10j-)(20j(
10)10j-(||20j10thN −+=+== ZZ
=−= 20j10 22.36∠-63.43° Ω To find , consider the circuit in Fig. (b). thV
j20 Ω 10 Ω
+
Vth
−
+ − 50∠30° V -j10 Ω
(b)
=°∠−
= )3050(10j20j
10j-thV -50∠30° V
=°∠
°∠==
43.63-36.223050-
th
thN Z
VI 2.236∠273.4° A
(b) To find , consider the circuit in Fig. (c). thZ
-j5 Ω
j10 ΩZth
8 Ω
(c)
=−+−
=−==5j810j)5j8)(10j(
)5j8(||10jthN ZZ 10∠26° Ω
To obtain , consider the circuit in Fig. (d). thV
-j5 Ω
Io
4∠0° A
+
Vth
−
j10 Ω8 Ω
(d) By current division,
5j832
)04(5j10j8
8o +
=°∠−+
=I
=+
==5j8
320j10j oth IV 33.92∠58° V
=°∠°∠
==2610
5892.33
th
thN Z
VI 3.392∠32° A
Chapter 10, Solution 56.
(a) To find , consider the circuit in Fig. (a). thZ
j4 Ω
-j2 ΩZth
6 Ω
(a)
4j62j4j
)2j-)(4j(6)2j-(||4j6thN −=
−+=+== ZZ
= 7.211∠-33.69° Ω By placing short circuit at terminals a-b, we obtain,
=NI 2∠0° A
=°∠°∠== )02()69.33-211.7(ththth IZV 14.422∠-33.69° V
(b) To find , consider the circuit in Fig. (b). thZ
j10 Ω
-j5 Ω60 ΩZth
30 Ω
(b)
2060||30 =
5j20)10j20)(5j-(
)10j20(||5j-thN ++
=+== ZZ
= 5.423∠-77.47° Ω
To find and , we transform the voltage source and combine the 30 Ω and 60 Ω resistors. The result is shown in Fig. (c).
thV NI
a
4∠45° A -j5 Ω
j10 Ω
20 ΩIN
b (c)
)454)(j2(52
)454(10j20
20N °∠−=°∠
+=I
= 3.578∠18.43° A
)43.18578.3()47.77-423.5(Nthth °∠°∠== IZV = 19.4∠-59° V
Chapter 10, Solution 57.
To find , consider the circuit in Fig. (a). thZ
5 Ω -j10 Ω 2 Ω
j20 ΩZth
(a)
10j5)10j5)(20j(
2)10j5(||20j2thN +−
+=−+== ZZ
=− 12j18= 21.633∠-33.7° Ω To find , consider the circuit in Fig. (b). thV
5 Ω -j10 Ω 2 Ω
+
Vth
−
+ − j20 Ω60∠120° V
(b)
)12060(2j1
4j)12060(
20j10j520j
th °∠+
=°∠+−
=V
= 107.3∠146.56° V
=°∠°∠
==7.33-633.21
56.1463.107
th
thN Z
VI 4.961∠-179.7° A
Chapter 10, Solution 58.
Consider the circuit in Fig. (a) to find . thZ
Zth
-j6 Ω
j10 Ω8 Ω
(a)
)j2(54j8
)6j8)(10j()6j8(||10jth +=
+−
=−=Z
= 11.18∠26.56° Ω Consider the circuit in Fig. (b) to find . thV
+
Vth
Io
5∠45° A
-j6 Ω
j10 Ω8 Ω
(b)
)455(2j43j4
)455(10j6j8
6j8o °∠
+−
=°∠+−
−=I
=+
°∠−==
)j2)(2()455)(3j4)(10j(
10j oth IV 55.9∠71.56° V
Chapter 10, Solution 59.
The frequency-domain equivalent circuit is shown in Fig. (a). Our goal is to find and
across the terminals of the capacitor as shown in Figs. (b) and (c). thV
thZ
3 Ω j Ω 3 Ω j Ω
Zth
b
a+
Vo
−
-j Ω+ − 5∠-60° A10∠-45° V +
−
(b) (a)
3 Ω j Ω Zth
+
Vth
−
10∠-45° V + −
+ −
+
Vo
−b
a
(d)
+ −
Vth5∠-60° A -j Ω
(c) From Fig. (b),
)3j1(103
j33j
j||3th +=+
==Z
From Fig.(c),
0j
60-53
45-10 thth =−°∠
+−°∠ VV
3j1301545-10
th −°∠−°∠
=V
From Fig. (d),
°∠−°∠=−
= 301545-10j
j-th
tho V
ZV
=oV 15.73∠247.9° V Therefore, =ov 15.73 cos(t + 247.9°) V Chapter 10, Solution 60.
(a) To find , consider the circuit in Fig. (a). thZ
10 Ω -j4 Ω
b
a
j5 Ω 4 Ω
(a)
Zth
)4j24j-(||4)5j||104j-(||4th ++=+=Z
== 2||4thZ 1.333 Ω To find , consider the circuit in Fig. (b). thV
+
Vth
−
j5 Ω 4 Ω
V2V1
+ − 4∠0° A20∠0° V
10 Ω -j4 Ω
(b) At node 1,
4j-5j1020 2111 VVVV −
+=−
205.2j)5.0j1( 21 =−+ VV (1)
At node 2,
44j-4 221 VVV
=−
+
16j)j1( 21 +−= VV (2)
Substituting (2) into (1) leads to
2)3j5.1(16j28 V−=−
333.5j83j5.1
16j282 +=
−−
=V
Therefore,
== 2th VV 9.615∠33.69° V
(b) To find , consider the circuit in Fig. (c). thZ
c dZth
j5 Ω 4 Ω
10 Ω -j4 Ω
(c)
+
+=+=j2
10j4||4j-)5j||104(||4j-thZ
=+=+= )4j6(64j-
)4j6(||4j-thZ 2.667 – j4 Ω
To find ,we will make use of the result in part (a). thV
)2j3()38(333.5j82 +=+=V )j5()38(16j16j)j1( 21 −+=+−= VV
=+=−= 8j31621th VVV 9.614∠56.31° V
Chapter 10, Solution 61. First, we need to find and across the 1 Ω resistor. thV thZ
4 Ω -j3 Ω j8 Ω 6 Ω
Zth
(a) From Fig. (a),
8.0j4.45j10
)8j6)(3j4()8j6(||)3j4(th −=
++−
=+−=Z
=thZ 4.472∠-10.3° Ω
4 Ω -j3 Ω j8 Ω 6 Ω
+
Vth
−
+ − 2 A -j16 V
(b) From Fig. (b),
8j62
3j416j- thth
+=+
−− VV
°∠=+−
= 45.43-93.204.0j22.056.2j92.3
thV
°∠°∠
=+
=43.8-46.5
45.43-93.201 th
tho Z
VV
°∠= 02.35-835.3oV Therefore, =ov 3.835 cos(4t – 35.02°) V
Chapter 10, Solution 62.
First, we transform the circuit to the frequency domain. 1,012)tcos(12 =ω°∠→
2jLjH2 =ω→
4j-Cj
1F
41
=ω
→
8j-Cj
1F
81
=ω
→
To find , consider the circuit in Fig. (a). thZ
3 Io
21
Io 4 Ω
-j8 Ω
Vx
-j4 Ω + −
1 V
j2 Ω Ix
(a)At node 1,
2j1
34j-4
xo
xx VI
VV −=++ , where
4- x
o
VI =
Thus, 2j
14
24j-
xxx VVV −=−
8.0j4.0x +=V At node 2,
2j1
8j-1
3 xox
VII
−+=+
83
j)5.0j75.0( xx −+= VI
425.0j1.0-x +=I
Ω°∠=−== 24.103-29.2229.2j5246.0-1
xth I
Z
To find , consider the circuit in Fig. (b). thV
3 Io
V2
12∠0° V
Io 4 Ω
-j8 Ω
1 2 +
Vth
−
V1
-j4 Ω+ −
j2 Ω
(b)At node 1,
2j4j-3
412 211
o1 VVV
IV −
++=−
, where 4
12 1o
V−=I
21 2j)j2(24 VV −+= (1)
At node 2,
8j-3
2j2
o21 V
IVV
=+−
21 3j)4j6(72 VV −+= (2)
From (1) and (2),
++
=
2
1
3j-4j62j-j2
7224
VV
6j5- +=∆ , 24j-2 =∆
°∠=∆∆
== 8.219-073.322th VV
Thus,
229.2j4754.1)8.219-073.3)(2(
22
thth
o −°∠
=+
= VZ
V
°∠=°∠°∠
= 3.163-3.25.56-673.28.219-146.6
oV
Therefore, =ov 2.3 cos(t – 163.3°) V
Chapter 10, Solution 63.
Transform the circuit to the frequency domain.
200,304)30t200cos(4 =ω°∠→°+ Ω==ω→ k2j)10)(200(jLjH10
Ω=×
=ω
→µ kj-)105)(200(j
1Cj
1F5 6-
NZ is found using the circuit in Fig. (a).
-j kΩ
2 kΩZN
j2 kΩ
(a)
Ω=++=+= k1j1j-2j||2j-NZ We find using the circuit in Fig. (b). NI
-j kΩ
j2 kΩ 2 kΩ4∠30° A IN
(b)
j12||2j += By the current division principle,
°∠=°∠−+
+= 75657.5)304(
jj1j1
NI
Therefore,
=Ni 5.657 cos(200t + 75°) A =NZ 1 kΩ
Chapter 10, Solution 64.
is obtained from the circuit in Fig. (a). NZ
ZN60 Ω 40 Ω
-j30 Ωj80 Ω
(a)
50j100)50j)(100(
50j||100)30j80j(||)4060(N +==−+=Z
=+= 40j20NZ 44.72∠63.43° Ω
To find , consider the circuit in Fig. (b). NI
IN
I2
I1
Is3∠60° A
j80 Ω
60 Ω 40 Ω
-j30 Ω
(b)
°∠= 603sI For mesh 1,
060100 s1 =− II °∠= 608.11I
For mesh 2,
080j)30j80j( s2 =−− II °∠= 608.42I
=−= 21N III 3∠60° A
Chapter 10, Solution 65.
2,05)t2cos(5 =ω°∠→ 8j)4)(2(jLjH4 ==ω→
2j-)4/1)(2(j
1Cj
1F
41
==ω
→
j-)2/1)(2(j
1Cj
1F
21
==ω
→
To find , consider the circuit in Fig. (a). NZ
2 Ω
ZN
-j2 Ω -j Ω
(a)
)10j2(131
3j2)2j2(j-
)2j2(||j-N −=−−
=−=Z
To find , consider the circuit in Fig. (b). NI
2 Ω+ −
IN-j2 Ω
5∠0° V
-j Ω
(b)
5jj-05
N =°∠
=I
The Norton equivalent of the circuit is shown in Fig. (c).
Io
ZN IN j8 Ω
(c)
Using current division,
94j210j50
8j)10j2)(131()5j)(10j2)(131(
8j NN
No +
+=
+−−
=+
= IZ
ZI
°∠=−= 47.77-05425294.0j1176.0oI Therefore, =oi 0.542 cos(2t – 77.47°) A
Chapter 10, Solution 66.
ω 10=
5j)5.0)(10(jLjH5.0 ==ω→
10j-)1010)(10(j
1Cj
1mF10 3- =
×=
ω→
Vx
j5 Ω 2 Vo +
Vo
−
-j10 Ω
1 A 10 Ω
(a)
To find , consider the circuit in Fig. (a). thZ
10j105j21 xx
o −+=+
VVV , where
10j10 x
o −V
10=V
2j2110j10-
5j10j1019
1 xxx
++
=→=−
+ VVV
=°∠°∠
===44.5095.21
135142.141
xthN
VZZ 0.67∠129.56° Ω
To find and , consider the circuit in Fig. (b). thV NI
2 Vo I
− ++
Vth
−
+
Vo
−
-j10 Ω
j5 Ω10 Ω
12∠0° V
-j2 A
(b)
012)2(5j)2j-)(10()5j10j10( o =−+−+− VI where )2j-)(10(o IV −= Thus,
20j188-)105j10( −=− I
105j10-20j188
++
=I
200105j)40j21(5j)2(5j oth −=+=+= IIVIV
076.2j802.11-200105j10-
)20j188(105jth +=−
++
=V
=thV 11.97∠170° V
=°∠°∠
==56.12967.0
17097.11
th
thN Z
VI 17.86∠40.44° A
Chapter 10, Solution 67.
Ω+=++
+−−
=++−== 079.1j243.116j20
)6j8(125j23
)5j13(10)6j8//(12)5j13//(10ZZ ThN
Ω+=∠++
=+=∠−
= 37.454j93.25)4560(6j20)6j8(V,44.21j78.13)4560(
5j2310V o
bo
a
A 09.9734.38ZV
I V, 599.11.433VVV o
Th
ThN
obaTh −∠==−∠=−=
Chapter 10, Solution 68.
10j1x10jLj H1 ==ω→
2j
201x10j
1Cj
1 F201
−==ω
→
We obtain VTh using the circuit below.
Io 4 Ω a + + + -j2 j10 Vo 6<0o Vo/3 - 4Io - - b
5.2j2j10j)2j(10j)2j//(10j −=
−−
=−
ooo I10j)5.2j(xI4V −=−= (1)
0V31I46 oo =++− (2)
Combining (1) and (2) gives
oooTho 19.5052.11
3/10j460jI10jVV,
3/10j46I −∠=
−−
=−==−
=
)19.50t10sin(52.11v o
Th −= To find RTh, we insert a 1-A source at terminals a-b, as shown below. Io 4 Ω a + + -j2 j10 Vo Vo/3 - 4Io -
1<0o
12V
I0V31I4 o
ooo −=→=+
10jV
2jV
I41 ooo +
−=+
Combining the two equations leads to
4766.1j2293.14.0j333.0
1Vo −=+
=
Ω−== 477.12293.11
VZ o
Th
Chapter 10, Solution 69.
This is an inverting op amp so that
=ω
==Cj1
R--
i
f
s
o
ZZ
VV
-jωRC
When and ms V=V RC1=ω ,
°∠==⋅⋅⋅= 90-VVj-VRCRC1
j- mmmoV
Therefore,
=°−ω= )90tsin(V)t(v mo - Vm cos(ωt) Chapter 10, Solution 70.
This may also be regarded as an inverting amplifier.
44 104,02)t104cos(2 ×=ω°∠→×
Ω=××
=ω
→ k5.2j-)1010)(104(j
1Cj
1nF10 9-4
i
f
s
o -ZZ
VV
=
where and Ω= k50iZ Ω−
== kj40
100j-)k5.2j-(||k100fZ .
Thus, j40
2j-
s
o
−=
VV
If , °∠= 02sV
°∠=°∠
°∠=
−= 57.88-1.0
43.1-01.4090-4
j404j-
oV
Therefore,
=)t(vo 0.1 cos(4x104 t – 88.57°) V
Chapter 10, Solution 71.
oo 308)30t2cos(8 ∠→+
0. Ω−==ω
→µ−
k1j10x5.0x2j
1Cj
1F5 6
At the inverting terminal,
)j6.0(308)j1.0(Vk2308
k10308V
k1j308V
ooo
oo
o +∠=+→∠
=∠−
+−
∠−
o
o 747.4283.9j1.0
)j6.0)(4j9282.6(V ∠=+
++=
vo(t) = 9.283cos(2t + 4.75o) V
Chapter 10, Solution 72.
44 10,04)t10cos(4 =ω°∠→
Ω==ω
→ k100j-)10)(10(j
1Cj
1nF1 9-4
Consider the circuit as shown below.
4∠0° V
Vo Vo
-j100 kΩ
50 kΩ
+ −
+− Io
100 kΩ
At the noninverting node,
5.0j14
100j-504
ooo
+=→=
−V
VV
A56.26-78.35mA)5.0j1)(100(
4k100
oo µ°∠=
+==
VI
Therefore,
=)t(io 35.78 cos(104 t – 26.56°) µA
Chapter 10, Solution 73.
As a voltage follower, o2 VV =
Ω=××
=ω
→= k-j20)1010)(105(j
1Cj1
nF10C 9-31
1
Ω=××
=ω
→= k-j10)1020)(105(j
1Cj1
nF20C 9-32
2
Consider the circuit in the frequency domain as shown below.
-j20 kΩ
Zin
Io
V1
V2Is
-j10 kΩ
20 kΩ10 kΩ
+ −
+− Vo
VS
At node 1,
2020j-10o1o11s VVVVVV −
+−
=−
o1s )j1()j3(2 VVV +−+= (1) At node 2,
10j-0
20oo1 −
=− VVV
o1 )2j1( VV += (2) Substituting (2) into (1) gives
os 6j2 VV = or so 31
-j VV =
so1 31
j32
)2j1( VVV
−=+=
s1s
s k10)j1)(31(
k10V
VVI
−=
−=
k30j1
s
s −=
VI
k)j1(15j1
k30
s
sin +=
−==
IV
Z
=inZ 21.21∠45° kΩ Chapter 10, Solution 74.
11i Cj
1R
ω+=Z ,
22f Cj
1R
ω+=Z
=
ω+
ω+
===
11
22
i
f
s
ov
Cj1
R
Cj1
R-ZZ
VV
A
ω+ω+
11
22
2
1
CRj1CRj1
CC
At , 0=ω =vA2
1
CC
As ω , ∞→ =vA1
2
RR
At 11CR
1=ω , =vA
+
+
j1CRCRj1
CC 1122
2
1
At 22CR
1=ω , =vA
++
22112
1
CRCRj1j1
CC
Chapter 10, Solution 75.
3102×=ω
Ω=××
=ω
→== k-j500)101)(102(j
1Cj1
nF1CC 9-31
21
Consider the circuit shown below.
100 kΩ
-j500 kΩ
+
Vo
−
-j500 kΩ
20 kΩ
V1
V2
100 kΩ20 kΩ
+ −
+−
VS
At node 1,
500j-100500j-211o1s VVVVVV −
+−
=−
2o1s 5j)5j2( VVVV −−+= (1) At node 2,
100500j-221 VVV
=−
21 )5j1( VV −= (2) But
2RRR o
o43
32
VVV =
+= (3)
From (2) and (3),
o1 )5j1(21
VV −⋅= (4)
Substituting (3) and (4) into (1),
ooos 21
5j)5j1)(5j2(21
VVVV −−−+⋅=
os )25j26(21
VV −⋅=
=−
=25j26
2
s
o
VV
0.0554∠43.88°
Chapter 10, Solution 76.
Let the voltage between the -jkΩ capacitor and the 10kΩ resistor be V1.
o1o
o1o11o
V6.0jV)6.0j1(302
k20VV
k10VV
k4jV302
+−=∠
→−
+−
=−
−∠ (1)
Also,
o1oo1 V)5j1(Vk2j
Vk10VV
+=→−
=−
(2)
Solving (2) into (1) yields
V 34.813123.03088.0j047.0V oo −∠=−=
Chapter 10, Solution 77.
Consider the circuit below.
1
2
+ −
VS
+
Vo
−
C1
C2 R2 R1
V1
V1
R3
−+
At node 1,
11
1s CjR
VVV
ω=−
111s )CRj1( VV ω+= (1) At node 2,
)(CjRR
0o12
2
o1
3
1 VVVVV
−ω+−
=−
ω+−= 32
2
31o1 RCj
RR
)( VVV
13223
o RCj)RR(1
1 VV
ω++= (2)
From (1) and (2),
ω++
ω+=
3223
2
11
so RRCjR
R1
CRj1V
V
=s
o
VV
)RRCjR()CRj1(RRCjRR
322311
32232
ω+ω+ω++
Chapter 10, Solution 78. 400,02)t400sin(2 =ω°∠→
Ω=×
=ω
→µ k5j-)105.0)(400(j
1Cj
1F5.0 6-
Ω=×
=ω
→µ k10j-)1025.0)(400(j
1Cj
1F25.0 6-
Consider the circuit as shown below.
20 kΩ
At node 1,
10 kΩ2∠0° V
10 kΩ -j5 kΩ
20 kΩ
V1 V2
-j10 kΩ40 kΩ
+ −
+− Vo
205j-10j-102 o12111 VVVVVV −
+−
+=−
o21 4j)6j3(4 VVV −−+= (1) At node 2,
105j221 VVV
=−−
21 )5.0j1( VV −= (2) But
oo2 31
402020
VVV =+
= (3)
From (2) and (3),
o1 )5.0j1(31
VV −⋅= (4)
Substituting (3) and (4) into (1) gives
oooo 61
j134
j)5.0j1(31
)6j3(4 VVVV
−=−−−⋅⋅+=
°∠=−
= 46.9945.3j6
24oV
Therefore, =)t(vo 3.945 sin(400t + 9.46°) V
Chapter 10, Solution 79.
1000,05)t1000cos(5 =ω°∠→
Ω=×
=ω
→µ k10j-)101.0)(1000(j
1Cj
1F1.0 6-
Ω=×
=ω
→µ k5j-)102.0)(1000(j
1Cj
1F2.0 6-
Consider the circuit shown below.
20 kΩ
+ − Vs = 5∠0° V
V1
-j5 kΩ
-j10 kΩ 40 kΩ
+Vo−
− +
10 kΩ −+
Since each stage is an inverter, we apply ii
fo
-V
ZZ
V = to each stage.
1o 15j-40-
VV =
(1) and
s1 10)10j-(||20-
VV =
(2) From (1) and (2),
°∠
−
= 0510j20
)10-j)(20(-10
8j-oV
°∠=+= 56.2678.35)j2(16oV
Therefore, =)t(vo 35.78 cos(1000t + 26.56°) V Chapter 10, Solution 80. 4
1000,60-4)60t1000cos( =ω°∠→°−
Ω=×
=ω
→µ k10j-)101.0)(1000(j
1Cj
1F1.0 6-
Ω=×
=ω
→µ k5j-)102.0)(1000(j
1Cj
1F2.0 6-
The two stages are inverters so that
+°∠⋅=
10j5-
5020
)60-4(10j-
20oo VV
o52
2j-)60-4()2j(
2j- V⋅+°∠⋅⋅=
°∠=+ 60-4)5j1( oV
°∠=+
°∠= 31.71-922.3
5j160-4
oV
Therefore, =)t(vo 3.922 cos(1000t – 71.31°) V
Chapter 10, Solution 81. The schematic is shown below. The pseudocomponent IPRINT is inserted to print the value of Io in the output. We click Analysis/Setup/AC Sweep and set Total Pts. = 1, Start Freq = 0.1592, and End Freq = 0.1592. Since we assume that w = 1. The output file includes: FREQ IM(V_PRINT1) IP(V_PRINT1) 1.592 E-01 1.465 E+00 7.959 E+01 Thus, Io = 1.465∠79.59o A
Chapter 10, Solution 82. The schematic is shown below. We insert PRINT to print Vo in the output file. For AC Sweep, we set Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592. After simulation, we print out the output file which includes: FREQ VM($N_0001) VP($N_0001) 1.592 E-01 7.684 E+00 5.019 E+01
which means that Vo = 7.684∠50.19o V
hapter 10, Solution 83.
he schematic is shown below. The frequency is
C
T 15.1592
10002/f =π
=πω=
When the circuit is saved and simulated, we obtain from the output file
REQ VM(1) VP(1)
hus, vo = 6.611cos(1000t – 159.2o) V
F1.592E+02 6.611E+00 -1.592E+02 T
hapter 10, Solution 84.
he schematic is shown below. We set PRINT to print Vo in the output file. In AC
FREQ VM($N_0003)
1.592 E-01 1.664 E+00 -1.646
amely, Vo = 1.664∠-146.4o V
C TSweep box, we set Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592. After simulation, we obtain the output file which includes: VP($N_0003) E+02 N
hapter 10, Solution 85.
C
The schematic is shown below. We let =ω rad/s so that L=1H and C=1F.
When the circuit is saved and simulated, we obtain from the output file
FREQ VM(1) VP(1)
From this, we conclude that
5.167228.2Vo −∠= V
1
1.591E-01 2.228E+00 -1.675E+02
Chapter 10, Solution 86.
e insert three pseudocomponent PRINTs at nodes 1, 2, and 3 to print V1, V2, and V3,
FREQ VM($N_0002)
1.592 E-01 6.000 E+01 3.000
FREQ VM($N_0003)
1.592 E-01 2.367 E+02 -8.483
Winto the output file. Assume that w = 1, we set Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592. After saving and simulating the circuit, we obtain the output file which includes: VP($N_0002) E+01 VP($N_0003) E+01
FREQ VM($N_0001)
1.592 E-01 1.082 E+02 1.254
herefore,
V1 = 60∠30o V
VP($N_0001) E+02 T
V2 = 236.7∠-84.83o V V3 = 108.2∠125.4o V
hapter 10, Solution 87.
he schematic is shown below. We insert three PRINTs at nodes 1, 2, and 3. We set otal Pts = 1, Start Freq = 0.1592, End Freq = 0.1592 in the AC Sweep box. After
VM($N_0004) VP($N_0004)
1.696
FREQ VM($N_0001) VP($N_0001)
-1.386
C TTsimulation, the output file includes: FREQ 1.592 E-01 1.591 E+01 E+02 1.592 E-01 5.172 E+00 E+02
FREQ VM($N_0003) VP($N_0003)
-1.524
ore,
∠ o V
1.592 E-01 2.270 E+00 E+02 Theref
V1 = 15.91 169.6 V2 = 5.172∠-138.6o V V3 = 2.27∠-152.4o V
hapter 10, Solution 88.
ow. We insert IPRINT and PRINT to print Io and Vo in the utput file. Since w = 4, f = w/2π = 0.6366, we set Total Pts = 1, Start Freq = 0.6366, nd End Freq = 0.6366 in the AC Sweep box. After simulation, the output file includes:
6.366 E-01 3.496 E+01 1.261
FREQ IM(V_PRINT2) IP _PRINT2)
6.366 E-01 8.912 E-01
C The schematic is shown beloa FREQ VM($N_0002) VP($N_0002) E+01 (V -8.870 E+01
Therefore, Vo = 34.96∠12.6o V, Io = 0.8912∠-88.7o A
vo = 34.96 cos(4t + 12.6o)V, io = 0.8912cos(4t - 88.7o )A
onsider the circuit below.
Chapter 10, Solution 89. C
At node 1,
in−
2
2
1
in
RVVV
=−0
in
R−
in1
22in R
R- VVV =+ (1)
At node 3,
Cj1R4in
3
in2
ω−
=− VVVV
V
C R1 R2 Vin
+
Iin
in
3
R3 R4
− +
−+
V1
2 4
3
2in4in CRj
-ω−
=+VV
VV (2)
rom (1) and (2),
F
in13
24in RCRj
R-- VVV
ω=+
1R
Thus,
in43
2
4
4inin RCRj
RR
VVV
Iω
=−
=
eq2
431
in
inin Lj
RRRCRj
ω=ω
==IV
Z
where 2
431eq R
CRRRL =
Chapter 10, Solution 90.
et
LRCj1
RC
1||R
ω+==Z
C
j4 ω
CjRCj1
Cj1
R3 ωω+
=ω
+=Z
onsider the circuit shown below.
R2Z4
+ Vo −Vi + −
Z3 R1
i21
2i
43
4o RR
RVV
ZZZ
V+
−+
=
21
2
i
o
RRR
CjRCj1
Cj1R
Cj1R
+−
ωω+
+ω+
ω+=
VV
21
22 RR
R)RCj1(RCj
RCj+
−ω++ω
ω=
21
2222
i
o
RRR
RC3jCR1RCj
+−
ω+ω−ω
=V
V
For and to be in phase, oV iVi
o
VV
must be purely real. This happens when
0CR1 222 =ω−
f2RC1
π==ω
or RC21
fπ
=
At this freque cy, n
21
2
i
ov RR
R31
+−==
VV
A
Chapter 10, Solution 91.
(a) Let
=2V voltage at the noninverting terminal of the op amp output=oV voltage of the op amp
Rk10 =Ω op =Z1Cj
LjRs ω+ω+=Z
s in Section 10.9, A
Cj
LjRR
R
o
o
ps
p
o
2
ω−ω++
=+
=ZZ
ZVV
)1LC(j)RR(CCR
2o
o
o
2
−ω++ωω
=VV
For this to be purely real,
LC1
01LC o2o =ω→=−ω
)102)(104.0(21
LC21
f9-3-o
××π=
π=
=of 180 kHz
(b) At oscillation,
o
o
oo
oo
o
2
RRR
)RR(CCR
+=
+ωω
=VV
This must be compensated for by
52080
12
ov =+==
VV
A
==→=+ o
o
o R4R51
RRR
40 kΩ
Chapter 10, Solution 92.
Let voltage at the noninverting terminal of the op amp =2V=oV output voltage of the op amp
os R=Z
)1LC(jRLRL
Lj1
CjR1
1R||
Cj1
||Lj 2p −ω+ωω
=
ω+ω+
=ω
ω=Z
As in Section 10.9,
)1LC(jRLRL
R
)1LC(jRLRL
2o
2
ps
p
o
2
−ω+ωω
+
−ω+ωω
=+
=ZZ
ZVV
)1LC(RjRLRRLRL
2ooo
2
−ω+ω+ωω
=VV
For this to be purely real,
LC21
f1LC o2o π
=→=ω
(a) At , oω=ω
oooo
o
o
2
RRR
LRRLRL
+=
ω+ωω
=VV
This must be compensated for by
11k100k1000
1RR
1o
f
2
ov =+=+==
VV
A
Hence,
==→=+
R10R111
RRR
oo
100 kΩ
(b) )102)(1010(2
1f
9-6-o××π
=
=of 1.125 MHz Chapter 10, Solution 93. As shown below, the impedance of the feedback is
jωL
1Cj1ω2Cj
1ω
ZT
ω+ω
ω=
21T Cj
1Lj||
Cj1
Z
)CLCCC(j
LC1
Cj-
LjCj-
Cj-
LjCj-
212
21
2
21
21T ω−+
ω−ω=
ω+ω+
ω
ω+ω
ω=Z
In order for to be real, the imaginary term must be zero; i.e. TZ
0CLCCC 212o21 =ω−+
T21
212o LC
1CLCCC
=+
=ω
To LC2
1f
π=
Chapter 10, Solution 94. If we select C
nF20C21 ==
nF102
CCC
CCC 1
21
21T ==
+=
Since T
o LC21
fπ
= ,
mH13.10)1010)(102500)(4(
1C)f2(
1L 9-62
T2 =
××π=
π=
=××π
=ω
=)1020)(1050)(2(
1C1
X 9-32
c 159 Ω
We may select and , say Ω= k20R i if RR ≥ Ω= k20R f . Thus,
== 21 CC 20 nF, =L 10.13 mH == if RR 20 kΩ Chapter 10, Solution 95. First, we find the feedback impedance.
C
L1L2
ZT
ω
+ωω=Cj
1Lj||Lj 21TZ
)1)LL(C(j)L1(CL
Cj
LjLj
Cj
LjLj
212
212
21
21
T −+ωω−ω
=
ω−ω+ω
ω−ωω
=Z
In order for to be real, the imaginary term must be zero; i.e. TZ
01)LL(C 212o =−+ω
)LL(C1
f221
oo +=π=ω
)LL(C21
f21
o +π=
Chapter 10, Solution 96.
(a) Consider the feedback portion of the circuit, as shown below.
V2V1
+ −
R
R
jωL
Vo jωL
21 LjLjR
LjRLj
VVVV 12 ωω+
=→ω+
ω= (1)
Applying KCL at node 1,
LjRRLj111o
ω++=
ω− VVVV
ω+
+ω=−LjR
1R1
Lj 11o VVV
ω+ω−ω
+=)LjR(RLRL2j
122
1o VV
(2) From (1) and (2),
2
22
o )LjR(RLRL2j
1Lj
LjRVV
ω+ω−ω
+
ωω+
=
RLjLRL2jRLjR 222
2
o
ωω−ω+ω+
=VV
RLjLR
3
1222
o
2
ωω−
+=
VV
=o
2
VV
)LRRL(j31
ω−ω+
(b) Since the ratio o
2
VV
must be real,
0L
RR
L
o
o =ω
−ω
LR
Lo
2
o ω=ω
LR
f2 oo =π=ω
L2R
fo π=
(c) When oω=ω
31
o
2 =VV
This must be compensated for by 3v =A . But
3RR
11
2v =+=A
12 R2R =
Chapter 11, Solution 1.
)t50cos(160)t(v = )9018030t50cos(2)30t50sin(20-)t(i °−°+°−=°−=
)60t50cos(20)t(i °+=
)60t50cos()t50cos()20)(160()t(i)t(v)t(p °+== [ ] W)60cos()60t100cos(1600)t(p °+°+=
=)t(p W)60t100cos(1600800 °++
)60cos()20)(160(21
)cos(IV21
P ivmm °=θ−θ=
=P W800
Chapter 11, Solution 2. First, transform the circuit to the frequency domain.
°∠→ 030)t500cos(30 , 500=ω 150jLjH3.0 =ω→
100j-)10)(20)(500(
j-Cj
1F20 6- ==
ω→µ
I
I1
I2
+ − 30∠0° V j150 Ω
-j100 Ω
200 Ω
2.0j-902.0150j
0301 =°−∠=
°∠=I
)t500sin(2.0)90t500cos(2.0)t(i1 =°−=
06.0j12.056.261342.0j2
3.0100j200030
2 +=°∠=−
=−
°∠=I
)56.25t500cos(1342.0)t(i2 °+=
°∠=−=+= 49.4-1844.014.0j12.021 III )35t500cos(1844.0)t(i °−=
For the voltage source,
])35t500cos(1844.0[])t500cos(30[)t(i)t(v)t(p °−×== At , s2t = )351000cos()1000cos(532.5p °−=
)935.0)(5624.0)(532.5(p = =p W91.2
For the inductor,
])t500sin(2.0[])t500cos(30[)t(i)t(v)t(p ×== At , s2t = )1000sin()1000cos(6p =
)8269.0)(5624.0)(6(p = =p W79.2
For the capacitor,
°∠== 63.44-42.13)100j-(2c IV )56.25t500cos(1342.0[])44.63500cos(42.13[)t(i)t(v)t(p °+×°−==
At , s2t = )56.261000cos()44.631000cos(18p °+°−=
)1329.0)(991.0)(18(p = =p W37.2
For the resistor,
°∠== 56.2584.26200 2R IV ])56.26t500cos(1342.0[])56.26t500cos(84.26[)t(i)t(v)t(p °+×°+==
At , s2t = )56.251000(cos602.3p 2 °+=
21329.0)(602.3(p = =p W0636.0
Chapter 11, Solution 3. 10 , °∠→°+ 3010)30t2cos( 2=ω
2jLjH1 =ω→
-j2Cj
1F25.0 =
ω→
4 Ω 2 ΩI I1
I2
+ − 10∠30° V j2 Ω -j2 Ω
2j22
)2j2)(2j()2j2(||2j +=
−=−
°∠=++°∠
= 565.11581.12j24
3010I
°∠=== 565.101581.1j22j
1 III
°∠=−
= 565.56236.22
2j22 II
For the source,
)565.11-581.1)(3010(21* °∠°∠== IVS
5.2j5.718.43905.7 +=°∠=S
The average power supplied by the source = W5.7 For the 4-Ω resistor, the average power absorbed is
=== )4()581.1(21
R21
P 22I W5
For the inductor,
5j)2j()236.2(21
21 2
L
2
2 === ZIS
The average power absorbed by the inductor = W0
For the 2-Ω resistor, the average power absorbed is
=== )2()581.1(21
R21
P 22
1I W5.2
For the capacitor,
5.2j-)2j-()581.1(21
21 2
c
2
1 === ZIS
The average power absorbed by the capacitor = W0
Chapter 11, Solution 4.
20 Ω 10 Ω
I2I1+ − -j10 Ω50 V j5 Ω
For mesh 1,
21 10j)10j20(50 II +−=
21 j)j2(5 II +−= (1) For mesh 2,
12 10j)10j5j10(0 II +−+=
12 2j)j2(0 II +−= (2) In matrix form,
−
−=
2
1
j22jjj2
05
II
4j5−=∆ , )j2(51 −=∆ , -j102 =∆
°∠=−−
=∆∆
= 1.12746.14j5
)j2(511I
°∠==∆∆
= 66.128562.1j4-5
j10-22I
For the source,
°∠== 12.1-65.4321 *
1IVS
The average power supplied =°= )1.12cos(65.43 W68.42 For the 20-Ω resistor,
== R21
P2
1I W48.30
For the inductor and capacitor, =P W0
For the 10-Ω resistor,
== R21
P2
2I W2.12
Chapter 11, Solution 5. Converting the circuit into the frequency domain, we get: 1 Ω 2 Ω
+ − j6
–j2 8∠–40˚
W4159.112
6828.1P
38.256828.1
2j26j)2j2(6j1
408I
21
1
==
°−∠=
−+−
+
°−∠=
Ω
Ω
P3H = P0.25F = 0
W097.522
258.2P
258.238.256828.12j26j
6jI
22
2
==
=°−∠−+
=
Ω
Ω
Chapter 11, Solution 6.
20 Ω 10 Ω
I2I1+ − -j10 Ω50 V j5 Ω
For mesh 1,
04)604(2j)2j4( o1 =+°∠−+ VI (1) )604(2 2o IV −°∠= (2)
For mesh 2, 04)604(2)j2( o2 =−°∠−− VI (3)
Substituting (2) into (3), 0)604(8608)j2( 22 =−°∠−°∠−− II
j106040
2 −°∠
=I
Hence,
j10608j-
j106040
6042o −°∠
=
−
°∠−°∠=V
Substituting this into (1),
−−
°∠=−
°∠+°∠=+
j10j14
)608j(j10
6032j608j)2j4( 1I
°∠=+
+°∠= 125.06498.2
8j21)14j1)(604(
1I
=== )4()498.2(21
R21
P 22
14 I W48.12
Chapter 11, Solution 7.
20 Ω 10 Ω
I2I1+ − -j10 Ω50 V j5 Ω
Applying KVL to the left-hand side of the circuit, oo 1.04208 VI +=°∠ (1)
Applying KCL to the right side of the circuit,
05j105j
8 11o =
−++
VVI
But, o11o 105j10
5j1010
VVVV−
=→−
=
Hence, 01050j
5j108 o
oo =+−
+V
VI
oo 025.0j VI = (2)
Substituting (2) into (1), )j1(1.0208 o +=°∠ V
j12080
o +°∠
=V
°∠== 25-2
1010
o1
VI
=
== )10(2
10021
R21
P2
1I W250
Chapter 11, Solution 8. We apply nodal analysis to the following circuit.
At node 1,
Io V2V1
6∠0° A 0.5 Io j10 Ω
-j20 Ω
I2
40 Ω
20j-10j6 211 VVV −
+= 21 120j VV −= (1)
At node 2,
405.0 2
oo
VII =+
But, j20-
21o
VVI
−=
Hence, 40j20-
)(5.1 221 VVV=
−
21 )j3(3 VV −= (2)
Substituting (1) into (2),
0j33360j 222 =+−− VVV
j6)-1(37360
j6360j
2 +=−
=V
j6)-1(379
402
2 +==V
I
=
== )40(379
21
R21
P2
2
2I W78.43
Chapter 11, Solution 9.
rmsV8)2)(4(V26
1V so ==
+=
=== mW1064
RV
P2o
10 mW4.6
The current through the 2 -kΩ resistor is
mA1k2
Vs =
== RIP 2
2 mW2 Similarly,
== RIP 26 mW6
Chapter 11, Solution 10.
No current flows through each of the resistors. Hence, for each resistor, =P W0 .
Chapter 11, Solution 11.
, , 377=ω 410R = -910200C ×=754.0)10200)(10)(377(RC -94 =×=ω
°=ω 02.37)RC(tan -1
Ω°∠=°∠+
= k37.02-375.637.02-)754.0(1
k10Z
2ab
mA)68t377cos(2)22t377sin(2)t(i °−=°+=
°∠= 68-2I
3
2-3
ab2rms 10)37.02-375.6(
2102
ZIS ×°∠
×
==
mVA37.02-751.12S °∠=
== )02.37cos(SP mW181.10
Chapter 11, Solution 12.
(a) We find using the circuit in Fig. (a). ThZ
Zth
8 Ω -j2 Ω
(a)
882.1j471.0)4j1(178
j28(8)(-j2)
-j2||8Th −=−=−
==Z
== *
ThL ZZ Ω+ 882.1j471.0
We find using the circuit in Fig. (b). ThV
Io +
Vth
−
-j2 Ω 4∠0° A 8 Ω
(b)
)04(2j8
2j-o °∠
−=I
j28j64-
I8 oTh −==V
=
==)471.0)(8(
6864
R8P
2
L
2
Thmax
VW99.15
(b) We obtain from the circuit in Fig. (c). ThZ
5 Ω -j3 Ω
Zth
j2 Ω
4 Ω
(c)
167.1j5.23j9
)3j4)(5(2j)3j4(||52jTh +=
−−
+=−+=Z
== *
ThL ZZ Ω− 167.1j5.2
Chapter 11, Solution 13.
(a) We find at the load terminals using the circuit in Fig. (a). ThZ
j100 Ω
-j40 ΩZth
80 Ω
(a)
6.1j2.51j6080
j100)(-j40)(80j100)(80||-j40Th −=
++
=+=Z
== *
ThL ZZ Ω+ 6.1j2.51
(b) We find at the load terminals using Fig. (b). ThV
j100 ΩIo
+
Vth
−
3∠20° A -j40 Ω80 Ω
(b)
6j8)203)(8(
)203(40j100j80
80o +
°∠=°∠
−+=I
6j8)2024)(40j(-
40j- oTh +°∠
== IV
=
⋅==
)2.51)(8(
241040
R8P
2
L
2
Thmax
VW5.22
From Fig.(d), we obtain V using the voltage division principle. Th
5 Ω -j3 Ω
+
Vth
−
+ − 10∠30° V
j2 Ω
4 Ω
(d)
°∠
−−
=°∠
−−
= 303
10j33j4
)3010(3j93j4
ThV
=
⋅==
)5.2)(8(3
10105
R8P
2
L
2
Thmax
VW389.1
Chapter 11, Solution 14.
I
+
VTh
_
16 Ω
–j10 Ω
j8 Ω
j24 Ω
10 Ω
ZTh 40∠90º A
Ω+==
Ω−=++−=+++++
+−=
∗ 3.2j245.8ZZ
3.2j245.87.7j245.810j8j1624j10)8j16)(24j10(10jZ
Th
Th
W6.456245.8)245.8x2(
2
V
245.8IP
V12.158j53.7166.6555.173
)8j16(40j8j1624j10
10)8j16(IV
2
2
2Th
2rmsmax
Th
===
+=°∠=
++++
=+=
Chapter 11, Solution 15. To find Z , insert a 1-A current source at the load terminals as shown in Fig. (a). Th
+
Vo
−
2 1 1 Ω -j Ω
2 Vo j Ω 1 A
(a)At node 1,
2oo2oo j
j-j1VV
VVVV=→
−=+ (1)
At node 2,
o2o2
o )j2(j1j-
21 VVVV
V +−=→−
=+ (2)
Substituting (1) into (2), 222 )j1()j)(j2(j1 VVV −=+−=
j11
2 −=V
5.0j5.02
j11
2Th +=
+==
VV
== *
ThL ZZ Ω− 5.0j5.0
We now obtain from Fig. (b). ThV
1 Ω -j Ω
+
Vth
−
+ − 12∠0° V
+
Vo
−
2 Vo j Ω
(b)
j112
2 ooo
VVV =
−+
j112-
o +=V
0)2j-( Thoo =+×− VVV
j1)2j1)(12(
j2)-(1 oTh ++
=+= VV
=
==)5.0)(8(
2512
R8P
2
L
2
Thmax
VW90
Chapter 11, Solution 16.
520/14
11F20/1,4H1,4 jxjCj
jLj −==→=→=ω
ωω
We find the Thevenin equivalent at the terminals of ZL. To find VTh, we use the circuit shown below. 0.5Vo 2Ω V1 4Ω V2 + + + 10<0o Vo -j5 j4 VTh
- - -
At node 1,
2121
111 25.0)2.01(5
425.0
5210 VjVVVV
jVV
−+=→−
++−
=−
(1)
At node 2,
)25.025.0(5.004
25.04 21
21
21 jVVjVVVV
+−+=→=+− (2)
Solving (1) and (2) leads to
2 6.1947 7.0796 9.4072 48.81oThV V j= = + = ∠ Chapter 11, Solution 17. We find ThR at terminals a-b following Fig. (a).
ba
j20 Ω
30 Ω
40 Ω
-j10 Ω (a)
10j40-j10))(40(
20j30)20j)(30(
)10j-(||4020j||30Th −+
+=+=Z
41.9j353.285.13j23.9Th −++=Z
Ω+= 44.4j583.11ThZ == *
ThL ZZ Ω− 44.4j583.11
We obtain from Fig. (b). ThV
I1 I2
+ VTh −
j20 Ω
30 Ω
40 Ω
-j10 Ω
j5 A
(b)
Using current division,
3.2j1.1-)5j(10j7020j30
1 +=++
=I
7.2j1.1)5j(10j7010j40
2 +=+−
=I
70j1010j30 12Th +=+= IIV
===)583.11)(8(
5000R8
PL
2
Th
max
VW96.53
Chapter 11, Solution 18. We find Z at terminals a-b as shown in the figure below. Th
a
b
Zth
40 Ω 80 Ω-j10 Ω
40 Ω
j20 Ω
j1080(80)(-j10)
20j20-j10)(||8040||4020jTh −++=++=Z
154.10j23.21Th +=Z
== *
ThL ZZ Ω− 15.10j23.21 Chapter 11, Solution 19. At the load terminals,
j9j)(6)(3
-j2)j3(||62j-Th ++
+=++=Z
561.1j049.2Th −=Z
Ω== 576.2R ThL Z
To get V , let Th 439.0j049.2)j3(||6 +=+=Z . By transforming the current sources, we obtain
756.1j196.8)04(Th +=°∠= ZV
===608.20258.70
R8P
L
2
Thmax
VW409.3
Chapter 11, Solution 20. Combine j20 Ω and -j10 Ω to get
-j20-j10||20j = To find , insert a 1-A current source at the terminals of , as shown in Fig. (a). ThZ LR
Io
-j20 Ω
4 Io
+ −V2V1
-j10 Ω
40 Ω
1 A
(a)At the supernode,
10j-20j-401 211 VVV
++=
21 4j)2j1(40 VV ++= (1)
Also, , where o21 4IVV +=40
- 1o
VI =
1.11.1 2
121
VVVV =→= (2)
Substituting (2) into (1),
22 4j1.1
)2j1(40 VV
+
+=
4.6j144
2 +=V
Ω−== 71.6j05.11
2Th
VZ
== ThLR Z Ω792.6
To find , consider the circuit in Fig. (b). ThV
+ − 120∠0° V
+
Vth
−
Io
-j20 Ω
4 Io
+ −V2V1
-j10 Ω
40 Ω
(b)At the supernode,
j10-j20-40120 211 VVV
+=−
21 4j)2j1(120 VV ++= (3)
Also, , where o21 4IVV +=40
120 1o
VI
−=
1.1122
1
+=
VV (4)
Substituting (4) into (3),
2)818.5j9091.0(82.21j09.109 V+=−
°∠=+−
== 92.43-893.18818.5j9091.082.21j09.109
2Th VV
===)792.6)(8(
)893.18(R8
P2
L
2
Thmax
VW569.6
Chapter 11, Solution 21. We find Z at terminals a-b, as shown in the figure below. Th
100 Ω -j10 Ω
40 Ω
b
a
Zth
j30 Ω50 Ω
])30j40(||10010j-[||50Th ++=Z
where 634.14j707.3130j140
)30j40)(100()30j40(||100 +=
++
=+
634.4j707.81)634.4j707.31)(50(
)634.4j707.31(||50Th ++
=+=Z
73.1j5.19Th +=Z
== ThLR Z Ω58.19
Chapter 11, Solution 22.
π<<= ttti 0,sin4)(
8)02
(1642sin
216sin161
00
22 =−=
−== ∫
ππππ
ππ tttdtI rms
A 828.28 ==rmsI
Chapter 11, Solution 23.
<<<<
=6t2,52t0,15
)t(v
[ ]6
550dt5dt15
61
V6
222
022
rms =+= ∫∫
=rmsV V574.9
Chapter 11, Solution 24.
, 2T =
<<<<
=2t15,-1t0,5
)t(v
[ ] 25]11[225
dt-5)(dt521
V2
121
022
rms =+=+= ∫∫
=rmsV V5
Chapter 11, Solution 25.
[ ]
266.33
32f
332]16016[
31
dt4dt0dt)4(31dt)t(f
T1f
rms
32
221
10
2T0
22rms
==
=++=
++−== ∫∫∫∫
Chapter 11, Solution 26.
, 4T =
<<<<
=4t2102t05
)t(v
[ ] 5.62]20050[41
dt)10(dt541
V4
222
022
rms =+=+= ∫∫
=rmsV V906.7
Chapter 11, Solution 27. , 5T = 5t0,t)t(i <<=
333.815
1253t
51
dtt51
I 50
35
022
rms ==⋅== ∫
=rmsI A887.2
Chapter 11, Solution 28.
[ ]∫∫ +=5
222
022
rms dt0dt)t4(51
V
533.8)8(1516
3t16
51
V 20
32rms ==⋅=
=rmsV V92.2
===2533.8
RV
P2rms W267.4
Chapter 11, Solution 29.
, 20T =
<<+<<−
=25t152t40-
15t5t220)t(i
[ ]∫∫ ++−=25
15215
522
eff dt2t)-40(dt)t220(201
I
+−++−= ∫∫
25
15
215
5
22eff dt400)t40t(dt)tt20100(
51I
+−+
+−= 2515
23
155
322
eff t400t203t
3t
t10t10051
I
332.33]33.8333.83[51
I2eff =+=
=effI A773.5
== RIP 2
eff W400 Chapter 11, Solution 30.
<<<<
=4t21-2t0t
)t(v
[ ] 1667.1238
41
dt-1)(dtt41
V4
222
022
rms =
+=+= ∫∫
=rmsV V08.1
Chapter 11, Solution 31.
6667.81634
21)4()2(
21)(
21 2
0
1
0
2
1
222 =
+=
−+== ∫ ∫ ∫ dtdttdttvV rms
V 944.2=rmsV
Chapter 11, Solution 32.
+= ∫∫
2
1
1
0
222rms dt0dt)t10(
21I
105t
50dtt50I 10
51
042
rms =⋅== ∫
=rmsI A162.3
Chapter 11, Solution 33.
<<<<−<<
=3t202t1t10201t010
)t(i
[ ]0dt)t1020(dt1031
I2
121
022
rms +−+= ∫∫
33.133)31)(100(100dt)tt44(100100I32
122
rms =+=+−+= ∫
==3
33.133Irms A667.6
Chapter 11, Solution 34.
[ ]
472.420f
20363t9
31
dt6dt)t3(31dt)t(f
T1f
rms
2
0
3
32
220
2T0
22rms
==
=
+=
+== ∫∫∫
Chapter 11, Solution 35.
[ ]∫∫∫∫∫ ++++=6
525
424
222
121
022
rms dt10dt20dt30dt20dt1061
V
67.466]1004001800400100[61
V2rms =++++=
=rmsV V6.21
Chapter 11, Solution 36.
(a) Irms = 10 A
(b) V 528.42916
234
222 =+=→
+= rmsrms VV (checked)
(c) A 055.92
3664 =+=rmsI
(d) V 528.42
16225
=+=rmsV
Chapter 11, Solution 37.
)302cos(6)10sin(48321oo ttiiii ++++=++=
A 487.9902
362
166432
22
12 ==++=++= rmsrmsrmsrms IIII
Chapter 11, Solution 38.
08.157j)5.0)(50)(2(jLjH5.0 =π=ω→
08.157j30jXR L +=+=Z
08.157j30)210( 2
*
2
−==
ZV
S
Apparent power ===160
)210( 2
S VA6.275
)19.79cos(36
08.157tancoscospf 1- °=
=θ=
=pf (lagging)1876.0
Chapter 11, Solution 39.
4j12)8j12)(4j(
)8j12(||4jT −−
=−=Z
°∠=+= 74.7456.4)11j3(4.0TZ
=°= )74.74cos(pf 2631.0
Chapter 11, Solution 40. At node 1,
)1(V4.0)6667.0j4.1(V60j92.103
50VV
30jV
20V30120
21
2111o
−−=+
→−
+=−∠
At node 2,
212221 )25.16(0401050
VjVjVVVV
++−=→−
+=−
(2)
Solving (1) and (2) leads to
V1 = 45.045 + j66.935, V2 = 9.423 + j9.193
(a) Ω−Ω == 4030 0 jj PP
W665.820/3.173||21 2
22
10 ====Ω RV
RVP rms
W6.034100/1.4603||21 2
2150 ==
−=Ω R
VVP
W86.8740/3514|30120|21 2
120 ==
−∠=Ω R
VPo
(b) 6092.10330120,3467.0944.22030120 1 jVjVI o
s
o
+=∠=−=−∠
=
VA 8.177||,3.1065.14221
==−== • SSjIVS s
(c ) pf = 142.5/177.8 = 0.8015 (leading). Chapter 11, Solution 41.
(a) -j6j
(-j2)(-j3)-j3||-j2)2j5j(||2j- ===−
°∠=−= 56.31-211.76j4TZ
=°= )-56.31cos(pf (leading)5547.0
(b) 52.1j64.03j4
)j4)(2j()j4(||2j +=
++
=+
°∠=++
=−+= 5.214793.044.0j64.144.0j64.0
)j52.1j64.0(||1Z
=°= )5.21cos(pf (lagging)9304.0
Chapter 11, Solution 42.
°=θ→θ== 683.30cos86.0pf
kVA798.9)683.30sin(
5sin
QSsinSQ =
°=
θ=→θ=
°∠=°∠×
==→= 683.30536.44220
683.3010798.9 3**
VS
IIVS
Peak current =× 536.442= A98.62 Apparent power = S = kVA798.9
Chapter 11, Solution 43.
(a) V 477.53021
29253
22
21
2 ==++=++= rmsrmsrmsrms VVVV
(b) W310/302
===R
VP rms
Chapter 11, Solution 44.
opf 46.49cos65.0 =→== θθ
kVA 385.32)7599.065.0(50)sin(cos jjjSS +=+=+= θθ
Thus, Average power = 32.5 kW, Reactive power = 38 kVAR
Chapter 11, Solution 45.
(a) V 9.4622002
60202
22 =→=+= rmsrms VV
AI rms 061.1125.125.01
22 ==+=
(b) W74.49== rmsrms IVP
Chapter 11, Solution 46.
(a) °∠=°∠°∠== 30-110)60-5.0)(30220(*IVS=S VA55j26.95 −
Apparent power =110 VA Real power = W26.95 Reactive power = VAR55 pf is leading because current leads voltage
(b) == *IVS °∠=°∠°∠ 151550)252.6)(10-250( =S VA2.401j2.1497 +
Apparent power =1550 VA Real power =1497 W2. Reactive power = VAR2.401 pf is lagging because current lags voltage
(c) °∠=°∠°∠== 15288)154.2)(0120(*IVS=S VA54.74j2.278 +
Apparent power = VA288 Real power = W2.278 Reactive power = VAR54.74 pf is lagging because current lags voltage
(d) °∠=°∠°∠== 135-1360)180-5.8)(45160(*IVS=S VA7.961j7.961- −
Apparent power =1360 VA Real power = W7.961- Reactive power = VAR7.961- pf is leading because current leads voltage
Chapter 11, Solution 47.
(a) , °∠= 10112V °∠= 50-4I
=°∠== 6022421 *IVS VA194j112+
Average power =112 W Reactive power =194 VAR
(b) , °∠= 0160V °∠= 4525I
=°∠== 45-20021 *IVS VA42.141j42.141 −
Average power =141 W42. Reactive power = VAR42.141-
(c) =°∠=°∠
== 3012830-50
)80( 2
*
2
ZV
S V64j51.90 A+
Average power = W51.90 Reactive power = VAR64
(d) =°∠== )45100)(100(2ZIS kVA071.7j071.7 +
Average power = kW071.7 Reactive power = kVAR071.7
Chapter 11, Solution 48.
(a) =−= jQPS V150j269 A−
(b) °=θ→=θ= 84.259.0cospf
31.4588)84.25sin(
2000sin
QSsinSQ =
°=
θ=→θ=
48.4129cosSP =θ=
=S VA2000j4129−
(c) 75.0600450
SQ
sinsinSQ ===θ→θ=
59.48=θ , 6614.0pf =
86.396)6614.0)(600(cosSP ==θ= =S VA450j9.396 +
(d) 121040
)220(S
22
===Z
V
8264.012101000
SP
coscosSP ===θ→θ=
°=θ 26.34
25.681sinSQ =θ=
=S VA2.681j1000+
Chapter 11, Solution 49.
(a) kVA))86.0(sin(cos86.04
j4 1-+=S
=S kVA373.2j4 +
(b) 6.0sincos8.026.1
SP
pf =θ→θ===
=θ−= sin2j6.1S kVA2.1j6.1 −
(c) VA)505.6)(20208(*
rmsrms °∠°∠== IVS=°∠= 70352.1S kVA2705.1j4624.0 +
(d) °∠
=−
==56.31-11.72
1440060j40)120( 2
*
2
ZV
S
=°∠= 56.317.199S VA16.166j77.110 + Chapter 11, Solution 50.
(a) ))8.0(sin(cos8.0
1000j1000jQP 1-−=−=S
750j1000−=S
But, *
2
rms
ZV
S =
23.23j98.30750j1000
)220( 22
rms* +=−
==S
VZ
=Z Ω− 23.23j98.30
(b) ZIS2
rms=
=+
== 22
rms)12(2000j1500
I
SZ Ω+ 89.13j42.10
(c) °∠=°∠
=== 60-6.1)604500)(2(
)120(2
222
rms*
SV
SV
Z
=°∠= 606.1Z Ω+ 386.1j8.0
Chapter 11, Solution 51.
(a) )6j8(||)5j10(2T +−+=Z
j1820j110
2j18
)6j8)(5j10(2T +
++=
++−
+=Z
°∠=+= 5.382188.8768.0j152.8TZ
=°= )5.382cos(pf (lagging)9956.0
(b) )5.382-188.8)(2(
)16(22
1 2
*
2
*
°∠===
ZV
IVS
°∠= 5.38263.15S
=θ= cosSP W56.15
(c) =θ= sinSQ VAR466.1
(d) == SS VA63.15
(e) =°∠= 382.563.15S VA466.1j56.15 +
Chapter 11, Solution 52.
749j4200SSSS500j1000S
2749j12009165.0x3000j4.0x3000S
1500j20006.08.0
2000j2000S
CBA
C
B
A
−=++=+=
−=−=
+=+=
(a) .leading9845.07494200
4200pf22=
+=
(b) °−∠=°∠
−=→= ∗∗ 11.5555.35
45120749j4200IIVS rmsrmsrms
Irms = 35.55∠–55.11˚ A.
Chapter 11, Solution 53. S = SA + SB + SC = 4000(0.8–j0.6) + 2400(0.6+j0.8) + 1000 + j500 = 5640 + j20 = 5640∠0.2˚
(a)
A8.2997.9388.2946.66x2I
8.2946.66
230120
2.05640V
SV
SSVSI
rmsrms
CA
rms
Brms
°∠=°∠=
°−∠=°∠°∠
==+
+=∗
(b) pf = cos(0.2˚) ≈ 1.0 lagging.
Chapter 11, Solution 54.
(a) ))8.0(sin(cos8.0
1000j1000jQP 1-−=−=S
750j1000−=S
But, *
2
rms
ZV
S =
23.23j98.30750j1000
)220( 22
rms* +=−
==S
VZ
=Z Ω− 23.23j98.30
(b) ZIS2
rms=
=+
== 22
rms)12(2000j1500
I
SZ Ω+ 89.13j42.10
(c) °∠=°∠
=== 60-6.1)604500)(2(
)120(2
222
rms*
SV
SV
Z
=°∠= 606.1Z Ω+ 386.1j8.0
Chapter 11, Solution 55. We apply mesh analysis to the following circuit.
40∠0° V rms I2+ −
j10 Ω
20 Ω
I3
I1+ −
-j20 Ω
50∠90° V rms
For mesh 1, 21 I20I)20j20(40 −−=
21 II)j1(2 −−= (1) For mesh 2, 12 I20I)10j20(50j- −+=
21 I)j2(I-2j5- ++= (2) Putting (1) and (2) in matrix form,
+
−=
2
1
II
j22-1-j1
j5-2
j1−=∆ , 3j41 −=∆ , 5j-12 −=∆
°∠=−=−−
=∆∆
= 13.8535.3)j7(21
j13j4
I 11
°∠=−=−−
=∆∆
= 56.31-605.33j2j15j1-
I 22
°∠=+=−−+=−= 8.66808.35.3j5.1)3j2()5.0j5.3(III 213
For the 40-V source,
=
−⋅== )j7(21
)40(-- *1IVS Vj20140- A+
For the capacitor, == c
2
1 ZIS Vj250- A For the resistor,
== R2
3IS V290 A For the inductor,
== L
2
2 ZIS VA130j For the j50-V source,
=+== )3j2)(50j(*2IVS Vj100150- A+
Chapter 11, Solution 56.
8.1j6.02j6
)2j-)(6(6||2j- −=
−=
j2.23.66||-j2)(4j3 +=++
The circuit is reduced to that shown below.
+
Vo
−
Io
2∠30° A 5 Ω
3.6 + j2.2 Ω
°∠=°∠++
= 47.0895.0)302(2.2j6.82.2j6.3
oI
°∠== 47.0875.45 oo IV
)30-)(247.0875.4(21
21 *
so °∠°∠⋅== IVS
=°∠= 17.0875.4S VA396.1j543.4 +
Chapter 11, Solution 57. Consider the circuit as shown below.
At node o,
+
V2
−
V1Vo
j2 Ω+ − 24∠0° V
4 Ω -j1 Ω 2 Ω
1 Ω 2 Vo
j-1424 1ooo VVVV −
+=−
1o 4j)4j5(24 VV −+= (1)
At node 1, 2j
2j-
1o
1o VV
VV=+
−
o1 )4j2( VV −= (2)
Substituting (2) into (1),
o)168j4j5(24 V−−+=
j41124-
o +=V ,
j411j4)-(-24)(2
1 +=V
The voltage across the dependent source is
o1o12 4)2)(2( VVVVV +=+=
4j11)4j6)(24-(
)44j2(j411
24-2 +
−=+−⋅
+=V
)2(21
21 *
o2*
2 VVIVS ==
)4j6(137576
j4-1124-
4j11)4j6)(24-(
−
=⋅+−
=S
=S VA82.16j23.25 −
Chapter 11, Solution 58.
4 kΩ
Ix
8 mA
-j3 kΩ j1 kΩ
10 kΩ
From the left portion of the circuit,
mA4.0500
2.0o ==I
mA820 o =I
From the right portion of the circuit,
mAj7
16)mA8(
3jj1044
x −=
−++=I
)1010(50
)1016(R 3
2-32
x ×⋅×
== IS
=S mVA2.51
Chapter 11, Solution 59. Consider the circuit below.
4 kΩ
Ix
8 mA
-j3 kΩ j1 kΩ
10 kΩ
30j40j20-50240
4 ooo
++=
−+
VVV
o)38.0j36.0(88 V+=
°∠=+
= 46.55-13.16838.0j36.0
88oV
°∠== 43.4541.8j20-
o1
VI
°∠=+
= 83.42-363.330j40
o2
VI
Reactive power in the inductor is
=⋅== )30j()363.3(21
21 2
L
2
2 ZIS VAR65.169j
Reactive power in the capacitor is
=⋅== )20j-()41.8(21
21 2
c2
1 ZIS VARj707.3-
Chapter 11, Solution 60.
15j20))8.0(sin(cos8.0
20j20S 1-1 +=+=
749.7j16))9.0(sin(cos9.0
16j16S 1-
2 +=+=
°∠=+=+= 29.32585.42749.22j36SSS 21
But o
*o V6IVS ==
==6S
Vo °∠ 29.32098.7
=°= )29.32cos(pf (lagging)8454.0
Chapter 11, Solution 61. Consider the network shown below.
I2
So
I1+
Vo
−
Io
S2
S3S1
kVA8.0j2.12 −=S
kVA937.1j4))9.0(sin(cos9.0
4j4 1-
3 +=+=S
Let kVA137.1j2.5324 +=+= SSS
But *2o4 2
1IVS =
104j74.2290100
10)137.1j2.5)(2(2 3
o
4*2 −=
°∠×+
==VS
I
104j74.222 +=I
Similarly, kVA2j2))707.0(sin(cos707.02
j2 1-1 −=−=S
But *1o1 2
1IVS =
40j40-100j
10)4j4(V2 3
o
1*1 −=
×−==
SI
j40-401 +=I
°∠=+=+= 83.96145j144-17.2621o III
*ooo IV
21
=S
VA)96.83-145)(90100(21
o °∠°∠⋅=S
=oS kVA862.0j2.7 −
Chapter 11, Solution 62. Consider the circuit below
0.2 + j0.04 Ω
+
V1
−
+
V2
−
+ −
I2
I1
I 0.3 + j0.15 Ω
Vs
25.11j15))8.0(sin(cos8.0
15j15 1-
2 −=−=S
But *
222 IVS =
12025.11j15
2
2*2
−==
VS
I
09375.0j125.02 +=I
)15.0j3.0(221 ++= IVV )15.0j3.0)(09375.0j125.0(1201 +++=V
0469.0j02.1201 +=V
843.4j10))9.0(sin(cos9.0
10j10 1-
1 +=+=S
But *
111 IVS =
°∠°∠
==02.002.12084.25111.11
1
1*1 V
SI
0405.0j0837.025.82-093.01 −=°∠=I
053.0j2087.021 +=+= III )04.0j2.0(1s ++= IVV
)04.0j2.0)(053.0j2087.0()0469.0j02.120(s ++++=V 0658.0j06.120s +=V
=sV V03.006.120 °∠ Chapter 11, Solution 63. Let S = . 321 SSS ++
929.6j12))866.0(sin(cos866.012
j12 1-1 −=−=S
916.9j16))85.0(sin(cos85.0
16j16 1-
2 +=+=S
20j1520j)6.0(sin(cos
)6.0)(20(1-3 +=+=S
*o2
1987.22j43 IVS =+=
11098.22j442*
o
+==
VS
I
=oI A27.58-4513.0 °∠
Chapter 11, Solution 64.
I2 I1
+ −
j12
Is
8 Ω 120∠0º V
Is + I2 = I1 or Is = I1 – I2
But,
333.3j83.20Ior
333.3j83.20120
400j2500
VS
IVI
923.6j615.412j8
120I
2
22
1
+=
−=−
==→
−=+
=
∗∗S =
Is = I1 – I2 = –16.22 – j10.256 = 19.19∠–147.69˚ A.
Chapter 11, Solution 65.
Ω=×
=ω
→= k-j1001010j-
Cj1
nF1C 9-4
At the noninverting terminal,
j14
j100-10004
ooo
+=→=
−°∠V
VV
°∠= 45-2
4oV
)45t10cos(2
4)t(v 4
o °−=
W1050
12
12
4R
VP 3
22rms
×
⋅==
=P W80 µ
Chapter 11, Solution 66. As an inverter,
)454(j34j4)(2--
si
fo °∠⋅
++
== VZZ
V
mAj3)j2)(4-(6
)45j4)(4(2-mA
2j6o
o +°∠+
=−
=V
I
The power absorbed by the 6-kΩ resistor is
36-
22
o 10610540420
21
R21
P ×××
××
⋅== I
=P mW96.0
Chapter 11, Solution 67.
51.02
11F1.0,6H3,2 jxjCj
jLj −==→=→=ω
ωω
42510
50)5//(10 jj
jj −=−
−=−
The frequency-domain version of the circuit is shown below. Z2=2-j4 Ω
Z1 =8+j6Ω I1 - + Io + + V Vo206.0 ∠ o - -
Ω= 123Z
(a) oo
jj
jI 87.1606.0
682052.05638.0
680206.0
1 −∠=++
=+
−∠=
mV 86.3618mVA 8.104.14)87.1606.0)(203.0(21
1* ooo
s jIVS ∠=+=+∠∠==
(b) ooooso j
jZV
IVZZ 7.990224.0)206.0(
)68(12)42(,
31
2 ∠=∠+−
−==−=V
mW 904.2)12()0224.0(5.0||21 22 === RIP o
A
Chapter 11, Solution 68.
Let S cLR SSS ++=
where 0jRI21
jQP 2oRRR +=+=S
LI21
j0jQP 2oLLL ω+=+=S
C1
I21
j0jQP 2occc ω⋅−=+=S
Hence, =S
ω−ω+
C1
LjRI21 2
o
Chapter 11, Solution 69.
(a) Given that 12j10 +=Z
°=θ→=θ 19.501012
tan
=θ= cospf 6402.0
(b) 09.354j12.295)12j10)(2(
)120(2
2
*
2
+=−
==Z
VS
The average power absorbed === )Re(P S W1.295
(c) For unity power factor, °=θ 01
09.354, which implies that the reactive power due
to the capacitor is Qc =
But 2VC21
X2V
Qc
2
c ω==
=π
=ω
= 22c
)120)(60)(2()09.354)(2(
VQ2
C F4.130 µ
Chapter 11, Solution 70.
6.0sin8.0cospf =θ→=θ= 528)6.0)(880(sinSQ ==θ=
If the power factor is to be unity, the reactive power due to the capacitor is
VAR528QQc ==
But 2c2
c
2rms
VQ2
CVC21
XV
Qω
=→ω==
=π
= 2)220)(50)(2()528)(2(
C F45.69 µ
Chapter 11, Solution 71.
67.666.0
508.08.0,6.0
,50,065.1067071.0150 22
2211 ======== SPQSQxQP
5067.66,065.106065.106 21 jSjS −=+=
9512.098.17cos,98.176.18106.56735.17221 ==∠=+=+= oo pfjSSS
058.56)098.17(tan735.172)tan(tan 21 =−=−= o
c PQ θθ
F 33.10120602
058.5622 µ
πω===
xxVQC
rms
c
Chapter 11, Solution 72.
(a) °==θ 54.40)76.0(cos-11
°==θ 84.25)9.0(cos-12
)tan(tanPQ 21c θ−θ=
kVAR])84.25tan()54.40tan()[40(Qc °−°= kVAR84.14Qc =
=π
=ω
= 22rms
c
)120)(60)(2(14840
VQ
C mF734.2
(b) , °=θ 54.401 °=θ 02
kVAR21.34kVAR]0)54.40tan()[40(Qc =−°=
=πω
= 22rms
c
)120)(60)(2(34210
VQ
C mF3.6
Chapter 11, Solution 73.
(a) kVA7j1022j15j10 +=+−=S
=+== 22 710S S kVA21.12
(b) 240
000,7j000,10** +==→=
VS
IIVS
=−= 167.29j667.41I A35-86.50 °∠
(c) °=
=θ 35107
tan 1-1 , °==θ 26.16)96.0(cos-1
2
])tan(16.26-)35tan([10]tantan[PQ 211c °°=θ−θ=
=cQ kVAR083.4
=π
=ω
= 22rms
c
)240)(60)(2(4083
VQ
C F03.188 µ
(d) , 222 jQP +=S kW10PP 12 ==
kVAR917.2083.47QQQ c12 =−=−=
kVA917.2j102 +=S
But *22 IVS =
2402917j000,102*
2
+==
VS
I
=−= 154.12j667.412I A16.26-4.43 °∠
Chapter 11, Solution 74.
(a) °==θ 87.36)8.0(cos-11
kVA308.0
24cos
PS
1
11 ==
θ=
kVAR18)6.0)(30(sinSQ 111 ==θ=
kVA18j241 +=S
°==θ 19.18)95.0(cos-12
kVA105.4295.0
40cos
PS
2
22 ==
θ=
kVAR144.13sinSQ 222 =θ=
kVA144.13j402 +=S
kVA144.31j6421 +=+= SSS
°=
=θ 95.2564144.31
tan 1-
=θ= cospf 8992.0
(b) , °=θ 95.252 °=θ 01
kVAR144.31]0)95.25tan([64]tantan[PQ 12c =−°=θ−θ=
=π
=ω
= 22rms
c
)120)(60)(2(144,31
VQ
C mF74.5
Chapter 11, Solution 75.
(a) VA59.323j75.5175j8
576050j80)240( 2
*1
2
1 −=+
=+
==ZV
S
VA91.208j13.3587j12
576070j120
)240( 2
2 +=−
=−
=S
VA96060
)240( 2
3 ==S
=++= 321 SSSS VA68.114j88.1835 −
(b) °=
=θ 574.388.183568.114
tan 1-
=θ= cospf 998.0
(c) ]0)574.3tan([88.35.18]tantan[PQ 12c −°=θ−θ=
VAR68.114Qc =
=π
=ω
= 22rms
c
)240)(50)(2(68.114
VQ
C F336.6 µ
Chapter 11, Solution 76.
The wattmeter reads the real power supplied by the current source. Consider the circuit below.
Vo4 Ω
8 Ω
-j3 Ω
j2 Ω12∠0° V + − 3∠30° A
8j23j412
303 ooo VVV+=
−−
+°∠
°∠=+=−+
= 19.86347.11322.11j7547.004.3j28.2
52.23j14.36oV
)30-3)(19.86347.11(21
21 *
oo °∠°∠⋅== IVS
°∠= 19.56021.17S
== )Re(P S W471.9
Chapter 11, Solution 77.
The wattmeter measures the power absorbed by the parallel combination of 0.1 F and 150 Ω.
°∠→ 0120)t2cos(120 , 2=ω 8jLjH4 =ω→
-j5Cj
1F1.0 =
ω→
Consider the following circuit.
j8 Ω6 Ω I
+ − 120∠0° V Z
5.4j5.15j15
(15)(-j5)-j5)(||15 −=
−==Z
°∠=−++
= 25.02-5.14)5.4j5.1()8j6(
120I
)5.4j5.1()5.14(21
21
21 22* −⋅=== ZIIVS
VA06.473j69.157 −=S
The wattmeter reads
== )Re(P S W69.157
Chapter 11, Solution 78.
4The wattmeter reads the power absorbed by the element to its right side.
°∠→ 02)t4cos(2 , =ω
4jLjH1 =ω→
-j3Cj
1F
121
=ω
→
Consider the following circuit.
10 Ω I
+ − 20∠0° V Z
3j4)3j-)(4(
4j53j-||44j5−
++=++=Z
08.2j44.6 +=Z
°∠=+
= 7.21-207.108.2j44.16
20I
)08.2j44.6()207.1(21
21 22
+⋅== ZIS
== )Re(P S W691.4
Chapter 11, Solution 79. The wattmeter reads the power supplied by the source and partly absorbed by the 40-Ω resistor.
20j10x500x100j
1Cj
1F500,j10x10x100jmH 10
,100
63 −==
ω→µ=→
=ω
−−
The frequency-domain circuit is shown below.
20 Io
I 40 j V1 V2 +1 2 Io 10<0o -j20 - At node 1,
21
21212121o
1
V)40j6(V)40j7(10jVV
20)VV(3
20VV
jVV
I240
V10
+−+−=
→−
+−
=−
+−
+=−
(1)
At node 2,
2122121 )19()20(02020
VjVjjVVV
jVV
+−+=→−
=−
+− (2)
Solving (1) and (2) yields V1 = 1.5568 –j4.1405
0703.22216.421,4141.08443.0
4010 1 jVISjVI −==+=
−= •
P = Re(S) = 4.222 W.
Chapter 11, Solution 80.
(a) ===4.6
110ZV
I A19.17
(b) 625.18904.6)110( 22
===Z
VS
°=θ→==θ 41.34825.0pfcos
≅=θ= 76.1559cosSP kW6.1
Chapter 11, Solution 81.
kWh consumed kWh77132464017 =−= The electricity bill is calculated as follows :
(a) Fixed charge = $12 (b) First 100 kWh at $0.16 per kWh = $16 (c) Next 200 kWh at $0.10 per kWh = $20 (d) The remaining energy (771 – 300) = 471 kWh
at $0.06 per kWh = $28.26. Adding (a) to (d) gives $ 26.76
Chapter 11, Solution 82.
(a) 0,000,5 11 == QP 171,17)82.0sin(cos000,30,600,2482.0000,30 1
22 ==== −QxP 171,17j600,29)QQ(j)PP(SSS 212121 +=+++=+=
AkV 22.34|S|S ==
(b) Q = 17.171 kVAR
(c ) 865.0220,34600,29
===SPpf
[ ] VAR 2833)9.0tan(cos)865.0tan(cos600,29
)tan(tanPQ11
21c
=−=
θ−θ=−−
(d) F 46.130240602
283322 µ
πω===
xxVQC
rms
c
Chapter 11, Solution 83.
(a) oooVIS 35840)258)(60210(21
21
∠=−∠∠== ∗
W1.68835cos840cos === oSP θ
(b) S = 840 VA (c) VAR 8.48135sin840sin === oS θQ (d) (lagging) 8191.035cos/ === oSPpf
Chapter 11, Solution 84.
(a) Maximum demand charge 000,72$30400,2 =×= Energy cost= 000,48$10200,104.0$ 3 =××Total charge = $ 000,120
(b) To obtain $120,000 from 1,200 MWh will require a flat rate of
=×
kWhper10200,1
000,120$3 kWhper10.0$
Chapter 11, Solution 85.
(a) 15 655.51015602 mH 3 jxxxj =→ −π
We apply mesh analysis as shown below. I1 + Ix 120<0o V 10Ω - In 30Ω Iz + 10Ω 120<0o V Iy - j5.655 Ω I2
For mesh x, 120 = 10 Ix - 10 Iz (1) For mesh y, 120 = (10+j5.655) Iy - (10+j5.655) Iz (2) For mesh z, 0 = -10 Ix –(10+j5.655) Iy + (50+j5.655) Iz (3) Solving (1) to (3) gives Ix =20, Iy =17.09-j5.142, Iz =8 Thus, I1 =Ix =20 A I2 =-Iy =-17.09+j5.142 = A 26.16385. o∠17 In =Iy - Ix =-2.091 –j5.142 = A 5.119907.5 o−∠
(b) 5.3085.1025)120(21,12002060)120(
21
21 jISxIS yx −===== ••
VA 5.3085.222521 jSS −=+=S
(c ) pf = P/S = 2225.5/2246.8 = 0.9905 Chapter 11, Solution 86. For maximum power transfer
*LThi
*ThL ZZZZZ ==→=
)104)(1012.4)(2(j75LjR -66L ××π+=ω+=Z
Ω+= 55.103j75LZ
=iZ Ω− 55.103j75 Chapter 11, Solution 87. jXR ±=Z
Ω=×
==→= k6.11050
80RR 3-
RR I
VIV
22222222
)6.1()3(RXXR −=−=→+= ZZ Ω= k5377.2X
°=
=
=θ 77.576.1
5377.2tan
RX
tan 1-1-
=θ= cospf 5333.0
Chapter 11, Solution 88.
(a) °∠=°∠= 55220)552)(110(S
=°=θ= )55cos(220cosSP W2.126
(b) == SS VA220 Chapter 11, Solution 89.
(a) Apparent power == S kVA12
kW36.9)78.0)(12(cosSP ==θ= kVAR51.7))78.0(sin(cos12sinSQ -1 ==θ=
=+= jQPS kVA51.7j36.9 +
(b) 3
22
**
2
10)51.7j36.9()210(
×+==→=
SV
ZZV
S
=Z Ω+ 6.27j398.34
Chapter 11, Solution 90 Original load :
kW2000P1 = , °=θ→=θ 79.3185.0cos 11
kVA94.2352cos
PS
1
11 =
θ=
kVAR5.1239sinSQ 111 =θ=
Additional load :
kW300P2 = , °=θ→=θ 87.368.0cos 22
kVA375cos
PS
2
22 =
θ=
kVAR225sinSQ 222 =θ=
Total load :
jQP)QQ(j)PP( 212121 +=+++=+= SSS kW23003002000P =+=
kVAR5.14642255.1239Q =+= The minimum operating pf for a 2300 kW load and not exceeding the kVA rating of the generator is
9775.094.2352
2300SP
cos1
===θ
or °=θ 177.12
The maximum load kVAR for this condition is
)177.12sin(94.2352sinSQ 1m °=θ= kVAR313.496Qm =
The capacitor must supply the difference between the total load kVAR ( i.e. Q ) and the permissible generator kVAR ( i.e. ). Thus, mQ
=−= mc QQQ kVAR2.968 Chapter 11, Solution 91
θ= cosSP
===θ=)15)(220(
2700SP
cospf 8182.0
3.1897)09.35sin()15(220sinSQ =°=θ=
When the power is raised to unity pf, °=θ 01 and Q 3.1897Qc ==
=π
=ω
= 22rms
c
)220)(60)(2(3.1897
VQ
C F104 µ
Chapter 11, Solution 92
(a) Apparent power drawn by the motor is
kVA8075.0
60cos
PSm ==
θ=
kVAR915.52)60()80(PSQ 2222
m =−=−= Total real power
kW8020060PPPP Lcm =++=++= Total reactive power
=+−=++= 020915.52QQQQ Lcm kVAR91.32 Total apparent power
=+= 22 QPS kVA51.86
(b) ===51.86
80SP
pf 9248.0
(c) ===550
86510VS
I A3.157
Chapter 11, Solution 93
(a) kW7285.3)7457.0)(5(P1 ==
kVA661.48.0
7285.3pfP
S 11 ===
kVAR796.2))8.0(sin(cosSQ -1
11 == kVA796.2j7285.31 +=S
kW2.1P2 = , VAR0Q2 =
kVA0j2.12 +=S
kW2.1)120)(10(P3 == , VAR0Q3 = kVA0j2.13 +=S
kVAR6.1Q4 = , 8.0sin6.0cos 44 =θ→=θ
kVA2sin
QS
4
44 =
θ=
kW2.1)6.0)(2(cosSP 444 ==θ=
kVA6.1j2.14 −=S
4321 SSSSS +++= kVA196.1j3285.7 +=S
Total real power = 7 kW3285. Total reactive power = 1 kVAR196.
(b) °=
=θ 27.93285.7196.1
tan 1-
=θ= cospf 987.0
Chapter 11, Solution 94
°=θ→=θ 57.457.0cos 11 kVA1000MVA1S1 == kW700cosSP 111 =θ=
kVAR14.714sinSQ 111 =θ= For improved pf,
°=θ→=θ 19.1895.0cos 22 kW700PP 12 ==
kVA84.73695.0
700cos
PS
2
22 ==
θ=
kVAR08.230sinSQ 222 =θ=
P1 = P2 = 700 kW
Q2
Q1
S1
S2
θ1 θ2
Qc
(a) Reactive power across the capacitor
kVAR06.48408.23014.714QQQ 21c =−=−= Cost of installing capacitors =×= 06.48430$ 80.521,14$
(b) Substation capacity released 21 SS −= kVA16.26384.7361000 =−=
Saving in cost of substation and distribution facilities
=×= 16.263120$ 20.579,31$
(c) Yes, because (a) is greater than (b). Additional system capacity obtained by using capacitors costs only 46% as much as new substation and distribution facilities.
Chapter 11, Solution 95
(a) Source impedance css XjR −=Z Load impedance 2LL XjR +=Z For maximum load transfer
LcLs*sL XX,RR ==→= ZZ
LC1
XX Lc ω=ω
→=
or f2LC1
π==ω
=××π
=π
=)1040)(1080(2
1LC2
1f
9-3-kHz814.2
(b) ===)10)(4(
)6.4(R4
VP
2
L
2s mW529 (since V is in rms) s
Chapter 11, Solution 96
ZTh
+ −
VTh ZL
(a) Hz300,V146VTh =
Ω+= 8j40ZTh
== *ThL ZZ Ω− 8j40
(b) ===)40)(8(
)146(R8
VP
2
Th
2
Th W61.66
Chapter 11, Solution 97
Ω+=+++= 22j2.100)20j100()j1.0)(2(ZT
22j2.100240
ZV
IT
s
+==
=+
=== 22
22
L2
)22()2.100()240)(100(
I100RIP W3.547
Chapter 12, Solution 1.
(a) If , then 400ab =V
=°∠= 30-3
400anV V30-231 °∠
=bnV V150-231 °∠ =cnV V270-231 °∠
(b) For the acb sequence,
°∠−°∠=−= 120V0V ppbnanab VVV
°∠=
−+= 30-3V
23
j21
1V ppabV
i.e. in the acb sequence, lags by 30°. abV anV Hence, if , then 400ab =V
=°∠= 303
400anV V30231 °∠
=bnV V150231 °∠ =cnV V90-231 °∠
Chapter 12, Solution 2.
Since phase c lags phase a by 120°, this is an acb sequence.
=°+°∠= )120(30160bnV V150160 °∠ Chapter 12, Solution 3. Since V leads by 120°, this is an bn cnV abc sequence.
=°+°∠= )120(130208anV V250208 °∠
Chapter 12, Solution 4.
=°∠= 120cabc VV V140208 °∠
=°∠= 120bcab VV V260208 °∠
=°∠°∠
=°∠
=303260208
303ab
an
VV V230120 °∠
=°∠= 120-anbn VV V110120 °∠
Chapter 12, Solution 5. This is an abc phase sequence.
°∠= 303anab VV
or =°∠°∠
=°∠
=3030420
303ab
an
VV V30-5.242 °∠
=°∠= 120-anbn VV V150-5.242 °∠
=°∠= 120ancn VV V905.242 °∠
Chapter 12, Solution 6.
°∠=+= 26.5618.115j10YZ The line currents are
=°∠
°∠==
26.5618.110220
Y
ana Z
VI A26.56-68.19 °∠
=°∠= 120-ab II A146.56-68.19 °∠
=°∠= 120ac II A93.4468.19 °∠
The line voltages are
=°∠= 303200abV V30381 °∠ =bcV V90-381 °∠ =caV V210-381 °∠
The load voltages are
=== anYaAN VZIV V0220 °∠ == bnBN VV V120-220 °∠ == cnCN VV V120220 °∠
Chapter 12, Solution 7. This is a balanced Y-Y system.
+ − 440∠0° V ZY = 6 − j8 Ω
Using the per-phase circuit shown above,
=−
°∠=
8j60440
aI A53.1344 °∠
=°∠= 120-ab II A66.87-44 °∠ =°∠= 120ac II A13.73144 °∠
Chapter 12, Solution 8. , V220VL = Ω+= 9j16YZ
°∠=+
=== 29.36-918.6)9j16(3
2203VV
Y
L
Y
pan ZZ
I
=LI A918.6
Chapter 12, Solution 9.
=+
°∠=
+=
15j200120
YL
ana ZZ
VI A36.87-8.4 °∠
=°∠= 120-ab II A156.87-8.4 °∠
=°∠= 120ac II A83.138.4 °∠
As a balanced system, =nI A0
Chapter 12, Solution 10. Since the neutral line is present, we can solve this problem on a per-phase basis.
For phase a,
°∠=−
°∠=
+= 36.5355.6
20j270220
2A
ana Z
VI
For phase b,
°∠=°∠
=+
= 120-1022
120-2202B
bnb Z
VI
For phase c,
°∠=+
°∠=
+= 97.3892.16
5j12120220
2C
cnc Z
VI
The current in the neutral line is
)-( cban IIII ++= or cban- IIII ++=
)78.16j173.2-()66.8j5-()9.3j263.5(- n ++−++=I
=−= 02.12j91.1nI A81-17.12 °∠
Chapter 12, Solution 11.
°∠°∠
=°∠
=°∠
=90-3
1022090-390-3
BCbcan
VVV
=anV V100127 °∠
=°∠= 120BCAB VV V130220 °∠
V110-220120-BCAC °∠=°∠= VV
If , then °∠= 6030bBI
°∠= 18030aAI , °∠= 60-30cCI
°∠=°∠°∠
=°∠
= 21032.1730-3
1803030-3
aAAB
II
°∠= 9032.17BCI , °∠= 30-32.17CAI
== CAAC -II A15032.17 °∠
BCBC VZI =
=°∠
°∠==
9032.170220
BC
BC
IV
Z Ω°∠ 80-7.12
Chapter 12, Solution 12. Convert the delta-load to a wye-load and apply per-phase analysis.
Ia
110∠0° V + −
ZY
Ω°∠== ∆ 45203Y
ZZ
=°∠°∠
=45200110
aI A45-5.5 °∠
=°∠= 120-ab II A165-5.5 °∠ =°∠= 120ac II A755.5 °∠
Chapter 12, Solution 13.
First we calculate the wye equivalent of the balanced load.
ZY = (1/3)Z∆ = 6+j5
Now we only need to calculate the line currents using the wye-wye circuits.
A07.58471.615j8120110I
A07.178471.615j8120110I
A93.61471.65j610j2
110I
c
b
a
°∠=+
°∠=
°∠=+
°−∠=
°−∠=+++
=
Chapter 12, Solution 14. We apply mesh analysis. Ω+ 2j1 A a + ZL 100 ZV 0o∠ L - I3 n I1 B C - -
100 + - + c I
V 120100 o∠ V 120o∠ Ω+= 1212 jZ L
2 b Ω+ 2j1 Ω+ 2j1 For mesh 1,
0)1212()21()1614(120100100 321 =+−+−++∠+− IjIjjIo or
6.861506.8650100)1212()21()1614( 321 jjIjIjIj −=−+=+−+−+ (1) For mesh 2,
0)1614()1212()21(120100120100 231 =+++−+−−∠−∠ IjIjjIoo or
2.1736.86506.8650)1212()1614()21( 321 jjjIjIjIj −=−+−−=+−+++− (2) For mesh 3,
0)3636()1212()1212( 321 =+++−+− IjIjIj (3) Solving (1) to (3) gives
016.124197.4,749.16098.10,3.19161.3 321 jIjIjI −−=−−=−−= A 3.9958.191
oaA II −∠==
A 8.159392.712
obB III ∠=−=
A 91.5856.192
ocC II ∠=−=
Chapter 12, Solution 15.
Convert the delta load, , to its equivalent wye load. ∆Z
10j83Ye −== ∆Z
Z
°∠=−
−+== 14.68-076.8
5j20)10j8)(5j12(
|| YeYp ZZZ
047.2j812.7p −=Z
047.1j812.8LpT −=+= ZZZ
°∠= 6.78-874.8TZ
We now use the per-phase equivalent circuit.
Lp
pa
VZZ
I+
= , where 3
210pV =
°∠=°∠
= 78.666.13)6.78-874.8(3
210aI
== aL II A66.13
Chapter 12, Solution 16.
(a) °∠=°+°∠== 15010)180-30(10- ACCA II This implies that °∠= 3010ABI
°∠= 90-10BCI
=°∠= 30-3ABa II A032.17 °∠ =bI A120-32.17 °∠ =cI A12032.17 °∠
(b) =°∠°∠
==∆ 30100110
AB
AB
IV
Z Ω°∠ 30-11
Chapter 12, Solution 17. Convert the ∆-connected load to a Y-connected load and use per-phase analysis.
Ia
+ −
ZL
Van ZY
4j33Y +== ∆Z
Z
°∠=+++°∠
=+
= 48.37-931.19)5.0j1()4j3(
0120
LY
ana ZZ
VI
But °∠= 30-3ABa II
=°∠
°∠=
30-348.37-931.19
ABI A18.37-51.11 °∠
=BCI A138.4-51.11 °∠ =CAI A101.651.11 °∠
)53.1315)(18.37-51.11(ABAB °∠°∠== ∆ZIV
=ABV V76.436.172 °∠
=BCV V85.24-6.172 °∠ =CAV V8.5416.172 °∠
Chapter 12, Solution 18.
°∠=°∠°∠=°∠= 901.762)303)(60440(303anAB VV
°∠=+=∆ 36.87159j12Z
=°∠°∠
==∆ 36.8715
901.762ABAB Z
VI A53.1381.50 °∠
=°∠= 120-ABBC II A66.87-81.50 °∠
=°∠= 120ABCA II A173.1381.50 °∠ Chapter 12, Solution 19.
°∠=+=∆ 18.4362.3110j30Z The phase currents are
=°∠
°∠==
∆ 18.4362.310173ab
AB ZV
I A18.43-47.5 °∠
=°∠= 120-ABBC II A138.43-47.5 °∠ =°∠= 120ABCA II A101.5747.5 °∠
The line currents are
°∠=−= 30-3ABCAABa IIII
=°∠= 48.43-347.5aI A48.43-474.9 °∠
=°∠= 120-ab II A168.43-474.9 °∠ =°∠= 120ac II A71.57474.9 °∠
Chapter 12, Solution 20.
°∠=+=∆ 36.87159j12Z The phase currents are
=°∠
°∠=
36.87150210
ABI A36.87-14 °∠
=°∠= 120-ABBC II A156.87-14 °∠ =°∠= 120ABCA II A83.1314 °∠
The line currents are
=°∠= 30-3ABa II A66.87-25.24 °∠ =°∠= 120-ab II A186.87-25.24 °∠
=°∠= 120ac II A53.1325.24 °∠
Chapter 12, Solution 21.
(a) )rms(A66.9896.1766.38806.12
1202308j10
120230IAC °−∠=°∠
°∠−=
+°∠−
=
(b)
A34.17110.31684.4j75.30220.11j024.14536.6j729.16
66.3896.1766.15896.178j100230
8j10120230IIIII ABBCBABCbB
°∠=+−=+−−−=
°−∠−°−∠=+
°∠−
+−∠
=−=+=
Chapter 12, Solution 22.
Convert the ∆-connected source to a Y-connected source.
°∠=°∠=°∠= 30-12030-3
20830-
3
VpanV
Convert the ∆-connected load to a Y-connected load.
j8)5j4)(6j4(
)5j4(||)6j4(3
||Y +−+
=−+== ∆ZZZ
2153.0j723.5 −=Z
Ia
+ −
ZL
Van Z
=−
°∠=
+=
2153.0j723.730120
L
ana ZZ
VI A28.4-53.15 °∠
=°∠= 120-ab II A148.4-53.15 °∠
=°∠= 120ac II A91.653.15 °∠
Chapter 12, Solution 23.
(a) oAB
AB ZVI
6025208∠
==∆
o
o
oo
ABa II 90411.146025
303208303 −∠=∠
−∠=−∠=
A 41.14|| == aL II
(b) kW 596.260cos25
3208)208(3cos321 =
==+= o
LL IVPPP θ
Chapter 12, Solution 24. Convert both the source and the load to their wye equivalents.
10j32.1730203Y +=°∠== ∆Z
Z
°∠=°∠= 02.24030-3ab
an
VV
We now use per-phase analysis.
Ia
+ −
1 + j Ω
Van 20∠30° Ω
=°∠
=+++
=3137.212.240
)10j32.17()j1(an
a
VI A31-24.11 °∠
=°∠= 120-ab II A151-24.11 °∠
=°∠= 120ac II A8924.11 °∠
But °∠= 30-3ABa II
=°∠°∠
=30-331-24.11
ABI A1-489.6 °∠
=°∠= 120-ABBC II A121-489.6 °∠
=°∠= 120ABCA II A119489.6 °∠
Chapter 12, Solution 25.
Convert the delta-connected source to an equivalent wye-connected source and consider the single-phase equivalent.
Ya 3
)3010(440Z
I°−°∠
=
where °°∠=−=−++= 78.24-32.146j138j102j3YZ
=°∠
°∠=
)24.78-32.14(320-440
aI A4.7874.17 °∠
=°∠= 120-ab II A115.22-74.17 °∠
=°∠= 120ac II A124.7874.17 °∠
Chapter 12, Solution 26. Transform the source to its wye equivalent.
°∠=°∠= 30-17.7230-3
VpanV
Now, use the per-phase equivalent circuit.
ZV
I anaA = , °∠=−= 32-3.2815j24Z
=°∠°∠
=32-3.2830-17.72
aAI A255.2 °∠
=°∠= 120-aAbB II A118-55.2 °∠
=°∠= 120aAcC II A12255.2 °∠
Chapter 12, Solution 27.
)15j20(310-220
330-
Y
aba +
°∠=
°∠=
ZV
I
=aI A46.87-081.5 °∠
=°∠= 120-ab II A166.87-081.5 °∠
=°∠= 120ac II A73.13081.5 °∠
Chapter 12, Solution 28. Let °∠= 0400abV
°∠=°∠°∠
=°∠
= 307.7)60-30(3
30-4003
30-
Y
ana Z
VI
== aLI I A7.7
°∠=°∠== 30-94.23030-3an
YaAN
VZIV
== ANpV V V9.230
Chapter 12, Solution 29.
, θ= cosIV3P pp 3V
V Lp = , pL II =
θ= cosIV3P LL
pL
L I05.20)6.0(3240
5000cosV3
PI ===
θ=
911.6)05.20(3
240I3
VIV
L
L
p
pY ====Z
°=θ→=θ 13.536.0cos
(leading)53.13-911.6Y °∠=Z
=YZ Ω− 53.5j15.4
83336.0
5000pfP
S ===
6667sinSQ =θ=
=S VA6667j5000−
Chapter 12, Solution 30.
Since this a balanced system, we can replace it by a per-phase equivalent, as shown below. + ZL Vp
-
3,33 *
2L
pp
pp
VVZVSS ===
kVA 454421.14530)208( 2
*
2o
op
L
ZVS ∠=
−∠==
kW 02.1cos == θSP
Chapter 12, Solution 31.
(a) kVA 5.78.0/6cos
,8.0cos,000,6 =====θ
θ Ppp
PSP
kVAR 5.4sin == θPp SQ
kVA 5.1318)5.46(33 jjSS p +=+== For delta-connected load, Vp = VL= 240 (rms). But
Ω+=+
==→= 608.4144.6,10)5.1318(
)240(3333
22*
*
2
jZxjS
VZZVS P
pp
p
p
(b) A 04.188.02403
6000cos3 ==→=xx
IIVP LLLp θ
(c ) We find C to bring the power factor to unity
F 2.207240602
4500kVA 5.4 22 µπω
===→==xxV
QCQQrms
cpc
Chapter 12, Solution 32.
θ∠= LLIV3S
3LL 1050IV3S ×=== S
==)440(3
5000IL A61.65
For a Y-connected load,
61.65II Lp == , 03.2543
4403
VV L
p ===
872.361.6503.254
IV
p
p===Z
θ∠= ZZ , °==θ 13.53)6.0(cos-1
)sinj)(cos872.3( θ+θ=Z
)8.0j6.0)(872.3( +=Z
=Z Ω+ 098.3j323.2
Chapter 12, Solution 33. θ∠= LLIV3S
LLIV3S == S
For a Y-connected load,
pL II = , pL V3V =
pp IV3S =
====)208)(3(
4800V3S
IIp
pL A69.7
=×== 2083V3V pL V3.360
Chapter 12, Solution 34.
3
2203
VV L
p ==
°∠=−
== 5873.6)16j10(3
200V
Y
pa Z
I
== pL II A73.6
°∠××=θ∠= 58-73.62203IV3 LLS
=S VA8.2174j1359−
Chapter 12, Solution 35.
(a) This is a balanced three-phase system and we can use per phase equivalent circuit. The delta-connected load is converted to its wye-connected equivalent
10203/)3060(31'' jjZZ y +=+== ∆
IL
+ 230 V Z’y Z’’y
-
5.55.13)1020//()1040(//' '' jjjZZZ yyy +=++==
A 953.561.145.55.13
230 jj
I L −=+
=
(b) kVA 368.1361.3* jIVS Ls +== (c ) pf = P/S = 0.9261
Chapter 12, Solution 36.
(a) S = 1 [0.75 + sin(cos-10.75) ] =0.75 + 0.6614 MVA
(b) 49.5252.5942003
10)6614.075.0(3
36
** jx
xjVSIIVp
ppp +=+
==→=S
kW 19.25)4()36.79(|| 22 === lpL RIP
(c) kV 2.709-4.443 kV 21.04381.4)4( o∠=−=++= jjIV pLsV Chapter 12, Solution 37.
206.0
12pfP
S ===
kVA16j1220S −=θ∠=θ∠=S
But θ∠= LLIV3S
=××
=20831020
I3
L A51.55
p
2
p3 ZIS = For a Y-connected load, pL II = .
2
3
2
L
p )51.55)(3(10)16j12(
I3
×−==
SZ
=pZ Ω− 731.1j298.1
Chapter 12, Solution 38.
As a balanced three-phase system, we can use the per-phase equivalent shown below.
14j100110
)12j9()2j1(0110
a +°∠
=+++°∠
=I
)12j9()1410(
)110(21
21
22
2
Y
2
ap +⋅+
⋅== ZIS
The complex power is
)12j9(296
)110(23
32
p +⋅⋅== SS
=S VA81.735j86.551 +
Chapter 12, Solution 39. Consider the system shown below.
I2
I1I3
5 Ω
5 Ω
-j6 Ω
10 Ω
j3 Ω
B C
A
8 Ω
4 Ω
b
a
−+
+ −
100∠-120°
100∠120° 100∠0°− +
5 Ω
c
For mesh 1,
321 )6j8(5)6j18(100 III −−−−= (1) For mesh 2,
312 10520120-100 III −−=°∠
321 24-120-20 III −+=°∠ (2)
For mesh 3, 321 )3j22(10)6j8(-0 III −+−−= (3)
To eliminate , start by multiplying (1) by 2, 2I
321 )12j16(10)12j36(200 III −−−−= (4) Subtracting (3) from (4),
31 )15j38()18j44(200 II −−−= (5) Multiplying (2) by 45 ,
321 5.2525.1-120-25 III −+=°∠ (6) Adding (1) and (6),
31 )6j5.10()6j75.16(65.21j5.87 II −−−=− (7) In matrix form, (5) and (7) become
+−+−
=
− 3
1
6j5.10-6j75.1615j38-18j44
65.12j5.87200
II
25.26j5.192 −=∆ , 2.935j25.9001 −=∆ , 6.1327j3.1103 −=∆
144.4j242.538.33-682.67.76-28.194
46.09-1.129811 −=°∠=
°∠°∠
=∆∆
=I
694.6j485.177.49-857.67.76-28.194
85.25-2.133233 −=°∠=
°∠°∠
=∆∆
=I
We obtain from (6), 2I
312 21
41
120-5 III ++°∠=
)347.3j7425.0()0359.1j3104.1()33.4j-2.5(2 −+−+−=I
713.8j4471.0-2 −=I The average power absorbed by the 8-Ω resistor is
W89.164)8(551.2j756.3)8(P22
311 =+=−= II The average power absorbed by the 4-Ω resistor is
W1.188)4()8571.6()4(P 22
32 === I
The average power absorbed by the 10-Ω resistor is
W12.78)10(019.2j1.9321-)10(P 22323 =−=−= II
Thus, the total real power absorbed by the load is
=++= 321 PPPP W1.431 Chapter 12, Solution 40. Transform the delta-connected load to its wye equivalent.
8j73Y +== ∆Z
Z
Using the per-phase equivalent circuit above,
°∠=+++
°∠= 46.75-567.8
)8j7()5.0j1(0100
aI
For a wye-connected load,
567.8II aap === I
)8j7()567.8)(3(3 2p
2
p +== ZIS
=== )7()567.8)(3()Re(P 2S k541.1 W Chapter 12, Solution 41.
kVA25.68.0
kW5pfP
S ===
But LLIV3S =
=××
==40031025.6
V3S
I3
LL A021.9
Chapter 12, Solution 42. The load determines the power factor.
°=θ→==θ 13.53333.13040
tan
(leading)6.0cospf =θ=
kVA6.9j2.7)8.0(6.02.7
j2.7 −=
−=S
But p
2
p3 ZIS =
80)40j30)(3(
10)6.9j2.7(3
3
p
2
p =−
×−==
ZS
I
A944.8Ip =
== pL II A944.8
=×
==)944.8(3
1012I3
SV
3
LL V6.774
Chapter 12, Solution 43.
p
2
p3 ZIS = , Lp II = for Y-connected loads
)047.2j812.7()66.13)(3( 2 −=S =S kVA145.1j373.4 −
Chapter 12, Solution 44. For a ∆-connected load,
Lp VV = , pL I3I =
LLIV3S =
273.31)240(3
10)512(V3
SI
322
LL =
×+==
At the source,
LLL'L ZIVV +=
)3j1)(273.31(0240'L ++°∠=V
819.93j273.271'L +=V
='LV V04.287
Also, at the source,
*L
'L
' 3 IVS = )273.31)(819.93j273.271(3' +=S
078.19273.271
819.93tan 1- =
=θ
=θ= cospf 9451.0
Chapter 12, Solution 45. θ∠= LLIV3S
LL V3
-I
θ∠=
S, kVA6.635
708.010450
pfP 3
=×
==S
A45-8344403
-)6.635(L °∠=
×θ∠
=I
At the source,
)2j5.0(0440 LL ++°∠= IV
)76062.2)(45-834(440L °∠°∠+=V °∠+= 137.1719440LV
7.885j1.1914L +=V =LV V.8324109.2 °∠
Chapter 12, Solution 46. For the wye-connected load,
pL II = , pL V3V = Zpp VI =
*
2
L*
2
p*pp
3333
ZV
ZV
IVS ===
W121100
)110( 2
*
2
L ===Z
VS
For the delta-connected load,
Lp VV = , pL I3I = , Zpp VI =
*
2
L*
2
p*pp
333
ZV
ZV
IVS ===
W363100
)110)(3( 2
==S
This shows that the delta-connected load will deliver three times more average
power than the wye-connected load. This is also evident from 3Y∆=
ZZ .
Chapter 12, Solution 47.
°==θ→= 87.36)8.0(cos(lagging)8.0pf -1 kVA150j20087.362501 +=°∠=S
°==θ→= 19.18-)95.0(cos(leading)95.0pf -1
kVA65.93j28519.81-3002 −=°∠=S
°==θ→= 0)1(cos0.1pf -1
kVA4503 =S
kVA45.37.93635.56j935321T °∠=+=++= SSSS
LLT IV3=S
=××
=)108.13(3
107.936I
3
3
L rmsA19.39
=°=θ= )45.3cos(cospf (lagging)9982.0
Chapter 12, Solution 48.
(a) We first convert the delta load to its equivalent wye load, as shown below. A A ZA 18-j12Ω 40+j15Ω ZC C B C 60Ω
ZB
B
923.1577.73118
)1218)(1540( jjjjZ A −=
+−+
=
105.752.203118
).1540(60 jjjZ B −=
++
=
3303.6992.83118
)1218(60 jjjZC −=
+−
=
The system becomes that shown below.
a 2+j3 A + 240<0o ZA - I1 - - ZB ZC 240<120o 240<-120o + + 2+j3 c I2 b B C 2+j3
We apply KVL to the loops. For mesh 1, 0)()2(120240240 21 =+−+++−∠+− lBBAl
o ZZIZZZI or
85.207360)105.1052.22()13.11097.32( 21 jIjIj +=+−+ (1) For mesh 2,
0)2()(120240120240 21 =++++−−∠−∠ CBllBoo ZZZIZZI
or
69.415)775.651.33()105.1052.22( 21 jIjIj −=+++− (2) Solving (1) and (2) gives
89.11165.15,328.575.23 21 jIjI −=−=
A 6.14281.10,A 64.1234.24 121o
bBo
aA IIIII −∠=−=−∠==
A 9.14127.192o
cC II ∠=−= (b) ooo
aS 64.126.5841)64.1234.24)(0240( ∠=∠∠= ooo
bS 6.224.2594)6.14281.10)(120240( ∠=∠−∠= ooo
bS 9.218.4624)9.14127.19)(120240( −∠=−∠∠= kVA 54.24.12kVA 55.0386.12 o
cba jSSSS ∠=+=++=
Chapter 12, Solution 49.
(a) For the delta-connected load, (rms) 220,1020 ==Ω+= Lpp VVjZ ,
kVA 56.26943.629045808)1020(
22033 2
*
2o
p
p jj
xZVS ∠=+=
−==
(b) For the wye-connected load, 3/,1020 Lpp VVjZ =Ω+= ,
kVA 56.26164.2)1020(3
22033 2
*
2o
p
p
jx
ZVS ∠=
−==
Chapter 12, Solution 50.
kVA 3kVA, 4.68.4)8.06.0(8 121 =+=+=+= SjjSSS Hence,
kVA 4.68.112 jSSS +=−=
But p
LLp
p
p
ZVSVV
ZVS *
2
2*
2
2.
3,3
=→==
Ω+=→+
== 34.8346.210)4.68.1(
2403
2
2
** jZ
xjSVZ p
Lp
Chapter 12, Solution 51. Apply mesh analysis to the circuit as shown below.
Za
i2
i1
Zc
Zb
+ −
− +
− +
150∠-120°
150∠0°
n 150∠120°
For mesh 1,
0)(150- 2b1ba =−++ IZIZZ
21 )9j12()j18(150 II +−+= (1) For mesh 2,
0)(120-150- 1b2cb =−++°∠ IZIZZ
12 )9j12()9j27(120-150 II +−+=°∠ (2) From (1) and (2),
+−−+
=
°∠ 2
1
9j279j12-9j12-j18
120-150150
II
27j414 −=∆ , 8.3583j9.37801 +=∆ , 2.1063j9.5792 −=∆
°∠=°∠°∠
=∆∆
= 47.256.123.73-88.414
43.475.520911I
°∠=°∠°∠
=∆∆
= 57.66-919.23.73-88.414
61.39-1.121122I
== 1a II A47.256.12 °∠
∆−
=∆∆−∆
=−=4647j3201-12
12b III
=°∠°∠
=3.73-88.414
235.443.5642bI A239.176.13 °∠
== 2c -II A122.34919.2 °∠
Chapter 12, Solution 52. Since the neutral line is present, we can solve this problem on a per-phase basis.
°∠=°∠°∠
== 6066020120120
AN
ana Z
VI
°∠=°∠°∠
== 040300120
BN
bnb Z
VI
°∠=°∠°∠
== 150-33040120-120
CN
cnc Z
VI
Thus,
cban- IIII ++= °∠+°∠+°∠= 150-304606- nI
)5.1j598.2-()4()196.5j3(- n −+++=I °∠=+= 4075.5696.3j405.4- nI
=nI A22075.5 °∠
Chapter 12, Solution 53.
3
250Vp =
Since we have the neutral line, we can use per-phase equivalent circuit for each phase.
=°∠
⋅°∠
=6040
13
0250aI A60-608.3 °∠
=°∠
⋅°∠
=45-60
13120-250
bI A75-406.2 °∠
=°∠
⋅°∠
=020
13120250
cI A120217.7 °∠
cban- IIII ++=
)25.6j609.3-()324.2j6227.0()125.3j804.1(- n ++−+−=I
=−= 801.0j1823.1nI A34.12- 428.1 °∠
Chapter 12, Solution 54. Consider the circuit shown below.
Iaa
B C
A
j50 Ω
-j50 Ω
50 Ω
IAB
c b
Vp∠0°
+−
Vp∠120° Vp∠-120°
+ −
− +
°∠×== 303100abAB VV
=°∠
==50
303100
AB
ABAB Z
VI A30464.3 °∠
=°∠°∠
==90-50
90-3100
BC
BCBC Z
VI A0464.3 °∠
=°∠
°∠==
90501503100
CA
CACA Z
VI A60464.3 °∠
Chapter 12, Solution 55. Consider the circuit shown below.
Iaa A
Ib
I1
I2
c b
220∠0°
+−
220∠120° 220∠-120°
+ −
− +
30 + j40100 – j120
B C
60 + j80
IcFor mesh 1,
0)120j100()40j160(0220120-220 21 =−−−+°∠−°∠ II
21 )6j5()2j8(120-1111 II −−−=°∠− (1) For mesh 2,
0)120j100()80j130(120-220120220 12 =−−−+°∠−°∠ II
21 )4j5.6()6j5(-12011120-11 II −+−=°∠−°∠ (2) From (1) and (2),
+
+−=
+
2
1
j4-6.5j65-j65-2j8
j19.053-526.9j5.16
II
15j55+=∆ , 35.99j04.311 −=∆ , 8.203j55.1012 −=∆
°∠=°∠°∠
=∆∆
= 87.91-8257.115.2601.57
72.65-08.10411I
°∠=°∠°∠
=∆∆
= 78.77-994.315.2601.5763.51-7.2272
2I
°∠== 87.91-8257.11a II
°∠=+−
=∆∆−∆
=−= 71.23-211.215j55
45.104j51.701212b III
°∠== 23.011994.3- 2c II
7.266j99.199)80j60()8257.1( 2AN
2
aA +=+== ZIS
6.586j9.488)120j100()211.2( 2BN
2
bB −=−== ZIS
1.638j6.478)40j30()994.3( 2CN
2
cC +=+== ZIS
=++= CBA SSSS V2.318j5.1167 A+ Chapter 12, Solution 56.
(a) Consider the circuit below.
For mesh 1,
0)(10j0440120-440 31 =−+°∠−°∠ II
°∠=+
=− 60-21.7610j
)866.0j5.1)(440(31 II (1)
For mesh 2,
0)(20120-440120440 32 =−+°∠−°∠ II
1.38j20
)732.1j)(440(23 ==− II (2)
For mesh 3, 05j)(20)(10j 32313 =−−+− IIIII
I2
I1
b −
+
+−
440∠-120°440∠120°
440∠0°
− +
BI3
20 Ω
j10 Ω
a A
-j5 Ω
C c
Substituting (1) and (2) into the equation for mesh 3 gives,
°∠=+
= 6042.152j5
j0.866)-1.5)(440(3I (3)
From (1),
°∠=+=°∠+= 3013266j315.11460-21.7631 II From (2),
°∠=+=−= 50.9493.1209.93j21.761.38j32 II
== 1a II A30132 °∠
=+=−= j27.9-38.10512b III A143.823.47 °∠
== 2c -II A230.99.120 °∠
(b) kVA08.58j)10j(2
31AB =−= IIS
kVA04.29)20(2
32BC =−= IIS
kVA-j116.16-j5)((152.42)-j5)( 22
3CA === IS
kVA08.58j04.29CABCAB −=++= SSSS Real power absorbed = 29 kW04.
(c) Total complex supplied by the source is =S kVA08.58j04.29 −
Chapter 12, Solution 57.
We apply mesh analysis to the circuit shown below. Ia
+ Va Ω+ 50j80 - I1 - - Ω+ 3020 j
Vc Vb + + Ib
I2 Ic
Ω− 4060 j
263.95165)3020()80100( 21 jVVIjIj ba +=−=+−+ (1) 53.190)1080()3020( 21 jVVIjIj cb −=−=−++− (2)
Solving (1) and (2) gives 722.19088.0,6084.08616.1 21 jIjI −=−= .
55.1304656.11136.1528.0,A 1.189585.1 121 bo
a jIIIII −∠=−−=−=−∠==
A 8.117947.12o
c II ∠=−= Chapter 12, Solution 58. The schematic is shown below. IPRINT is inserted in the neutral line to measure thecurrent through the line. In the AC Sweep box, we select Total Ptss = 1, Start Freq0.1592, and End Freq. = 0.1592. After simulation, the output file includes
FREQ IM(V_PRINT4) IP(V_PRINT4) 1.592 E–01 1.078 E+01 –8.997 E+01
A o
. =
i.e. In = 10.78∠–89.97° A
Chapter 12, Solution 59. The schematic is shown below. In the AC Sweep box, we set Total Pts = 1, Start Freq = 60, and End Freq = 60. After simulation, we obtain an output file which includes
FREQ VM(1) VP(1) 6.000 E+01 2.206 E+02 –3.456 E+01 FREQ VM(2) VP(2) 6.000 E+01 2.141 E+02 –8.149 E+01 FREQ VM(3) VP(3) 6.000 E+01 4.991 E+01 –5.059 E+01
i.e. VAN = 220.6∠–34.56°, VBN = 214.1∠–81.49°, VCN = 49.91∠–50.59° V
hapter 12, Solution 60.
he schematic is shown below. IPRINT is inserted to give Io. We select Total Pts = 1,
FREQ IM(V_PRINT4) IP(V_PRINT4)
.592 E–01 1.421 E+00 –1.355 E+02
om which, Io = 1.421∠–135.5° A
C TStart Freq = 0.1592, and End Freq = 0.1592 in the AC Sweep box. Upon simulation, the output file includes
1
fr
Chapter 12, Solution 61. The schematic is shown below. Pseudocomponents IPRINT and PRINT are inserted to measure IaA and VBN. In the AC Sweep box, we set Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592. Once the circuit is simulated, we get an output file which includes
FREQ VM(2) VP(2) 1.592 E–01 2.308 E+02 –1.334 E+02 FREQ IM(V_PRINT2) IP(V_PRINT2) 1.592 E–01 1.115 E+01 3.699 E+01
from which IaA = 11.15∠37° A, VBN = 230.8∠–133.4° V
Chapter 12, Solution 62. Because of the delta-connected source involved, we follow Example 12.12. In the AC Sweep box, we type Total Pts = 1, Start Freq = 60, and End Freq = 60. After simulation, the output file includes
FREQ IM(V_PRINT2) IP(V_PRINT2) 6.000 E+01 5.960 E+00 –9.141 E+01 FREQ IM(V_PRINT1) IP(V_PRINT1) 6.000 E+01 7.333 E+07 1.200 E+02
From which Iab = 7.333x107∠120° A, IbB = 5.96∠–91.41° A
Chapter 12, Solution 63.
Let F 0333.0X1 C and H, 20X/ L that so 1 =====ω
ωω
The schematic is shown below..
.
When the file is saved and run, we obtain an output file which includes the following: FREQ IM(V_PRINT1)IP(V_PRINT1) 1.592E-01 1.867E+01 1.589E+02 FREQ IM(V_PRINT2)IP(V_PRINT2) 1.592E-01 1.238E+01 1.441E+02 From the output file, the required currents are:
A 1.14438.12 A, 9.15867.18 oAC
oaA II ∠=∠=
Chapter 12, Solution 64. We follow Example 12.12. In the AC Sweep box we type Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592. After simulation the output file includes
FREQ IM(V_PRINT1) IP(V_PRINT1) 1.592 E–01 4.710 E+00 7.138 E+01 FREQ IM(V_PRINT2) IP(V_PRINT2) 1.592 E–01 6.781 E+07 –1.426 E+02 FREQ IM(V_PRINT3) IP(V_PRINT3) 1.592 E–01 3.898 E+00 –5.076 E+00 FREQ IM(V_PRINT4) IP(V_PRINT4) 1.592 E–01 3.547 E+00 6.157 E+01 FREQ IM(V_PRINT5) IP(V_PRINT5) 1.592 E–01 1.357 E+00 9.781 E+01 FREQ IM(V_PRINT6) IP(V_PRINT6) 1.592 E–01 3.831 E+00 –1.649 E+02
from this we obtain
IaA = 4.71∠71.38° A, IbB = 6.781∠–142.6° A, IcC = 3.898∠–5.08° A
IAB = 3.547∠61.57° A, IAC = 1.357∠97.81° A, IBC = 3.831∠–164.9° A
Chapter 12, Solution 65. Due to the delta-connected source, we follow Example 12.12. We type Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592. The schematic is shown below. After it is saved and simulated, we obtain an output file which includes
FREQ IM(V_PRINT1) IP(V_PRINT1) 1.592 E–01 6.581 E+00 9.866 E+01 FREQ IM(V_PRINT2) IP(V_PRINT2) 1.592 E–01 1.140 E+01 –1.113 E+02 FREQ IM(V_PRINT3) IP(V_PRINT3) 1.592 E–01 6.581 E+00 3.866 E+01
Thus, IaA = 6.581∠98.66° A, IbB = 11.4∠–111.3 A, IcC = 6.581∠38.66° A
Chapter 12, Solution 66. Chapter 12, Solution 66.
(a) ===3
2083
VV L
p V120 (a)
===3
2083
VV L
p V120
(b) Because the load is unbalanced, we have an unbalanced three-phase system. Assuming an abc sequence,
(b) Because the load is unbalanced, we have an unbalanced three-phase system. Assuming an abc sequence,
A05.248
01201 °∠=
°∠=I
A120-340
120-1202 °∠=
°∠=I
A120260
1201203 °∠=
°∠=I
A05.248
01201 °∠=
°∠=I
A120-340
120-1202 °∠=
°∠=I
A120260
1201203 °∠=
°∠=I
++
−+=++=
23
j5.0-)2(23
j0.5-)3(5.2- 321N IIII
++
−+=++=
23
j5.0-)2(23
j0.5-)3(5.2- 321N IIII
A90866.0866.0j23
jN °∠===I
Hence, =1I A5.2 , =2I A3 , =3I A2 , =NI A866.0
(c) === )48()5.2(RIP 2
1211 W300
=== )40()3(RIP 22
222 W360
=== )60()2(RIP 23
233 W240
(d) =++= 321T PPPP W900
Chapter 12, Solution 67.
(a) The power to the motor is kW221)85.0)(260(cosSPT ==θ=
The motor power per phase is
kW67.73P31
P Tp ==
Hence, the wattmeter readings are as follows:
=+= 2467.73Wa kW67.97 =+= 1567.73Wb kW67.88 =+= 967.73Wc kW67.83
(b) The motor load is balanced so that 0IN = .
For the lighting loads,
A200120
000,24Ia ==
A125120
000,15Ib ==
A75120000,9
Ic ==
If we let
A02000Iaa °∠=°∠=I A120-125b °∠=I
A12075c °∠=I Then,
cbaN- IIII ++=
++
−+=
23
j0.5-)75(23
j5.0-)125(200- NI
A602.86100- N −=I
=NI A3.132 Chapter 12, Solution 68.
(a) === )4.8)(330(3IV3S LL VA4801
(b) SP
cospfcosSP =θ=→θ=
==24.4801
4500pf 9372.0
(c) For a wye-connected load,
== Lp II A4.8
(d) ===3
3303
VV L
p V53.190
Chapter 12, Solution 69.
MVA 8.0SMVA, 323.15.1)661.075.0(2SMVA, 72.096.0)6.08.0(2.1 321 =−=−=+=+= jjjS
9833.03153.326.3MVA, 603.026.3321 ===−=++=
SPpfjSSSS
MVA 1379.0)99.0tan(cos)9833.0[tan(cos26.3)tan(tan 11 =−=−= −−
newoldc PQ
mF 28106.6602
101379.031
62
6
==xxx
xxC
π
Chapter 12, Solution 70.
8004001200PPP 21T =−=+=
1600-1200400-PPQ 12T =−=−=
°=θ→===θ -63.43-28001600-
PQ
tanT
T
=θ= cospf (leading)4472.0
406
240IV
ZL
Lp ===
=pZ Ω°∠ 63.43-40
Chapter 12, Solution 71.
(a) If , °∠= 0208abV °∠= 120-208bcV , °∠= 120208caV ,
°∠=°∠
== 04.1020
0208
Ab
abAB Z
VI
°∠=°∠°∠
== 75-708.1445-210
120-208
BC
bcBC Z
VI
°∠=°∠°∠
== 97.381622.6213
120208
CA
caCA Z
VI
°∠−°∠=−= 97.381604.10CAABaA III
867.15j055.24.10aA −+=I °∠= 51.87-171.20aAI
°∠−°∠=−= 75-708.1497.8316BCCAcC III
°∠= 101.0364.30cCI
)cos(PaAab IVaAab1 θ−θ= IV
=°+°= )87.510cos()171.20)(208(P1 W2590
)cos(PcCcb IVcCcb2 θ−θ= IV
But °∠== 60208- bccb VV
=°−°= )03.10160cos()64.30)(208(P2 W4808
(b) W17.7398PPP 21T =+=
VAR25.3840)PP(3Q 12T =−= =+= TTT jQPS VA25.3840j17.7398 +
== TTS S V8335 A Chapter 12, Solution 72. From Problem 12.11,
V130220AB °∠=V and A18030aA °∠=I
=°−°= )180130cos()30)(220(P1 W4242
°∠== 190220- BCCB VV °∠= 60-30cCI
=°+°= )60190cos()30)(220(P2 W2257-
Chapter 12, Solution 73. Consider the circuit as shown below.
I1
I2
Z
Z
240∠-120° V − +
Ia
Ib
+ − 240∠-60° V
Z
Ic
°∠=+= 71.5762.3130j10Z
°∠=°∠
°∠= 131.57-59.7
71.5762.3160-240
aI
°∠=°∠°∠
= 191.57-59.771.5762.31120-240
bI
0120-24060-240c =°∠−°∠+ZI
°∠=°∠
= 108.4359.771.5731.62
240-cI
°∠=−= 101.57-146.13ca1 III °∠=+= 138.43146.13cb2 III
[ ] [ ] =°∠°∠== ).57101146.13)(60-240(ReReP *
111 IV W2360
[ ] [ ] =°∠°∠== )138.43-146.13)(120-240(ReReP *222 IV W8.632-
Chapter 12, Solution 74. Consider the circuit shown below.
Z = 60 − j30 Ω
For mesh 1,
− + 208∠-60° V I2
I1+ − 208∠0° V
Z
Z
212208 IZIZ −= For mesh 2,
21 2-60-208- IZIZ +=°∠
In matrix form,
=
°∠ 2
1
2--2
60-208-208
II
ZZZZ
23Z=∆ , Z)866.0j5.1)(208(1 +=∆ , Z)732.1j)(208(2 =∆
°∠=−+
=∆∆
= 56.56789.1)30j60)(3(
)866.0j5.1)(208(11I
°∠=−
=∆∆
= 116.5679.1)30j60)(3()732.1j)(208(2
2I
[ ] [ ] =°∠== )56.56-789.1)(208(ReReP *
111 IV W98.208
[ ] [ ] =°∠°∠== )63.4479.1))(60-208(Re)-(ReP *222 IV W65.371
Chapter 12, Solution 75.
(a) ===60012
RV
I mA20
(b) ===600120
RV
I mA200
Chapter 12, Solution 76. If both appliances have the same power rating, P,
sVP
I =
For the 120-V appliance, 120
PI1 = .
For the 240-V appliance, 240P
I2 = .
Power loss =
=appliance -V240 for the
240RP
appliance -V120 for the120
RP
R2
2
2
2
2I
Since 22 2401
1201
> , the losses in the 120-V appliance are higher.
Chapter 12, Solution 77. , lineloadTg PPPP −−= 85.0pf =
But 3060pf3600cos3600PT =×=θ=
=−−= )80)(3(25003060Pg W320 Chapter 12, Solution 78.
°=θ→==θ 79.3185.06051
cos 11
kVAR61.31)5268.0)(60(sinSQ 111 ==θ=
kW51PP 12 ==
°=θ→=θ 19.1895.0cos 22
kVA68.53cos
PS
2
22 =
θ=
kVAR759.16sinSQ 222 =θ=
kVAR851.14759.1661.3QQQ 21c =−=−=
For each load,
kVAR95.43
QQ c
1c ==
=π
=ω
= 221c
)440)(60)(2(4950
VQ
C F82.67 µ
Chapter 12, Solution 79. Consider the per-phase equivalent circuit below.
Ia
n
a
Van + −
2 ΩA
ZY = 12 + j5 Ω
N
=+
°∠=
+=
5j140255
2Y
ana Z
VI A19.65-15.17 °∠
Thus,
=°∠= 120-ab II A139.65-15.17 °∠ =°∠= 120ac II A100.3515.17 °∠
=°∠°∠== )22.6213)(19.65-15.17(YaAN ZIV V2.97223 °∠
Thus,
=°∠= 120-ANBN VV V117.63-223 °∠ =°∠= 120ANCN VV V122.97223 °∠
Chapter 12, Solution 80.
)7071.07071.0(8)]83.0sin(cos83.0[6 21
321 jSjSSSS −+++=++= − kVA 31.26368.10 2SjS +−= (1)
But
kVA j18.28724.383 VA )6.08.0)(6.84)(208(33 +=+=∠= jIVS LL θ (2) From (1) and (2),
kVA 28.5676.246.20746.132 ∠=+= jS Thus, the unknown load is 24.76 kVA at 0.5551 pf lagging.
Chapter 12, Solution 82.
p
p
ZVjjS *
2
21 3SkVA, 240320)6.08.0(400 =+=+=
For the delta-connected load, V pL V=
kVA 93.8427.1053810)2400(3
2
2 jj
xS +=−
=
MVA 0829.13737.121 jSSS +=+=
Let I = I1 + I2 be the total line current. For I1,
3,3 1
*1
Lpp
VVIVS ==
735.5798.76,)2400(310)240320(
3 1
31
1* jIxj
VS
IL
−=+
==
For I2, convert the load to wye.
76.2891.273303810
24003032 jj
II oop −=−∠
+=−∠=
5.34735021 jIII −=+=
kV 372.5||kV 405.1185.5)63(2400 =→+=++=+= slineLs VjjIVVV
Chapter 12, Solution 83.
kVA 80SkVA, 135.60135.60)707.0707.0(95.0746120 21 =+=+= jjxxS
kVA 135.60135.14021 jSSS +=+=
A 42.1834803
1049.1523
||3||But 3
===→=xx
VSIIVSL
LLL
Chapter 12, Solution 84. We first find the magnitude of the various currents.
For the motor,
A248.53440
4000V3
SI
LL ===
For the capacitor,
A091.4440
1800VQ
IL
cC ===
For the lighting,
V2543
440Vp ==
A15.3254800
VP
Ip
LiLi ===
Consider the figure below.
Ia I1a
-jXC
R In
ILi
I3
I2
Ic
+
Vab
−Ib
IC
b
c
n
If , °∠= 0VpanV °∠= 30V3 pabV °∠= 120VpcnV
°∠== 120091.4X-j C
abC
VI
)30(091.4ab1 °+θ∠==
∆ZV
I
where θ °== 95.43)72.0(cos-1
°∠= 73.95249.51I
°∠= 46.05-249.52I
°∠= 193.95249.53I
°∠== 12015.3R
cnLi
VI
Thus,
°∠+°∠=+= 120091.473.95249.5C1a III =aI A93.96608.8 °∠
°∠−°∠=−= 120091.446.05-249.5C2b III
=bI A52.16-271.9 °∠
°∠+°∠=+= 12015.3193.95249.5Li3c III =cI A167.6827.6 °∠
== Lin -II A60-15.3 °∠
Chapter 12, Solution 85. Let Z RY =
V56.1383
2403
VV L
p ===
RV
kW92
27IVP
2p
pp ====
Ω=== 133.29000
)56.138(P
VR
22p
Thus, =YZ Ω133.2
Chapter 12, Solution 86. Consider the circuit shown below.
1 Ω
B
N
A
b
n
I1
I2+ − 120∠0° V rms 15 + j4 Ω
24 – j2 Ω
1 Ω
a
+ − 120∠0° V rms
1 Ω
For the two meshes,
21)2j26(120 II −−= (1)
12)4j17(120 II −+= (2)
In matrix form,
+
−=
2
1
4j171-1-2j26
120120
II
70j449 +=∆ , )4j18)(120(1 +=∆ , )2j27)(120(2 −=∆
°∠=°∠°∠×
=∆∆
= 3.6787.48.8642.454
12.5344.1812011I
°∠=°∠
°∠×=
∆∆
= 13.1-15.78.8642.454
4.24-07.2712022I
== 1aA II A3.6787.4 °∠ == 2bB -II A166.915.7 °∠
∆∆−∆
=−= 1212nN III
=+−
=70j449
)6j9)(120(nNI A42.55-856.2 °∠
Chapter 12, Solution 87. 85.18j)5010)(60)(2(jLjmH50L 3- =π=ω→=Consider the circuit below.
1 Ω
30 Ω
20 ΩI1
I2+ −
2 Ω
+ −
1 Ω
115 V
15 + j18.85 Ω
115 V
Applying KVl to the three meshes, we obtain
11520223 321 =−− III (1) 11530332- 321 =−+ III (2)
0)85.18j65(3020- 321 =++− III (3) In matrix form,
=
+ 0115115
j18.856530-20-30-332-20-2-23
3
2
1
III
232,14j775,12 +=∆ , )8.659j1975)(115(1 +=∆
)3.471j1825)(115(2 +=∆ , )1450)(115(3 =∆
°∠=°∠°∠×
=∆∆
= 29.62-52.1248.0919214
47.18208211511I
°∠=°∠
°∠×=
∆∆
= 33.61-33.1148.0919124
14.489.188411522I
=+
−=
∆∆−∆
=−=75.231,14j775,12
)5.188j-150)(115(1212n III A176.6-448.1 °∠
=°∠== )29.6252.12)(115(IVS *
111 VA6.711j1252 +
=°∠== )33.6133.1)(115(*222 IVS V2.721j1085 A+
Chapter 13, Solution 1. For coil 1, L1 – M12 + M13 = 6 – 4 + 2 = 4
For coil 2, L2 – M21 – M23 = 8 – 4 – 5 = – 1
For coil 3, L3 + M31 – M32 = 10 + 2 – 5 = 7
LT = 4 – 1 + 7 = 10H
or LT = L1 + L2 + L3 – 2M12 – 2M23 + 2M12 LT = 6 + 8 + 10 = 10H Chapter 13, Solution 2. L = L1 + L2 + L3 + 2M12 – 2M23 2M31
= 10 + 12 +8 + 2x6 – 2x6 –2x4 = 22H Chapter 13, Solution 3. L1 + L2 + 2M = 250 mH (1) L1 + L2 – 2M = 150 mH (2)
Adding (1) and (2),
2L1 + 2L2 = 400 mH
But, L1 = 3L2,, or 8L2 + 400, and L2 = 50 mH
L1 = 3L2 = 150 mH From (2), 150 + 50 – 2M = 150 leads to M = 25 mH
k = M/ 150x50/5.2LL 21 = = 0.2887
Chapter 13, Solution 4. (a) For the series connection shown in Figure (a), the current I enters each coil from its dotted terminal. Therefore, the mutually induced voltages have the same sign as the self-induced voltages. Thus,
Leq = L1 + L2 + 2M
L2
L1
L2 L1
I1 I2
Is
Vs+–
Leq (a) (b) (b) For the parallel coil, consider Figure (b).
Is = I1 + I2 and Zeq = Vs/Is
Applying KVL to each branch gives,
Vs = jωL1I1 + jωMI2 (1)
Vs = jωMI1 + jω L2I2 (2)
or
ωωωω
=
2
1
2
1
s
s
II
LjMjMjLj
VV
∆ = –ω2L1L2 + ω2M2, ∆1 = jωVs(L2 – M), ∆2 = jωVs(L1 – M)
I1 = ∆1/∆, and I2 = ∆2/∆
Is = I1 + I2 = (∆1 + ∆2)/∆ = jω(L1 + L2 – 2M)Vs/( –ω2(L1L2 – M))
Zeq = Vs/Is = jω(L1L2 – M)/[jω(L1 + L2 – 2M)] = jωLeq
i.e., Leq = (L1L2 – M)/(L1 + L2 – 2M)
Chapter 13, Solution 5.
(a) If the coils are connected in series,
=++=++= 60x25)5.0(26025M2LLL 21 123.7 mH
(b) If they are connected in parallel,
=−+−
=−+−
= mH 36.19x26025
36.1960x25M2LL
MLLL2
21
221 24.31 mH
Chapter 13, Solution 6.
V1 = (R1 + jωL1)I1 – jωMI2
V2 = –jωMI1 + (R2 + jωL2)I2
Chapter 13, Solution 7. Applying KVL to the loop,
20∠30° = I(–j6 + j8 + j12 + 10 – j4x2) = I(10 + j6)
where I is the loop current.
I = 20∠30°/(10 + j6)
Vo = I(j12 + 10 – j4) = I(10 + j8)
= 20∠30°(10 + j8)/(10 + j6) = 22∠37.66° V Chapter 13, Solution 8. Consider the current as shown below.
j 6 j 4
1 Ω
-j3
j2
10 + – I2I1
4 Ω
+ Vo –
For mesh 1, 10 = (1 + j6)I1 + j2I2 (1)
For mesh 2, 0 = (4 + j4 – j3)I2 + j2I1 0 = j2I1 +(4 + j)I2 (2)
In matrix form,
+
+=
2
1
II
j42j2j6j1
010
∆ = 2 + j25, and ∆2 = –j20
I2 = ∆2/∆ = –j20/(2 + j25)
Vo = –j3I2 = –60/(2 + j25) = 2.392∠94.57°
Chapter 13, Solution 9. Consider the circuit below.
2 Ω -j1 2 Ω
For loop 1,
-j2V+ – j 4 j 4
8∠30o + – I2I1
8∠30° = (2 + j4)I1 – jI2 (1)
For loop 2, ((j4 + 2 – j)I2 – jI1 + (–j2) = 0 or I1 = (3 – j2)i2 – 2 (2) Substituting (2) into (1), 8∠30° + (2 + j4)2 = (14 + j7)I2
I2 = (10.928 + j12)/(14 + j7) = 1.037∠21.12°
Vx = 2I2 = 2.074∠21.12°
Chapter 13, Solution 10. Consider the circuit below.
Io
jωL jωL
1/jωC
I2I1
jωM
Iin∠0o
221 LLLkM == = L, I1 = Iin∠0°, I2 = Io
Io(jωL + R + 1/(jωC)) – jωLIin – (1/(jωC))Iin = 0 Io = j Iin(ωL – 1/(ωC)) /(R + jωL + 1/(jωC))
Chapter 13, Solution 11. Consider the circuit below.
I1
jωL1R1
R2
jωL2 1/jωC
jωM
V2
+ –
V1 + –
I3
I2
For mesh 1, V1 = I1(R1 + 1/(jωC)) – I2(1/jωC)) –R1I3 For mesh 2,
0 = –I1(1/(jωC)) + (jωL1 + jωL2 + (1/(jωC)) – j2ωM)I2 – jωL1I3 + jωMI3
For mesh 3, –V2 = –R1I1 – jω(L1 – M)I2 + (R1 + R2 + jωL1)I3
or V2 = R1I1 + jω(L1 – M)I2 – (R1 + R2 + jωL1)I3
Chapter 13, Solution 12.
Let .1=ω j4 j2 • + j6 j8 j10 1V - I1 I2
• Applying KVL to the loops,
21 481 IjIj += (1)
21 1840 IjIj += (2) Solving (1) and (2) gives I1 = -j0.1406. Thus
H 111.711
11
==→==jI
LjLI
Z eqeq
We can also use the equivalent T-section for the transform to find the equivalent inductance. Chapter 13, Solution 13. We replace the coupled inductance with an equivalent T-section and use series and parallel combinations to calculate Z. Assuming that ,1=ω
10,101020,81018 21 ===−=−==−=−= MLMLLMLL cba The equivalent circuit is shown below:
12Ω j8Ω j10Ω 2 Ω j10Ω -j6Ω Z j4Ω Z=12 +j8 + j14//(2 + j4) = 13.195 + j11.244Ω Chapter 13, Solution 14. To obtain VTh, convert the current source to a voltage source as shown below.
5 Ω -j3 Ω
+ – 8 V +
VTh –
2 Ω j8 Ω
a
b
j10 V + –
j6 Ω
I
j2 Note that the two coils are connected series aiding.
ωL = ωL1 + ωL2 – 2ωM
jωL = j6 + j8 – j4 = j10 Thus, –j10 + (5 + j10 – j3 + 2)I + 8 = 0
I = (– 8 + j10)/ (7 + j7) But, –j10 + (5 + j6)I – j2I + VTh = 0
VTh = j10 – (5 + j4)I = j10 – (5 + j4)(–8 + j10)/(7 + j7)
VTh = 5.349∠34.11° To obtain ZTh, we set all the sources to zero and insert a 1-A current source at the terminals a–b as shown below.
I2I1 1 A
5 Ω j6 Ω j8 Ω -j3 Ωa
j2
2 Ω
+ Vo –
b Clearly, we now have only a super mesh to analyze.
(5 + j6)I1 – j2I2 + (2 + j8 – j3)I2 – j2I1 = 0 (5 + j4)I1 + (2 + j3)I2 = 0 (1) But, I2 – I1 = 1 or I2 = I1 – 1 (2) Substituting (2) into (1), (5 + j4)I1 +(2 + j3)(1 + I1) = 0
I1 = –(2 + j3)/(7 + j7) Now, ((5 + j6)I1 – j2I1 + Vo = 0
Vo = –(5 + j4)I1 = (5 + j4)(2 + j3)/(7 + j7) = (–2 + j23)/(7 + j7) = 2.332∠50°
ZTh = Vo/1 = 2.332∠50° ohms
Chapter 13, Solution 15.
To obtain IN, short-circuit a–b as shown in Figure (a).
I1
I2 I1
20 Ω
j10 Ω
j20 Ω a
j5
1 + –
20 Ω
j10 Ω
j20 Ω a
j5
IN60∠30o
+ –
I2
b b (a) (b)For mesh 1,
60∠30° = (20 + j10)I1 + j5I2 – j10I2 or 12∠30° = (4 + j2)I1 – jI2 (1) For mesh 2,
0 = (j20 + j10)I2 – j5I1 – j10I1 or I1 = 2I2 (2) Substituting (2) into (1), 12∠30° = (8 + j3)I2
IN = I2 = 12∠30°/(8 + j3) = 1.404∠9.44° A To find ZN, we set all the sources to zero and insert a 1-volt voltage source at terminals a–b as shown in Figure (b). For mesh 1, 1 = I1(j10 + j20 – j5x2) + j5I2 1 = j20I1 + j5I2 (3) For mesh 2, 0 = (20 + j10)I2 + j5I1 – j10I1 = (4 + j2)I2 – jI1 or I2 = jI1/(4 + j2) (4) Substituting (4) into (3), 1 = j20I1 + j(j5)I1/(4 + j2) = (–1 + j20.5)I1
I1 = 1/(–1 + j20.5)
ZN = 1/I1 = (–1 + j20.5) ohms
Chapter 13, Solution 16. To find IN, we short-circuit a-b.
jΩ 8Ω -j2Ω a
••
+ j4Ω j6Ω I2 IN 80 V Io0∠ 1 - b
80)28(0)428(80 2121 =−+→=−+−+− jIIjjIIjj (1)
2112 606 IIjIIj =→=− (2) Solving (1) and (2) leads to
A 91.126246.1362.0584.11148
802
oN j
jII −∠=−=
+==
To find ZN, insert a 1-A current source at terminals a-b. Transforming the current source to voltage source gives the circuit below.
jΩ 8Ω -j2Ω 2Ω a
•• +
j4Ω j6Ω I2 2V I1 - b
28)28(0 2
121 jjIIjIIj+
=→−+= (3)
0)62(2 12 =−++ jIIj (4)
Solving (3) and (4) leads to I2 = -0.1055 +j0.2975, Vab=-j6I2 = 1.7853 +0.6332
Ω∠== oabN 53.19894.1
1V
Z
Chapter 13, Solution 17.
Z = -j6 // Zo where
7.15j5213.045j2j30j
14420jZo +=++−
+=
Ω−=+−
−= 7.9j1989.0
Z6jxZ6j
Zo
o
Chapter 13, Solution 18. Let 5105.0,20,5.1 2121 ====== xLLkMLLω We replace the transformer by its equivalent T-section.
5,25520,1055)( 11 −=−==+=+==+=−−= MLMLLMLL cba We find ZTh using the circuit below. -j4 j10 j25 j2 -j5 ZTh 4+j6
Ω+=++
+=++= 12.29215.274
)4(627)6//()4(27 jj
jjjjjjZTh
We find VTh by looking at the circuit below.
-j4 j10 j25 j2 + -j5 + VTh 120<0o
4+j6 - -
V 22.4637.61)120(64
4 oTh jj
jV −∠=+++
=
Chapter 13, Solution 19.
Let H 652540)(.1 1 =+=−−== MLLaω
25LH, 552530 C2 −=−==+=+= MMLLb
Thus, the T-section is as shown below. j65Ω j55Ω -j25Ω
Chapter 13, Solution 20. Transform the current source to a voltage source as shown below.
4 Ω
20∠0o + –
8 Ω
j12 + –
I3
j10 j10
-j5 I2I1
k=0.5
k = M/ 21LL or M = k 21LL
ωM = k 21 LL ωω = 0.5(10) = 5 For mesh 1, j12 = (4 + j10 – j5)I1 + j5I2 + j5I2 = (4 + j5)I1 + j10I2 (1) For mesh 2, 0 = 20 + (8 + j10 – j5)I2 + j5I1 + j5I1 –20 = +j10I1 + (8 + j5)I2 (2)
From (1) and (2),
++++
=
2
1
II
5j810j10j5j4
2012j
∆ = 107 + j60, ∆1 = –60 –j296, ∆2 = 40 – j100
I1 = ∆1/∆ = 2.462∠72.18° A
I2 = ∆2/∆ = 0.878∠–97.48° A
I3 = I1 – I2 = 3.329∠74.89° A
i1 = 2.462 cos(1000t + 72.18°) A
i2 = 0.878 cos(1000t – 97.48°) A
At t = 2 ms, 1000t = 2 rad = 114.6°
i1 = 0.9736cos(114.6° + 143.09°) = –2.445
i2 = 2.53cos(114.6° + 153.61°) = –0.8391 The total energy stored in the coupled coils is
w = 0.5L1i12 + 0.5L2i2
2 – Mi1i2 Since ωL1 = 10 and ω = 1000, L1 = L2 = 10 mH, M = 0.5L1 = 5mH
w = 0.5(10)(–2.445)2 + 0.5(10)(–0.8391)2 – 5(–2.445)(–0.8391)
w = 43.67 mJ Chapter 13, Solution 21. For mesh 1, 36∠30° = (7 + j6)I1 – (2 + j)I2 (1)
For mesh 2, 0 = (6 + j3 – j4)I2 – 2I1 jI1 = –(2 + j)I1 + (6 – j)I2 (2)
Placing (1) and (2) into matrix form,
−−−−−+
=
°∠
2
1
II
j6j2j26j7
03036
∆ = 48 + j35 = 59.41∠36.1°, ∆1 = (6 – j)36∠30° = 219∠20.54°
∆2 = (2 + j)36∠30° = 80.5∠56.56°, I1 = ∆1/∆ = 3.69∠–15.56°, I2 = ∆2/∆ = 1.355∠20.46°
Power absorbed fy the 4-ohm resistor, = 0.5(I2)24 = 2(1.355)2 = 3.672 watts
Chapter 13, Solution 22. With more complex mutually coupled circuits, it may be easier to show the effects of the coupling as sources in terms of currents that enter or leave the dot side of the coil. Figure 13.85 then becomes,
Ix
Ib
Ia
-j50
j30Ib
j60
j20Ia
j10Ia
j80
j30Ic
j40 j10Ib
+ − + − − +
− +
I3
50∠0° V
j20Ic
Io
I2I1
− +
− +
+ −
100 Ω Note the following,
Ia = I1 – I3 Ib = I2 – I1 Ic = I3 – I2
and Io = I3
Now all we need to do is to write the mesh equations and to solve for Io. Loop # 1,
-50 + j20(I3 – I2) j 40(I1 – I3) + j10(I2 – I1) – j30(I3 – I2) + j80(I1 – I2) – j10(I1 – I3) = 0
j100I1 – j60I2 – j40I3 = 50 Multiplying everything by (1/j10) yields 10I1 – 6I2 – 4I3 = - j5 (1)
Loop # 2, j10(I1 – I3) + j80(I2–I1) + j30(I3–I2) – j30(I2 – I1) + j60(I2 – I3) – j20(I1 – I3) + 100I2 = 0
-j60I1 + (100 + j80)I2 – j20I3 = 0 (2)
Loop # 3,
-j50I3 +j20(I1 –I3) +j60(I3 –I2) +j30(I2 –I1) –j10(I2 –I1) +j40(I3 –I1) –j20(I3 –I2) = 0 -j40I1 – j20I2 + j10I3 = 0 Multiplying by (1/j10) yields, -4I1 – 2I2 + I3 = 0 (3) Multiplying (2) by (1/j20) yields -3I1 + (4 – j5)I2 – I3 = 0 (4) Multiplying (3) by (1/4) yields -I1 – 0.5I2 – 0.25I3 = 0 (5) Multiplying (4) by (-1/3) yields I1 – ((4/3) – j(5/3))I2 + (1/3)I3 = -j0.5 (7) Multiplying [(6)+(5)] by 12 yields (-22 + j20)I2 + 7I3 = 0 (8) Multiplying [(5)+(7)] by 20 yields -22I2 – 3I3 = -j10 (9) (8) leads to I2 = -7I3/(-22 + j20) = 0.2355∠42.3o = (0.17418+j0.15849)I3 (10) (9) leads to I3 = (j10 – 22I2)/3, substituting (1) into this equation produces, I3 = j3.333 + (-1.2273 – j1.1623)I3 or I3 = Io = 1.3040∠63o amp.
Chapter 13, Solution 23. ω = 10
0.5 H converts to jωL1 = j5 ohms
1 H converts to jωL2 = j10 ohms
0.2 H converts to jωM = j2 ohms
25 mF converts to 1/(jωC) = 1/(10x25x10-3) = –j4 ohms
The frequency-domain equivalent circuit is shown below.
j10
I2I1
+ −
–j4
j5 j2
5 Ω 12∠0°
For mesh 1, 12 = (j5 – j4)I1 + j2I2 – (–j4)I2 –j2 = I1 + 6I2 (1) For mesh 2, 0 = (5 + j10)I2 + j2I1 –(–j4)I1 0 = (5 + j10)I2 + j6I1 (2) From (1), I1 = –j12 – 6I2 Substituting this into (2) produces,
I2 = 72/(–5 + j26) = 2.7194∠–100.89°
I1 = –j12 – 6 I2 = –j12 – 163.17∠–100.89 = 5.068∠52.54°
Hence, i1 = 5.068cos(10t + 52.54°) A, i2 = 2.719cos(10t – 100.89°) A.
At t = 15 ms, 10t = 10x15x10-3 0.15 rad = 8.59°
i1 = 5.068cos(61.13°) = 2.446
i2 = 2.719cos(–92.3°) = –0.1089
w = 0.5(5)(2.446)2 + 0.5(1)(–0.1089)2 – (0.2)(2.446)(–0.1089) = 15.02 J Chapter 13, Solution 24. (a) k = M/ 21LL = 1/ 2x4 = 0.3535 (b) ω = 4 1/4 F leads to 1/(jωC) = –j/(4x0.25) = –j 1||(–j) = –j/(1 – j) = 0.5(1 – j) 1 H produces jωM = j4 4 H produces j16 2 H becomes j8
j4
j8
j16
2 Ω
I2I1
+ −
0.5(1–j) 12∠0°
12 = (2 + j16)I1 + j4I2 or 6 = (1 + j8)I1 + j2I2 (1) 0 = (j8 + 0.5 – j0.5)I2 + j4I1 or I1 = (0.5 + j7.5)I2/(–j4) (2) Substituting (2) into (1),
24 = (–11.5 – j51.5)I2 or I2 = –24/(11.5 + j51.5) = –0.455∠–77.41°
Vo = I2(0.5)(1 – j) = 0.3217∠57.59°
vo = 321.7cos(4t + 57.6°) mV (c) From (2), I1 = (0.5 + j7.5)I2/(–j4) = 0.855∠–81.21°
i1 = 0.885cos(4t – 81.21°) A, i2 = –0.455cos(4t – 77.41°) A At t = 2s,
4t = 8 rad = 98.37°
i1 = 0.885cos(98.37° – 81.21°) = 0.8169
i2 = –0.455cos(98.37° – 77.41°) = –0.4249
w = 0.5L1i12 + 0.5L2i2
2 + Mi1i2
= 0.5(4)(0.8169)2 + 0.5(2)(–.4249)2 + (1)(0.1869)(–0.4249) = 1.168 J
Chapter 13, Solution 25. m = k 21LL = 0.5 H
We transform the circuit to frequency domain as shown below.
12sin2t converts to 12∠0°, ω = 2
0.5 F converts to 1/(jωC) = –j
2 H becomes jωL = j4 j1
–j1
b
a
10 Ω
2 ΩIo 4 Ω
j2j2
1 Ω
+ −
j4 12∠0° Applying the concept of reflected impedance,
Zab = (2 – j)||(1 + j2 + (1)2/(j2 + 3 + j4))
= (2 – j)||(1 + j2 + (3/45) – j6/45)
= (2 – j)||(1 + j2 + (3/45) – j6/45)
= (2 – j)||(1.0667 + j1.8667)
=(2 – j)(1.0667 + j1.8667)/(3.0667 + j0.8667) = 1.5085∠17.9° ohms
Io = 12∠0°/(Zab + 4) = 12/(5.4355 + j0.4636) = 2.2∠–4.88°
io = 2.2sin(2t – 4.88°) A
Chapter 13, Solution 26.
M = k 21LL ωM = k 21 LL ωω = 0.6 40x20 = 17 The frequency-domain equivalent circuit is shown below.
j17
Io–j30
I2I1
50 Ω
j40j20
+ −
10 Ω 200∠60°
For mesh 1, 200∠60° = (50 – j30 + j20)I1 + j17I2 = (50 – j10)I1 + j17I2 (1)
For mesh 2, 0 = (10 + j40)I2 + j17I1 (2)
In matrix form,
+
−=
°∠
2
1
II
40j1017j17j10j50
060200
∆ = 900 + j100, ∆1 = 2000∠60°(1 + j4) = 8246.2∠136°, ∆2 = 3400∠–30°
I2 = ∆2/∆ = 3.755∠–36.34° Io = I2 = 3.755∠–36.34° A Switching the dot on the winding on the right only reverses the direction of Io. This can be seen by looking at the resulting value of ∆2 which now becomes 3400∠150°. Thus, Io = 3.755∠143.66° A
Chapter 13, Solution 27.
Zin = –j4 + j5 + 9/(12 + j6) = 0.6 + j.07 = 0.922∠49.4° I1 = 12∠0°/0.922∠49.4° = 13∠–49.4° A
Chapter 13, Solution 28.
We find ZTh by replacing the 20-ohm load with a unit source as shown below.
j10Ω 8 -jX Ω
•• + j12Ω j15Ω I2 1V - I1 For mesh 1, 0 21 10)128( IjIjjX −+−= (1) For mesh 2, (2) jIIIjIj 1.05.1010151 2112 −=→=−+ Substituting (2) into (1) leads to
XjjXjI
5.18121.08.02.1
2 −+++−
=
XjXjj
IZTh 1.08.02.1
5.18121
2 −−−+
=−
=
6247275.108.0)1.02.1(
)5.18(1220|| 2
22
22
−+=→+−
−+== XX
X
XZTh
Solving the quadratic equation yields X = 6.425
Chapter 13, Solution 29.
30 mH becomes jωL = j30x10-3x103 = j30 50 mH becomes j50 Let X = ωM Using the concept of reflected impedance,
Zin = 10 + j30 + X2/(20 + j50) I1 = V/Zin = 165/(10 + j30 + X2/(20 + j50)) p = 0.5|I1|2(10) = 320 leads to |I1|2 = 64 or |I1| = 8 8 = |165(20 + j50)/(X2 + (10 + j30)(20 + j50))|
= |165(20 + j50)/(X2 – 1300 + j1100)|
or 64 = 27225(400 + 2500)/((X2 – 1300)2 + 1,210,000) (X2 – 1300)2 + 1,210,000 = 1,233,633 X = 33.86 or 38.13
If X = 38.127 = ωM
M = 38.127 mH k = M/ 21LL = 38.127/ 50x30 = 0.984
j38.127
I2I1
10 Ω
j50j30
+ −
20 Ω 165∠0°
165 = (10 + j30)I1 – j38.127I2 (1) 0 = (20 + j50)I2 – j38.127I1 (2)
In matrix form,
+−
−+=
2
1
II
50j20127.38j127.38j30j10
0165
∆ = 154 + j1100 = 1110.73∠82.03°, ∆1 = 888.5∠68.2°, ∆2 = j6291 I1 = ∆1/∆ = 8∠–13.81°, I2 = ∆2/∆ = 5.664∠7.97° i1 = 8cos(1000t – 13.83°), i2 = 5.664cos(1000t + 7.97°)
At t = 1.5 ms, 1000t = 1.5 rad = 85.94°
i1 = 8cos(85.94° – 13.83°) = 2.457 i2 = 5.664cos(85.94° + 7.97°) = –0.3862 w = 0.5L1i1
2 + 0.5L2i22 + Mi1i2
= 0.5(30)(2.547)2 + 0.5(50)(–0.3862)2 – 38.127(2.547)(–0.3862)
= 130.51 mJ
Chapter 13, Solution 30.
(a) Zin = j40 + 25 + j30 + (10)2/(8 + j20 – j6) = 25 + j70 + 100/(8 + j14) = (28.08 + j64.62) ohms (b) jωLa = j30 – j10 = j20, jωLb = j20 – j10 = j10, jωLc = j10 Thus the Thevenin Equivalent of the linear transformer is shown below.
j40 j20 j10 25 Ω 8 Ω
Zin
j10
–j6
Zin = j40 + 25 + j20 + j10||(8 + j4) = 25 + j60 + j10(8 + j4)/(8 + j14) = (28.08 + j64.62) ohms
Chapter 13, Solution 31.
(a) La = L1 – M = 10 H Lb = L2 – M = 15 H Lc = M = 5 H (b) L1L2 – M2 = 300 – 25 = 275
LA = (L1L2 – M2)/(L1 – M) = 275/15 = 18.33 H
LB = (L1L2 – M2)/(L1 – M) = 275/10 = 27.5 H
LC = (L1L2 – M2)/M = 275/5 = 55 H Chapter 13, Solution 32. We first find Zin for the second stage using the concept of reflected impedance.
Zin’
LBLb
R Zin’ = jωLb + ω2Mb
2/(R + jωLb) = (jωLbR - ω2Lb2 + ω2Mb
2)/(R + jωLb) (1) For the first stage, we have the circuit below.
Zin
LALa Zin’
Zin = jωLa + ω2Ma2/(jωLa + Zin)
= (–ω2La
2 + ω2Ma2 + jωLaZin)/( jωLa + Zin) (2)
Substituting (1) into (2) gives,
=
b
2b
22b
2b
a
b
2b
22b
2b
a2a
22a
2
LjRMLRLj
Lj
LjR)MLRLj(
LjML
ω+ω+ω−ω
+ω
ω+ω+ω−ω
ω+ω+ω−
–Rω2La
2 + ω2Ma2R – jω3LbLa + jω3LbMa
2 + jωLa(jωLbR – ω2Lb2 + ω2Mb
2) =
jωRLa –ω2LaLb + jωLbR – ω2La2 + ω2Mb
2
ω2R(La2 + LaLb – Ma
2) + jω3(La2Lb + LaLb
2 – LaMb2 – LbMa
2) Zin =
ω2(LaLb +Lb2 – Mb
2) – jωR(La +Lb) Chapter 13, Solution 33.
Zin = 10 + j12 + (15)2/(20 + j 40 – j5) = 10 + j12 + 225/(20 + j35)
= 10 + j12 + 225(20 – j35)/(400 + 1225)
= (12.769 + j7.154) ohms Chapter 13, Solution 34. Insert a 1-V voltage source at the input as shown below. j6Ω 1Ω 8Ω
•• + j12Ω j10Ω j4Ω 1<0o V I1 I2 - -j2Ω
For loop 1, 21 4)101(1 IjIj −+= (1)
For loop 2,
21112 )32(062)21048(0 IjjIIjIjIjjj ++−=→−+−++= (2) Solving (1) and (2) leads to I1=0.019 –j0.1068
Ω∠=+== ojI
Z 91.79219.9077.96154.11
1
Alternatively, an easier way to obtain Z is to replace the transformer with its equivalent T circuit and use series/parallel impedance combinations. This leads to exactly the same result. Chapter 13, Solution 35.
For mesh 1, 21 2)410(16 IjIj ++= (1) For mesh 2, 321 12)2630(20 IjIjIj −++= (2) For mesh 3, 32 )115(120 IjIj ++−= (3) We may use MATLAB to solve (1) to (3) and obtain
A 41.214754.15385.03736.11
ojI −∠=−= A 85.1340775.00549.00547.02
ojI −∠=−−= A 41.110077.00721.00268.03
ojI −∠=−−= Chapter 13, Solution 36.
Following the two rules in section 13.5, we obtain the following: (a) V2/V1 = –n, I2/I1 = –1/n (n = V2/V1) (b) V2/V1 = –n, I2/I1 = –1/n (c) V2/V1 = n, I2/I1 = 1/n (d) V2/V1 = n, I2/I1 = –1/n
Chapter 13, Solution 37.
(a) 5480
2400
1
2 ===VVn
(b) A 17.104480
000,50000,50 1222111 ==→==== IVISVIS
(c ) A 83.202400
000,502 ==I
Chapter 13, Solution 38.
Zin = Zp + ZL/n2, n = v2/v1 = 230/2300 = 0.1
v2 = 230 V, s2 = v2I2*
I2
* = s2/v2 = 17.391∠–53.13° or I2 = 17.391∠53.13° A
ZL = v2/I2 = 230∠0°/17.391∠53.13° = 13.235∠–53.13°
Zin = 2∠10° + 1323.5∠–53.13°
= 1.97 + j0.3473 + 794.1 – j1058.8
Zin = 1.324∠–53.05° kohms Chapter 13, Solution 39. Referred to the high-voltage side,
ZL = (1200/240)2(0.8∠10°) = 20∠10°
Zin = 60∠–30° + 20∠10° = 76.4122∠–20.31°
I1 = 1200/Zin = 1200/76.4122∠–20.31° = 15.7∠20.31° A
Since S = I1v1 = I2v2, I2 = I1v1/v2
= (1200/240)( 15.7∠20.31°) = 78.5∠20.31° A
Chapter 13, Solution 40.
V 60)240(41nVV
VV
n,41
2000500
NN
n 121
2
1
2 ===→====
W30012
60R
VP22===
Chapter 13, Solution 41. We reflect the 2-ohm resistor to the primary side.
Zin = 10 + 2/n2, n = –1/3
Since both I1 and I2 enter the dotted terminals, Zin = 10 + 18 = 28 ohms
I1 = 14∠0°/28 = 0.5 A and I2 = I1/n = 0.5/(–1/3) = –1.5 A Chapter 13, Solution 42. 10Ω 1:4 -j50Ω
+ •• + + + V1 V2 20 VΩ o 120<0o V I1 - - - - I2
Applying mesh analysis,
11 VI10120 += (1)
22 VI)50j20(0 +−= (2)
At the terminals of the transformer,
121
2 V4V4nVV
=→== (3)
211
2 I4I41
n1
II
−=→−=−= (4)
Substituting (3) and (4) into (1) gives 120 22 V25.0I40 +−= (5)
Solving (2) and (5) yields 6877.0j4756.2I2 −−=
V 52.1539.51I20V o
2o ∠=−= Chapter 13, Solution 43. Transform the two current sources to voltage sources, as shown below.
+
v1
−
+
v2
−
1 : 4
12V+ –
10 Ω
20 V + – I2I1
12 Ω Using mesh analysis, –20 + 10I1 + v1 = 0 20 = v1 + 10I1 (1) 12 + 12I2 – v2 = 0 or 12 = v2 – 12I2 (2) At the transformer terminal, v2 = nv1 = 4v1 (3) I1 = nI2 = 4I2 (4) Substituting (3) and (4) into (1) and (2), we get, 20 = v1 + 40I2 (5) 12 = 4v1 – 12I2 (6) Solving (5) and (6) gives v1 = 4.186 V and v2 = 4v = 16.744 V
Chapter 13, Solution 44.
We can apply the superposition theorem. Let i1 = i1’ + i1” and i2 = i2’ + i2” where the single prime is due to the DC source and the double prime is due to the AC source. Since we are looking for the steady-state values of i1 and i2,
i1’ = i2’ = 0.
For the AC source, consider the circuit below.
+–i2”
+
v1
−
1 : n
+
v2
−
R
i1”
Vn∠0°
v2/v1 = –n, I2”/I1” = –1/n
But v2 = vm, v1 = –vm/n or I1” = vm/(Rn)
I2” = –I1”/n = –vm/(Rn2)
Hence, i1(t) = (vm/Rn)cosωt A, and i2(t) = (–vm/(n2R))cosωt A Chapter 13, Solution 45. 48 Ω
+ −
Z4∠–90˚
4j8Cj8ZL −=
ω−= , n = 1/3
°−∠=°−∠
°−∠=
−+°−∠
=
−===
3.7303193.07.1628.125
90436j7248
904I
36j72Z9n
ZZ L2L
We now have some choices, we can go ahead and calculate the current in the second loop and calculate the power delivered to the 8-ohm resistor directly or we can merely say that the power delivered to the equivalent resistor in the primary side must be the same as the power delivered to the 8-ohm resistor. Therefore,
=== −Ω 7210x5098.072
2IP 3
28 36.71 mW
The student is encouraged to calculate the current in the secondary and calculate the power delivered to the 8-ohm resistor to verify that the above is correct. Chapter 13, Solution 46. (a) Reflecting the secondary circuit to the primary, we have the circuit shown below.
Zin
+ −
+ − 16∠60° I1
10∠30°/(–n) = –5∠30°
Zin = 10 + j16 + (1/4)(12 – j8) = 13 + j14 –16∠60° + ZinI1 – 5∠30° = 0 or I1 = (16∠60° + 5∠30°)/(13 + j14) Hence, I1 = 1.072∠5.88° A, and I2 = –0.5I1 = 0.536∠185.88° A
(b) Switching a dot will not effect Zin but will effect I1 and I2.
I1 = (16∠60° – 5∠30°)/(13 + j14) = 0.625 ∠25 A and I2 = 0.5I1 = 0.3125∠25° A
Chapter 13, Solution 47.
0.02 F becomes 1/(jωC) = 1/(j5x0.02) = –j10 We apply mesh analysis to the circuit shown below.
–j10
+
vo
−
I3
+
v1
−
+
v2
−
3 : 1 10 Ω
10∠0° + – I2I1
2 Ω For mesh 1, 10 = 10I1 – 10I3 + v1 (1) For mesh 2, v2 = 2I2 = vo (2) For mesh 3, 0 = (10 – j10)I3 – 10I1 + v2 – v1 (3) At the terminals, v2 = nv1 = v1/3 (4) I1 = nI2 = I2/3 (5) From (2) and (4), v1 = 6I2 (6) Substituting this into (1), 10 = 10I1 – 10I3 (7) Substituting (4) and (6) into (3) yields 0 = –10I1 – 4I2 + 10(1 – j)I3 (8) From (5), (7), and (8)
=
−−−−
−
0100
III
10j10410106100333.01
3
2
1
I2 = 33.93j20
100j1002
−−−
=∆∆
= 1.482∠32.9°
vo = 2I2 = 2.963∠32.9° V (a) Switching the dot on the secondary side effects only equations (4) and (5). v2 = –v1/3 (9) I1 = –I2/3 (10) From (2) and (9), v1 = –6I2 Substituting this into (1),
10 = 10I1 – 10I3 – 6I2 = (23 – j5)I1 (11) Substituting (9) and (10) into (3),
0 = –10I1 + 4I2 + 10(1 – j)I3 (12)
From (10) to (12), we get
=
−−−−
0100
III
10j10410106100333.01
3
2
1
I2 = 33.93j20
100j1002
+−−
=∆∆
= 1.482∠–147.1°
vo = 2I2 = 2.963∠–147.1° V Chapter 13, Solution 48. We apply mesh analysis. 8Ω 2:1 10Ω + + •• + V1 V2 I1 - j6Ω 100∠0o V - I2 - Ix -j4Ω
121 4)48(100 VIjIj +−−= (1)
212 4)210(0 VIjIj +−+= (2)
But
211
2 221 VVn
VV
=→== (3)
211
2 5.021 IInI
I−=→−=−= (4)
Substituting (3) and (4) into (1) and (2), we obtain
22 2)24(100 VIj +−−= (1)a
22)410(0 VIj ++= (2)a
Solving (1)a and (2)a leads to I2 = -3.5503 +j1.4793
A 4.157923.15.0 221o
x IIII ∠==+= Chapter 13, Solution 49.
101F 201,2 j
Cj−=→=
ωω
Ix -j10 2 Ω I1 1:3 I2 1 2 + • + + V1 V2 6Ω 12<0o V - - - •
At node 1,
2111211 2.0)2.01(212
10212 VjjVII
jVVV
−++=→+−−
=−
(1)
At node 2,
212221
2 )6.01(6.060610
VjVjIVjVVI +−+=→=
−−
+ (2)
At the terminals of the transformer, 1212 31,3 IIV −=−=V
Substituting these in (1) and (2),
)4.23(60),8.01(612 1212 jVIjVI ++=++−= Adding these gives V1=1.829 –j1.463 and
ox j
VjVVI 34.51937.0
104
10121 ∠=
−=
−−
=
A )34.512cos(937.0 o
x ti += Chapter 13, Solution 50.
The value of Zin is not effected by the location of the dots since n2 is involved.
Zin’ = (6 – j10)/(n’)2, n’ = 1/4 Zin’ = 16(6 – j10) = 96 – j160 Zin = 8 + j12 + (Zin’ + 24)/n2, n = 5 Zin = 8 + j12 + (120 – j160)/25 = 8 + j12 + 4.8 – j6.4 Zin = (12.8 + j5.6) ohms
Chapter 13, Solution 51. Let Z3 = 36 +j18, where Z3 is reflected to the middle circuit.
ZR’ = ZL/n2 = (12 + j2)/4 = 3 + j0.5 Zin = 5 – j2 + ZR’ = (8 – j1.5) ohms I1 = 24∠0°/ZTh = 24∠0°/(8 – j1.5) = 24∠0°/8.14∠–10.62° = 8.95∠10.62° A
Chapter 13, Solution 52. For maximum power transfer,
40 = ZL/n2 = 10/n2 or n2 = 10/40 which yields n = 1/2 = 0.5 I = 120/(40 + 40) = 3/2 p = I2R = (9/4)x40 = 90 watts.
Chapter 13, Solution 53.
(a) The Thevenin equivalent to the left of the transformer is shown below. 8 Ω
20 V
+ −
The reflected load impedance is ZL’ = ZL/n2 = 200/n2. For maximum power transfer, 8 = 200/n2 produces n = 5.
(b) If n = 10, ZL’ = 200/10 = 2 and I = 20/(8 + 2) = 2
p = I2ZL’ = (2)2(2) = 8 watts. Chapter 13, Solution 54. (a)
I1
+ −
ZTh
ZL/n2 VS
For maximum power transfer,
ZTh = ZL/n2, or n2 = ZL/ZTh = 8/128
n = 0.25 (b) I1 = VTh/(ZTh + ZL/n2) = 10/(128 + 128) = 39.06 mA (c) v2 = I2ZL = 156.24x8 mV = 1.25 V But v2 = nv1 therefore v1 = v2/n = 4(1.25) = 5 V Chapter 13, Solution 55. We reflect Zs to the primary side.
ZR = (500 – j200)/n2 = 5 – j2, Zin = Zp + ZR = 3 + j4 + 5 – j2 = 8 + j2
I1 = 120∠0°/(8 + j2) = 14.552∠–14.04°
I2I1
+ −
1 : n Zp Vp Zs
Since both currents enter the dotted terminals as shown above,
I2 = –(1/n)I1 = –1.4552∠–14.04° = 1.4552∠166° S2 = |I2|2Zs = (1.4552)(500 – j200) P2 = Re(S2) = (1.4552)2(500) = 1054 watts
Chapter 13, Solution 56. We apply mesh analysis to the circuit as shown below.
+
v1
−
5 Ω
+
v2
− I2I1
+ −
1 : 2 2 Ω 10 Ω 46V For mesh 1, 46 = 7I1 – 5I2 + v1 (1) For mesh 2, v2 = 15I2 – 5I1 (2) At the terminals of the transformer, v2 = nv1 = 2v1 (3) I1 = nI2 = 2I2 (4) Substituting (3) and (4) into (1) and (2), 46 = 9I2 + v1 (5) v1 = 2.5I2 (6) Combining (5) and (6), 46 = 11.5I2 or I2 = 4 P10 = 0.5I2
2(10) = 80 watts.
Chapter 13, Solution 57.
(a) ZL = j3||(12 – j6) = j3(12 – j6)/(12 – j3) = (12 + j54)/17
Reflecting this to the primary side gives Zin = 2 + ZL/n2 = 2 + (3 + j13.5)/17 = 2.3168∠20.04° I1 = vs/Zin = 60∠90°/2.3168∠20.04° = 25.9∠69.96° A(rms) I2 = I1/n = 12.95∠69.96° A(rms)
(b) 60∠90° = 2I1 + v1 or v1 = j60 –2I1 = j60 – 51.8∠69.96°
v1 = 21.06∠147.44° V(rms) v2 = nv1 = 42.12∠147.44° V(rms) vo = v2 = 42.12∠147.44° V(rms)
(c) S = vsI1
* = (60∠90°)(25.9∠–69.96°) = 1554∠20.04° VA Chapter 13, Solution 58. Consider the circuit below. 20 Ω
+
vo
−
I3
+
v1
−
+
v2
−
1 : 5 20 Ω
80∠0° + – I2I1
100 Ω
For mesh1, 80 = 20I1 – 20I3 + v1 (1) For mesh 2, v2 = 100I2 (2)
For mesh 3, 0 = 40I3 – 20I1 which leads to I1 = 2I3 (3) At the transformer terminals, v2 = –nv1 = –5v1 (4) I1 = –nI2 = –5I2 (5) From (2) and (4), –5v1 = 100I2 or v1 = –20I2 (6) Substituting (3), (5), and (6) into (1),
4 = I1 – I2 – I 3 = I1 – (I1/(–5)) – I1/2 = (7/10)I1 I1 = 40/7, I2 = –8/7, I3 = 20/7
p20(the one between 1 and 3) = 0.5(20)(I1 – I3)2 = 10(20/7)2 = 81.63 watts p20(at the top of the circuit) = 0.5(20)I3
2 = 81.63 watts p100 = 0.5(100)I2
2 = 65.31 watts Chapter 13, Solution 59. We apply nodal analysis to the circuit below. 2 Ω
I3
v2 v1 I1
4 Ω
+
v1
−
+
v2
−
2 : 1 8 Ω
20∠0° + –
I2
20 = 8I1 + V1 (1) V1 = 2I3 + V2 (2)
V2 = 4I2 (3) At the transformer terminals, v2 = 0.5v1 (4) I1 = 0.5I2 (5) Solving (1) to (5) gives I1 = 0.833 A, I2 = 1.667 A, I3 = 3.333 A V1 = 13.33 V, V2 = 6.667 V.
P8Ω = 0.5(8)|(20 – V1)/8|2 = 2.778 W P2Ω = 0.5(2)I3
2 = 11.11 W, P4Ω = 0.5V22/4 = 5.556 W
Chapter 13, Solution 60. (a) Transferring the 40-ohm load to the middle circuit,
ZL’ = 40/(n’)2 = 10 ohms where n’ = 2 10||(5 + 10) = 6 ohms We transfer this to the primary side. Zin = 4 + 6/n2 = 4 + 96 = 100 ohms, where n = 0.25 I1 = 120/100 = 1.2 A and I2 = I1/n = 4.8 A
I2’
120∠0° + –
5 Ω
10 Ω
I1
+
v1
−
+
v2
−
1 : 4 4 Ω I2 10 Ω
Using current division, I2’ = (10/25)I2 = 1.92 and I3 = I2’/n’ = 0.96 A (b) p = 0.5(I3)2(40) = 18.432 watts
Chapter 13, Solution 61. We reflect the 160-ohm load to the middle circuit.
ZR = ZL/n2 = 160/(4/3)2 = 90 ohms, where n = 4/3
Io’
24∠0° + –
14 Ω
60 Ω
I1
+
v1
−
+
vo
−
1 : 5 2 Ω Io
90 Ω
14 + 60||90 = 14 + 36 = 50 ohms
We reflect this to the primary side.
ZR’ = ZL’/(n’)2 = 50/52 = 2 ohms when n’ = 5 I1 = 24/(2 + 2) = 6A 24 = 2I1 + v1 or v1 = 24 – 2I1 = 12 V vo = –nv1 = –60 V, Io = –I1 /n1 = –6/5 = –1.2 Io‘ = [60/(60 + 90)]Io = –0.48A I2 = –Io’/n = 0.48/(4/3) = 0.36 A
Chapter 13, Solution 62. (a) Reflect the load to the middle circuit.
ZL’ = 8 – j20 + (18 + j45)/32 = 10 – j15 We now reflect this to the primary circuit so that Zin = 6 + j4 + (10 – j15)/n2 = 7.6 + j1.6 = 7.767∠11.89°, where n = 5/2 = 2.5 I1 = 40/Zin = 40/7.767∠11.89° = 5.15∠–11.89° S = 0.5vsI1
* = (20∠0°)(5.15∠11.89°) = 103∠11.89° VA
(b) I2 = –I1/n, n = 2.5
I3 = –I2/n’, n = 3
I3 = I1/(nn’) = 5.15∠–11.89°/(2.5x3) = 0.6867∠–11.89°
p = 0.5|I2|2(18) = 9(0.6867)2 = 4.244 watts Chapter 13, Solution 63. Reflecting the (9 + j18)-ohm load to the middle circuit gives,
Zin’ = 7 – j6 + (9 + j18)/(n’)2 = 7 – j6 + 1 + j12 = 8 + j4 when n’ = 3 Reflecting this to the primary side, Zin = 1 + Zin’/n2 = 1 + 2 – j = 3 – j, where n = 2 I1 = 12∠0°/(3 – j) = 12/3.162∠–18.43° = 3.795∠18.43A I2 = I1/n = 1.8975∠18.43° A I3 = –I2/n2 = 632.5∠161.57° mA
Chapter 13, Solution 64. We find ZTh at the terminals of Z by considering the circuit below. 10Ω 1:n + •• + V1 V2 + – – I1 I2 1 V – 20Ω
For mesh 1, 02030 121 =++ VII (1) For mesh 2, 12020 221 =++ VII (2)
At the terminals, nIInVV 1
212 , −==
Substituting these in (1) and (2) leads to
1)1(20,0)3020( 1212 =+−=+− nVInVIn Solving these gives
5.72040301204030
1 2
222 =+−==→
+−= nn
IZ
nnI Th
Solving the quadratic equation yields n=0.5 or 0.8333 Chapter 13, Solution 65.
40Ω 10Ω I1 1:2 I2 50Ω I2 1:3 I3 1 • + •• 2 + + + + 200 V V3 V4 (rms) V1 V2 - - 20 Ω- - - • At node 1,
1411411 1025.025.1200
4010200 IVVIVVV
+−=→+−
=− (1)
At node 2,
3413441 403
2040IVVIVVV
+=→+=− (2)
At the terminals of the first transformer,
121
2 22 VVVV
−=→−= (3)
211
2 22/1 IIII
−=→−= (4)
For the middle loop,
223322 50050 IVVVIV −=→=++− (5) At the terminals of the second transformer,
343
4 33 VVVV
=→= (6)
322
3 33/1 IIII
−=→−= (7)
We have seven equations and seven unknowns. Combining (1) and (2) leads to
314 50105.3200 IIV ++= But from (4) and (7), 3321 6)3(22 IIII =−−=−= . Hence
34 1105.3200 IV += (8) From (5), (6), (3), and (7),
3122224 45061503)50(3 IVIVIVV +−=−=−= Substituting for V1 in (2) gives
433344 21019450)403(6 VIIIVV =→++−= (9)
Substituting (9) into (8) yields
87.14452.13200 44 =→= VV
W05.1120
42
==VP
Chapter 13, Solution 66.
v1 = 420 V (1) v2 = 120I2 (2) v1/v2 = 1/4 or v2 = 4v1 (3) I1/I2 = 4 or I1 = 4 I2 (4)
Combining (2) and (4),
v2 = 120[(1/4)I1] = 30 I1 4v1 = 30I1 4(420) = 1680 = 30I1 or I1 = 56 A
Chapter 13, Solution 67.
(a) V 1604004.04.04.0
112
2
21
2
1 ===→=+
= xVVN
NNVV
(b) A 25.311605000000,5 2222 ==→== IVIS
(c ) A 5.12400
5000000,5 21112 ==→=== IVISS
Chapter 13, Solution 68. This is a step-up transformer.
I1
I2
N12 – j6
+
v1
−
+
v2
−
N2 + −
10 + j40 20∠30°
For the primary circuit, 20∠30° = (2 – j6)I1 + v1 (1) For the secondary circuit, v2 = (10 + j40)I2 (2)
At the autotransformer terminals,
v1/v2 = N1/(N1 + N2) = 200/280 = 5/7,
thus v2 = 7v1/5 (3) Also, I1/I2 = 7/5 or I2 = 5I1/7 (4)
Substituting (3) and (4) into (2), v1 = (10 + j40)25I1/49 Substituting that into (1) gives 20∠30° = (7.102 + j14.408)I1 I1 = 20∠30°/16.063∠63.76° = 1.245∠–33.76° A I2 = 5I1/7 = 0.8893∠–33.76° A Io = I1 – I2 = [(5/7) – 1]I1 = –2I1/7 = 0.3557∠146.2° A p = |I2|2R = (0.8893)2(10) = 7.51 watts
Chapter 13, Solution 69. We can find the Thevenin equivalent.
I1
I2
75 Ω
+
VTh
−
N2
j125 Ω
+
v1
−
+
v2
−
N1 + − 120∠0°
I1 = I2 = 0
As a step up transformer, v1/v2 = N1/(N1 + N2) = 600/800 = 3/4
v2 = 4v1/3 = 4(120)/3 = 160∠0° rms = VTh. To find ZTh, connect a 1-V source at the secondary terminals. We now have a step-down transformer.
I2
I1
75 Ω j125 Ω
+
v2
−
+
v1
−
+ − 1∠0° V
v1 = 1V, v2 =I2(75 + j125)
But v1/v2 = (N1 + N2)/N1 = 800/200 which leads to v1 = 4v2 = 1
and v2 = 0.25
I1/I2 = 200/800 = 1/4 which leads to I2 = 4I1
Hence 0.25 = 4I1(75 + j125) or I1 = 1/[16(75 + j125)
ZTh = 1/I1 = 16(75 + j125) Therefore, ZL = ZTh
* = (1.2 – j2) kΩ Since VTh is rms, p = (|VTh|/2)2/RL = (80)2/1200 = 5.333 watts
Chapter 13, Solution 70. This is a step-down transformer.
I1
I2
+
v2
−
+
v1
−
+ −
30 + j12 120∠0° 20 – j40
I1/I2 = N2/(N1 + N2) = 200/1200 = 1/6, or I1 = I2/6 (1) v1/v2 = (N2 + N2)/N2 = 6, or v1 = 6v2 (2)
For the primary loop, 120 = (30 + j12)I1 + v1 (3) For the secondary loop, v2 = (20 – j40)I2 (4)
Substituting (1) and (2) into (3),
120 = (30 + j12)( I2/6) + 6v2 and substituting (4) into this yields
120 = (49 – j38)I2 or I2 = 1.935∠37.79° p = |I2|2(20) = 74.9 watts.
Chapter 13, Solution 71.
Zin = V1/I1 But V1I1 = V2I2, or V2 = I2ZL and I1/I2 = N2/(N1 + N2) Hence V1 = V2I2/I1 = ZL(I2/I1)I2 = ZL(I2/I1)2I1
V1/I1 = ZL[(N1 + N2)/N2] 2 Zin = [1 + (N1/N2)] 2ZL
Chapter 13, Solution 72.
(a) Consider just one phase at a time.
1:n A
B 20MVA
Load
C
a b c
n = VL/ )312470/(7200V3 Lp = = 1/3
(b) The load carried by each transformer is 60/3 = 20 MVA.
Hence ILp = 20 MVA/12.47 k = 1604 A ILs = 20 MVA/7.2 k = 2778 A
(c) The current in incoming line a, b, c is
85.1603x3I3 Lp = = 2778 A
Current in each outgoing line A, B, C is 2778/(n 3 ) = 4812 A
Chapter 13, Solution 73.
(a) This is a three-phase ∆-Y transformer. (b) VLs = nvLp/ 3 = 450/(3 3 ) = 86.6 V, where n = 1/3
As a Y-Y system, we can use per phase equivalent circuit. Ia = Van/ZY = 86.6∠0°/(8 – j6) = 8.66∠36.87° Ic = Ia∠120° = 8.66∠156.87° A ILp = n 3 ILs I1 = (1/3) 3 (8.66∠36.87°) = 5∠36.87° I2 = I1∠–120° = 5∠–83.13° A
(c) p = 3|Ia|2(8) = 3(8.66)2(8) = 1.8 kw.
Chapter 13, Solution 74.
(a) This is a ∆-∆ connection. (b) The easy way is to consider just one phase.
1:n = 4:1 or n = 1/4
n = V2/V1 which leads to V2 = nV1 = 0.25(2400) = 600
i.e. VLp = 2400 V and VLs = 600 V
S = p/cosθ = 120/0.8 kVA = 150 kVA
pL = p/3 = 120/3 = 40 kw 4:1
Ipp
IL
Ips
ILs VLp VLs
But pLs = VpsIps
For the ∆-load, IL = 3 Ip and VL = Vp Hence, Ips = 40,000/600 = 66.67 A
ILs = 3 Ips = 3 x66.67 = 115.48 A
(c) Similarly, for the primary side
ppp = VppIpp = pps or Ipp = 40,000/2400 = 16.667 A
and ILp = 3 Ip = 28.87 A (d) Since S = 150 kVA therefore Sp = S/3 = 50 kVA
Chapter 13, Solution 75.
(a) n = VLs/( 3 VLp) 4500/(900 3 ) = 2.887 (b) S = 3 VLsILs or ILs = 120,000/(900 3 ) = 76.98 A ILs = ILp/(n 3 ) = 76.98/(2.887 3 ) = 15.395 A
Chapter 13, Solution 76. (a) At the load, VL = 240 V = VAB
VAN = VL/ 3 = 138.56 V
Since S = 3 VLIL then IL = 60,000/(240 3 ) = 144.34 A
1:n 0.05 Ω j0.1 Ω
0.05 Ω j0.1 Ω
0.05 Ω
A
B
Balanced
Load 60kVA 0.85pf leading C
240V j0.1 Ω2640V
(b) Let VAN = |VAN|∠0° = 138.56∠0° cosθ = pf = 0.85 or θ = 31.79° IAA’ = IL∠θ = 144.34∠31.79° VA’N’ = ZIAA’ + VAN = 138.56∠0° + (0.05 + j0.1)(144.34∠31.79°) = 138.03∠6.69°
VLs = VA’N’ 3 = 137.8 3 = 238.7 V (c) For Y-∆ connections,
n = 3 VLs/Vps = 3 x238.7/2640 = 0.1569
fLp = nILs/ 3 = 0.1569x144.34/ 3 = 13.05 A Chapter 13, Solution 77.
(a) This is a single phase transformer. V1 = 13.2 kV, V2 = 120 V
n = V2/V1 = 120/13,200 = 1/110, therefore n = 110 (b) P = VI or I = P/V = 100/120 = 0.8333 A
I1 = nI2 = 0.8333/110 = 7.576 mA
Chapter 13, Solution 78. The schematic is shown below.
k = 21LL/M = 3x6/1 = 0.2357 In the AC Sweep box, set Total Pts = 1, Start Freq = 0.1592 and End Freq = 0.1592. After simulation, the output file includes FREQ IM(V_PRINT1) IP(V_PRINT1) 1.592 E–01 4.253 E+00 –8.526 E+00 FREQ IM(V_PRINT2) IP(V_PRINT2) 1.592 E–01 1.564 E+00 2.749 E+01 From this, I1 = 4.253∠–8.53° A, I2 = 1.564∠27.49° A The power absorbed by the 4-ohm resistor = 0.5|I|2R = 0.5(1.564)2x4
= 4.892 watts
Chapter 13, Solution 79. The schematic is shown below.
k1 = 5000/15 = 0.2121, k2 = 8000/10 = 0.1118 In the AC Sweep box, we type Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592. After the circuit is saved and simulated, the output includes FREQ IM(V_PRINT1) IP(V_PRINT1) 1.592 E–01 4.068 E–01 –7.786 E+01 FREQ IM(V_PRINT2) IP(V_PRINT2) 1.592 E–01 1.306 E+00 –6.801 E+01 FREQ IM(V_PRINT3) IP(V_PRINT3) 1.592 E–01 1.336 E+00 –5.492 E+01 Thus, I1 = 1.306∠–68.01° A, I2 = 406.8∠–77.86° mA, I3 = 1.336∠–54.92° A
Chapter 13, Solution 80. The schematic is shown below.
k1 = 80x40/10 = 0.1768, k2 = 60x40/20 = 0.482
k3 = 60x80/30 = 0.433 In the AC Sweep box, we set Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592. After the simulation, we obtain the output file which includes FREQ IM(V_PRINT1) IP(V_PRINT1) 1.592 E–01 1.304 E+00 6.292 E+01 i.e. Io = 1.304∠62.92° A
Chapter 13, Solution 81. The schematic is shown below.
k1 = 8x4/2 = 0.3535, k2 = 8x2/1 = 0.25 In the AC Sweep box, we let Total Pts = 1, Start Freq = 100, and End Freq = 100. After simulation, the output file includes FREQ IM(V_PRINT1) IP(V_PRINT1) 1.000 E+02 1.0448 E–01 1.396 E+01 FREQ IM(V_PRINT2) IP(V_PRINT2) 1.000 E+02 2.954 E–02 –1.438 E+02 FREQ IM(V_PRINT3) IP(V_PRINT3) 1.000 E+02 2.088 E–01 2.440 E+01
i.e. I1 = 104.5∠13.96° mA, I2 = 29.54∠–143.8° mA,
I3 = 208.8∠24.4° mA.
Chapter 13, Solution 82. The schematic is shown below. In the AC Sweep box, we type Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592. After simulation, we obtain the output file which includes FREQ IM(V_PRINT1) IP(V_PRINT1) 1.592 E–01 1.955 E+01 8.332 E+01 FREQ IM(V_PRINT2) IP(V_PRINT2) 1.592 E–01 6.847 E+01 4.640 E+01 FREQ IM(V_PRINT3) IP(V_PRINT3) 1.592 E–01 4.434 E–01 –9.260 E+01
i.e. V1 = 19.55∠83.32° V, V2 = 68.47∠46.4° V,
Io = 443.4∠–92.6° mA.
hapter 13, Solution 83.
he schematic is shown below. In the AC Sweep box, we set Total Pts = 1, Start Freq
FREQ IM(V_PRINT1) IP(V_PRINT1)
FREQ VM($N_0001) VP($N_0001)
i.e. iX = 1.08∠33.91° A
C T= 0.1592, and End Freq = 0.1592. After simulation, the output file includes 1.592 E–01 1.080 E+00 3.391 E+01 1.592 E–01 1.514 E+01 –3.421 E+01
, Vx = 15.14∠–34.21° V.
Chapter 13, Solution 84.
he schematic is shown below. We set Total Pts = 1, Start Freq = 0.1592, and End
FREQ IM(V_PRINT1) IP(V_PRINT1)
FREQ IM(V_PRINT2) IP(V_PRINT2)
FREQ IM(V_PRINT3) IP(V_PRINT3)
i.e. I1 = 4.028∠–52.38° A
TFreq = 0.1592. After simulation, the output file includes 1.592 E–01 4.028 E+00 –5.238 E+01 1.592 E–01 2.019 E+00 –5.211 E+01 1.592 E–01 1.338 E+00 –5.220 E+01
, I2 = 2.019∠–52.11° A,
I3 = 1.338∠–52.2° A.
Chapter 13, Solution 85.
For maximum power transfer,
Z1 = ZL/n2 or n2 = ZL/Z1 = 8/7200 = 1/900
n = 1/30 = N2/N1. Thus N2 = N1/30 = 3000/30 = 100 turns.
Z1
+ − ZL/n2
VS
Chapter 13, Solution 86.
n = N2/N1 = 48/2400 = 1/50 ZTh = ZL/n2 = 3/(1/50)2 = 7.5 kΩ
Chapter 13, Solution 87. ZTh = ZL/n2 or n = 300/75Z/Z ThL = = 0.5 Chapter 13, Solution 88. n = V2/V1 = I1/I2 or I2 = I1/n = 2.5/0.1 = 25 A
p = IV = 25x12.6 = 315 watts Chapter 13, Solution 89. n = V2/V1 = 120/240 = 0.5
S = I1V1 or I1 = S/V1 = 10x103/240 = 41.67 A S = I2V2 or I2 = S/V2 = 104/120 = 83.33 A
Chapter 13, Solution 90.
(a) n = V2/V1 = 240/2400 = 0.1 (b) n = N2/N1 or N2 = nN1 = 0.1(250) = 25 turns (c) S = I1V1 or I1 = S/V1 = 4x103/2400 = 1.6667 A S = I2V2 or I2 = S/V2 = 4x104/240 = 16.667 A
Chapter 13, Solution 91.
(a) The kVA rating is S = VI = 25,000x75 = 1875 kVA (b) Since S1 = S2 = V2I2 and I2 = 1875x103/240 = 7812 A
Chapter 13, Solution 92.
(a) V2/V1 = N2/N1 = n, V2 = (N2/N1)V1 = (28/1200)4800 = 112 V (b) I2 = V2/R = 112/10 = 11.2 A and I1 = nI2, n = 28/1200
I1 = (28/1200)11.2 = 261.3 mA (c) p = |I2|2R = (11.2)2(10) = 1254 watts.
Chapter 13, Solution 93. (a) For an input of 110 V, the primary winding must be connected in parallel, with series-aiding on the secondary. The coils must be series-opposing to give 12 V. Thus the connections are shown below.
12 V 110 V
(b) To get 220 V on the primary side, the coils are connected in series, with series-aiding on the secondary side. The coils must be connected series-aiding to give 50 V. Thus, the connections are shown below.
50 V
220 V
Chapter 13, Solution 94. V2/V1 = 110/440 = 1/4 = I1/I2 There are four ways of hooking up the transformer as an auto-transformer. However it is clear that there are only two outcomes.
V2
V1
V2
V1
V2
V1
V2
V1
(1) (2) (3) (4) (1) and (2) produce the same results and (3) and (4) also produce the same results. Therefore, we will only consider Figure (1) and (3). (a) For Figure (3), V1/V2 = 550/V2 = (440 – 110)/440 = 330/440 Thus, V2 = 550x440/330 = 733.4 V (not the desired result) (b) For Figure (1), V1/V2 = 550/V2 = (440 + 110)/440 = 550/440 Thus, V2 = 550x440/550 = 440 V (the desired result) Chapter 13, Solution 95.
(a) n = Vs/Vp = 120/7200 = 1/60 (b) Is = 10x120/144 = 1200/144
S = VpIp = VsIs
Ip = VsIs/Vp = (1/60)x1200/144 = 139 mA
Chapter 14, Solution 1.
RCj1RCj
Cj1RR
)(i
o
ω+ω
=ω+
==ωVV
H
=ω)(H0
0
j1j
ωω+ωω
, where RC1
0 =ω
20
0
)(1)(H
ωω+
ωω=ω= H
ωω
−π
=ω∠=φ0
1-tan2
)(H
This is a highpass filter. The frequency response is the same as that for P.P.14.1 except that RC10 =ω . Thus, the sketches of H and φ are shown below.
H
ωω0 = 1/RC
1 0.7071
0
0
90°
φ
ωω0 = 1/RC
45°
Chapter 14, Solution 2.
=ω+
=ω+
=ωRLj1
1LjR
R)(H
0j11ωω+
, where LR
0 =ω
20 )(1
1)(H
ωω+=ω= H
ωω
=ω∠=φ0
1-tan-)(H
The frequency response is identical to the response in Example 14.1 except that
LR0 =ω . Hence the response is shown below.
φ
H
ω ω0 = R/L
0.7071 1
0
ω
ω0 = R/L
-45°
-90°
0°
Chapter 14, Solution 3.
(a) The Thevenin impedance across the second capacitor where V is taken is o
sRC1R
RsC1||RRTh ++=+=Z
sRC1sC1RsC1 i
iTh +=
+=
VVV
ZTh
sC1
+
Vo
−
+ −
VTh
)sC1)(sRC1(sC1sC1
Th
iTh
Tho Z
VV
ZV
++=⋅
+=
))sRC1(sRCsRC1)(sRC1(1
)sRC1)(sC1(1
)sThi
o
++++=
++==
ZVV
H(
=)s(H1sRC3CRs
1222 ++
(b) 08.01080)102)(1040(RC -3-63 =×=××=
There are no zeros and the poles are at
==RC0.383-
s1 4.787-
==RC2.617-
s2 32.712-
Chapter 14, Solution 4.
(a) RCj1
RCj
1||R
ω+=
ω
)RCj1(LjRR
RCj1R
Lj
RCj1R
)(i
o
ω+ω+=
ω++ω
ω+==ω
VV
H
=ω)(HLjRRLC-
R2 ω++ω
(b) )LjR(Cj1
)LjR(CjCj1LjR
LjR)(
ω+ω+ω+ω
=ω+ω+
ω+=ωH
=ω)(HRCjLC1
RCjLC-2
2
ω+ω−ω+ω
Chapter 14, Solution 5.
(a) Cj1LjR
Cj1)(
i
o
ω+ω+ω
==ωVV
H
=ω)(HLCRCj1
12ω−ω+
(b) RCj1
RCj
1||R
ω+=
ω
)RCj1(LjR)RCj1(Lj
)RCj1(RLjLj
)(i
o
ω+ω+ω+ω
=ω++ω
ω==ω
VV
H
=ω)(HRLCLjR
RLCLj2
2
ω−ω+ω−ω
Chapter 14, Solution 6.
(a) Using current division,
Cj1LjRR
)(i
o
ω+ω+==ω
II
H
)25.0)(10()25.0)(20(j1)25.0)(20(j
LCRCj1RCj
)( 22 ω−ω+ω
=ω−ω+
ω=ωH
=ω)(H 25.25j15j
ω−ω+ω
(b) We apply nodal analysis to the circuit below.
1/jωC+ −
IoVx
0.5 Vx
R Is jωL
Cj1Lj5.0
Rxxx
s ω+ω−
+=VVV
I
But )Cj1Lj(2Cj1Lj
5.0ox
xo ω+ω=→
ω+ω= IV
VI
Cj1Lj5.0
R1
x
s
ω+ω+=
VI
)Cj1Lj(21
R1
)Cj1Lj(2 o
s
ω+ω+=
ω+ωII
1R
)Cj1Lj(2
o
s +ω+ω
=II
)LC1(2RCjRCj
R)Cj1Lj(211)( 2
s
o
ω−+ωω
=ω+ω+
==ωII
H
)25.01(2jj
)( 2ω−+ωω
=ωH
=ω)(H 25.0j2j
ω−ω+ω
Chapter 14, Solution 7.
(a) Hlog2005.0 10=
Hlog105.2 10-3 =×
== × -3105.210H 005773.1
(b) Hlog206.2- 10=Hlog0.31- 10=
== -0.3110H 4898.0
(c) Hlog207.104 10=Hlog235.5 10=
== 235.510H 510718.1 ×
Chapter 14, Solution 8.
(a) 05.0H === 05.0log20H 10dB 26.02- , =φ °0
(b) 125H =
== 125log20H 10dB 41.94 , =φ °0
(c) °∠=+
= 43.63472.4j2
10j)1(H
== 472.4log20H 10dB 01.13 , =φ °43.63
(d) °∠=−=+
++
= 23.55-254.47.1j9.3j2
6j1
3)1(H
== 254.4log20H 10dB 577.12 , =φ °23.55-
Chapter 14, Solution 9.
)10j1)(j1(1
)(ω+ω+
=ωH
10/j1log20j1log20-H 1010dB ω+−ω+= )10/(tan)(tan- -1-1 ω−ω=φ
The magnitude and phase plots are shown below.
HdB
0.1
-40
10/j11
log20 10 ω+
ω+ j11
log20 10
-20
10 1 100 ω
1
j11
arg
φ
-135°
10/j1arg
ω+
ω+
-90°
-180°
0.1 10 1 100 ω-45°
Chapter 14, Solution 10.
ω+ω
=ω+ω
=ω
5j1j1
10)j5(j
50)j(H
ωj1log20
ω+
5j1
1log20
20 log1
-40
0.1
10
1 100 ω
20
-20
HdB 40
φ
-135°
5/j11argω+
ωj1arg
-90°
-180°
0.1 10
1
100 ω -45°
Chapter 14, Solution 11.
)2j1(j)10j1(5
)(ω+ωω+
=ωH
2j1log20jlog2010j1log205log20H 10101010dB ω+−ω−ω++=
2tan10tan90- -1-1 ω−ω+°=φ
The magnitude and phase plots are shown below.
-40
0.1 10 1 100 ω
20 14
-20
HdB 40 34
φ
-45°
45°
ω1001 10 0.1
-90°
90°
Chapter 14, Solution 12.
201.0log20,)10/1(
)1(1.0)( −=++
=ωωω
jjjwT
The plots are shown below. |T| (db) 20 0 ω 0.1 1 10 100 -20 -40 arg T 90o
0 ω 0.1 1 10 100 -90o
Chapter 14, Solution 13.
)10j1()j()j1)(101(
)j10()j(j1
)( 22 ω+ωω+
=ω+ω
ω+=ωG
10j1log20jlog40j1log2020-G 101010dB ω+−ω−ω++=
10tantan-180 -1-1 ω−ω+°=φ The magnitude and phase plots are shown below.
GdB 40
20
0.1 1 10 100 ω-20
-40
φ
90°
0.1 1 10 100 ω-90°
-180°
Chapter 14, Solution 14.
ω
+ω
+ω
ω+=ω 2
5j
2510j
1j
j12550
)(H
ω−ω++= jlog20j1log202log20H 101010dB
2
10 )5j(52j1log20 ω+ω+−
ω−ω
−ω+°=φ51
2510tantan90- 2
1-1-
The magnitude and phase plots are shown below.
φ
-90°
-40
0.1 10 1 100 ω
20 6
40 26
-20
HdB
-180°
0.1 10 1 100 ω
90°
Chapter 14, Solution 15.
)10j1)(2j1()j1(2
)j10)(j2()j1(40
)(ω+ω+
ω+=
ω+ω+ω+
=ωH
10j1log202j1log20j1log202log20H 10101010dB ω+−ω+−ω++=
10tan2tantan -1-1-1 ω−ω−ω=φ
The magnitude and phase plots are shown below.
φ
-45°
6
40
-40
0.1 10 1 100 ω
20
-20
HdB
90°
-90°
0.1 10 1 100 ω
45°
Chapter 14, Solution 16.
2
10j1)j1(100
j)(G
ω+ω+
ω=ω
GdB
-20
ωjlog20
10jlog40 ω
−
φ
-90° 2
10j1
1arg
ω+
ω+ j11arg
arg(jω)
-180°
0.1 10 1 100ω
90°
20 log(1/100) -60
-40
0.1 10 1 100
ω
20
Chapter 14, Solution 17.
2)2j1)(j1(j)41(
)(ω+ω+ω
=ωG
2j1log40j1log20jlog204-20logG 10101010dB ω+−ω+−ω+=
2tan2tan--90 -1-1 ω−ω°=φ
The magnitude and phase plots are shown below.
GdB
-20
20
ω-12
1001 10 0.1
-40
φ
-90°
-180°
0.1 10 1 100 ω
90°
Chapter 14, Solution 18.
)10j1)(5j1(j50)2j1(4
)(2
ω+ω+ωω+
=ωG
ω−ω++= jlog202j1log40504log20G 101010dB
10j1log205j1log20 1010 ω+−ω+− where 94.21-504log1020 =
10tan5tan2tan290- -1-1-1 ω−ω−ω+°=φ
The magnitude and phase plots are shown below.
GdB
-20
180°
0.1 10 1 100 ω
90°
-90°
φ
-60
-40
0.1 10 1 100 ω
20
Chapter 14, Solution 19.
)10010j1(100j
)( 2ω−ω+ω
=ωH
10010j1log20100log20jlog20H 2
101010dB ω−ω+−−ω=
ω−ω
−°=φ1001
10tan90 2
1-
The magnitude and phase plots are shown below.
φ
-90°
20
-60
-40
0.1 10 1 100 ω
40
-20
HdB
-180°
0.1 10 1 100 ω
90°
Chapter 14, Solution 20.
)10j1)(j1()j1(10
)(2
ω+ω+ω−ω+
=ωN
2
101010dB j1log2010j1log20j1log2020N ω−ω++ω+−ω+−=
10tantan1
tan 1-1-2
1- ω−ω−
ω−ω
=φ
The magnitude and phase plots are shown below.
40
180°
0.1 10 1 100 ω
90°
-90°
φ
20
-40
0.1 10 1 100 ω-20
NdB
Chapter 14, Solution 21.
)10010j1)(10j1(100)j1(j
)( 2ω−ω+ω+ω+ω
=ωT
100log20j1log20jlog20T 101010dB −ω++ω=
10010j1log2010j1log20 21010 ω−ω+−ω+−
ω−ω
−ω−ω+°=φ1001
10tan10tantan90 2
1-1-1-
The magnitude and phase plots are shown below.
180°
-180°
0.1 10 1 100 ω
90°
-90°
φ
-40
-60
0.1 10 1 100 ω
20
-20
TdB
Chapter 14, Solution 22. 10kklog2020 10 =→=
A zero of slope at dec/dB20+ 2j12 ω+→=ω
A pole of slope t dec/dB20- a20j1
120
ω+→=ω
A pole of slope t dec/dB20- a100j1
1100
ω+→=ω
Hence,
)100j1)(20j1()2j1(10
)(ω+ω+
ω+=ωH
=ω)(H)j100)(j20(
)j2(104
ω+ω+ω+
Chapter 14, Solution 23.
A zero of slope at the origindec/dB20+ ω→ j
A pole of slope t dec/dB20- a1j1
11
ω+→=ω
A pole of slope at dec/dB40- 2)10j1(1
10ω+
→=ω
Hence,
2)10j1)(j1(j
)(ω+ω+
ω=ωH
=ω)(H 2)j10)(j1(j100
ω+ω+ω
Chapter 14, Solution 24. The phase plot is decomposed as shown below.
φ
-45°
ω+ 100/j11
arg
)10/j1(arg ω+
)j(arg ω
100
90°
-90°
0.1 10 1 1000 ω
45°
)j100(j)j10)(10(k
)100j1(j)10j1(k
)(ω+ωω+′
=ω+ωω+′
=ωG
where k′ is a constant since arg 0k =′ .
Hence, =ω)(G)j100(j
)j10(kω+ωω+
, where 10k constant isk ′=
Chapter 14, Solution 25.
s/krad5)101)(1040(
1LC1
6-3-0 =××
==ω
==ω R)( 0Z Ωk2
ω−
ω+=ω
C4
L4
jR)4(0
00Z
××
−×⋅×
+=ω)101)(105(
41040
4105
j2000)4( 6-33-
3
0Z
)5400050(j2000)4( 0 −+=ωZ
=ω )4( 0Z Ω− k75.0j2
ω−
ω+=ω
C2
L2
jR)2(0
00Z
××
−××
+=ω)101)(105(
2)1040(
2)105(
j2000)2( 6-33-
3
0Z
)52000100(j2000)4( 0 −+=ωZ
=ω )2( 0Z Ω− k3.0j2
ω−ω+=ω
C21
L2jR)2(0
00Z
××
−××+=ω)101)(105)(2(
1)1040)(105)(2(j2000)2( 6-3
3-30Z
=ω )2( 0Z Ω+ k3.0j2
ω−ω+=ω
C41
L4jR)4(0
00Z
××
−××+=ω)101)(105)(4(
1)1040)(105)(4(j2000)4( 6-3
3-30Z
=ω )4( 0Z Ω+ k75.0j2
Chapter 14, Solution 26.
(a) kHz 51.2210101052
12
139===
−− xxxLCfo
ππ
(b) krad/s 101010
1003 === −xL
RB
(c ) 142.14101.0
101050
1013
36
====−
xx
RL
LCRL
Q oω
Chapter 14, Solution 27. At resonance,
Ω== 10RZ , LC1
0 =ω
LR
B = and R
LB
Q 00 ω=
ω=
Hence,
H1650
)80)(10(RQL
0==
ω=
F25)16()50(
1L
1C 22
0µ==
ω=
s/rad625.01610
LR
B ===
Therefore, =R Ω10 , =L H16 , =C F25 µ , =B s/rad625.0
Chapter 14, Solution 28. Let = 10R . Ω
H5.02010
BR
L ===
F2)5.0()1000(
1L
1C 22
0µ==
ω=
5020
1000B
Q 0 ==ω
=
Therefore, if then Ω= 10R =L H5.0 , =C F2 µ , =Q 50
Chapter 14, Solution 29.
jω 1/jω
1 Z
jω
ω+ω
+ω
+ω=j1
jj1
jZ
2
2
1j1
jω+ω+ω
+
ω−ω=Z
Since and are in phase, )t(v )t(i
211
0)Im(ω+ω
+ω
−ω==Z
0124 =−ω+ω
618.02
411-2 =+±
=ω
=ω s/rad7861.0
Chapter 14, Solution 30. Select = 10R . Ω
mH5H05.0)20)(10(
10Q
RL
0===
ω=
F2.0)05.0)(100(
1L
1C 2
0==
ω=
s/rad5.0)2.0)(10(
1RC1
B ===
Therefore, if then Ω= 10R =L mH5 , =C F2.0 , =B s/rad5.0
Chapter 14, Solution 31.
ω=→ω= L
LX
LLX
rad/s 10x796.810x40
10x6.5x10x10x2X
RLRB 6
3
36
L=
π=
ω==
Chapter 14, Solution 32. Since 10Q > ,
2B
01 −ω=ω , 2B
02 +ω=ω
=×
=ω
=120
106Q
B6
0 s/krad50
=−=ω 025.061 s/rad10975.5 6×
=+=ω 025.062 s/rad10025.6 6×
Chapter 14, Solution 33.
pF 84.5610x40x10x6.5x2
80Rf2
QCRCQ36o
o =π
=π
=→ω=
H 21.1480x10x6.5x2
10x40Qf2
RLL
RQ6
3
ooµ=
π=
π=→
ω=
Chapter 14, Solution 34.
(a) krad/s 443.110x60x10x8
1LC1
63o ===ω−−
(b) rad/s 33.310x60x10x5
1RC1B
63===
−
(c) 9.43210x60x10x5x10x443.1RCQ 633
o ==ω= −
Chapter 14, Solution 35. At resonance,
=×
==→= 3-10251
Y1
RR1
Y Ω40
=×
=ω
=→ω=)40)(10200(
80R
QCRCQ 3
00 F10 µ
=××
=ω
=→=ω)1010)(104(
1C
1L
LC1
6-1020
0 H5.2 µ
=×
=ω
=80
10200Q
B3
0 s/krad5.2
=−=−ω=ω 5.22002B
01 s/krad5.197
=+=+ω=ω 5.22002B
01 s/krad5.202
Chapter 14, Solution 36.
s/rad5000LC1
0 ==ω
==ω→=ω R)(R1
)( 00 ZY Ωk2
kS75.18j5.0L
4C
4j
R1
)4(0
00 −=
ω−
ω+=ωY
=−
=ω01875.0j0005.0
1)4( 0Z Ω+ 3.53j4212.1
kS5.7j5.0L
2C
2j
R1
)2(0
00 −=
ω−
ω+=ωY
=−
=ω0075.0j0005.0
1)2( 0Z Ω+ 74.132j85.8
kS5.7j5.0C2
1L2j
R1
)2(0
00 +=
ω−ω+=ωY
=ω )2( 0Z Ω− 74.132j85.8
kS75.18j5.0C4
1L4j
R1
)4(0
00 +=
ω−ω+=ωY
=ω )4( 0Z Ω− 3.53j4212.1
Chapter 14, Solution 37.
22 )C
1L(R
)C
1L(jRLRjCL
LjCj
1R
)Cj
1R(Lj)
Cj1R//(LjZ
ω−ω+
ω−ω+
ω+
=ω+
ω+
ω+ω
=ω
+ω=
1)LCCR(0)
C1L(R
C1L
CLLR
)ZIm( 22222
2
=+ω→=
ω−ω+
ω−ω+ω
=
Thus,
22CRLC
1
+=ω
Chapter 14, Solution 38.
222 LRLjR
CjCjLjR
1Y
ω+ω−
+ω=ω+ω+
At resonance, , i.e. 0)Im( =Y
0LR
LC 22
02
00 =
ω+ω
−ω
CL
LR 220
2 =ω+
2
3-6-3-2
2
0 104050
)1010)(1040(1
LR
LC1
×−
××=−=ω
=ω0 s/rad4841
Chapter 14, Solution 39.
(a) s/krad810x)8690(2)ff(2B 31212 π=−π=−π=ω−ω=
π=π=ω+ω=ω 17610x)88(2)(21 3
21o
nF89.1910x2x10x8
1BR1C
RC1B 33 =
π==→=
(b) H4.16410x89.19x)176(
1C
1LLC1
92o
2o =π
=ω
=→=ω−
(c ) s/krad9.552176o =π=ω (d) s/krad13.258B =π=
(e) 228
176B
Q o =ππ
=ω
=
Chapter 14, Solution 40.
(a) L = 5 + 10 = 15 mH
===ω−− 630
10x20x10x15
1LC1 1.8257 k rad/sec
==ω= −633
0 10x20x10x25x10x8257.1RCQ 912.8
===−63 10x2010x25
1RC1B 2 rad
(b) To increase B by 100% means that B’ = 4.
==′
=′4x10x25
1BR1C 3 10 µF
Since F10CC
C
21
21 µ=+
′ CC = and C1 = 20 µF, we then obtain C2 = 20 µF.
Therefore, to increase the bandwidth, we merely add another 20 µF in series with the first one.
Chapter 14, Solution 41.
(a) This is a series RLC circuit.
Ω=+= 862R , H1L = , F4.0C =
===ω4.0
1LC1
0 s/rad5811.1
==ω
=8
5811.1R
LQ 0 1976.0
==LR
B s/rad8
(b) This is a parallel RLC circuit.
F263)6)(3(
F6andF3 µ=+
→µµ
F2C µ= , Ω= k2R , mH20L =
=××
==ω)1020)(102(
1LC1
3-6-0 s/krad5
=××
×=
ω=
)1020)(105(102
LR
Q 3-3
3
020
=××
==)102)(102(
1RC1
B 6-3 s/krad250
Chapter 14, Solution 42.
(a) )LjR(||)Cj1(in ω+ω=Z
RCjLC1LjR
Cj1
LjR
CjLjR
2in ω+ω−ω+
=
ω+ω+
ωω+
=Z
22222
2
in CR)LC1()RCjLC1)(LjR(
ω+ω−ω−ω−ω+
=Z
At resonance, Im( 0)in =Z , i.e.
CR)LC1(L0 22 ω−ω−ω= CRLLC 22 −=ω
=−
=ωLC
CRL 2
0 LR
C1 2
−
(b) )Cj1R(||Ljin ω+ω=Z
RCj)LC1()RCj1(Lj
Cj1LjR)Cj1R(Lj
2in ω+ω−ω+ω
=ω+ω+ω+ω
=Z
22222
22
in CR)LC1(]RCj)LC1[()LjRLC-(
ω+ω−ω−ω−ω+ω
=Z
At resonance, Im( 0)in =Z , i.e.
LCR)LC1(L0 2232 ω+ω−ω=
1)CRLC( 222 =−ω
=ω0 22CRLC1−
(c) )Cj1Lj(||Rin ω+ω=Z
RCj)LC1()LC1(R
Cj1LjR)Cj1Lj(R
2
2
in ω+ω−ω−
=ω+ω+ω+ω
=Z
22222
22
in CR)LC1(]RCj)LC1)[(LC1(R
ω+ω−ω−ω−ω−
=Z
At resonance, Im( 0)in =Z , i.e.
RC)LC1(R0 2 ωω−= 0LC1 2 =ω−
=ω0 LC1
Chapter 14, Solution 43. Consider the circuit below.
1/jωC
R1 Zin
jωL R2
(a) )Cj1R(||)Lj||R( 21in ω+ω=Z
ω
+
ω+ω
=Cj
1R||
LjRLjR
21
1inZ
LjRLjR
Cj1
R
Cj1
RLjR
LRj
1
12
21
1
in
ω+ω
+ω
+
ω
+⋅ω+
ω
=Z
12
21
21in LCR)CRj1)(LjR(
)CRj1(LRjω−ω+ω+
ω+ω=Z
)CRRL(jLCRLCRRLRjLCRR-
2122
12
1
1212
in +ω+ω−ω−ω+ω
=Z
221
222
21
21
2122
12
11212
in )CRRL()LCRLCRR()]CRRL(jLCRLCRR)[LRjLCRR-(
+ω+ω−ω−+ω−ω−ω−ω+ω
=Z
At resonance, Im( 0)in =Z , i.e.
)LCRLCRR(LR)CRRL(LCRR0 22
12
1121213 ω−ω−ω++ω=
CLRLRLCRR0 221
321
222
21
3 ω−ω+ω= LC1CR0 222
22 ω−+ω=
1)CRLC( 222
2 =−ω
222
0 CRLC1−
=ω
26-26-0 )109()1.0()109)(02.0(1
×−×=ω
=ω0 s/krad357.2 (b) At , s/krad357.20 =ω=ω
14.47j)1020)(10357.2(jLj -33 =××=ω
0212.0j9996.014.47j1
14.47jLj||R1 +=
+=ω
14.47j1.0)109)(10357.2(j
11.0
Cj1
R 6-32 −=××
+=ω
+
)Cj1R(||)Lj||R()( 210in ω+ω=ωZ
)14.47j1.0()0212.0j9996.0()14.47j1.0)(0212.0j9996.0(
)( 0in −++−+
=ωZ
=ω )( 0inZ Ω1
Chapter 14, Solution 44. We find the input impedance of the circuit shown below.
jω(2/3) 1/jωZ
1
1/jωC
Cj111
j2j
31ω+
+ω+ω
=Z
, 1=ω
2
2
C1jCC
-j0.5jC1
jCj-j1.5
1++
+=+
++=Z
)t(v and i are in phase when Z is purely real, i.e. )t(
1)1C(C1
C-0.50 2
2 =−→+
+= or =C F1
Ω=→=+
= 221
C1C1
2
2
ZZ
20)10)(2( === IZV
V)tsin(20)t(v = , i.e. =oV V20
Chapter 14, Solution 45.
(a) ω+
ω=ω
j1j
j||1 , ω+
=ω+
ω=
ω j11
j11j1
j1
||1
Transform the current source gives the circuit below.
ω+ j11
ω+ωj1
j
1 I
j1jω+
ω +
−
+
Vo
−
IVω+
ω⋅
ω+ω
+ω+
+
ω+=
j1j
j1j
j11
1
j11
o
==ωI
VH o)( 2)j1(2
jω+
ω
(b) 2)j1(21
)1(+
=H
==2)2(2
1)1(H 25.0
Chapter 14, Solution 46.
(a) This is an RLC series circuit.
nF26.1110x10x)10x15x2(
1L
1CLC1
323o
2o =π
=ω
=→=ω−
(b) Z = R, I = V/Z = 120/20 = 6 A
(c ) 12.471520
10x10x10x15x2R
LQ
33o =π=
π=
ω=
−
Chapter 14, Solution 47.
RLj1
1LjR
R)(
i
o
ω+=
ω+==ω
VV
H
1)0(H = and 0)(H =∞ showing that this circuit is a lowpass filter.
At the corner frequency, 2
1)(H c =ω , i.e.
RL
1
RL
1
12
1 c
2c
ω=→
ω
+
= or LR
c =ω
Hence,
cc f2LR
π==ω
=××
⋅π
=⋅π
= 3-
3
c 1021010
21
LR
21
f kHz796
Chapter 14, Solution 48.
Cj1
||RLj
Cj1
||R)(
ω+ω
ω=ωH
Cj1RCjR
Lj
Cj1RCjR
)(
ω+ω
+ω
ω+ω
=ωH
=ω)(HRLCLjR
R2ω−ω+
1)0(H = and showing that 0)(H =∞ this circuit is a lowpass filter.
Chapter 14, Solution 49.
At dc, 224
)0( =H = .
Hence, 2
2)0(H
21
)(H ==ω
2c1004
42
2ω+
=
2.081004 c
2c =ω→=ω+
10j12
20j24
)2(H+
=+
=
199.01012
)2(H ==
In dB, =)2(Hlog1020 14.023-
== 10-tan)2(Harg -1 °84.3- Chapter 14, Solution 50.
LjR
Lj)(
i
o
ω+ω
==ωVV
H
0)0(H = and showing that 1)(H =∞ this circuit is a highpass filter.
LR
1
LR
1
12
1)(
c2
c
c ω=→
ω+
==ωH
or cc f2LR
π==ω
=⋅π
=⋅π
=1.0
20021
LR
21
fc Hz3.318
Chapter 14, Solution 51.
RC1j
jRCj1
RCj)(
+ωω
=ω+
ω=ω′H (from Eq. 14.52)
This has a unity passband gain, i.e. 1)(H =∞ .
50RC1
c =ω=
ω+ω
=ω′=ωj50
10j)(10)( HH ^
=ω)(Hω+ωj50
10j
Chapter 14, Problem 52.
Design an RL lowpass filter that uses a 40-mH coil and has a cut-off frequency of 5 kHz.
Chapter 14, Solution 53.
cc f2LR
π==ω
=×π=π= )1040)(10)(2(Lf2R -35
c Ωk13.25 Chapter 14, Solution 54.
311 1020f2 ×π=π=ω
322 1022f2 ×π=π=ω
3
12 102B ×π=ω−ω=
3120 1021
2×π=
ω+ω=ω
=ππ
=ω
=221
BQ 0 5.11
C1
LLC1
20
0 ω=→=ω
=××π
=)1080()1021(
1L 12-23 H872.2
BLRLR
B =→=
=×π= )872.2)(102(R 3 Ωk045.18
Chapter 14, Solution 55.
Ω=π
=π
=→=π=ω−
k3.26510x300x10x2x2
1Cf2
1RRC1f2
123ccc
Chapter 14, Solution 56.
s/krad10)104.0)(1025(
1LC1
63o =××
==ω−−
s/krad4.01025
10LR
B 3- =×
==
==4.0
10Q 25
s/krad8.92.0102Bo1 =−=−ω=ω or kHz56.12
8.91 =
π=f
s/krad2.102.0102Bo2 =+=+ω=ω or kHz62.12
2.10f2 =
π=
Therefore,
kHz62.1fkHz56.1 <<
Chapter 14, Solution 57.
(a) From Eq 14.54,
LC1
LRss
LRs
LCssRC1sRC
sC1sLR
R)s(2
2++
=++
=++
=H
Since LR
B = and LC1
0 =ω ,
=)s(H 20
2 sBssB
ω++
(b) From Eq. 14.56,
LC1
LR
ss
LC1
s
sC1
sLR
sC1
sL)s(
2
2
++
+=
++
+=H
=)s(H 20
2
20
2
sBss
ω++ω+
Chapter 14, Solution 58.
(a) Consider the circuit below.
I I1
R1/sC+
Vo
−
+ −
R 1/sC
Vs
sC2
R
sC1
RsC1
RsC1
R||sC1
R)s(+
++=
++=Z
)sRC2(sCsRC1
R)s(+
++=Z
)sRC2(sCCRssRC31
)s(222
+++
=Z
ZV
I s=
)sRC2(RsC2sC1 s
1 +=
+=
ZV
II
222s
1o CRssRC31)sRC2(sC
sRC2R
R+++
⋅+
==V
IV
222s
o
CRssRC31sRC
)s(++
==VV
H
++=
222
CR1
sRC3
s
sRC3
31
)s(H
Thus, 2220 CR
1=ω or ==ω
RC1
0 s/rad1
==RC3
B s/rad3
(b) Similarly,
sLR2)sLR(R
sL)sLR(||RsL)s(++
+=++=Z
sLR2LssRL3R
)s(222
+++
=Z
ZV
I s= , )sLR2(
RsLR2
R s1 +
=+
=Z
VII
222s
1o LssRL3RsLR2
sLR2sLR
sL++
+⋅
+=⋅=
VIV
2
22
222s
o
LR
sLR3
s
sLR3
31
LssRL3RsRL
)s(++
=++
==VV
H
Thus, ==ωLR
0 s/rad1
==LR3
B s/rad3
Chapter 14, Solution 59.
(a) =×
==ω)1040)(1.0(
1LC1
12-0 s/rad105.0 6×
(b) 43
1021.0
102LR
B ×=×
==
250102105.0
BQ 4
60 =
××
=ω
=
As a high Q circuit,
=−=−ω=ω )150(102B 4
01 s/krad490
=+=+ω=ω )150(102B 4
02 s/krad510
(c) As seen in part (b), =Q 250
Chapter 14, Solution 60. Consider the circuit below.
Ro
+
Vo
−
+ − R
1/sC
Vi
sL
sC1sLR)sC1sL(R
sC1
sL||R)s(+++
=
+=Z
LCssRC1)LCs1(R
)s( 2
2
+++
=Z
LCRsRLCRsCsRRR)LCs1(R
R 2o
2oo
2
oi
o
+++++
=+
==Z
ZVV
H
LCssRC1)LCs1(R
RR 2
2
ooin +++
+=+= ZZ
LCssRC1LCRsRLCRsCsRRR
2
2o
2oo
in ++++++
=Z
ω= js
RCjLC1LCRRLCRCRRjR
2
2o
2oo
in ω+ω−ω−+ω−ω+
=Z
222
2o
2o
2o
in )RC()LC1()RCjLC1)(CRRjLCRLCRRR(
ω+ω−ω−ω−ω+ω−ω−+
=Z
0)Im( in =Z implies that
0)LC1(CRR]LCRLCRRR[RC- 2
o2
o2
o =ω−ω+ω−ω−+ω
0LCRRLCRLCRRR o2
o2
o2
o =ω+−ω−ω−+ RLCR2 =ω
=××
==ω)104)(101(
1LC1
6-3-0 s/krad811.15
LCRLCRRCRRjR)LC1(R
2o
2oo
2
ω−ω−+ω+ω−
=H
RRR
)0(HHo
max +==
or o
oo
2o
2
max RRR
)RR(LCCRRjRR
LC1Rlim)(HH
+=
+−ω
+ω+
−ω=∞=
∞→ω
At and ω , 1ω 2 mzxH2
1=H
CRRj)RR(LCRR)LC1(R
)RR(2R
oo2
o
2
o ω++ω−+ω−
=+
2o
2o
2o
2o
))RR(LCRR()CRR()LC1)(RR(
21
+ω−++ω
ω−+=
28-226-
-92
)10410()10(96)1041(10
21
×⋅ω−+ω×
×⋅ω−=
21
)10410()10(96)1041(10
028-226-
-92
−×⋅ω−+ω×
×⋅ω−=
0)10410()1096()2)(10410( 2-822-6-82 =×⋅ω−+ω×−×⋅ω−
2-822-62-82 )10410()1096()10410)(2( ×⋅ω−+ω×=×⋅ω−
0)10410()1096( 2-822-6 =×⋅ω−−ω×
010010092.8106.1 2-74-15 =+ω×−ω×
01025.610058.5 1684 =×+×−ω
××
=ω 8
82
101471.2109109.2
Hence,
s/krad653.141 =ω
s/krad061.172 =ω
=−=ω−ω= 653.14061.17B 12 s/krad408.2 Chapter 14, Solution 61.
(a) iCj1RCj1
VVω+
ω=+ , oVV =−
Since , −+ = VV
oiRCj11
VV =ω+
==ωi
o)(VV
HRCj1
1ω+
(b) iCj1RR
VVω+
=+ , oVV =−
Since , −+ = VV
oiRCj1RCj
VV =ω+
ω
==ωi
o)(VV
HRCj1
RCjω+
ω
Chapter 14, Solution 62. This is a highpass filter.
RCj11
RCj1RCj
)(ω−
=ω+
ω=ωH
ωω−=ω
cj11
)(H , )1000(2RC1
c π==ω
f1000j11
ffj11
)(c −
=−
=ωH
(a) i
o
5j11
)Hz200f(VV
H =−
==
=−
=5j1
mV120oV mV53.23
(b) i
o
5.0j11
)kHz2f(VV
H =−
==
=−
=5.0j1
mV120oV m3.107 V
(c) i
o
1.0j11
)kHz10f(VV
H =−
==
=−
=1.0j1
mV120oV m4.119 V
Chapter 14, Solution 63.
For an active highpass filter,
ii
fiRsC1
RsC)s(H
+−= (1)
But
10/s1s10)s(H
+−= (2)
Comparing (1) and (2) leads to:
Ω==→= M10C10R10RC
iffi
Ω==→= k100C
1.0R1.0RCi
iii
Chapter 14, Solution 64.
ff
f
fff CRj1
RCj1
||RZω+
=ω
=
i
ii
iii Cj
CRj1Cj
1RZ
ωω+
=ω
+=
Hence,
===ωi
f
i
o -)(
ZZ
VV
H)CRj1)(CRj1(
CRj-
iiff
if
ω+ω+ω
This is a bandpass filter. )(ωH is similar to the product of the transfer function of a lowpass filter and a highpass filter.
Chapter 14, Solution 65.
ii RCj1RCj
Cj1RR
VVVω+
ω=
ω+=+
ofi
i
RRR
VV+
=−
Since , −+ = VV
iofi
i
RCj1RCj
RRR
VVω+
ω=
+
==ωi
o)(VV
H
ω+
ω
+
RCj1RCj
RR
1i
f
It is evident that as ω , the gain is ∞→i
f
RR
+1 and that the corner frequency is RC1
.
Chapter 14, Solution 66.
(a) Proof
(b) When 3241 RRRR = ,
CR1ss
RRR
)s(243
4
+⋅
+=H
(c) When ∞→3R ,
CR1sCR1-
)s(2
1
+=H
Chapter 14, Solution 67.
fii
f R4R41
RR
gain DC =→==
s/rad)500(2CR
1frequencyCorner
ffc π==ω=
If we select , then Ω= k20R f Ω= k80R i and
nF915.15)1020)(500)(2(
1C 3 =
×π=
Therefore, if =fR Ωk20 , then =iR Ωk80 and =C nF915.15
Chapter 14, Solution 68.
ifi
f R5RRR
5gainfrequency High =→==
s/rad)200(2CR
1frequencyCorner
iic π==ω=
If we select , then Ω= k20R i Ω= k100fR and
nF8.39)1020)(200)(2(
1C 3 =
×π=
Therefore, if =iR Ωk20 , then =fR Ωk100 and =C nF8.39
Chapter 14, Solution 69. This is a highpass filter with f 2c = kHz.
RC1
f2 cc =π=ω
3c 104
1f2
1RC
×π=
π=
Hz108 may be regarded as high frequency. Hence the high-frequency gain is
410
RRf −
=−
or R5.2Rf =
If we let =R Ωk10 , then =fR Ωk25 , and =×π
= 41040001
C nF96.7 .
Chapter 14, Solution 70.
(a) )YYY(YYY
YY)s()s(
)s(321421
21
i
o
+++==
VV
H
where 11
1 GR1
Y == , 22
2 GR1
Y == , 13 sCY = , 24 sCY = .
=)s(H)sCGG(sCGG
GG
121221
21
+++
(b) 1GGGG
)0(H21
21 == , 0)(H =∞
showing that this circuit is a lowpass filter. Chapter 14, Solution 71. L , Ω= 50R , mH F1C40= µ=
)1040(KK
1LKK
L 3-
f
m
f
m ×⋅=→=′
mf KK25 = (1)
fm
-6
fm KK10
1KK
CC =→=′
mf
6
K1
K10 = (2)
Substituting (1) into (2),
ff
6
K251
K10 =
=fK -3102.0 ×
== fm K25K -3105×
Chapter 14, Solution 72.
CLLC
KKLC
CL 2f2
f ′′=→=′′
8-
-6-32f 104
)2)(1()1020)(104(
K ×=××
=
=fK -4102×
LC
CL
KKCL
CL 2
m2m ⋅
′′
=→=′′
3-
3-
-62m 105.2
)104)(2()1020)(1(
K ×=××
=
=mK -2105×
Chapter 14, Solution 73.
=×==′ )10800)(12(RKR 3m ΩM6.9
=×==′ )1040(1000800
LKK
L 6-
f
m F32 µ
=×
==′ )1000)(800(10300
KKC
C-9
fmpF375.0
Chapter 14, Solution 74.
Ω=== 300100x3RK'R 1m1
Ω=== k 1100x10RK'R 2m2
H 200)2(1010L
KK
'L 6
2
f
m µ===
nF 110101
KKC'C 8fm
===
Chapter 14, Solution 75.
Ω=== 20010x20RK'R m
H 400)4(1010L
KK
'L5f
m µ===
F 110x101
KKC'C
5fmµ===
Chapter 14, Solution 76.
Ω==→== 5010
10x50R10x50RK'R3
33
m
mH 101010x10x10LH 10L
KK
'L3
66
f
m ==→µ== −
mF 4010x10x10x40CKK
CFp 40'C 6312
fm==→== −
Chapter 14, Solution 77. L and C are needed before scaling.
H25
10BR
LLR
B ===→=
F5.312)2)(1600(
1L
1C
LC1
20
0 µ==ω
=→=ω
(a) ===′ )2)(600(LKL m H1200
=×
==′ 60010125.3
KC
C-4
mF5208.0 µ
(b) ===′ 3f 10
2KL
L mH2
=×
==′ 3
-4
f 1010125.3
KC
C nF5.312
(c) ===′ 5f
m
10)2)(400(
LKK
L mH8
=×
==′ )10)(400(10125.3
KKC
C 5
-4
fmpF81.7
Chapter 14, Solution 78.
Ω===′ k1)1)(1000(RKR m
H1.0)1(1010
LKK
L 4
3
f
m ===′
F1.0)10)(10(
1KK
CC 43
fmµ===′
The new circuit is shown below.
1 kΩ
1 kΩ 0.1 µF0.1 H1 kΩ +
Vx
−
I
Chapter 14, Solution 79.
(a) Insert a 1-V source at the input terminals.
Ro
+ −
3Vo
+
Vo
−
Io V2V1
sL + −
R 1/sC
1 V
There is a supernode.
sC1sLR1 21
+=
− VV (1)
But (2) o12o21 33 VVVVVV −=→+=
Also, sC1sLsLsC1sL
sL 2o2o +
=→+
=VV
VV (3)
Combining (2) and (3)
oo12 sLsC1sL
3 VVVV+
=−=
12
2
o LCs41LCs
VV+
= (4)
Substituting (3) and (4) into (1) gives
12o1
LCs41sC
sLR1
VVV
+==
−
12
2
121 LCs41sRCLCs41
LCs41sRC
1 VVV+
++=
++=
sRCLCs41LCs41
2
2
1 +++
=V
)sRCLCs41(RsRC
R1
21
o ++=
−=
VI
sCLCs4sRC11 2
oin
++==
IZ
sC1
RsL4in ++=Z (5)
When , , 5R = 2L = 1.0C = ,
=)s(inZs
105s8 ++
At resonance,
C1
L40)Im( in ω−ω==Z
or ===ω)2)(1.0(2
1LC21
0 s/rad118.1
(b) After scaling,
RKR m→′ Ω→Ω 404 Ω→Ω 505
H2.0)2(10010
LKK
Lf
m ===′
4-
fm10
)100)(10(1.0
KKC
C ===′
From (5),
=)s(inZs
1050s8.0
4
++
===ω)10)(2.0(2
1LC21
4-0 s/rad8.111
Chapter 14, Solution 80.
(a) Ω===′ 400)2)(200(RKR m
mH2010
)1)(200(K
LKL 4
f
m ===′
F25.0)10)(200(
5.0KK
CC 4
fmµ===′
The new circuit is shown below.
20 mHa
400 Ω
Ix
0.25 µF 0.5 Ix
b
(b) Insert a 1-A source at the terminals a-b.
V2V1
b
a sL
R
Ix
1/(sC) 1 A 0.5 Ix
At node 1,
sL
sC1 211
VVV
−+= (1)
At node 2,
R
5.0sL
2x
21 VI
VV=+
−
But, I . 1x sC V=
RsC5.0
sL2
121 V
VVV
=+−
(2)
Solving (1) and (2),
1sCR5.0LCs
RsL21 ++
+=V
1sCR5.0LCsRsL
1 21
Th +++
==V
Z
At , 410=ω
1)400)(1025.0)(10j(5.0)1025.0)(1020()10j(400)1020)(10j(
6-46-3-24
3-4
Th +×+××+×
=Z
200j6005.0j5.0
200j400Th −=
++
=Z
=ThZ ohms435.18-5.632 °∠
Chapter 14, Solution 81. (a)
1GR)LGRC(jLCRLjZtoleadswhich
LjR1)LjR)(CjG(
LjR1CjG
Z1
2 +++ω+ω−
+ω=
ω++ω+ω+
=ω+
+ω+=
LC1GR
CG
LRj
LCR
Cj
)(Z2 +
+
+ω+ω−
+ω
=ω (1)
We compare this with the given impedance:
25001j2)1j(1000)(Z 2 ++ω+ω−
+ω=ω (2)
Comparing (1) and (2) shows that
LR1 R/LmF, 1C1000C1
=→==→=
1CG2CG
LR
==→=+ mS
L4.0RR10
1R10LC
1GR25013
3==→
+=
+=
−
−
Thus, R = 0.4Ω, L = 0.4 H, C = 1 mF, G = 1 mS
(b) By frequency-scaling, Kf =1000. R’ = 0.4 Ω, G’ = 1 mS
mH4.010
4.0KL'L
3f=== , F1
1010
KC'C 3
3
fµ===
−
−
Chapter 14, Solution 82.
fmKK
CC =′
2001
200K c
f ==ωω′
=
50002001
101
K1
CC
K 6-f
m =⋅=⋅′
=
==′ RKR m 5 kΩ, thus, ==′ if R2R Ωk10
Chapter 14, Solution 83.
pF 1.010x100
10CKK
1'CF1 5
6
fm===→µ
−
pF 5.0'CF5 =→µ
Ω=Ω==→Ω M 1k 10x100RK'Rk 10 m
Ω=→Ω M 2'Rk 20 Chapter 14, Solution 84.
The schematic is shown below. A voltage marker is inserted to measure vo. In the AC sweep box, we select Total Points = 50, Start Frequency = 1, and End Frequency = 1000. After saving and simulation, we obtain the magnitude and phase plots in the probe menu as shown below.
Chapter 14, Solution 85. We let A so that Vo
s 01I ∠= .VI/ oso = The schematic is shown below. The circuit is simulated for 100 < f < 10 kHz.
Chapter 14, Solution 86.
The schematic is shown below. A current marker is inserted to measure I. We set Total Points = 101, start Frequency = 1, and End Frequency = 10 kHz in the AC sweep box. After simulation, the magnitude and phase plots are obtained in the Probe menu as shown below.
Chapter 14, Solution 87.
The schematic is shown below. In the AC Sweep box, we set Total Points = 50, Start Frequency = 1, and End Frequency = 100. After simulation, we obtain the magnitude response as shown below. It is evident from the response that the circuit represents a high-pass filter.
Chapter 14, Solution 88. Chapter 14, Solution 88.
The schematic is shown below. We insert a voltage marker to measure Vo. In the AC Sweep box, we set Total Points = 101, Start Frequency = 1, and End Frequency = 100. After simulation, we obtain the magnitude and phase plots of Vo as shown below.
The schematic is shown below. We insert a voltage marker to measure Vo. In the AC Sweep box, we set Total Points = 101, Start Frequency = 1, and End Frequency = 100. After simulation, we obtain the magnitude and phase plots of Vo as shown below.
Chapter 14, Solution 89.
The schematic is shown below. In the AC Sweep box, we type Total Points = 101, Start Frequency = 100, and End Frequency = 1 k. After simulation, the magnitude plot of the response Vo is obtained as shown below.
Chapter 14, Solution 90.
The schematic is shown below. In the AC Sweep box, we set Total Points = 1001, Start Frequency = 1, and End Frequency = 100k. After simulation, we obtain the magnitude plot of the response as shown below. The response shows that the circuit is a high-pass filter.
Chapter 14, Solution 91.
The schematic is shown below. In the AC Sweep box, we set Total Points = 1001, Start Frequency = 1, and End Frequency = 100k. After simulation, we obtain the magnitude plot of the response as shown below. The response shows that the circuit is a high-pass filter.
Chapter 14, Solution 92.
The schematic is shown below. We type Total Points = 101, Start Frequency = 1, and End Frequency = 100 in the AC Sweep box. After simulating the circuit, the magnitude plot of the frequency response is shown below.
hapter 14, Solution 93.
C
R
L
C
2
2
0 LR
LC1
21
f −π
=
610
10240400
L
7
6- =×
=28810
)10120)(10240(1
LC1 16
12-6- =××
= R
,
Since LC1
L<<
R
=π
=π
≅224
10LC2
1f
8
0 kHz938
If R is reduced to 40 Ω, LC1
LR<< .
The result remains the same. Chapter 14, Solution 94.
RC1
c =ω
We make R and C as small as possible. To achieve this, we connect 1.8 kΩ and 3.3 kΩ in parallel so that
Ω=+
= k 164.13.38.13.3x8.1R
We place the 10-pF and 30-pF capacitors in series so that
C = (10x30)/40 = 7.5 pF Hence,
rad/s 10x55.11410x5.7x10x164.1
1RC1 6
123c ===ω−
Chapter 14, Solution 95.
(a) LC2
1f0 π=
When , pF360C =
MHz541.0)10360)(10240(2
1f
12-6-0 =××π
=
When , pF40C =
MHz624.1)1040)(10240(2
1f
12-6-0 =××π
=
Therefore, the frequency range is MHz624.1fMHz541.0 0 <<
(b) RfL2
Qπ
=
At , MHz541.0f0 =
=××π
=12
)10240)(10541.0)(2(Q
-66
98.67
At , MHz624.1f0 =
=××π
=12
)10240)(10624.1)(2(Q
-66
1.204
Chapter 14, Solution 96.
C2 C1
+
Vo
−
VoV1
+ −
Ri
RL
L
Vi
Z1Z2
22
L
2L1 CsR1
RsC
1||R
+==Z
+++
=+=2L
2L2
L
11
12 CsR1
LCRsRsL||
sC1
)sL(||sC1
ZZ
2L
2L2
L
1
2L
2L2
L
12
CsR1LCRsRsL
sC1
CsR1LCRsRsL
sC1
+++
+
+++
⋅=Z
21L3
1L12
2L
2L2
L2 CLCRsCsRLCsCsR1
LCRsRsL++++
++=Z
ii2
21 R
VZ
ZV
+=
i1
1
22
21
1
1o sLRsL
VZ
ZZ
ZV
ZZ
V+
⋅+
=+
=
sLR 1
1
22
2
i
o
+⋅
+=
ZZ
ZZ
VV
where
=+ 22
2
RZZ
21Li3
1Li1i2
2Lii2L2
L
2L2
L
CLCRRsCRsRLCRsCRsRRLCRsRsLLCRsRsL
+++++++++
and 2L
2L
L
1
1
LCRssLRR
sL ++=
+ZZ
Therefore,
=i
o
VV
)LCRssLR)(CLCRRsCRsRLCRsCRsRRLCRsRsL(
)LCRsRsL(R
2L2
L21Li3
1Li1i2
2Lii2L2
L
2L2
LL
+++
++++++++
where s . ω= j
Chapter 14, Solution 97.
C2 C1
+
Vo
−
VoV1
+ −
Ri
RL
L
Vi
Z2 Z1
2L
2L
2L sC1sLR
)sC1R(sLsC
1R||sL
+++
=
+=Z , ω= js
i1i
1 sC1RV
ZZ
V++
=
i1i2L
L1
2L
Lo sC1RsC1R
RsC1R
RV
ZZ
VV++
⋅+
=+
=
)sC1sLR)(sC1R()sC1R(sL)sC1R(sL
sC1RR
)(2L1i2L
2L
2L
L
i
o
++++++
⋅+
==ωVV
H
=ω)(H)1CsR(LCs)1CsRLCs)(1CsR(
CCLRs
2L12
2L22
1i
21L3
+++++
where s . ω= j
Chapter 14, Solution 98.
π=−π=−π=ω−ω= 44)432454(2)ff(2B 1212
)44)(20(QBf2 00 π==π=ω
==ππ
= )22)(20(2
)44)(20(f0 Hz440
Chapter 14, Solution 99.
Cf21
C1
Xc π=
ω=
π=
××π=
π=
2010
)105)(102)(2(1
Xf21
C-9
36c
Lf2LXL π=ω=
π×
=×π
=π
=4103
)102)(2(300
f2X
L-4
6L
=
π⋅
π×
π
=π
=
2010
4103
2
1LC2
1f
9-4-0 MHz826.1
=
×π
== 4-1034
)100(LR
B s/rad10188.4 6×
Chapter 14, Solution 100.
RC1
f2 cc =π=ω
=××π
=π
=)105.0)(1020)(2(
1Cf2
1R 6-3
cΩ91.15
Chapter 14, Solution 101.
RC1
f2 cc =π=ω
=×π
=π
=)1010)(15)(2(
1Cf2
1R 6-
cΩk061.1
Chapter 14, Solution 102.
(a) When and 0R s = ∞=LR , we have a low-pass filter.
RC1
f2 cc =π=ω
=××π
=π
=)1040)(104)(2(
1RC21
f 9-3c Hz7.994
(b) We obtain across the capacitor. ThR)RR(||RR sLTh +=
Ω=+= k5.2)14(||5R Th
)1040)(105.2)(2(1
CR21
f 9-3Th
c ××π=
π=
=cf kHz59.1
Chapter 14, Solution 103.
sC1||RRR
)(12
2
i
o
+==ω
VV
H , ω= js
)sC1(RR)sC1R(R
sC1R)sC1(R
R
R)(
12
12
1
12
2
++
=
++
=ωH
=ω)(H21
12
sCRR)sCR1(R
++
Chapter 14, Solution 104.
The schematic is shown below. We click Analysis/Setup/AC Sweep and enter Total Points = 1001, Start Frequency = 100, and End Frequency = 100 k. After simulation, we obtain the magnitude plot of the response as shown.
Chapter 15, Solution 1.
(a) 2
ee)atcosh(
at-at +=
[ ] =
++
−=
as1
as1
21
)atcosh(L 22 ass−
(b) 2
ee)atsinh(
at-at −=
[ ] =
+−
−=
as1
as1
21
)atsinh(L 22 asa−
Chapter 15, Solution 2.
(a) )sin()tsin()cos()tcos()t(f θω−θω= [ ] [ ])tsin()sin()tcos()cos()s(F ωθ−ωθ= LL
=)s(F 22s)sin()cos(s
ω+θω−θ
(b) )sin()tcos()cos()tsin()t(f θω+θω=
[ ] [ ])tsin()cos()tcos()sin()s(F ωθ+ωθ= LL
=)s(F 22s)cos()sin(s
ω+θω−θ
Chapter 15, Solution 3.
(a) [ ] =)t(u)t3cos(e-2tL9)2s(
2s2 ++
+
(b) [ ] =)t(u)t4sin(e-2tL16)2s(
42 ++
(c) Since [ ] 22 ass
)atcosh(−
=L
[ ] =)t(u)t2cosh(e-3tL4)3s(
3s2 −+
+
(d) Since [ ] 22 asa
)atsinh(−
=L
[ =)t(u)tsinh(e-4tL ]1)4s(
12 −+
(e) [ ]4)1s(
2)t2sin(e 2
t-
++=L
If )s(F)t(f →←
)s(Fdsd-
)t(ft →←
Thus, [ ] ( )[ ]1-2t- 4)1s(2dsd-
)t2sin(et ++=L
)1s(2)4)1s((
222 +⋅
++=
[ ] =)t2sin(et -tL 22 )4)1s(()1s(4++
+
Chapter 15, Solution 4.
(a) 16s
se6e4s
s6)s(G 2
ss
22 +=
+=
−−
(b) 3s
e5s2)s(F
s2
2 ++=
−
Chapter 15, Solution 5.
(a) [ ]4s
)30sin(2)30cos(s)30t2cos( 2 +
°−°=°+L
[ ]
+−°
=°+4s
1)30cos(sdsd
)30t2cos(t 22
22L
( )
+
−=
1-2 4s1s23
dsd
dsd
( ) ( )
+
−−+=
2-21-2 4s1s23s24s
23
dsd
( )
( ) ( ) ( ) ( )32
2
222222 4s
1s23)s8(
4s
23s2
4s
1s232
4s
2s-23
+
−
++
−+
−
−+
=
( ) ( )32
2
22 4s
1s23)s8(
4ss32s3s3-
+
−
++
−+−=
( ) ( )32
23
32
2
4ss8s34
4s)42)(ss3(-3
+
−+
+
++=
[ ] =°+ )30t2cos(t 2L ( ) 32
32
4ss3s6s3128
+
+−−
(b) [ ] =+
⋅= 5t-4
)2s(!4
30et30L 5)2s(720+
(c) =−⋅−=
δ− )01s(4s2
)t(dtd
4)t(ut2 2L s4s22 −
(d) )t(ue2)t(ue2 -t1)--(t =
[ ] =)t(ue2 1)--(tL1s
e2+
(e) Using the scaling property,
[ ] =⋅⋅=⋅⋅=s2
125
)21(s1
211
5)2t(u5Ls5
(f) [ ] =+
=31s
6)t(ue6 3t-L
1s318+
(g) Let f . Then, )t()t( δ= 1)s(F = .
−′−−=
=
δ −− )0(fs)0(fs)s(Fs)t(fdtd
)t(dtd 2n1nn
n
n
n
n
LL
−⋅−⋅−⋅=
=
δ −− 0s0s1s)t(fdtd
)t(dtd 2n1nn
n
n
n
n
LL
=
δ )t(dtd
n
n
L ns
Chapter 15, Solution 6.
(a) [ ] =−δ )1t(2L -se2
(b) [ ] =− )2t(u10L 2s-es
10
(c) [ ] =+ )t(u)4t(Ls4
s12 +
(d) [ ] [ ] =−=− )4t(uee2)4t(ue2 4)--(t-4-t LL)1s(e
e24
-4s
+
Chapter 15, Solution 7.
(a) Since [ ] 22 4ss
)t4cos(+
=L , we use the linearity and shift properties to
obtain [ ] =−− )1t(u))1t(4cos(10L16ses10
2
s-
+
(b) Since [ ] 32
s2
t =L , [ ]s1
)t(u =L ,
[ ] 3t2-2
)2s(2
et+
=L , and [ ]s
e)3t(u
s3-
=−L
[ ] =−+ )3t(u)t(uet t2-2Ls
e)2s(
2 -3s
3 ++
Chapter 15, Solution 8.
(a) [ ]t3-t2- e10e4)t2(u6)t3(2 −++δL
3s10
2s4
2s1
21
631
2+
−+
+⋅⋅+⋅=
=3s
102s
4s6
32
+−
+++
(b) )1t(ue)1t(ue)1t()1t(uet -t-t-t −+−−=−
)1t(uee)1t(uee)1t()1t(uet -11)--(t-11)--(t-t −+−−=−
[ ] =+
++
=−1s
ee)1s(
ee)1t(uet
-s-1
2
-s-1t-L
1se
)1s(e 1)-(s
2
1)-(s
++
+
++
(c) [ ] =−− )1t(u))1t(2cos(L4s
es2
-s
+
(d) Since )t4sin()t4cos()4sin()4cos()t4sin())t(4sin( =π−π=π− )t(u))t(4sin()t(u)t4sin( π−π−=π−
[ ][ ])t(u)t(u)t4sin( π−−L
[ ] [ ])t(u))t(4sin()t(u)t4sin( π−π−−= LL
=+
−+
=π
16se4
16s4
2
s-
2 )e1(16s
4 s-2
π−⋅+
Chapter 15, Solution 9.
(a) )2t(u2)2t(u)2t()2t(u)4t()t(f −−−−=−−=
=)s(F 2
-2s
2
-2s
se2
se
−
(b) )1t(uee2)1t(ue2)t(g 1)--4(t-4-4t −=−=
=)s(G)4s(e
e24
-s
+
(c) )t(u)1t2cos(5)t(h −=
)Bsin()Asin()Bcos()Acos()BAcos( +=−
)1sin()t2sin()1cos()t2cos()1t2cos( +=−
)t(u)t2sin()1sin(5)t(u)t2cos()1cos(5)t(h +=
4s2
)1sin(54s
s)1cos(5)s(H 22 +
⋅++
⋅=
=)s(H4s
415.84ss702.2
22 ++
+
(d) )4t(u6)2t(u6)t(p −−−=
=)s(P 4s-2s- es6
es6
−
Chapter 15, Solution 10.
(a) By taking the derivative in the time domain, )tsin(et)tcos()ee-t()t(g -t-t-t −+=
)tsin(et)tcos(et)tcos(e)t(g -t-t-t −−=
++
+
++
++
+++
=1)1s(
1dsd
1)1s(1s
dsd
11)(s1s)s(G 222
=++
+−
+++
−++
+= 2222
2
2 )2s2(s22s
)2s2(s2ss
2s2s1s
)s(G 22
2
)2s2(s2)(ss+++
(b) By applying the time differentiation property,
)0(f)s(sF)s(G −= where , f)tcos(et)t(f -t= 0)0( =
=+++
=
++
+⋅= 22
2
2 )2s2(s2s)(s)(s
1)1s(1s
dsd-
)s()s(G 22
2
)2s2(s2)(ss+++
Chapter 15, Solution 11.
(a) Since [ ] 22 ass
)atcosh(−
=L
=−+
+=
4)1s()1s(6
)s(F 2 3s2s)1s(6
2 −++
(b) Since [ ] 22 asa
)atsinh(−
=L
[ ]12s4s
1216)2s(
)4)(3()t4sinh(e3 22
2t-
−+=
−+=L
[ ] [ ]1-22t- )12s4s(12dsd-
)t4sinh(e3t)s(F −+=⋅= L
=−++= 2-2 )12s4s)(4s2)(12()s(F 22 )12s4s()2s(24
−++
(c) )ee(21
)tcosh( t-t +⋅=
)2t(u)ee(21
e8)t(f t-t3t- −+⋅⋅=
= )2t(ue4)2t(ue4 -4t-2t −+− = )2t(uee4)2t(uee4 2)--4(t-82)--2(t-4 −+− [ ] [ ])t(ueee4)2t(uee4 -2-2s-42)--2(t-4 LL ⋅=−
[ ]2s
e4)2t(uee4
4)-(2s2)-2(t-4-
+=−
+
L
Similarly, [ ]4s
e4)2t(uee4
8)-(2s2)-4(t-8-
+=−
+
L
Therefore,
=+
++
=++
4se4
2se4
)s(F8)-(2s4)-(2s [ ]
8s6s)e8e16(s)e4e4(e
2
-22-226)-(2s
++++++
Chapter 15, Solution 12.
)2s(2
)2s(s2
2
2s
)1t(22)1t(222)1t(2
e)2s(3s
2s11
2se
2see
)2s(ee)s(f
)1t(uee)1t(uee)1t()1t(uete)t(f
+−+−−
−−
−
−−−−−−−−−
+
+=
++
+=
++
+=
−+−−=−=
Chapter 15, Solution 13.
(a) )()( sFdsdt −→←tf
If f(t) = cost, then 22
2
2 )1()12()1)(1()( and
1)(
++−+
=+
=s
ssssFdsd
sssF
22
2
)1(1)cos(
+−+
=sssttL
(b) Let f(t) = e-t sin t.
221
1)1(1)( 22 ++
=++
=sss
sF
22
2
)22()22)(1()0)(22(
+++−++
=ss
sssdsdF
22 )22()1(2)sin(++
+=−=−
sss
dsdFtte tL
(c ) ∫∞
→←s
dssFttf )()(
Let 22)( then ,sin)(β
ββ+
==s
sFttf
sssds
stt
ss
ββ
πββ
ββ
ββ 11122 tantan
2tan1sin −−∞−
∞
=−==+
=
∫L
Chapter 15, Solution 14.
<<−<<
=2t1t5101t0t5
)t(f
We may write f as )t(
[ ] [ ])2t(u)1t(u)t510()1t(u)t(ut5)t(f −−−−+−−= )2t(u)2t(5)1t(u)1t(10)t(ut5 −−+−−−=
s2-2
s-22 e
s5
es10
s5
)s(F +−=
=)s(F )ee21(s5 s2-s-2 +−
Chapter 15, Solution 15. [ ])2t(u)1t(u)1t(u)t(u10)t(f −+−−−−=
=
+−=s
ee
s2
s1
10)s(Fs2-
s- 2s- )e1(s
10−
Chapter 15, Solution 16.
)4t(u5)3t(u3)1t(u3)t(u5)t(f −−−+−−=
=)s(F [ ]s4-s3-s- e5e3e35s1
−+−
Chapter 15, Solution 17.
><<<<
<
=
3t03t111t0t
0t0
)t(f2
[ ] [ ])3t(u)1t(u1)1t(u)t(ut)t(f 2 −−−+−−=
)3t(u)1t(u)1t(u)1t2-()1t(u)1t()t(ut 22 −−−+−++−−−= )3t(u)1t(u)1t(2)1t(u)1t()t(ut 22 −−−−−−−−=
=)s(Fs
ee
s2
)e1(s2 s-3
s-2
s-3 −−−
Chapter 15, Solution 18.
(a) [ ] [ ])3t(u)2t(u3)2t(u)1t(u2)1t(u)t(u)t(g −−−+−−−+−−= )3t(u3)2t(u)1t(u)t(u −−−+−+=
=)s(G )e3ee1(s1 s3-s2-s- −++
(b) [ ] [ ])3t(u)1t(u2)1t(u)t(ut2)t(h −−−+−−=
[ ])4t(u)3t(u)t28( −−−−+ )3t(u2)1t(u2)1t(u2)1t(u)1t(2)t(ut2 −−−+−−−−−=
)4t(u)4t(2)3t(u2)3t(u)3t(2 −−+−+−−− )4t(u)4t(2)3t(u)3t(2)1t(u)1t(2)t(ut2 −−+−−−−−−=
=+−−= s4-2
s3-2
s-2 e
s2
es2
)e1(s2
)s(H )eee1(s2 s4-s3-s-2 +−−
Chapter 15, Solution 19.
Since [ ] 1)t( =δL and , 2T = =)s(F s2-e11−
Chapter 15, Solution 20.
Let 1t0),tsin()t(g1 <<π= [ ])1t(u)t(u)tsin( −−π= )1t(u)tsin()t(u)tsin( −π−π=
Note that sin( )tsin(-)tsin())1t( π=π−π=−π . So, 1)-u(t1))-t(sin(u(t)t)sin()t(g1 π+π=
)e1(s
)s(G s-221 +
π+π
=
=−
= 2s-1
e1)s(G
)s(G)e1)(s(
)e1(2s-22
-s
−π++π
Chapter 15, Solution 21.
π= 2T
Let [ ])1t(u)t(u2t
1)t(f1 −−
π−=
)1t(u21
1)1t(u)1t(21
)t(u2t
)t(u)t(f1 −
π−−−−
π+
π−=
[ ] [ ]
2
-s-ss-
2
s-
21 s2e1-se)12-(2
s1
e21
1-s2
es2
1s1
)s(Fπ
+++π+π=⋅
π++
π+
π−=
=−
= Ts-1
e1)s(F
)s(F[ ] [ ]
)e1(s2e1-se)12-(2
s2-2
-s-s
π−π+++π+π
Chapter 15, Solution 22.
(a) Let 1t0,t2)t(g1 <<= [ ])1t(u)t(ut2 −−= )1t(u2)1t(u)1t(2)t(ut2 −+−−−=
s-
2
-s
21 es2
se2
s2
)s(G +−=
1T,e1
)s(G)s(G sT-
1 =−
=
=)s(G)e1(s
)ese1(2s-2
-s-s
−+−
(b) Let h , where is the periodic triangular wave. )t(uh 0 += 0h
Let h be within its first period, i.e. 1 0h
<<−<<
=2t1t241t0t2
)t(h1
)2t(u)2t(2)1t(ut2)1t(u4)1t(ut2)t(ut2)t(h1 −−−−−−+−−=
)2t(u)2t(2)1t(u)1t(4)t(ut2)t(h1 −−−−−−=
2s-
22
-2ss-
221 )e1(s2
se2
es4
s2
)s(H −=−−=
)e1()e1(
s2
)s(H 2s-
2-s
20 −−
=
=)s(H)e1(
)e1(s2
s1
2s-
2-s
2 −−
+
Chapter 15, Solution 23.
(a) Let
<<<<
=2t11-1t01
)t(f1
[ ] [ ])2t(u)1t(u)1t(u)t(u)t(f1 −−−−−−=
)2t(u)1t(u2)t(u)t(f1 −+−−=
2s-2s-s-1 )e1(
s1
)ee21(s1
)s(F −=+−=
2T,)e1(
)s(F)s(F sT-
1 =−
=
=)s(F)e1(s
)e1(2s-
2-s
−−
(b) Let [ ] )2t(ut)t(ut)2t(u)t(ut)t(h 222
1 −−=−−= )2t(u4)2t(u)2t(4)2t(u)2t()t(ut)t(h 22
1 −−−−−−−−=
2s-2s-2
2s-31 e
s4
es4
)e1(s2
)s(H −−−=
2T,)e1(
)s(H)s(H Ts-
1 =−
=
=)s(H)e1(s
)ss(es4)e1(22s-3
2-2s-2s
−+−−
Chapter 15, Solution 24.
(a) 5s6s
ss10lim)s(sFlim)0(f 2
4
ss +++
==∞→∞→
==++
+
∞→ 010
s5
s6
s1
s110
lim432
3
s= ∞
=+++
==∞→→ 5s6s
ss10lim)s(sFlim)(f 2
4
0s0s0
(b) =+−
+==
∞→∞→ 6s4sss
lim)s(sFlim)0(f 2
2
ss1
The complex poles are not in the left-half plane.
)(f ∞ existnotdoes
(c) )5s2s)(2s)(1s(
s7s2lim)s(sFlim)0(f 2
3
ss +++++
==∞→∞→
==
++
+
+
+
∞→ 10
s5
s21
s21
s11
s7
s2
lim
2
3
s= 0
==++++
+==∞
→→ 100
)5s2s)(2s)(1s(s7s2
lim)s(sFlim)(f 2
3
0s0s0
Chapter 15, Solution 25.
(a) )4s)(2s()3s)(1s)(8(
lim)s(sFlim)0(fss ++
++==
∞→∞→
=
+
+
+
+
∞→
s41
s21
s31
s11)8(
lims
= 8
===∞→→ )4)(2(
)3)(1)(8(lim)s(sFlim)(f
0s0s3
(b) 1s
)1s(s6lim)s(sFlim)0(f 4ss −
−==
∞→∞→
==−
−=
∞→ 10
s1
1
s1
s1
6lim)0(f
4
42
s0
All poles are not in the left-half plane.
)(f ∞ existnotdoes Chapter 15, Solution 26.
(a) =++
+==
∞→∞→ 6s4ss3s
lim)s(sFlim)0(f 23
3
ss1
Two poles are not in the left-half plane.
)(f ∞ existnotdoes
(b) )4s2s)(2s(
ss2slim)s(sFlim)0(f 2
23
ss ++−+−
==∞→∞→
=
++
−
+−
∞→
2
2
s
s4
s21
s21
s1
s21
lim= 1
One pole is not in the left-half plane.
)(f ∞ existnotdoes Chapter 15, Solution 27.
(a) =)t(f -te2)t(u +
(b) 4s
113
4s11)4s(3
)s(G+
−=+
−+=
=)t(g -4te11)t(3 −δ
(c) 3s
B1s
A)3s)(1s(
4)s(H
++
+=
++=
2A = , -2B =
3s2
1s2
)s(H+
−+
=
=)t(h -3t-t e2e2 −
(d) 4s
C)2s(
B2s
A)4s()2s(
12)s(J 22 +
++
++
=++
=
62
12B == , 3
)-2(12
C 2 ==
2)2s(C)4s(B)4s()2s(A12 ++++++=
Equating coefficients :
2s : -3-CACA0 ==→+=1s : 6-2ABBA2C4BA60 ==→+=++=0s : 121224-24C4B4A812 =++=++=
4s3
)2s(6
2s3-
)s(J 2 ++
++
+=
=)t(j -2t-2t-4t et6e3e3 +−
Chapter 15, Solution 28. (a)
)t(ut)e4e2()t(f
5s4
3s2
5s2
)4(2
3s2
)2(2
)s(F
t5t3 −− +−=
++
+−
=+−−
++
−
=
(b)
( ) )t(ut2sine5.1t2cose2e2)t(h5s2s
1s21s
2)s(H
1C;2B;2A3CB,3CB;BA;11CA5
CA5s)CBA2(s)BA()1s)(CBs()5s2s(A11s3
5s2sCBs
1sA
)5s2s)(1s(11s3)s(H
ttt2
22
22
−−− +−=→++
+−+
+=
=−==→−==+−−==+
++++++=+++++=+
++
++
+=
+++
+=
Chapter 15, Solution 29.
2222
2222
3)2s(3
32
3)2s()2s(2
s2)s(V
6Band2A26s2BsAs26s8s2;3)2s(
BAss2)s(V
++−
++
+−=
−=−=→+=++++++
++=
v(t) = 0t,t3sine32t3cose2)t(u t2t2 ≥−− −−2
Chapter 15, Solution 30.
(a) 22222213)2s(
332
3)2s()2s(2
3)2s(2)2s(2)s(H
+++
++
+=
++
++=
t3sine32t3cose2)t(h t2t2
1−− +=
(b) )5s2s(
DCs)1s(
B)1s(
A)5s2s()1s(
4s)s( 2222
22
++
++
++
+=
+++
+=H
22222 )1s(D)1s(Cs)5s2s(B)5s2s)(1s(A4s ++++++++++=+
or
)1s2s(D)ss2s(C)5s2s(B)5s7s3s(A4s 2232232 ++++++++++++=+ Equating coefficients:
ACCA0:s3 −=→+=
DBADC2BA31:s2 ++=+++=
2/1C,2/1A2A4D2B2A6D2CB2A70:s =−=→+=++=+++=
4/1D,4/5B1B4A4DB5A54:constant ==→++=++=
++
−++
++
+−
=
++
++
++
+−
=)2)1s(
1)1s(2)1s(
5)1s(
241
)5s2s(1s2
)1s(5
)1s(2
41)s(H 222222
Hence,
( ) )t(ut2sine5.0t2cose2te5e241)t(h tttt
2−−−− −++−=
(c )
++
+=
++
+=
++
+= −−
−
)3s(1
)1s(1e
21
)3s(B
)1s(Ae
)3s)(1s(e)2s()s(H ss
s3
( ) )1t(uee21)t(h )1t(3)1t(
3 −+= −−−−
Chapter 15, Solution 31.
(a) 3s
C2s
B1s
A)3s)(2s)(1s(
s10)s(F
++
++
+=
+++=
-5210-
)1s()s(FA 1-s ==+= =
201-20-
)2s()s(FB -2s ==+= =
-15230-
)3s()s(FC 3-s ==+= =
3s15
2s20
1s5-
)s(F+
−+
++
=
=)t(f -3t-2t-t e15e20e5- −+
(b) 323
2
)2s(D
)2s(C
2sB
1sA
)2s)(1s(1s4s2
)s(F+
++
++
++
=++++
=
-1)1s()s(FA 1-s =+= =
-1)2s()s(FD -2s3 =+= =
)4s4s)(1s(B)4s4s)(2s(A1s4s2 222 +++++++=++
)1s(D)2s)(1s(C +++++ Equating coefficients :
3s : 1 -ABBA0 ==→+=2s : 3 A2CCACB5A62 =−=→+=++=1s : DC3A4DC3B8A124 ++=+++=
-1A-2DDA64 =−=→++=0s : 116-4D2C4ADC2B4A81 =−+=++=+++=
32 )2s(1
)2s(3
2s1
1s1-
)s(F+
−+
++
++
=
2t-
22t-2t-t- e
2t
et3ee-f(t) −++=
=)t(f 2t-2
t- e2t
t31e-
−++
(c) 5s2s
CBs2s
A)5s2s)(2s(
1s)s(F 22 ++
++
+=
++++
=
51-
)2s()s(FA -2s =+= =
)2s(C)s2s(B)5s2s(A1s 22 ++++++=+
Equating coefficients :
2s : 51
-ABBA0 ==→+=
1s : 1CC0CB2A21 =→+=++=0s : 121-C2A51 =+=+=
222222 2)1s(54
2)1s()1s(51
2s51-
2)1s(1s51
2s51-
)s(F++
++++
++
=+++⋅
++
=
=)t(f )t2sin(e4.0)t2cos(e2.0e0.2- -t-t-2t ++
Chapter 15, Solution 32.
(a) 4s
C2s
BsA
)4s)(2s(s)3s)(1s(8
)s(F+
++
+=++++
=
3)4)(2(
(8)(3)s)s(FA 0s === =
2)-4(
(8)(-1))2s()s(FB -2s ==+= =
3(-2))-4(
)(8)(-1)(-3)4s()s(FC -4s ==+= =
4s3
2s2
s3
)s(F+
++
+=
=)t(f -4t-2t e3e2)t(u3 ++
(b) 22
2
)2s(C
2sB
1sA
)2s)(1s(4s2s
)s(F+
++
++
=+++−
=
)1s(C)2s3s(B)4s4s(A4s2s 222 +++++++=+−
Equating coefficients :
2s : A1BBA1 −=→+=1s : CA3CB3A42- ++=++= 0s : -6B2B-CB2A44 =→−=++=
7B1A =−= -12A--5C ==
2)2s(12
2s6
1s7
)s(F+
−+
−+
=
=)t(f -2t-t e)t21(6e7 +−
(c) 5s4s
CBs3s
A)5s4s)(3s(
1s)s(F 22
2
+++
++
=+++
+=
)3s(C)s3s(B)5s4s(A1s 222 ++++++=+
Equating coefficients :
2s : A1BBA1 −=→+=1s : -3CACA3CB3A40 =+→++=++=0s : 5A2A-9C3A51 =→+=+=
-4A1B =−= -83A-C =−=
1)2s()2s(4
3s5
1)2s(8s4
3s5
)s(F 22 +++
−+
=++
+−
+=
=)t(f )tcos(e4e5 -2t-3t −
Chapter 15, Solution 33.
(a) 1s
C1sBAs
)1s)(1s(6
1s)1s(6
)s(F 224 ++
++
=++
=−−
=
)1s(C)1s(B)ss(A6 22 +++++=
Equating coefficients :
2s : -CACA0 =→+=1s : C -ABBA0 ==→+=0s : 3B2BCB6 =→=+=
-3A = , 3B = , 3C =
1s3
1s3s-
1s3
1s33s-
1s3
)s(F 222 ++
++
+=
++
++
=
=)t(f )tcos(3)tsin(3e3 -t −+
(b) 1s
es)s(F 2
s-
+=
π
=)t(f )t(u)tcos( π−π−
(c) 323 )1s(D
)1s(C
1sB
sA
)1s(s8
)s(F+
++
++
+=+
=
8A = , -8D =
sD)ss(C)ss2s(B)1s3s3s(A8 22323 +++++++++=
Equating coefficients :
3s : -ABBA0 =→+=2s : B-ACCACB2A30 ==→+=++=1s : -ADDADCBA30 =→+=+++=0s : 8D,8C,8B,8A −=−=−==
32 )1s(8
)1s(8
1s8
s8
)s(F+
−+
−+
−=
=)t(f [ ] )t(uet5.0ete18 -t2-t-t −−−
Chapter 15, Solution 34.
(a) 4s
311
4s34s
10)s(F 22
2
+−=
+−+
+=
=)t(f )t2sin(5.1)t(11 −δ
(b) )4s)(2s(
e4e)s(G
-2s-s
+++
=
Let 4s
B2s
A)4s)(2s(
1+
++
=++
21A = 21B =
++
++
++
+=
4s1
2s1
e24s
12s
12
e)s(G 2s-
-s
=)t(g [ ] [ ] )2t(uee2)1t(uee5.0 2)--4(t2)--2(t-1)-4(t-1)-2(t −−+−−
(c) Let 4s
C3s
BsA
)4s)(3s(s1s
++
++=
+++
121A = , 32B = , 43-C =
2s-e
4s43
3s32
s1
121
)s(H
+−
++⋅=
=)t(h )2t(ue43
e32
121 2)-4(t-2)-3(t- −
−+
Chapter 15, Solution 35.
(a) Let 2s
B1s
A)2s)(1s(
3s)s(G
++
+=
+++
=
2A = , -1B =
2t-t- ee2)t(g
2s1
1s2
)s(G −=→+
−+
=
)6t(u)6t(g)t(f)s(Ge)s(F -6s −−=→=
=)t(f [ ] )6t(uee2 6)--2(t6)--(t −−
(b) Let 4s
B1s
A)4s)(1s(
1)s(G
++
+=
++=
31A = , 31-B =
)4s(31
)1s(31
)s(G+
−+
=
[ ]4t-t- ee31
)t(g −=
)s(Ge)s(G4)s(F -2t−=
)2t(u)2t(g)t(u)t(g4)t(f −−−=
=)t(f [ ] [ ] )2t(uee31
)t(uee34 2)-4(t-2)-(t-4t-t- −−−−
(c) Let 4sCBs
3sA
)4s)(3s(s
)s(G 22 ++
++
=++
=
133-A =
)3s(C)s3s(B)4s(As 22 +++++=
Equating coefficients :
2s : -ABBA0 =→+=1s : CB31 +=0s : C3A40 +=
133-A = , 133B = , 134C =
4s4s3
3s3-
)s(G13 2 ++
++
=
)t2sin(2)t2cos(3e-3)t(g13 -3t ++=
)s(Ge)s(F -s=
)1t(u)1t(g)t(f −−=
=)t(f [ ] )1t(u))1t(2sin(2))1t(2cos(3e3-131 1)-3(t- −−+−+
Chapter 15, Solution 36.
(a) 3s
D2s
CsB
sA
)3s)(2s(s1
)s(X 22 ++
+++=
++=
61B = , 41C = , 91-D =
)s2s(D)s3s(C)6s5s(B)s6s5s(A1 2323223 +++++++++=
Equating coefficients :
3s : DCA0 ++= 2s : CBA3D2C3BA50 ++=+++= 1s : B5A60 +=0s : 61BB61 =→=
365-B65-A ==
3s91
2s41
s61
s365-
)s(X 2 +−
+++=
=)t(x 3t-2t- e91
e41
t61
)t(u36
5-−++
(b) 22 )1s(C
1sB
sA
)1s(s1
)s(Y+
++
+=+
=
1A = , 1-C =
sC)ss(B)1s2s(A1 22 +++++=
Equating coefficients :
2s : -ABBA0 =→+=1s : -ACCACBA20 =→+=++=0s : -1C-1,B,A1 ===
2)1s(1
1s1
s1
)s(Y+
−+
−=
=)t(y -t-t ete)t(u −−
(c) 10s6s
DCs1s
BsA
)s(Z 2 +++
++
+=
101A = , 51-B =
)ss(D)ss(C)s10s6s(B)10s16s7s(A1 2232323 ++++++++++=
Equating coefficients :
3s : CBA0 ++= 2s : DB5A6DCB6A70 ++=+++= 1s : A 2-BB5A10DB10A160 =→+=++=0s : 101AA101 =→=
101A = , 51--2AB == , 101AC == , 104
A4D ==
10s6s4s
1s2
s1
)s(Z10 2 +++
++
−=
1)3s(1
1)3s(3s
1s2
s1
)s(Z10 22 +++
+++
++
−=
=)t(z [ ] )t(u)tsin(e)tcos(ee211.0 -3t-3t-t ++−
Chapter 15, Solution 37.
(a) Let 4sCBs
sA
)4s(s12
)s(P 22 ++
+=+
=
3412s)s(PA 0s === =
sCsB)4s(A12 22 +++=
Equating coefficients :
0s : 3AA412 =→=1s : C0 =2s : -3-ABBA0 ==→+=
4ss3
s3
)s(P 2 +−=
)t2cos(3)t(u3)t(p −=
)s(Pe)s(F -2s=
=)t(f [ ] )2t(u))2t(2cos(13 −−−
(b) Let 9sDCs
1sBAs
)9s)(1s(1s2
)s(G 2222 ++
+++
=++
+=
)1s(D)ss(C)9s(B)s9s(A1s2 2323 +++++++=+
Equating coefficients :
3s : -ACCA0 =→+=
2s : -BDDB0 =→+=1s : 82-C,82AA8CA92 ==→=+= 0s : 81-D,81BB8DB91 ==→=+=
++
−
++
=9s1s2
81
1s1s2
81
)s(G 22
9s1
81
9ss
41
1s1
81
1ss
41
)s(G 2222 +⋅−
+⋅−
+⋅+
+⋅=
=)t(g )t3sin(241
)t3cos(41
)tsin(81
)tcos(41
−−+
(c) Let 13s4s
117s369
13s4ss9
)s(H 22
2
+++
−=++
=
2222 3)2s(3
153)2s(
2s369)s(H
++⋅−
+++
⋅−=
=)t(h )t3sin(e15)t3cos(e36)t(9 t2-t2- −−δ
Chapter 15, Solution 38.
(a) 26s10s
26s626s10s26s10s
s4s)s(F 2
2
2
2
++−−++
=++
+=
26s10s26s6
1)s(F 2 +++
−=
2222 1)5s(4
1)5s()5s(6
1)s(F++
+++
+−=
=)t(f )t5sin(e4)t5cos(e6)t( -t-t +−δ
(b) 29s4s
CBssA
)29s4s(s29s7s5
)s(F 22
2
+++
+=++++
=
sCsB)29s4s(A29s7s5 222 ++++=++
Equating coefficients :
0s : 1AA2929 =→=1s : 3A47CCA47 =−=→+=2s : 4 A5BBA5 =−=→+=
1A = , 4B = , 3C =
22222 5)2s(5
5)2s()2s(4
s1
29s4s3s4
s1
)s(F++
−++
++=
+++
+=
=)t(f )t5sin(e)t5cos(e4)t(u -2t-2t −+
Chapter 15, Solution 39.
(a) 20s4s
DCs17s2s
BAs)20s4s)(17s2s(
1s4s2)s(F 2222
23
+++
+++
+=
++++++
=
)20s4s(B)s20s4s(A1s4s 22323 +++++=++
)17s2s(D)s17s2s(C 223 ++++++Equating coefficients :
3s : CA2 +=2s : DC2BA44 +++= 1s : D2C17B4A200 +++= 0s : D17B201 +=
Solving these equations (Matlab works well with 4 unknowns),
-1.6A = , -17.8B = , 6.3C = , 21D =
20s4s21s6.3
17s2s17.81.6s-
)s(F 22 +++
+++
−=
22222222 4)2s()4((3.45)
4)2s()2(3.6)(s
4)1s()4((-4.05)
4)1s(1)(-1.6)(s
)s(F++
++++
+++
++++
=
=)t(f )t4sin(e45.3)t4cos(e6.3)t4sin(e05.4)t4cos(e6.1- -2t-2t-t-t ++−
(b) 3s6s
DCs9sBAs
)3s6s)(9s(4s
)s(F 2222
2
+++
+++
=+++
+=
)9s(D)s9s(C)3s6s(B)s3s6s(A4s 232232 +++++++++=+
Equating coefficients :
3s : -ACCA0 =→+=2s : DBA61 ++= 1s : A -CBC6B6C9B6A30 ==→+=++=0s : D9B34 +=
Solving these equations,
121A = , 121B = , 121-C = , 125D =
3s6s5s-
9s1s
)s(F12 22 +++
+++
=
5.449--0.551,2
12-366-03s6s2 =
±→=++
Let 449.5s
F551.0s
E3s6s
5s-)s(G 2 +
++
=++
+=
133.1449.5s5s-
E -0.551s =+
+= =
2.133- 551.0s5s-
F -5.449s =++
= =
449.5s133.2
551.0s133.1
)s(G+
−+
=
449.5s133.2
551.0s133.1
3s3
31
3ss
)s(F12 2222 +−
++
+⋅+
+=
=)t(f -5.449t-0.551t e1778.0e0944.0)t3sin(02778.0)t3cos(08333.0 −++
Chapter 15, Solution 40.
Let 5s2s
CBs2s
A)5s2s)(2s(
13s7s4)s( 22
2
++
++
+=
+++
++=H
)2s(C)s2s(B)5s2s(A13s7s4 222 ++++++=++
Equating coefficients gives:
BA4:s2 +=
1CCB2A27:s −=→++=
1B 3,Aor 15A5C2A513:constant ===→+=
222 2)1s(2)1s(
2s3
5s2s1s
2s3)s(H
++
−++
+=
++
−+
+=
Hence,
)t2sinsinAt2coscosA(ee3t2sinet2cosee3)t(h tt2ttt2 α−α+=−+= −−−−−
where o45,2A1sinA,1cosA =α=→=α=α Thus,
[ ] )t(ue3)45t2cos(e2)t(h t2ot −− ++= Chapter 15, Solution 41. Let y(t) = f(t)*h(t). For 0 < t < 1, f )t( λ− )(λh 0 t 1 2 λ
2t0
t
0
2 t22d4)1()t(y =λ=λλ= ∫
For 1 <t<3, )t(f λ− )(h λ 0 1 t 2
4t2t8)28(2d)48)(1(d4)1()t(y 2t1
2t0
2t
1
1
0−−=λ−λ+λ=λλ−+λλ= ∫∫
For 3 < t < 4 h )(λ )t(f λ− 0 1 t-2 2 3 t 4 λ
222t
2
2t
2 t2t163228d)48()t(y +−=λ−λ=λλλ−= −−∫
Thus,
<<+<<−
<<
=
otherwise 0,4 t 3 ,2t16t-32
3 t 1 ,42t-8t1 t 0 ,t2
)t(y 2
2
2
Chapter 15, Solution 42.
(a) For , 1t0 << )t(f1 λ− and f )(2 λ overlap from 0 to t, as shown in Fig. (a).
2t
2d))(1()t(f)t(f)t(y
2t0
2t
021 =λ
=λλ=∗= ∫
1
λt-1 t 1 0
1
λt-1 t 1
f2(λ)f1(t - λ)
0
(a) (b) For 1 , 2t << )t(f1 λ− and f )(2 λ overlap as shown in Fig. (b).
2t
t2
d))(1()t(y2
11t
21
1t−=
λ=λλ= −−
∫
For , there is no overlap. 2t > Therefore,
=)t(y
<<−<<
otherwise,02t1,2tt1t0,2t
2
2
(b) For , the two functions overlap as shown in Fig. (c). 1t0 <<
td)1)(1()t(f)t(f)t(yt
021 =λ=∗= ∫
1
λt-1 t 1 0
1
λt-1 t 1
f2(λ)f1(t - λ)
0
(c) (d)
For 1 , the functions overlap as shown in Fig. (d). 2t <<
t2d)1)(1()t(y 11t
1
1t−=λ=λ= −−
∫ For , there is no overlap. 2t > Therefore,
=)t(y
<<−<<
otherwise,02t1,t21t0,t
(c) For , there is no overlap. For -1-t < 0t1 << , f )t(1 λ− and overlap
as shown in Fig. (e). )(f2 λ
t1-
2t
1-21 2d)1)(1()t(f)t(f)t(y
λ+λ
=λ+λ=∗= ∫
22 )1t(
21
)1t2t(21
)t(y +=++=
-1
1
λ t 1 0-1
1
λt 1
f1(λ)f2(t - λ)
0
(e) (f) For , the functions overlap as shown in Fig. (f). 1t0 <<
∫∫ λλ−+λ+λ=t
0
0
1-d)1)(1(d)1)(1()t(y
t0
201-
2
22)t(y
λ
−λ+
λ+λ
=
)tt21(21
)t(y 2−+=
For , the two functions overlap. 1t >
∫∫ λλ−+λ+λ=1
0
0
1-d)1)(1(d)1)(1()t(y
121
121
221
)t(y 10
2
=−+=
λ
−λ+=
Therefore,
=)t(y
><<++<<++
<
1t1,1t0),1t2t-(5.00t1-),1t2t(5.0
-1t,0
2
2
Chapter 15, Solution 43.
(a) For , 1t0 << )t(x λ− and )(h λ overlap as shown in Fig. (a).
2t
2d))(1()t(h)t(x)t(y
2t0
2t
0=
λ=λλ=∗= ∫
1
λt-1 t 1
h(λ)
x(t - λ)
0
1
λ t-1 t 1 0
(a) (b) For 1 , 2t << )t(x λ− and )(h λ overlap as shown in Fig. (b).
1t2t21-
2d)1)(1(d))(1()t(y 2t
11
1t
2t
1
1
1t−+=λ+
λ=λ+λλ= −−
∫∫
For , there is a complete overlap so that 2t >
1)1t(td)1)(1()t(y t1t
t
1t=−−=λ=λ= −−
∫
Therefore,
=)t(y
><<−+<<
otherwise,02t,1
2t1,1t2)2t(-1t0,2t
2
2
(b) For , the two functions overlap as shown in Fig. (c). 0t >
t0
-t
0- e-2de2)1()t(h)t(x)t(y λλ =λ=∗= ∫
2
1
h(λ) = 2e-λ
λ0 t
x(t-λ)
(c) Therefore,
=)t(y 0t),e1(2 -t >−
(c) For , 0t1- << )t(x λ− and )(h λ overlap as shown in Fig. (d).
21t0
21t
0)1t(
21
2d))(1()t(h)t(x)t(y +=
λ=λλ=∗= +
+∫
2-1 t+1
1
λ t-1 t 1
h(λ)x(t - λ)
0
(d) For , 1t0 << )t(x λ− and )(h λ overlap as shown in Fig. (e).
∫∫ +λλ−+λλ=
1t
1
1
0d)2)(1(d))(1()t(y
21
tt21-
22
2)t(y 21t
1
210
2
++=
λ
−λ+λ
= +
t 2t+1
1
λt-1-1 10
(e) For 1 , 2t << )t(x λ− and )(h λ overlap as shown in Fig. (f).
∫∫ λλ−+λλ=−
2
1
1
1td)2)(1(d))(1()t(y
21
tt21-
22
2)t(y 22
1
21
1t
2
++=
λ
−λ+λ
= −
1
t 2 λt-1 10 t+1
(f) For , 3t2 << )t(x λ− and )(h λ overlap as shown in Fig. (g).
221t
22
1tt
21
t329
22d)2)(1()t(y +−=
λ
−λ=λλ−= −−∫
1
t2 λ t-110 t+1
(g)
Therefore,
=)t(y
<<+−<<++<<++
otherwise,03t2,29t3)2t(2t0,21t)2t(-0t1-,21t)2t(
2
2
2
Chapter 15, Solution 44.
(a) For , 1t0 << )t(x λ− and )(h λ overlap as shown in Fig. (a).
td)1)(1()t(h)t(x)t(yt
0=λ=∗= ∫
0
h(λ)1x(t - λ)
-1
t 2 λt-1 1
(a) For 1 , 2t << )t(x λ− and )(h λ overlap as shown in Fig. (b).
t23d)1)(1-(d)1)(1()t(y t1
11t
t
1
1
1t−=λ−λ=λ+λ= −−
∫∫ For , 3t2 << )t(x λ− and )(h λ overlap as shown in Fig. (c).
3t-d)-1)(1()t(y 21t
2
1t−=λ=λ= −−
∫
0 λ1 t-1 2 t
-1
1
0 2 λ1 t-1 t
-1
1
(b) (c) Therefore,
=)t(y
<<−<<−<<
otherwise,03t2,3t2t1,t231t0,t
(b) For , there is no overlap. For 2t < 3t2 << , f )t(1 λ− and overlap,
as shown in Fig. (d). )(f2 λ
∫ λλ−=∗=t
221 d)t)(1()t(f)t(f)t(y
2t22t
2t
2t2
2
+−=
λ−λ=
t-1 t
1
1 5432 λ
f2(λ) f1(t - λ)
0
(d)
0 1
1
2 3 4 5tt-1 λ (e)
For , 5t3 << )t(f1 λ− and )(f2 λ overlap as shown in Fig. (e).
21
2td)t)(1()t(y t
1t
2t
1t=
λ
−λ=λλ−= −−∫
For , the functions overlap as shown in Fig. (f). 6t5 <<
12t5t21-
2td)t)(1()t(y 25
1t
25
1t−+=
λ
−λ=λλ−= −−∫
0 1 λ
1
2 3 4 5 t t-1
(f) Therefore,
=)t(y
<<−+<<<<+−
otherwise,06t5,12t5)2t(-5t3,213t2,2t2)2t(
2
2
Chapter 15, Solution 45.
(a) t
t
0)(fd)t()(f)t()t(f =λλ=λλ−δλ=δ∗ ∫
)t(f)t()t(f =δ∗
(b) ∫ λλ−λ=∗t
0d)t(u)(f)t(u)t(f
Since
>λ<λ
=λ−t0t1
)t(u
∫ λλ=∗t
od)(f)t(u)t(f
Alternatively,
s
)s(F)t(u)t(f =∗L
∫ λλ=∗=
−t
o1 d)(f)t(u)t(f
s)s(F
L
Chapter 15, Solution 46.
(a) Let ∫ λλ−=∗=t
0 1221 d)t(x)t(x)t(x)t(x)t(y For , 3t0 << )t(x1 λ− and )(x 2 λ overlap as shown in Fig. (a).
)ee(4dee4dee4)t(y t2-t-t
0-t-t
0)-(t-2- −=λ=λ= ∫∫ λλλ
4
3t0 λ
(a) For , the two functions overlap as shown in Fig. (b). 3t >
( ) e1(e4e-e4dee4)t(y 3-t-30
-t-3
0)-(t-2- −==λ= λλλ∫
λ
4
3 t0
(b)
x1(t-λ)
x2(λ)
)
Therefore,
=)t(y
>−<<−3t),e1(e4
3t0),ee(43-t-
t-2t-
(b) For 1 , 2t << )(x1 λ and )t(x 2 λ− overlap as shown in Fig. (c).
1td)1)(1()t(x)t(x)t(y t1
t
121 −=λ=λ=∗= ∫
1 3
1
λt-1 t 2
x1(λ)x2(t - λ)
0
(c) For , the two functions overlap completely. 3t2 <<
1)1t(td)1)(1()t(y t1t
t
1t=−−=λ=λ= −−
∫ For , the two functions overlap as shown in Fig. (d). 4t3 <<
t4d)1)(1()t(y 31t
3
1t−=λ=λ= −−
∫
1 3
1
λ t-1 t20
(d)Therefore,
=)t(y
<<−<<<<−
otherwise,04t3,t43t2,12t1,1t
(c) For , 0t1- << )t(x1 λ− and )(x 2 λ overlap as shown in Fig. (e).
∫ λ=∗= λt
1-)-(t-
21 de4)1()t(x)t(x)t(y
[ ]1)(t-t
1-t- e14dee4)t(y +λ −=λ= ∫
λ1-1 t-1
1
x1(t - λ)
x2(λ)
(e) For , 1t0 <<
∫∫ λ+λ= λλt
0)-(t-0
1-)-(t- de4)1-(de4)1()t(y
4e4e8ee4ee4)t(y 1)-(t-tt0
-t01-
-t −−=−= +λλ For , the two functions overlap completely. 1t >
∫∫ λ+λ= λλ1
0)-(t-0
1-)-(t- de4)1-(de4)1()t(y
1)-(t1)-(t-t10
-t01-
-t e4e4e8ee4ee4)t(y −+λλ −−=−= Therefore,
=)t(y[ ]
>−−<<−−<<−
−+
+
+
1t,e4e4e81t0,4e4e80t1-,e14
1)(t-1)(t-t-
1)(t-t-
1)-(t
Chapter 15, Solution 47.
)tcos()t(f)t(f 21 ==
[ ] ∫ λλ−λ=t
0211- d)tcos()cos()s(F)s(FL
[ ])BAcos()BAcos(21
)Bcos()Acos( −++=
[ ] ∫ λλ−+=t
0211- d)]2tcos()t[cos(
21)s(F)s(FL
[ ] t0
t021
1-
2-)2tsin(
21
)tcos(21
)s(F)s(Fλ−
⋅+λ⋅=L
[ ] =)s(F)s(F 21
1-L )tsin(5.0)tcos(t5.0 + Chapter 15, Solution 48.
(a) Let 222 2)1s(2
5s2s2
)s(++
=++
=G
)t2sin(e)t(g -t=
)s(G)s(G)s(F =
[ ] λλ−λ== ∫ d)t(g)(g)s(G)s(G)t(f
t
01-L
∫ λλ−λ= λ−λt
0)t(-- d))t(2sin(e)2sin(e)t(f
[ ])BAcos()BAcos(21
)Bsin()Asin( +−−=
[ ]∫ λλ−−= λt
0-t- d))2t(2cos()t2cos(ee
21
)t(f
∫∫ λλ−−λ= λλt
02-
-tt
02-
-t
d)4t2cos(e2e
de)t2cos(2e
)t(f
[ ]∫ λλ+λ−⋅= λλ t
02-
-tt0
-2-t
d)4sin()t2sin()4cos()t2cos(e2e
2-e
)t2cos(2
e)t(f
∫ λλ−+= λt
02-
-tt2-t- d)4cos(e)t2cos(
2e
)1e-()t2cos(e41
)t(f
∫ λλ− λt
02-
-t
d)4sin(e)t2sin(2e
)e1()t2cos(e41
)t(f t2-t- −=
( ) t0
-2-t
)4sin(4)2cos(4-164
e)t2cos(
2e
λ−λ+
−λ
( ) t0
-2-t
)4cos(4)2sin(4-164
e)t2sin(
2e
λ+λ+
−λ
=)t(f )t4cos()t2cos(20e
)t2cos(20e
)t2cos(4
e)t2cos(
2e -3t-t-3t-t
+−−
)t2sin(10e
)t4sin()t2cos(10e -t-3t
++
)t4cos()t2sin(10e
)t4sin()t2sin(20e -t-t
−+
(b) Let 1s
2)s(X
+= ,
4ss
)s(Y+
=
)t(ue2)t(x -t= , )t(u)t2cos()t(y =
)s(Y)s(X)s(F =
[ ] ∫∞ λλ−λ==
01- d)t(x)(y)s(Y)s(X)t(f L
∫ λ⋅λ= λ−t
0)(t- de2)2cos()t(f
( ) t0
t- )2sin(2)2cos(41
ee2)t(f λ+λ
+⋅=
λ
( )[ ]1)t2sin(2)t2cos(ee52
)t(f tt- −+=
=)t(f t-e52
)t2sin(54
)t2cos(52
−+
Chapter 15, Solution 49.
Let x(t) = u(t) – u(t-1) and y(t) = h(t)*x(t).
[ ]
+−
=
−
+==
−−
−−−
)2s(s)e1(4)
se
s1()s(X)s(H)t(y
s1
s11 L
2s4LL
But
+−=
++=
+ 2s1
s1
21
2sB
sA
)2s(s1
++−
+−=
−−
2se
se
2s1
s12)s(Y
ss
)1t(u]e1[4)t(u]e1[2)t(y )1t(2t2 −−−−= −−−
Chapter 15, Solution 50.
Take the Laplace transform of each term.
[ ] [ ]4s
s3)s(V10)0(v)s(Vs2)0(v)0(vs)s(Vs 2
2
+=+−+′−−
4ss3
)s(V102)s(Vs22s)s(Vs 22
+=+−++−
4ss7s
4ss3
s)s(V)10s2s( 2
3
22
++
=+
+=++
10s2sDCs
4sBAs
)10s2s)(4s(s7s
)s(V 2222
3
+++
+++
=+++
+=
)4s(D)s4s(C)10s2s(B)s10s2s(As7s 232233 +++++++++=+
Equating coefficients :
3s : A1CCA1 −=→+=2s : DBA20 ++= 1s : 4B2A6C4B2A107 ++=++=
0s : B-2.5DD4B100 =→+=
Solving these equations yields
269
A = , 2612
B = , 2617
C = , 2630-
D =
++−
+++
=10s2s
30s174s
12s9261
)s(V 22
++
−++
+⋅+
+⋅+
+= 222222 3)1s(
473)1s(
1s17
4s2
64s
s9261
)s(V
=)t(v )t3sin(e7847
)t3cos(e2617
)t2sin(266
)t2cos(269 t-t- −++
Chapter 15, Solution 51.
Taking the Laplace transform of the differential equation yields
[ ] [ ]1s
10)s(V6)]0(v)s(sV5)0('v)0(sv)s(Vs2+
=+−+−−
or ( ))3s)(2s)(1s(
24s16s2)s(V1s
10104s2)s(V6s5s2
2+++
++=→
+=−−−++
Let 3C,0B,5A,3s
C2s
B1s
A)s( −===+
++
++
=V
Hence,
)t(u)e3e5()t(v t3t −− −=
Chapter 15, Solution 52.
Take the Laplace transform of each term. [ ] [ ] 01)s(I2)0(i)s(Is3)0(i)0(is)s(Is2 =++−+′−−
0133s)s(I)2s3s( 2 =+−−−++
2sB
1sA
)2s)(1s(5s
)s(I+
++
=++
+=
4A = , 3-B =
2s3
1s4
)s(I+
−+
=
=)t(i )t(u)e3e4( -2t-t −
Chapter 15, Solution 53.
Take the Laplace transform of each term.
[ ] [ ]4s
s)s(V6)0(y)s(Ys5)0(y)0(ys)s(Ys 2
2
+=+−+′−−
4ss
54s)s(Y)6s5s( 22
+=−−−++
4s)4s)(9s(s
4ss
9s)s(Y)6s5s( 2
2
22
++++
=+
++=++
4sDCs
3sB
2sA
)4s)(3s)(2s(36s5s9s
)s(Y 22
23
++
++
++
=+++
+++=
427
)s(Y)2s(A -2s =+= = , 1375-
)s(Y)3s(B -3s =+= =
When , 0s =
265
D4D
3B
2A
)4)(3)(2(36
=→++=
When , 1s =
521C
5D
5C
4B
3A
)5)(12(546
=→+++=+
Thus,4s
265s5213s
13752s427
)s(Y 2 ++⋅
++
−+
=
=)t(y )t2sin(525
)t2cos(521
e1375
e4
27 3t-2t- ++−
Chapter 15, Solution 54.
Taking the Laplace transform of the differential equation gives
[ ] [ ]3s
5)s(V2)0(v)s(Vs3)0(v)0(vs)s(Vs2
+=+−+′−−
3ss2
13s
5)s(V)2s3s( 2
+−
=−+
=++
)3s)(2s)(1s(s2
)2s3s)(3s(s2
)s(V 2 +++−
=+++
−=
3sC
2sB
1sA
)s(V+
++
++
=
23A = , -4B = , 25C =
3s25
2s4
1s23
)s(V+
++
−+
=
=)t(v )t(u)e5.2e4e5.1( -3t-2t-t +−
Chapter 15, Solution 55.
Take the Laplace transform of each term. [ ] [ ])0(y)0(ys)s(Ys6)0(y)0(ys)0(ys)s(Ys 223 ′−−+′′−′−−
[ ] 22 2)1s(1s)0(y)s(Ys8++
+=−+
Setting the initial conditions to zero gives
5s2s1s
)s(Y)s8s6s( 223
+++
=++
5s2sEDs
4sC
2sB
sA
)5s2s)(4s)(2s(s)1s(
)s(Y 22 +++
++
++
+=++++
+=
401
A = , 201
B = , 104
3-C = ,
653-
D = , 657-
E =
22 2)1s(7s3
651
4s1
1043
2s1
201
s1
401
)s(Y++
+⋅−
+⋅−
+⋅+⋅=
2222 2)1s(4
651
2)1s()1s(3
651
4s1
1043
2s1
201
s1
401
)s(Y++
⋅−++
+⋅−
+⋅−
+⋅+⋅=
=)t(y )t2sin(e652
)t2cos(e653
e104
3e
201
)t(u401 t-t-4t-2t- −−−+
Chapter 15, Solution 56.
Taking the Laplace transform of each term we get:
[ ] 0)s(Vs
12)0(v)s(Vs4 =+−
8)s(Vs
12s4 =
+
3ss2
12s4s8
)s(V 22 +=
+=
=)t(v ( )t3cos2
Chapter 15, Solution 57.
Take the Laplace transform of each term.
[ ]4s
s)s(Y
s9
)0(y)s(Ys 2 +=+−
4s4ss
4ss
1)s(Ys
9s2
2
2
2
+++
=+
+=
+
9sDCs
4sBAs
)9s)(4s(s4ss
)s(Y 2222
23
++
+++
=++
++=
)4s(D)s4s(C)9s(B)s9s(As4ss 232323 +++++++=++
Equating coefficients :
0s : D4B90 +=1s : C4A94 +=2s : DB1 +=3s : CA1 +=
Solving these equations gives
0A = , 54-B = , 1C = , 59D =
9s59
9ss
4s54-
9s59s
4s54-
)s(Y 22222 ++
++
+=
++
++
=
=)t(y )t3sin(6.0)t3cos()t2sin(0.4- ++
Chapter 15, Solution 58.
We take the Laplace transform of each term and obtain
10s6sse)s(Ve)s(V
s10)]0(v)s(sV[)s(V6 2
s2s2
++=→=+−+
−−
1)3s(e3e)3s()s(V
2
s2s2
++
−+=
−−
Hence, 3( 2) 3( 2)( ) cos( 2) 3 sin( 2) ( 2)t tv t e t e t u t− − − − = − − − −
Chapter 15, Solution 59.
Take the Laplace transform of each term of the integrodifferential equation.
[ ]2s
6)s(Y
s3
)s(Y4)0(y)s(Ys+
=++−
−+
=++ 12s
6s)s(Y)3s4s( 2
)3s)(2s)(1s(s)s4(
)3s4s)(2s()s4(s
)s(Y 2 +++−
=+++
−=
3sC
2sB
1sA
)s(Y+
++
++
=
5.2A = , 6B = , -10.5C =
3s5.10
2s6
1s5.2
)s(Y+
−+
++
=
=)t(y -3t-2t-t e5.10e6e5.2 −+
Chapter 15, Solution 60.
Take the Laplace transform of each term of the integrodifferential equation.
[ ]16s
4s4
)s(Xs3
)s(X5)0(x)s(Xs2 2 +=+++−
16s64s36s4s2
16ss4
4s2)s(X)3s5s2( 2
23
22
+−+−
=+
+−=++
)16s)(5.1s)(1s(32s18s2s
)16s)(3s5s2(64s36s4s2
)s(X 2
23
22
23
+++−+−
=+++−+−
=
16sDCs
5.1sB
1sA
)s(X 2 ++
++
++
=
-6.235)s(X)1s(A -1s =+= =
329.7)s(X)5.1s(B -1.5s =+= =
When , 0s =
2579.0D16D
5.1B
A(1.5)(16)
32-=→++=
)16s16ss(B)24s16s5.1s(A32s18s2s 232323 +++++++=−+−
)5.1s5.2s(D)s5.1s5.2s(C 223 ++++++
Equating coefficients of the s terms, 3
0935.0-CCBA1 =→++=
16s0.25790.0935s-
5.1s329.7
1s6.235-
)s(X 2 ++
++
++
=
=)t(x )t4sin(0645.0)t4cos(0935.0e329.7e6.235- -1.5t-t +−+
Chapter 16, Solution 1.
Consider the s-domain form of the circuit which is shown below.
I(s)
+ −
1
1/s 1/s
s
222 )23()21s(1
1ss1
s1s1s1
)s(I++
=++
=++
=
= t
23sine
32)t(i 2t-
=)t(i A)t866.0(sine155.1 -0.5t
Chapter 16, Solution 2.
8/s s
s4 2
+ −
+
Vx
−
4
V)t(u)e2e24(v
38j
34s
125.0
38j
34s
125.0s25.016
)8s8s3(s2s16V
s32s16)8s8s3(V
0VsV)s4s2(s
)32s16()8s4(V
0
s84
0V2
0Vs
s4V
t)9428.0j3333.1(t)9428.0j3333.1(x
2x
2x
x2
x2
x
xxx
−−+− ++−=
−+
−+
++
−+−=
++
+−=
+=++
=++++
−+
=+
−+
−+
−
vx = Vt3
22sine2
6t3
22cose)t(u4 3/t43/t4
−
− −−
Chapter 16, Solution 3. s
5/s 1/2 +
Vo
−
1/8
Current division leads to:
)625.0s(165
s16105
s81
21
21
s5
81Vo +
=+
=
++=
vo(t) = ( ) V)t(ue13125. t625.0−−0
Chapter 16, Solution 4.
The s-domain form of the circuit is shown below.
6 s
10/s1/(s + 1) + −
+
Vo(s)
−
Using voltage division,
+++
=
+++
=1s
110s6s
101s
1s106s
s10)s(V 2o
10s6sCBs
1sA
)10s6s)(1s(10
)s(V 22o +++
++
=+++
=
)1s(C)ss(B)10s6s(A10 22 ++++++=
Equating coefficients :
2s : -ABBA0 =→+=1s : A5-CCA5CBA60 =→+=++=0s : -10C-2,B,2AA5CA1010 ===→=+=
22222o 1)3s(4
1)3s()3s(2
1s2
10s6s10s2
1s2
)s(V++
−++
+−
+=
+++
−+
=
=)t(vo V)tsin(e4)tcos(e2e2 -3t-3t-t −−
Chapter 16, Solution 5.
s2
2s1+
2
Io
s
( )
( ) A)t(ut3229.1sin7559.0e
or
A)t(ueee3779.0eee3779.0e)t(i
3229.1j5.0s)646.2j)(3229.1j5.1(
)3229.1j5.0(
3229.1j5.0s)646.2j)(3229.1j5.1(
)3229.1j5.0(
2s1
)3229.1j5.0s)(3229.1j5.0s)(2s(s
2VsI
)3229.1j5.0s)(3229.1j5.0s)(2s(s2
2sss2
2s1
2s
21
s1
12s
1V
t2
t3229.1j2/t90t3229.1j2/t90t2o
22
2o
2
−=
++=
−+++
+−
+++
−−−−
++
=
−++++==
−++++=
+++=
+++=
−
−°−−°−−
Chapter 16, Solution 6.
2
2s5+
Io
10/s
s
Use current division.
t3sine35t3cose5)t(i
3)1s(5
3)1s()1s(5
10s2ss5
2s5
s102s
2sI
tto
22222o
−− −=
++−
++
+=
++=
+++
+=
Chapter 16, Solution 7. The s-domain version of the circuit is shown below. 1/s 1 Ix + 2s
1
2+s
– Z
2
2
2 s211s2s2
s21s21
s2s1
)s2(s1
1s2//s11Z
+
++=
++=
++=+=
)5.0ss(CBs
)1s(A
)5.0ss)(1s(1s2
1s2s2s21x
1s2
ZVI 22
2
2
2x
++
++
+=
+++
+=
++
++
==
)1s(C)ss(B)5.0ss(A1s2 222 ++++++=+
BA2:s2 +=
2CC2CBA0:s −=→+=++=
-4B ,6A3 0.5A or CA5.01:constant ==→=+=
222x866.0)5.0s()5.0s(4
1s6
75.0)5.0s(2s4
1s6I
++
+−
+=
++
+−
+=
[ ] A)t(ut866.0cose46)t(i t5.0x
−−=
Chapter 16, Solution 8.
(a) )1s(s
1s5.1ss22)s21(
s1)s21//(1
s1Z
2
+++
=++
+=++=
(b) )1s(s2
2s3s3
s11
1s1
21
Z1 2
+++
=+
++=
2s3s3)1s(s2Z 2 ++
+=
Chapter 16, Solution 9.
(a) The s-domain form of the circuit is shown in Fig. (a).
=+++
=+=s1s2)s1s(2
)s1s(||2Zin 1s2s)1s(2
2
2
+++
1
1
2 s 2/s 1/s
s
2
(a) (b)(b) The s-domain equivalent circuit is shown in Fig. (b).
2s3)2s(2
s23)s21(2
)s21(||2++
=++
=+
2s36s5
)s21(||21++
=++
=
++
+
++
⋅=
++
=
2s36s5
s
2s36s5
s
2s36s5
||sZin 6s7s3)6s5(s
2 +++
Chapter 16, Solution 10. To find ZTh, consider the circuit below. 1/s Vx + 1V 2 Vo 2Vo - Applying KCL gives
s/12VV21 x
o +=+
But xo Vs/12
2V+
= . Hence
s3)1s2(V
s/12V
s/12V41 x
xx +−=→
+=
++
s3)1s2(
1VZ x
Th+
−==
To find VTh, consider the circuit below. 1/s Vy +
1
2+s
2 Vo 2Vo
- Applying KCL gives
)1s(34V
2VV2
1s2
oo
o +−=→=+
+
But 0Vs1V2V ooy =+•+−
)1s(s3)2s(4
s2s
)1s(34)
s21(VVV oyTh +
+−=
+
+−=+==
Chapter 16, Solution 11. The s-domain form of the circuit is shown below.
4/s s
I2I1+ −
+ −
2 4/(s + 2) 1/s
Write the mesh equations.
21 I2Is4
2s1
−
+= (1)
21 I)2s(I-22s
4-++=
+ (2)
Put equations (1) and (2) into matrix form.
+
+=
+ 2
1
II
2s2-2-s42
2)(s4-s1
)4s2s(s2 2 ++=∆ ,
)2s(s4s4s2
1 ++−
=∆ , s6-
2 =∆
4s2sCBs
2sA
)4s2s)(2s()4s4s(21
I 22
21
1 +++
++
=++++−⋅
=∆∆
=
)2s(C)s2s(B)4s2s(A)4s4s(21 222 ++++++=+−⋅
Equating coefficients :
2s : BA21 += 1s : CB2A22- ++=
0s : C2A42 += Solving these equations leads to A 2= , 23-B = , -3C =
221 )3()1s(3s23-
2s2
I++−
++
=
22221 )3()1s(3
323-
)3()1s()1s(
23-
2s2
I++
⋅++++
⋅++
=
=)t(i1 [ ] A)t(u)t732.1sin(866.0)t732.1cos(e5.1e2 -t-2t −−
2222
2 )3()1s(3-
)4s2s(2s
s6-I
++=
++⋅=
∆∆
=
== )t3sin(e33-
)t(i t-2 A)t(u)t732.1sin(e1.732- -t
Chapter 16, Solution 12. We apply nodal analysis to the s-domain form of the circuit below.
Vo
10/(s + 1) + −
1/(2s) 4
s
3/s
oo
osV2
4V
s3
s
V1s
10
+=+−
+
1s15s1510
151s
10V)ss25.01( o
2
+++
=++
=++
1s25.0sCBs
1sA
)1s25.0s)(1s(25s15
V 22o +++
++
=+++
+=
740
V)1s(A 1-so =+= =
)1s(C)ss(B)1s25.0s(A25s15 22 ++++++=+
Equating coefficients :
2s : -ABBA0 =→+=1s : C-0.75ACBA25.015 +=++= 0s : CA25 +=
740A = , 740-B = , 7135C =
43
21
s
23
32
7155
43
21
s
21
s
740
1s1
740
43
21
s
7135
s740-
1s740
V 222o
+
+
⋅+
+
+
+−
+=
+
+
++
+=
+
−= t
23
sine)3)(7()2)(155(
t23
cose740
e740
)t(v 2t-2t-t-o
=)t(vo V)t866.0sin(e57.25)t866.0cos(e714.5e714.5 2-t2-t-t +−
Chapter 16, Solution 13.
Consider the following circuit.
Vo
1/(s + 2)
1/s 2s
Io
2 1
Applying KCL at node o,
ooo V
1s21s
s12V
1s2V
2s1
++
=+
++
=+
)2s)(1s(1s2
Vo +++
=
2sB
1sA
)2s)(1s(1
1s2V
I oo +
++
=++
=+
=
1A = , -1B =
2s1
1s1
Io +−
+=
=)t(io ( ) A)t(uee -2t-t −
Chapter 16, Solution 14. We first find the initial conditions from the circuit in Fig. (a).
1 Ω 4 Ω
io
+
vc(0)
−
+ −
5 V
(a)
A5)0(io =− , V0)0(vc =−
We now incorporate these conditions in the s-domain circuit as shown in Fig.(b).
2s 5/s
Io
Vo
+ −
1 4
15/s 4/s
(b)At node o,
0s440V
s5
s2V
1s15V ooo =
+−
+++−
oV)1s(4
ss2
11
s5
s15
+++=−
o
2
o
22
V)1s(s42s6s5
V)1s(s4
s2s2s4s4s
10+++
=+
++++=
2s6s5)1s(40
V 2o +++
=
s5
)4.0s2.1s(s)1s(4
s5
s2V
I 2o
o +++
+=+=
4.0s2.1sCBs
sA
s5
I 2o +++
++=
sCsB)4.0s2.1s(A)1s(4 s2 ++++=+
Equating coefficients :
0s : 10AA4.04 =→=1s : -84-1.2ACCA2.14 =+=→+=2s : -10-ABBA0 ==→+=
4.0s2.1s8s10
s10
s5
I 2o +++
−+=
2222o 2.0)6.0s()2.0(10
2.0)6.0s()6.0s(10
s15
I++
−++
+−=
=)t(io ( )[ ] A)t(u)t2.0sin()t2.0cos(e1015 0.6t- −−
Chapter 16, Solution 15.
First we need to transform the circuit into the s-domain.
2s5+
Vo
+ −
+ − 5/s
s/4
+ Vx −
10 3Vx
2s5VV
2s5VV,But
2ss5V120V)40ss2(0
2ss5sVVs2V120V40
010
2s5V
s/50V
4/sV3V
xoox
xo2
oo2
xo
ooxo
++=→
+−=
+−−++==
+−++−
=+−
+−
+−
We can now solve for Vx.
)40s5.0s)(2s()20s(5V
2s)20s(10V)40s5.0s(2
02s
s5V1202s
5V)40ss2(
2
2x
2x
2
xx2
−++
+−=
++
−=−+
=+
−−
++++
Chapter 16, Solution 16. We first need to find the initial conditions. For 0t < , the circuit is shown in Fig. (a). To dc, the capacitor acts like an open circuit and the inductor acts like a short circuit.
1 Ω
+ −Vo
1 F
Vo/2 + − 1 H io
+ −
2 Ω
3 V
(a)
Hence,
A1-33-
i)0(i oL === , V1-vo =
V5.221-
)1-)(2(-)0(vc =
−=
We now incorporate the initial conditions for as shown in Fig. (b). 0t >
I2I1 + −
s
2.5/s + −
1
+ −Vo
1/s
Vo/2 + −
Io
− +
2
-1 V
5/(s + 2)
(b)For mesh 1,
02
Vs5.2
Is1
Is1
22s
5- o21 =++−
+++
But, 2oo IIV ==
s5.2
2s5
Is1
21
Is1
2 21 −+
=
−+
+ (1)
For mesh 2,
0s5.2
2V
1Is1
Is1
s1 o12 =−−+−
++
1s5.2
Is1
s21
Is1
- 21 −=
+++ (2)
Put (1) and (2) in matrix form.
−
−+
=
++
−+
1s5.2
s5.2
2s5
I
I
s1
s21
s1-
s1
21
s1
2
2
1
s3
2s2 ++=∆ , )2s(s
5s4
2-2 +++=∆
3s2s2CBs
2sA
)3s2s2)(2s(132s-
II 22
22
2o +++
++
=+++
+=
∆∆
==
)2s(C)s2s(B)3s2s2(A132s- 222 ++++++=+
Equating coefficients :
2s : B A22- +=1s : CB2A20 ++=0s : C2A313 +=
Solving these equations leads to
7143.0A = , , -3.429B = 429.5C =
5.1ss714.2s7145.1
2s7143.0
3s2s2429.5s429.3
2s7143.0
I 22o ++−
−+
=++
−−
+=
25.1)5.0s()25.1)(194.3(
25.1)5.0s()5.0s(7145.1
2s7143.0
I 22o +++
+++
−+
=
=)t(io [ ] A)t(u)t25.1sin(e194.3)t25.1cos(e7145.1e7143.0 -0.5t-0.5t-2t +−
Chapter 16, Solution 17. We apply mesh analysis to the s-domain form of the circuit as shown below.
2/(s+1)
I3
+ −
1/s
I2I1
4
s
1 1
For mesh 3,
0IsIs1
Is1
s1s
2213 =−−
++
+ (1)
For the supermesh,
0Iss1
I)s1(Is1
1 321 =
+−++
+ (2)
But (3) 4II 21 −= Substituting (3) into (1) and (2) leads to
+=
+−
++s1
14Is1
sIs1
s2 32 (4)
1s2
s4-
Is1
sIs1
s- 32 +−=
++
+ (5)
Adding (4) and (5) gives
1s2
4I2 2 +−=
1s1
2I2 +−=
== )t(i)t(i 2o A)t(u)e2( -t− Chapter 16, Solution 18.
vs(t) = 3u(t) – 3u(t–1) or Vs = )e1(s3
se
s3 s
s−
−−=−
1 Ω
+
Vo
−
+ − 1/s
Vs 2 Ω
V)]1t(u)e22()t(u)e22[()t(v
)e1(5.1s
2s2)e1(
)5.1s(s3V
VV)5.1s(02
VsV
1VV
)1t(5.1t5.1o
sso
soo
oso
−−−−=
−
+−=−
+=
=→=++−
−−−
−−
+
Chapter 16, Solution 19. We incorporate the initial conditions in the s-domain circuit as shown below.
I
1/s
− +
2 I V1 Vo
1/s
s
+ −
2
2 4/(s + 2)
At the supernode,
o11 sV
s1
sV
22
V)2s(4++=+
−+
o1 Vss1
Vs1
21
22s
2++
+=++
(1)
But and I2VV 1o +=s
1VI 1 +=
2s2Vs
s)2s(s2V
Vs
)1V(2VV oo
11
1o +−
=+−
=→+
+= (2)
Substituting (2) into (1)
oo Vs2s
2V
2ss
s1s2
s1
22s
2+
+−
+
+
=−++
oVs2s1s2
)2s(s)1s2(2
s1
22s
2
+
++
=++
+−++
o
22
V2s
1s4s2s9s2
)2s(ss9s2
+++
=++
=++
732.3sB
2679.0sA
1s4s9s2
V 2o ++
+=
+++
=
443.2A = , 4434.0-B =
732.3s4434.0
2679.0s443.2
Vo +−
+=
Therefore,
=)t(vo V)t(u)e4434.0e443.2( -3.732t-0.2679t −
Chapter 16, Solution 20. We incorporate the initial conditions and transform the current source to a voltage source as shown.
1 s
+ −1/s 2/s
Vo
+ −
1
1/(s + 1) 1/s
At the main non-reference node, KCL gives
s1
sV
1V
s11Vs2)1s(1 ooo ++=
+−−+
s1s
V)s1s)(1s(Vs21s
soo
++++=−−
+
oV)s12s2(2s
1s1s
s++=−
+−
+
)1s2s2)(1s(1s4s2-
V 2
2
o +++−−
=
5.0ssCBs
1sA
)5.0ss)(1s(5.0s2s-
V 22o +++
++
=+++
−−=
1V)1s(A 1-so =+= =
)1s(C)ss(B)5.0ss(A5.0s2s- 222 ++++++=−−
Equating coefficients : 2s : -2BBA1- =→+=1s : -1CCBA2- =→++=0s : -0.515.0CA5.00.5- =−=+=
222o )5.0()5.0s()5.0s(2
1s1
5.0ss1s2
1s1
V++
+−
+=
+++
−+
=
=)t(vo [ ] V)t(u)2tcos(e2e 2-t-t −
Chapter 16, Solution 21. The s-domain version of the circuit is shown below. 1 s
V1 Vo + 2/s 2 1/s
At n
s
s1
10−
At n
1 −s
V
Subs
10 =
=Vo
=10
cons
=Vo
Taki
(vo
10/
-
ode 1,
ooo VsVsVs
sVVV
)12
()1(102
2
11
1−++=→+
−= (1)
ode 2,
)12
(2
21 ++=→+= ssVVsVVV
oooo (2)
tituting (2) into (1) gives
ooo VsssVsVsss )5.12()12
()12/)(1( 22
2 ++=−++++
5.12)5.12(10
22 +++
+=++ ss
CBssA
sss
CsBsssA ++++ 22 )5.12( BAs +=0:2 CAs += 20:
-40/3C -20/3,B ,3/205.110:tant ===→= AA
++
−+++
−=
+++
− 22222 7071.0)1(7071.0414.1
7071.0)1(11
320
5.1221
320
sss
ssss
s
ng the inverse Laplace tranform finally yields
[ ] V)t(ut7071.0sine414.1t7071.0cose1320)t tt −− −−=
Chapter 16, Solution 22. The s-domain version of the circuit is shown below. 4s V1 V2
1s +
12 1 2 3/s
At node 1,
s4V
s411V
1s12
s4VV
1V
1s12 2
1211 −
+=
+→
−+=
+ (1)
At node 2,
++=→+=
−1s2s
34VVV
3s
2V
s4VV 2
212221 (2)
Substituting (2) into (1),
222
2 V23s
37s
34
s41
s4111s2s
34V
1s12
++=
−
+
++=
+
)89s
47s(
CBs)1s(
A
)89s
47s)(1s(
9V22
2++
++
+=
+++=
)1s(C)ss(B)89s
47s(A9 22 ++++++=
Equating coefficients:
BA0:s2 +=
A43CCA
43CBA
470:s −=→+=++=
-18C -24,B ,24AA83CA
899:constant ===→=+=
6423)
87s(
3
6423)
87s(
)8/7s(24)1s(
24
)89s
47s(
18s24)1s(
24V222
2++
+++
+−
+=
++
+−
+=
Taking the inverse of this produces:
[ ] )t(u)t5995.0sin(e004.5)t5995.0cos(e24e24)t(v t875.0t875.0t2
−−− +−= Similarly,
)89s
47s(
FEs)1s(
D
)89s
47s)(1s(
1s2s349
V22
2
1++
++
+=
+++
++
=
)1s(F)ss(E)89s
47s(D1s2s
349 222 ++++++=
++
Equating coefficients:
ED12:s2 +=
D436FFD
436or FED
4718:s −=→+=++=
0F 4,E ,8DD833or FD
899:constant ===→=+=
6423)
87s(
2/7
6423)
87s(
)8/7s(4)1s(
8
)89s
47s(
s4)1s(
8V222
1++
−++
++
+=
+++
+=
Thus, [ ] )t(u)t5995.0sin(e838.5)t5995.0cos(e4e8)t(v t875.0t875.0t
1−−− −+=
Chapter 16, Solution 23.
The s-domain form of the circuit with the initial conditions is shown below. V
I
1/sC sL R -2/s
4/s 5C
At the non-reference node,
sCVsLV
RV
C5s2
s4
++=++
++=+
LC1
RCs
ss
CVs
sC56 2
LC1RCssC6s5
V 2 +++
=
But 88010
1RC1
== , 208041
LC1
==
22222 2)4s()2)(230(
2)4s()4s(5
20s8s480s5
V++
+++
+=
+++
=
=)t(v V)t2sin(e230)t2cos(e5 -4t-4t +
)20s8s(s4480s5
sLV
I 2 +++
==
20s8sCBs
sA
)20s8s(s120s25.1
I 22 +++
+=++
+=
6A = , -6B = , -46.75C =
22222 2)4s()2)(375.11(
2)4s()4s(6
s6
20s8s75.46s6
s6
I++
−++
+−=
+++
−=
=)t(i 0t),t2sin(e375.11)t2cos(e6)t(u6 -4t-4t >−−
Chapter 16, Solution 24. At t = 0-, the circuit is equivalent to that shown below. + 9A 4Ω 5Ω vo -
20)9(54
4x5)0(vo =+
=
For t > 0, we have the Laplace transform of the circuit as shown below after transforming the current source to a voltage source. 4Ω 16Ω Vo + 36V 10A 2/s 5
Ω
- Applying KCL gives
8.12B,2.7A,5.0s
BsA
)5.0s(ss206.3V
5V
2sV
1020
V36o
ooo −==+
+=++
=→+=+−
Thus,
[ ] )t(ue8.122.7)t(v t5.0o
−−=
Chapter 16, Solution 25. For , the circuit in the s-domain is shown below. 0t >
s 6 I
Applying KVL,
+
V
−
+ −
(2s)/(s2 + 16)
+ −
9/s
2/s
0s2
Is9
s616ss2
2 =+
++++−
)16s)(9s6s(32s4
I 22
2
++++
=
)16s()3s(s288s36
s2
s2I
s9V 22
2
+++
+=+=
16sEDs
)3s(C
3sB
sA
s2
22 ++
++
++
++=
)s48s16s3s(B)144s96s25s6s(A288s36 2342342 ++++++++=+
)s9s6s(E)s9s6s(D)s16s(C 232343 ++++++++
Equating coefficients : 0s : A144288 = (1) 1s : E9C16B48A960 +++= (2) 2s : E6D9B16A2536 +++= (3) 3s : ED6CB3A60 ++++= (4) 4s : (5) DBA0 ++=
Solving equations (1), (2), (3), (4) and (5) gives
2A = , B , C7984.1-= 16.8-= , D 2016.0-= , 765.2E =
16s)4)(6912.0(
16ss2016.0
)3s(16.8
3s7984.1
s4
)s(V 222 ++
+−
+−
+−=
=)t(v V)t4sin(6912.0)t4cos(2016.0et16.8e7984.1)t(u4 -3t-3t +−−−
Chapter 16, Solution 26.
Consider the op-amp circuit below.
R2
+Vo −
0
+ −
R1
1/sC
−+
Vs
At node 0,
sC)V0(R
V0R
0Vo
2
o
1
s −+−
=−
( )o2
1s V-sCR1
RV
+=
211s
o
RRCsR1-
VV
+=
But 21020
RR
2
1 == , 1)1050)(1020(CR 6-31 =××=
So, 2s
1-VV
s
o
+=
)5s(3Ve3V s
-5ts +=→=
5)2)(s(s3-
Vo ++=
5sB
2sA
5)2)(s(s3
V- o ++
+=
++=
1A = , -1B =
2s1
5s1
Vo +−
+=
=)t(vo ( ) )t(uee -2t-5t −
Chapter 16, Solution 27.
Consider the following circuit.
For mesh 1,
I2
2s
I1+ −
10/(s + 3) 1
2s s
1
221 IsII)s21(3s
10−−+=
+
21 I)s1(I)s21(3s
10+−+=
+ (1)
For mesh 2,
112 IsII)s22(0 −−+=
21 I)1s(2I)s1(-0 +++= (2) (1) and (2) in matrix form,
++++
=
+
2
1
II
)1s(2)1s(-)1s(-1s2
0)3s(10
1s4s3 2 ++=∆
3s)1s(20
1 ++
=∆
3s)1s(10
2 ++
=∆
Thus
=∆∆
= 11I )1s4s3)(3s(
)1s(202 ++++
=∆∆
= 22I
)1s4s3)(3s()1s(10
2 ++++
2I1=
Chapter 16, Solution 28.
Consider the circuit shown below.
VI1 +
−
1
2s s I2
s 6/s
For mesh 1,
21 IsI)s21(s6
++=
For mesh 2,
21 I)s2(Is0 ++=
21 Is2
1-I
+=
Substituting (2) into (1) gives
2
2
22 Is
2)5s(s-IsI
s2
12s)-(1s6 ++
=+
++=
or 2s5s
6-I 22 ++=
+
o
−2
(1)
(2)
4.561)0.438)(s(s12-
2s5s12-
I2V 22o ++=
++==
Since the roots of s are -0.438 and -4.561, 02s52 =++
561.4sB
438.0sA
Vo ++
+=
-2.914.123
12-A == , 91.2
4.123-12-
B ==
561.4s91.2
0.438s2.91-
)s(Vo ++
+=
=)t(vo [ ] V)t(uee91.2 t438.0-4.561t −
Chapter 16, Solution 29.
Consider the following circuit.
1 : 2Io
+ −
10/(s + 1)
1
4/s 8
Let 1s2
8s48)s4)(8(
s4
||8ZL +=
+==
When this is reflected to the primary side,
2n,nZ
1Z 2L
in =+=
1s23s2
1s22
1Zin ++
=+
+=
3s21s2
1s10
Z1
1s10
Iin
o ++
⋅+
=⋅+
=
5.1sB
1sA
)5.1s)(1s(5s10
Io ++
+=
+++
=
-10A = , 20B =
5.1s20
1s10-
)s(Io ++
+=
=)t(io [ ] A)t(uee210 t-1.5t −−
Chapter 16, Solution 30.
)s(X)s(H)s(Y = , 1s3
1231s
4)s(X
+=
+=
22
2
)1s3(34s8
34
)1s3(s12
)s(Y++
−=+
=
22 )31s(1
274
)31s(s
98
34
)s(Y+
⋅−+
⋅−=
Let 2)31s(s
98-
)s(+
⋅=G
Using the time differentiation property,
+=⋅= 3t-3t-3t- eet31-
98-
)et(dtd
98-
)t(g
3t-3t- e
98
et278
)t(g −=
Hence,
3t-3t-3t- et274
e98
et278
)t(u34
)t(y −−+=
=)t(y 3t-3t- et274
e98
)t(u34
+−
Chapter 16, Solution 31.
s1
)s(X)t(u)t(x =→=
4ss10
)s(Y)t2cos(10)t(y 2 +=→=
==)s(X)s(Y
)s(H4s
s102
2
+
Chapter 16, Solution 32.
(a) )s(X)s(H)s(Y =
s1
5s4s3s
2 ⋅++
+=
5s4s
CBssA
)5s4s(s3s
22 +++
+=++
+=
CsBs)5s4s(A3s 22 ++++=+
Equating coefficients :
0s : 53AA53 =→= 1s : 57- A41CCA41 =−=→+= 2s : 53--ABBA0 ==→+=
5s4s7s3
51
s53
)s(Y 2 +++
⋅−=
1)2s(1)2s(3
51
s6.0
)s(Y 2 ++++
⋅−=
=)t(y [ ] )t(u)tsin(e2.0)tcos(e6.06.0 -2t-2t −−
(b) 22t-
)2s(6
)s(Xet6)t(x+
=→=
22 )2s(6
5s4s3s
)s(X)s(H)s(Y+
⋅++
+==
5s4sDCs
)2s(B
2sA
)5s4s()2s()3s(6
)s(Y 2222 +++
++
++
=+++
+=
Equating coefficients :
3s : (1) -ACCA0 =→+=2s : DBA2DC4BA60 ++=+++= (2) 1s : D4B4A9D4C4B4A136 ++=+++= (3) 0s : BA2D4B5A1018 +=++= (4)
Solving (1), (2), (3), and (4) gives
6A = , 6B = , 6-C = , -18D =
1)2s(18s6
)2s(6
2s6
)s(Y 22 +++
−+
++
=
1)2s(6
1)2s()2s(6
)2s(6
2s6
)s(Y 222 ++−
+++
−+
++
=
=)t(y [ ] )t(u)tsin(e6)tcos(e6et6e6 -2t-2t-2t-2t −−+
Chapter 16, Solution 33.
)s(X)s(Y
)s(H = , s1
)s(X =
16)2s()4)(3(
16)2s(s2
)3s(21
s4
)s(Y 22 ++−
++−
++=
== )s(Ys)s(H20s4s
s1220s4s
s2)3s(2
s4 22
2
++−
++−
++
Chapter 16, Solution 34.
Consider the following circuit.
+
Vo(s)
−
Vo
10/s
2
4+ −
Vs
s
Using nodal analysis,
s10V
4V
2sVV ooos +=
+−
o2
os V)s2s(101
)2s(41
1V10s
41
2s1
)2s(V
++++=
+++
+=
( ) o2
s V30s9s2201
V ++=
=s
o
VV
30s9s220
2 ++
Chapter 16, Solution 35.
Consider the following circuit.
+
Vo
−
I V1
+ −
2/s
2I
s
Vs 3
At node 1,
3sV
II2 1
+=+ , where
s2VV
I 1s −=
3sV
s2VV
3 11s
+=
−⋅
1s1 V
2s3
V2s3
3sV
−=+
s1 V2s3
V2s3
3s1
=
++
s21 V2s9s3
)3s(s3V
+++
=
s21o V2s9s3
s9V
3s3
V++
=+
=
==s
o
VV
)s(H2s9s3
s92 ++
Chapter 16, Solution 36.
From the previous problem,
s21 V
2s9s3s3
3sV
I3++
=+
=
s2 V2s9s3
sI
++=
But o
2
s Vs9
2s9s3V
++=
2
2
3 9 23 9 2 9 9
ooVs s sI V
s s s+ +
= ⋅+ +
=
==I
V)s(H o 9
Chapter 16, Solution 37.
(a) Consider the circuit shown below.
3 2s
+
Vx
−
2/s + −
I1 I2+ −
Vs 4Vx
For loop 1,
21s Is2
Is2
3V −
+= (1)
For loop 2,
0Is2
Is2
s2V4 12x =−
++
But,
−=s2
)II(V 21x
So, 0Is2
Is2
s2)II(s8
1221 =−
++−
21 Is2s6
Is6-
0
−+= (2)
In matrix form, (1) and (2) become
−
+=
2
1s
II
s2s6s6-s2-s23
0V
−
−
+=∆
s2
s6
s2s6
s2
3
4s6s
18−−=∆
s1 Vs2s6
−=∆ , s2 Vs6
=∆
s1
1 Vs64s18
)s2s6(I
−−−
=∆∆
=
=−−
−=
32s9ss3
VI
s
1
9s2s33s
2
2
−+−
(b) ∆∆
= 22I
∆∆−∆
=−= 2121x s
2)II(
s2
V
∆=
∆−−
= ssx
V4-)s6s2s6(Vs2V
==s
s
x
2
4V-Vs6
VI
2s3-
Chapter 16, Solution 38.
(a) Consider the following circuit.
1
+
Vo
−
s Is
1/s
VoV1
1/s+ −
Vs
1 Io
At node 1,
sVV
Vs1
VV o11
1s −+=
−
o1s Vs1
Vs1
s1V −
++= (1)
At node o,
oooo1 V)1s(VVs
sVV
+=+=−
o
21 V)1ss(V ++= (2)
Substituting (2) into (1)
oo2
s Vs1V)1ss)(s11s(V −++++=
o23
s V)2s3s2s(V +++=
==s
o1 V
V)s(H
2s3s2s1
23 +++
(b) o
2o
231ss V)1ss(V)2s3s2s(VVI ++−+++=−=
o23
s V)1s2ss(I +++=
==s
o2 I
V)s(H
1s2ss1
23 +++
(c) 1
VI o
o =
==== )s(HIV
II
)s(H 2s
o
s
o3 1s2ss
123 +++
(d). ==== )s(HVV
VI
)s( 1s
o
s
o4H
2s3s2s1
23 +++
Chapter 16, Solution 39.
Consider the circuit below.
+
Vo
−
Io
+ −
1/sC
R Vb
Va−+
Vs
Since no current enters the op amp, flows through both R and C. oI
+=sC1
RI-V oo
sCI-
VVV osba ===
=+
==sC1
sC1RVV
)s(Hs
o 1sRC +
Chapter 16, Solution 40.
(a) LRs
LRsLR
RVV
)s(Hs
o
+=
+==
=)t(h )t(ueLR LRt-
(b) s1)s(V)t(u)t(v ss =→=
LRsB
sA
)LRs(sLR
VLRs
LRV so +
+=+
=+
=
1A = , -1B =
LRs1
s1
Vo +−=
=−= )t(ue)t(u)t(v L-Rt
o )t(u)e1( L-Rt− Chapter 16, Solution 41.
)s(X)s(H)s(Y =
1s2
)s(H)t(ue2)t(h t-
+=→=
s5)s(X)s(V)t(u5)t(v ii ==→=
1sB
sA
)1s(s10
)s(Y+
+=+
=
10A = , -10B =
1s10
s10
)s(Y+
−=
=)t(y )t(u)e1(10 -t−
Chapter 16, Solution 42.
)s(X)s(Y)s(Ys2 =+
)s(X)s(Y)1s2( =+
)21s(21
1s21
)s(X)s(Y
)s(H+
=+
==
=)t(h )t(ue5.0 2-t
Chapter 16, Solution 43.
i(t)
+ −
1Ω
1F u(t)
1H
First select the inductor current iL and the capacitor voltage vC to be the state variables. Applying KVL we get: '
CC vi;0'ivi)t(u ==+++−
Thus,
)t(uivi
iv
C'
'C
+−−=
=
Finally we get,
[ ] [ ] )t(u0i
v10)t(i;)t(u
10
iv
1110
iv CCC +
=
+
−−
=
′
′
Chapter 16, Solution 44. 1/8 F 1H
)t(u4 2Ω
+ −
+
vx
−
4Ω
First select the inductor current iL and the capacitor voltage vC to be the state variables. Applying KCL we get:
LCxxLC
'C
C
'C
Cx
x'L
xL'C
'Cx
L
i3333.1v3333.0vor;v2i4v2
vv
8v
4vv
v)t(u4i
v4i8vor;08
v2
vi
+=−+=+=+=
−=
−==++−
LC
'L
LCLCL'C
i3333.1v3333.0)t(u4i
i666.2v3333.1i333.5v3333.1i8v
−−=
+−=−−=
Now we can write the state equations.
=
+
−−
−=
L
Cx
L
C'L
'C
iv
3333.13333.0
v;)t(u40
iv
3333.13333.0666.23333.1
iv
Chapter 16, Solution 45.
First select the inductor current iL (current flowing left to right) and the capacitor voltage vC (voltage positive on the left and negative on the right) to be the state variables. Applying KCL we get:
2o'L
oL'CL
o'C
vvi
v2i4vor0i2
v4
v
−=
+==++−
1Co vvv +−=
21C
'L
1CL'C
vvvi
v2v2i4v
−+−=
+−=
[ ] [ ]
+
−=
−+
−−
=
′
′
)t(v)t(v
01vi
10)t(v;)t(v)t(v
0211
vi
2410
vi
2
1
C
Lo
2
1
C
L
C
L
Chapter 16, Solution 46.
First select the inductor current iL (left to right) and the capacitor voltage vC to be the state variables. Letting vo = vC and applying KCL we get:
sC'L
sLC'Cs
C'CL
vvi
iiv25.0vor0i4
vvi
+−=
++−==−++−
Thus,
+
=
+
−
−=
s
s
L
Co
s
s'L
'C
'L
'C
iv
0000
iv
01
)t(v;iv
0110
iv
01125.0
iv
Chapter 16, Solution 47.
First select the inductor current iL (left to right) and the capacitor voltage vC (+ on the left) to be the state variables.
Letting i1 = 4
v'C and i2 = iL and applying KVL we get:
Loop 1:
1CL'CL
'C
C1 v2v2i4vor0i4
v2vv +−==
−++−
Loop 2:
21C21CL
L'L
2'L
'C
L
vvvv2
v2v2i4i2i
or0vi4
vi2
−+−=−+−
+−=
=++
−
1CL1CL
1 v5.0v5.0i4
v2v2i4i +−=
+−=
+
−=
−+
−−
=
′
′
)t(v)t(v
0005.0
vi
015.01
)t(i)t(i
;)t(v)t(v
0211
vi
2410
vi
2
1
C
L
2
1
2
1
C
L
C
L
Chapter 16, Solution 48.
Let x1 = y(t). Thus, )t(zx4x3yxandxyx 21'22
''1 +−−=′′===
This gives our state equations.
[ ] [ ] )t(z0xx
01)t(y;)t(z10
xx
4310
xx
2
1
2
1'2
'1 +
=
+
−−
=
Chapter 16, Solution 49.
zxyorzyzxxand)t(yxLet 2'''
121 +=−=−== Thus, z3x5x6zz2z)zx(5x6zyx 21
''21
''2 −−−=−+++−−=−′′=
This now leads to our state equations,
[ ] [ ] )t(z0xx
01)t(y;)t(z3
1xx
5610
xx
2
1
2
1'2
'1 +
=
−
+
−−
=
Chapter 16, Solution 50.
Let x1 = y(t), x2 = .xxand,x '23
'1 =
Thus, )t(zx6x11x6x 321
"3 +−−−=
We can now write our state equations.
[ ] [ ] )t(z0xxx
001)t(y;)t(z100
xxx
6116100010
xxx
3
2
1
3
2
1
'3
'2
'1
+
=
+
−−−=
Chapter 16, Solution 51.
We transform the state equations into the s-domain and solve using Laplace transforms.
+=−
s1B)s(AX)0(x)s(sX
Assume the initial conditions are zero.
+++
=
−+=
=−
−
)s/2(0
4s24s
8s4s1
s1
20
s244s
)s(X
s1B)s(X)AsI(
2
1
222222
221
2)2s(2
2)2s()2s(
s1
2)2s(4s
s1
8s4s4s
s1
)8s4s(s8)s(X)s(Y
++
−+
++
+−+=
++
−−+=
++
−−+=
++==
y(t) = ( )( ) )t(ut2sint2cose1 t2 +− −
Chapter 16, Solution 52.
Assume that the initial conditions are zero. Using Laplace transforms we get,
+−+
++=
+−
+=
−
s/4s/3
2s214s
10s6s1
s/2s/1
0411
4s212s
)s(X 2
1
222211)3s(8.1s8.0
s8.0
)1)3s((s8s3X
++
−−+=
++
+=
2222 1)3s(16.
1)3s(3s8.0
s8.0
+++
++
+−=
)t(u)tsine6.0tcose8.08.0()t(x t3t3
1−− +−=
222221)3s(4.4s4.1
s4.1
1)3s((s14s4X
++
−−+=
++
+=
2222 1)3s(12.0
1)3s(3s4.1
s4.1
++−
++
+−=
)t(u)tsine2.0tcose4.14.1()t(x t3t3
2−− −−=
)t(u)tsine8.0tcose4.44.2(
)t(u2)t(x2)t(x2)t(yt3t3
211−− −+−=
+−−=
)t(u)tsine6.0tcose8.02.1()t(u2)t(x)t(y t3t3
12−− +−−=−=
Chapter 16, Solution 53.
If is the voltage across R, applying KCL at the non-reference node gives oV
oo
oo
s VsL1
sCR1
sLV
VsCRV
I
++=++=
RLCsRsLIsRL
sL1
sCR1
IV 2
sso ++
=++
=
RsLRLCsIsL
RV
I 2so
o ++==
LC1RCssRCs
RsLRLCssL
II
)s(H 22s
o
++=
++==
The roots
LC1
)RC2(1
RC21-
s 22,1 −±=
both lie in the left half plane since R, L, and C are positive quantities. Thus, the circuit is stable.
Chapter 16, Solution 54.
(a) 1s
3)s(H1 += ,
4s1
)s(H2 +=
)4s)(1s(3
)s(H)s(H)s(H 21 ++==
[ ]
++
+==
4sB
1sA
)s(H)t(h 1-1- LL
1A = , 1-B = =)t(h )t(u)ee( -4t-t −
(b) Since the poles of H(s) all lie in the left half s-plane, the system is stable.
Chapter 16, Solution 55.
Let be the voltage at the output of the first op amp. 1oV
sRC1
RsC1
VV
s
1o −=
−= ,
sRC1
VV
1o
o −=
222s
o
CRs1
VV
)s(H ==
22CRt
)t(h =
∞=
∞→)t(hlim
t, i.e. the output is unbounded.
Hence, the circuit is unstable.
Chapter 16, Solution 56.
LCs1sL
sC1
sL
sC1
sL
sC1
||sL 2+=
+
⋅=
RsLRLCssL
LCs1sL
R
LCs1sL
VV
2
2
2
1
2
++=
++
+=
LC1
RC1
ss
RC1
s
VV
21
2
+⋅+
⋅=
Comparing this with the given transfer function,
RC1
2 = , LC1
6 =
If , Ω= k1R ==R21
C F500 µ
==C61
L H3.333
Chapter 16, Solution 57. The circuit in the s-domain is shown below.
+
Vx
−
V1
+ −
R1
C R2
L
Vi
Z
CsRLCs1sLR
sC1sLR)sLR()sC1(
)sLR(||sC1
Z2
22
2
22 ++
+=
+++⋅
=+=
i1
1 VZR
ZV
+=
i12
21
2
2o V
ZRZ
sLRR
VsLR
RV
+⋅
+=
+=
CsRLCs1sLR
R
CsRLCs1sLR
sLRR
ZRZ
sLRR
VV
22
21
22
2
2
2
12
2
i
o
+++
+
+++
⋅+
=+
⋅+
=
sLRRCRsRLCRsR
VV
212112
2
i
o
++++=
LCRRR
CR1
LR
ss
LCRR
VV
1
21
1
22
1
2
i
o
++
++
=
Comparing this with the given transfer function,
LCRR
51
2= CR
1L
R6
1
2 += LCR
RR25
1
21 +=
Since and , Ω= 4R1 Ω= 1R 2
201
LCLC41
5 =→= (1)
C41
L1
6 += (2)
201
LCLC45
25 =→=
Substituting (1) into (2),
01C24C80C41
C206 2 =+−→+=
Thus, 201
,41
C =
When 41
C = , 51
C201
L == .
When 201
C = , 1C20
1L == .
Therefore, there are two possible solutions. =C F25.0 =L H2.0 or =C F05.0 =L H1
Chapter 16, Solution 58.
We apply KCL at the noninverting terminal at the op amp. )YY)(V0(Y)0V( 21o3s −−=−
o21s3 V)YY(-VY +=
21
3
s
o
YYY-
VV
+=
Let , 11 sCY = 12 R1Y = , 23 sCY =
11
12
11
2
s
o
CR1sCsC-
R1sCsC-
VV
+=
+=
Comparing this with the given transfer function,
1CC
1
2 = , 10CR
1
11=
If , Ω= k1R1
=== 421 101
CC F100 µ
Chapter 16, Solution 59. Consider the circuit shown below. We notice that o3 VV = and o32 VVV == .
Y4
Y3
Y1
V1
V2Y2
+ −
−+ Vo
Vin
At node 1, 4o12o111in Y)VV(Y)VV(Y)VV( −+−=−
)YY(V)YYY(VYV 42o42111in +−++= (1) At node 2,
3o2o1 Y)0V(Y)VV( −=−
o3221 V)YY(YV +=
o2
321 V
YYY
V+
= (2)
Substituting (2) into (1),
)YY(VV)YYY(Y
YYYV 42oo421
2
321in +−++⋅
+=
)YYYYYYYYYYYYYY(VYYV 42
2243323142
2221o21in −−+++++=
43323121
21
in
o
YYYYYYYYYY
VV
+++=
1Y and must be resistive, while and must be capacitive. 2Y 3Y 4Y
Let 1
1 R1
Y = , 2
2 R1
Y = , 13 sCY = , 24 sCY =
212
2
1
1
1
21
21
in
o
CCsRsC
RsC
RR1
RR1
VV
+++=
2121221
212
2121
in
o
CCRR1
CRRRR
ss
CCRR1
VV
+
+⋅+
=
Choose R , then Ω= k11
6
212110
CCRR1
= and 100CRRR
221
21 =R
+
We have three equations and four unknowns. Thus, there is a family of solutions. One such solution is
=2R Ωk1 , C =1 nF50 , =2C F20 µ
Chapter 16, Solution 60. With the following MATLAB codes, the Bode plots are generated as shown below. num=[1 1]; den= [1 5 6]; bode(num,den);
Chapter 16, Solution 61. We use the following codes to obtain the Bode plots below. num=[1 4]; den= [1 6 11 6]; bode(num,den);
Chapter 16, Solution 62. The following codes are used to obtain the Bode plots below. num=[1 1]; den= [1 0.5 1]; bode(num,den);
Chapter 16, Solution 63. We use the following commands to obtain the unit step as shown below. num=[1 2]; den= [1 4 3]; step(num,den);
Chapter 16, Solution 64. With the following commands, we obtain the response as shown below. t=0:0.01:5; x=10*exp(-t); num=4; den= [1 5 6]; y=lsim(num,den,x,t); plot(t,y)
Chapter 16, Solution 65. We obtain the response below using the following commands. t=0:0.01:5; x=1 + 3*exp(-2*t); num=[1 0]; den= [1 6 11 6]; y=lsim(num,den,x,t); plot(t,y)
Chapter 16, Solution 66. We obtain the response below using the following MATLAB commands. t=0:0.01:5; x=5*exp(-3*t); num=1; den= [1 1 4]; y=lsim(num,den,x,t); plot(t,y)
Chapter 16, Solution 67.
Using the result of Practice Problem 16.14,
)YYY(YYYYY-
VV
321432
21
i
o
+++=
When , 11 sCY = F5.0C1 µ=
12 R
1Y = , Ω= k10R1
23 YY = , 24 sCY = , F1C2 µ=
)RsC2(RsC1RsC-
)R2sC(sCR1RsC-
VV
1112
11
11221
11
i
o
++=
++=
1RC2sRCCsRsC-
VV
122121
211
i
o
+⋅+=
1)10)(1010(1)2(s)10)(1010)(110(0.5s)10)(1010(0.5s-
VV
36-236-6-2
3-6
i
o
+××+×××××
=
42i
o
102s400ss100-
VV
×++=
Therefore,
=a 100- , =b 400 , =c 4102× Chapter 16, Solution 68.
(a) Let 3s
)1s(K)s(Y
++
=
Ks31
)s11(Klim
3s)1s(K
lim)(Yss
=++
=++
=∞∞→∞→
i.e. . K25.0 =
Hence, Y =)s()3s(4
1s++
(b) Consider the circuit shown below.
I
Vs = 8 V + −
t = 0
YS
s8V)t(u8V ss =→=
)3s(s)1s(2
3s1s
s48
)s(V)s(YZV
I ss
++
=++
⋅===
3sB
sA
I+
+=
32A = , 34-B =
=)t(i [ ] A)t(ue4231 3t-−
Chapter 16, Solution 69.
The gyrator is equivalent to two cascaded inverting amplifiers. Let be the voltage at the output of the first op amp.
1V
ii1 -VVRR-
V ==
i1o VsCR
1V
RsC1-
V ==
CsRV
RV
I 2oo
o ==
CsRIV 2
o
o =
CRLwhen,sLIV 2
o
o ==
Chapter 17, Solution 1.
(a) This is periodic with ω = π which leads to T = 2π/ω = 2. (b) y(t) is not periodic although sin t and 4 cos 2πt are independently periodic. (c) Since sin A cos B = 0.5[sin(A + B) + sin(A – B)],
g(t) = sin 3t cos 4t = 0.5[sin 7t + sin(–t)] = –0.5 sin t + 0.5 sin7t which is harmonic or periodic with the fundamental frequency
ω = 1 or T = 2π/ω = 2π. (d) h(t) = cos 2 t = 0.5(1 + cos 2t). Since the sum of a periodic function and a constant is also periodic, h(t) is periodic. ω = 2 or T = 2π/ω = π. (e) The frequency ratio 0.6|0.4 = 1.5 makes z(t) periodic.
ω = 0.2π or T = 2π/ω = 10. (f) p(t) = 10 is not periodic. (g) g(t) is not periodic.
Chapter 17, Solution 2.
(a) The frequency ratio is 6/5 = 1.2. The highest common factor is 1. ω = 1 = 2π/T or T = 2π.
(b) ω = 2 or T = 2π/ω = π. (c) f3(t) = 4 sin2 600π t = (4/2)(1 – cos 1200π t)
ω = 1200π or T = 2π/ω = 2π/(1200π) = 1/600. (d) f4(t) = ej10t = cos 10t + jsin 10t. ω = 10 or T = 2π/ω = 0.2π.
Chapter 17, Solution 3.
T = 4, ωo = 2π/T = π/2 g(t) = 5, 0 < t < 1 10, 1 < t < 2 0, 2 < t < 4
ao = (1/T) = 0.25[ + ] = ∫T
0dt)t(g ∫
1
0dt5 ∫
2
1dt10 3.75
an = (2/T) = (2/4)[ ∫ ωT
0 o dt)tncos()t(g ∫π1
0dt)t
2ncos(5 + ∫
π2
1dt)t
2ncos(10 ]
= 0.5[1
0
t2
nsinn25 ππ
+ 2
1
t2
nsinn2 ππ
10 ] = (–1/(nπ))5 sin(nπ/2)
an = (5/(nπ))(–1)(n+1)/2, n = odd 0, n = even
bn = (2/T) = (2/4)[ ∫ ωT
0 o dt)tnsin()t(g ∫π1
0dt)t
2nsin(5 + ∫
π2
1dt)t
2nsin(10 ]
= 0.5[1
0
t2
ncosn
5x2 ππ
− – 2
1
t2
ncosn
10x2 ππ
] = (5/(nπ))[3 – 2 cos nπ + cos(nπ/2)]
Chapter 17, Solution 4.
f(t) = 10 – 5t, 0 < t < 2, T = 2, ωo = 2π/T = π
ao = (1/T) = (1/2) = ∫T
0dt)t(f ∫ −
2
0dt)t510(
2
0
2 )]2/t5(t10[5.0 − = 5
an = (2/T) = (2/2) ∫ ωT
0 o dt)tncos()t(f ∫ π−2
0dt)tncos()t510(
= – ∫ π2
0dt)tncos()10( ∫ π
2
0dt)tncos()t5(
= 2
022 tncos
n5
ππ
− + 2
0
tnsinn
t5π
π = [–5/(n2π2)](cos 2nπ – 1) = 0
bn = (2/2) ∫ π−2
0dt)tnsin()t510(
= – ∫ π2
0dt)tnsin()10( ∫ π
2
0dt)tnsin()t5(
= 2
022 tnsin
n5
ππ
− + 2
0
tncosn
t5π
π = 0 + [10/(nπ)](cos 2nπ) = 10/(nπ)
Hence f(t) = )tnsin(n110
1nπ
π+ ∑
∞
=
5
Chapter 17, Solution 5.
1T/2,2T =π=ωπ=
5.0]x2x1[21dt)t(z
T1a
T
0o −=π−π
π== ∫
∫∫∫π
π
ππ
ππ
=π
−π
=π
−π
=ω=2
20
0
T
0on 0ntsin
n2nt..sin
n1ntdtcos21ntdtcos11dtncos)t(z
T2a
∫∫∫π
π
ππ
ππ
=
=π=
π+
π−=
π−
π=ω=
22
00
T
0on
evenn 0,
oddn,n6
ntcosn2ntcos
n1ntdtsin21ntdtsin11dtncos)t(z
T2b
Thus,
ntsinn65.0)t(z
oddn1n∑∞
== π
+−=
Chapter 17, Solution 6.
.0a,functionoddanisthisSince
326)1x21x4(
21dt)t(y
21a
22,2T
n
20o
o
=
==+==
π=π
=ω=
∫
∑
∫∫∫
∞
==
=
=π
ππ
+=
=π−π
=π−π
−π−π
=
π−ππ
−−ππ
−=π
π−π
π−
=
π+π=ω=
oddn1n
evenn,0
oddn,n4
21
10
21
10
20 on
)tnsin(n143)t(y
))ncos(1(n2))ncos(1(
n2))ncos(1(
n4
))ncos()n2(cos(n2)1)n(cos(
n4)tncos(
n2)tncos(
n4
dt)tnsin(2dt)tnsin(4dt)tnsin()t(y22b
Chapter 17, Solution 7.
0a,6
T/2,12T 0 =π
=π=ω=
∫∫∫ π−+π=ω=−
10
4
4
2
T
0on ]dt6/tncos)10(dt6/tncos10[
61dtncos)t(f
T2a
[ ]3/n5sin3/nsin3/n2sin2n106/tnsin
n106/tnsin
n10 10
44
2 π−π+ππ
=ππ
−ππ
= −
∫∫∫ π−+π=ω=−
10
4
4
2
T
0on ]dt6/tnsin)10(dt6/tnsin10[
61dtnsin)t(f
T2b
[ ]3/n2sin23/ncos3/n5cosn106/ntncos
n106/tncos
n10 10
44
2 π−π+ππ
=ππ
+ππ
−= −
( )∑∞
=π+π=
1nnn 6/tnsinb6/tncosa)t(f
where an and bn are defined above.
Chapter 17, Solution 8.
π=π=ω=<<+= T/22,T1, t 1- ),t1(2)t(f o
2ttdt)1t(221dt)t(f
T1a
1
1
1
1
2T
0o =+=+==
−−∫∫
0tnsinn1tnsin
nttncos
n12tdtncos)1t(2
22dtncos)t(f
T2a
1
122
1
1
T
0on =
π
π+π
π+π
π=π+=ω=
−−∫∫
ππ
−=
π
π−π
π−π
π−=π+=ω=
−−∫∫ ncos
n4tncos
n1tncos
nttnsin
n12tdtnsin)1t(2
22dtnsin)t(f
T2b
1
122
1
1
T
0on
∑∞
=π
−π
−=1n
ntncos
n)1(42)t(f
Chapter 17, Solution 9. f(t) is an even function, bn=0.
4//2,8 ππω === TT
183.3104/sin)4(4
1004/cos1082)(1 2
0
2
00
===
+== ∫∫ π
ππ
π tdttdttfT
aT
o
[ ]dtntntdttntdtntfT
aT
on ∫∫∫ −++=+==2
0
2
0
2/
0
4/)1(cos4/)1(cos5]04/cos4/cos10[840cos)(4 ππππω
For n = 1,
102/sin25]12/[cos52
0
2
01 =
+=+= ∫ tdttdtta ππ
π
For n>1,
2)1(sin
)1(20
2)1(sin
)1(20
4)1(sin
)1(20
4)1(sin
)1(20 2
0
−−
++
+=
−−
++
+=
nn
nn
nn
tnn
anπ
ππ
ππ
ππ
π
0sin102sin420,3662.62/sin20sin10
32 =+==+= ππ
ππ
ππ
ππ
aa
Thus,
3213210 0,0,362.6,10,183.3 bbbaaaa ======= Chapter 17, Solution 10.
π=π=ω= T/2,2T o
π−−
π−=
−+==
π−π−π−π−ω−∫ ∫ ∫ 2
1
tjn10
tjnT
0
10
21
tjntjntojnn jn
e2jn
e421dte)2(dte4
21dte)t(h
T1c
[ ] ,even n ,0
oddn ,n
j6]6ncos6[
n2je2e24e4
n2jc jnn2jnj
n
=
=π
−=−ππ
=+−−π
= π−π−π−
Thus,
tjn
oddnn
en
6j)t(f π∞
=−∞=∑
π−
=
Chapter 17, Solution 11.
2/T/2,4T o π=π=ω=
∫ ∫ ∫
++==−
π−π−ω−T
0
01
10
2/tjn2/tjntojnn dte)1(dte)1t(
41dte)t(y
T1c
π−
π−−π−
π−= π−
−π−
π− 10
2/tjn01
2/tjn22
2/tjnn e
jn2e
jn2)12/tjn(
4/ne
41c
π
+π
−π
+−ππ
+π
−π
π−ππjn2e
jn2e
jn2)12/jn(e
n4
jn2
n4
41 2/jn2/jn2/jn
2222=
But
2/nsinj2/nsinj2/ncose,2/nsinj2/nsinj2/ncose 2/jn2/jn π−=π−π=π=π+π= π−π
[ ]2/nsinn2/nsin)12/jn(j1n
1c 22n ππ+π−π+π
=
[ ] 2/tjn22
ne2/nsinn2/nsin)12/jn(j1
n1)t(y π
∞
−∞=ππ+π−π+
π= ∑
Chapter 17, Solution 12.
A voltage source has a periodic waveform defined over its period as v(t) = t(2π - t) V, for all 0 < t < 2π Find the Fourier series for this voltage. v(t) = 2π t – t2, 0 < t < 2π, T = 2π, ωo = 2π/T = 1 ao =
(1/T) dt)tt2(21dt)t(f
2
0
2T
0 ∫∫π
−ππ
=3
2)3/21(24)3/tt(
21 23
2
032 π
=−ππ
=−ππ
= π
an = π
π
+π
π=−π∫
2
02
T
0
2 )ntsin(n
t2)ntcos(n21dt)ntcos()tt2(
T2
[ ] π+−
π−
2
0
223 )ntsin(tn)ntsin(2)ntcos(nt2
n1
232 n4)n2cos(n4
n1)11(
n2 −
=πππ
−−=
bn = ∫∫ −π
=− dt)ntsin()tnt2(1dt)ntsin()tnt2(T2 2T
0
2
ππ −+
π−−
π=
2
0
22302 ))ntcos(tn)ntcos(2)ntsin(nt2(
n1))ntcos(nt)nt(sin(
n1n2
0n
4n4
=π
+π−
=
Hence, f(t) = ∑∞
=
−π
1n2
2
)ntcos(n4
32
Chapter 17, Solution 13. T = 2π, ωo = 1
ao = (1/T) dttsin10[21dt)t(h
0
T
0 ∫∫π
π= + ]dt)tsin(20
2
∫π
ππ−
[ ]π
=π−−−π
= π
π
π 30)tcos(20tcos1021 2
0
an = (2/T) ∫ ωT
0 o dt)tncos()t(h
= [2/(2π)]
π−+∫ ∫
π π
π0
2dt)ntcos()tsin(20dt)ntcos(tsin10
Since sin A cos B = 0.5[sin(A + B) + sin(A – B)] sin t cos nt = 0.5[sin((n + 1)t) + sin((1 – n))t] sin(t – π) = sin t cos π – cost sin π = –sin t sin(t – π)cos(nt) = –sin(t)cos(nt)
an =
−++−−++
π ∫ ∫π π
π0
2dt)]t]n1sin([)t]n1[sin([20dt)]t]n1sin([)t]n1[sin([10
21
=
−−
+++
+
−−
−++
−π
π
π
π 2
0 n1)t]n1cos([2
n1)t]n1cos([2
n1)t]n1cos([
n1)t]n1cos([5
an =
−π−
−+
π+−
−+
+π n1)]n1cos([3
n1)]n1cos([3
n13
n135
But, [1/(1+n)] + [1/(1-n)] = 1/(1–n2)
cos([n–1]π) = cos([n+1]π) = cos π cos nπ – sin π sin nπ = –cos nπ
an = (5/π)[(6/(1–n2)) + (6 cos(nπ)/(1–n2))]
= [30/(π(1–n2))](1 + cos nπ) = [–60/(π(n–1))], n = even = 0, n = odd
bn = (2/T) ∫ ωT
0 o dttnsin)t(h
= [2/(2π)][ + ∫π
0dtntsintsin10 ∫
π
π−
2dtntsin)tsin(20
But, sin A sin B = 0.5[cos(A–B) – cos(A+B)] sin t sin nt = 0.5[cos([1–n]t) – cos([1+n]t)] bn = (5/π)[(sin([1–n]t)/(1–n)) – (sin([1+n]t)/ π+
0)]n1(
+ [(2sin([1-n]t)/(1-n)) – (2sin([1+n]t)/ π
π+ 2)]n1(
=
+π+
+−
π−−
π n1)]n1sin([
n1)]n1sin([5 = 0
Thus, h(t) = ∑∞
= −π−
π 1k2 )1k4(
)kt2cos(6030
Chapter 17, Solution 14. Since cos(A + B) = cos A cos B – sin A sin B.
f(t) = ∑∞
=
π
+−π
++
1n33 )nt2sin()4/nsin(
1n10)nt2cos()4/ncos(
1n102
Chapter 17, Solution 15.
(a) Dcos ωt + Esin ωt = A cos(ωt - θ) where A = 22 ED + , θ = tan-1(E/D)
A = 622 n1
)1n(+
+16 , θ = tan-1((n2+1)/(4n3))
f(t) = ∑+10∞
=
−
+−+
+1n3
21
622 n41ntannt10cos
n1
)1n(16
(b) Dcos ωt + Esin ωt = A sin(ωt + θ)
where A = 22 ED + , θ = tan-1(D/E)
f(t) = ∑+10∞
=
−
+
+++1n
2
31
622 1nn4tannt10sin
n1
)1n(16
Chapter 17, Solution 16.
If v2(t) is shifted by 1 along the vertical axis, we obtain v2*(t) shown below, i.e.
v2*(t) = v2(t) + 1.
1
v2*(t)
2
t
-2 -1 0 1 2 3 4 5
Comparing v2*(t) with v1(t) shows that
v2
*(t) = 2v1((t + to)/2) where (t + to)/2 = 0 at t = -1 or to = 1 Hence v2
*(t) = 2v1((t + 1)/2) But v2
*(t) = v2(t) + 1 v2(t) + 1 = 2v1((t+1)/2) v2(t) = -1 + 2v1((t+1)/2)
= -1 + 1
+
+π+
+
π+
+
ππ
−2
1t5cos251
21t3cos
91
21tcos8
2
v2(t) =
+
π
+π
+
π
+π
+
π
+π
π−
25
2t5cos
251
23
2t3cos
91
22tcos8
2
v2(t) =
+
π
+
π
+
π
π 2t5sin
251
2t3sin
91
2tsin8
2−
Chapter 17, Solution 17. We replace t by –t in each case and see if the function remains unchanged.
(a) 1 – t, neither odd nor even.
(b) t2 – 1, even
(c) cos nπ(-t) sin nπ(-t) = - cos nπt sin nπt, odd
(d) sin2 n(-t) = (-sin πt)2 = sin2 πt, even
(e) et, neither odd nor even.
Chapter 17, Solution 18.
(a) T = 2 leads to ωo = 2π/T = π
f1(-t) = -f1(t), showing that f1(t) is odd and half-wave symmetric.
(b) T = 3 leads to ωo = 2π/3 f2(t) = f2(-t), showing that f2(t) is even.
(c) T = 4 leads to ωo = π/2
f3(t) is even and half-wave symmetric. Chapter 17, Solution 19. This is a half-wave even symmetric function.
ao = 0 = bn, ωo = 2π/T π/2
an = ∫ ω
−
2/T
0 o dt)tncos(Tt41
T4
= [4/(nπ)2](1 − cos nπ) = 8/(n2π2), n = odd
= 0, n = even
f (t) = ∑∞
=
π
π oddn22 2
tncosn18
Chapter 17, Solution 20. This is an even function.
bn = 0, T = 6, ω = 2π/6 = π/3
ao =
−= ∫∫∫
3
2
2
1
2/T
0dt4dt)4t4(
62dt)t(f
T2
=
−+− )23(4)t4t2(
31 2
1
2 = 2
an = ∫ π4/T
0dt)3/tncos()t(f
T4
= (4/6)[ + ] ∫ π−2
1dt)3/tncos()4t4( ∫ π
3
2dt)3/tncos(4
= 3
2
2
122 3
tnsinn3
616
3tnsin
n3
3tnsin
nt3
3tncos
n9
616
π
π+
π
π−
π
π+
π
π
= [24/(n2π2)][cos(2nπ/3) − cos(nπ/3)]
Thus f(t) = ∑∞
=
π
π−
π
π+
1n22 3
tncos3ncos
3n2cos
n1242
At t = 2,
f(2) = 2 + (24/π2)[(cos(2π/3) − cos(π/3))cos(2π/3)
+ (1/4)(cos(4π/3) − cos(2π/3))cos(4π/3) + (1/9)(cos(2π) − cos(π))cos(2π) + -----]
= 2 + 2.432(0.5 + 0 + 0.2222 + -----)
f(2) = 3.756 Chapter 17, Solution 21. This is an even function.
bn = 0, T = 4, ωo = 2π/T = π/2. f(t) = 2 − 2t, 0 < t < 1 = 0, 1 < t < 2
ao = ∫
−=−
1
0
1
0
2
2ttdt)t1(2
42 = 0.5
an = ∫∫
π
−=ω1
0
2/T
0 o dt2
tncos)t1(244dt)tncos()t(f
T4
= [8/(π2n2)][1 − cos(nπ/2)]
f(t) = ∑∞
=
π
π
−π
+1n
22 2tncos
2ncos1
n8
21
Chapter 17, Solution 22.
Calculate the Fourier coefficients for the function in Fig. 16.54. f(t)
4
t
-5 -4 -3 -2 -1 0 1 2 3 4 5
Figure 17.61 For Prob. 17.22
This is an even function, therefore bn = 0. In addition, T=4 and ωo = π/2.
ao = === ∫∫1
0
21
0
2T
0ttdt4
42dt)t(f
T2 1
an = ∫∫ π=ω1
0
2T
0 o dt)2/tncos(t444dt)ntcos()t(f
T4
1
022 )2/tnsin(
nt2)2/tncos(
n44
π
π+π
π=
an = )2/nsin(n8)1)2/n(cos(
n16
22 ππ
+−ππ
Chapter 17, Solution 23. f(t) is an odd function.
f(t) = t, −1< t < 1 ao = 0 = an, T = 2, ωo = 2π/T = π
bn = ∫∫ π=ω1
0
2/T
0 o dt)tnsin(t24dt)tnsin()t(f
T4
= [ ]1022 )tncos(tn)tnsin(
n2
ππ−ππ
= −[2/(nπ)]cos(nπ) = 2(−1)n+1/(nπ)
f(t) = ∑∞
=
+
π−
π 1n
1n
)tnsin(n)1(2
Chapter 17, Solution 24.
(a) This is an odd function.
ao = 0 = an, T = 2π, ωo = 2π/T = 1
bn = ∫ ω2/T
0 o dt)ntsin()t(fT4
f(t) = 1 + t/π, 0 < t < π
bn = ∫π
π+π 0
dt)ntsin()/t1(24
= π
π−
π+−
π 02 )ntcos(
nt)ntsin(
n1)ntcos(
n12
= [2/(nπ)][1 − 2cos(nπ)] = [2/(nπ)][1 + 2(−1)n+1]
a2 = 0, b2 = [2/(2π)][1 + 2(−1)] = −1/π = −0.3183
(b) ωn = nωo = 10 or n = 10
a10 = 0, b10 = [2/(10π)][1 − cos(10π)] = −1/(5π)
Thus the magnitude is A10 = 2
102 ba10+ = 1/(5π) = 0.06366
and the phase is φ10 = tan−1(bn/an) = −90°
(c) f(t) = ∑ π ∞
=
π−π1n
)ntsin()]ncos(21[n2
f(π/2) = ∑∞
=
ππ−π1n
)2/nsin()]ncos(21[n2
π
For n = 1, f1 = (2/π)(1 + 2) = 6/π For n = 2, f2 = 0 For n = 3, f3 = [2/(3π)][1 − 2cos(3π)]sin(3π/2) = −6/(3π) For n = 4, f4 = 0 For n = 5, f5 = 6/(5π), ----
Thus, f(π/2) = 6/π − 6/(3π) + 6/(5π) − 6/(7π) ---------
= (6/π)[1 − 1/3 + 1/5 − 1/7 + --------] f(π/2) ≅ 1.3824 which is within 8% of the exact value of 1.5.
(d) From part (c) f(π/2) = 1.5 = (6/π)[1 − 1/3 + 1/5 − 1/7 + - - -] (3/2)(π/6) = [1 − 1/3 + 1/5 − 1/7 + - - -] or π/4 = 1 − 1/3 + 1/5 − 1/7 + - - - Chapter 17, Solution 25. This is an odd function since f(−t) = −f(t).
ao = 0 = an, T = 3, ωo = 2π/3.
b n = ∫∫ π=ω1
0
2/T
0 o dt)3/nt2sin(t34dt)tnsin()t(f
T4
= 1
022 3
nt2cosn2t3
3nt2sin
n49
34
π
π−
π
π
=
π
π−
π
π 3n2cos
n23
3n2sin
n49
34
22
f(t) = ∑∞
=
π
π
π−
π
π1n22 3
t2sin3n2cos
n2
3n2sin
n3
Chapter 17, Solution 26. T = 4, ωo = 2π/T = π/2
ao =
++= ∫∫∫∫
4
3
3
1
1
0
T
0dt1dt2dt1
41dt)t(f
T1 = 1
an = dt)tncos()t(fT
T
0 o∫ ω2
an =
π+π+π ∫∫∫
4
3
3
2
2
1dt)2/tncos(1dt)2/tncos(2dt)2/tncos(1
42
=
ππ
+π
π+
ππ
4
3
3
2
2
1 2tnsin
n2
2tnsin
n4
2tnsin
n22
=
π
−π
π 2nsin
2n3sin
n4
bn = ∫ ωT
0 o dt)tnsin()t(fT2
=
π
+π
+π
∫∫∫4
3
3
2
2
1dt
2tnsin1dt
2tnsin2dt
2tnsin1
42
=
ππ
−π
π−
ππ
−4
3
3
2
2
1 2tncos
n2
2tncos
n4
2tncos
n22
= [ ]1)ncos(n4
−ππ
Hence f(t) =
[ ]∑∞
=
π−π+ππ−ππ
+1n
)2/tnsin()1)n(cos()2/tncos())2/nsin()2/n3(sin(n41
Chapter 17, Solution 27. (a) odd symmetry. (b) ao = 0 = an, T = 4, ωo = 2π/T = π/2 f(t) = t, 0 < t < 1 = 0, 1 < t < 2
bn = 1
022
1
0 2tncos
nt2
2tnsin
n4dt
2tnsint
4
π
π−
ππ
=π
∫4
= 02
ncosn2
2nsin
n4
22 −π
π−
ππ
= 4(−1)(n−1)/2/(n2π2), n = odd
−2(−1)n/2/(nπ), n = even
a3 = 0, b3 = 4(−1)/(9π2) = 0.045 (c) b1 = 4/π2, b2 = 1/π, b3 = −4/(9π2), b4 = −1/(2π), b5 = (25π2)
Frms = ( )∑ ++ 2n
2n
2o ba
21a
Frms
2 = 0.5Σbn2 = [1/(2π2)][(16/π2) + 1 + (16/(8π2)) + (1/4) + (16/(625π2))]
= (1/19.729)(2.6211 + 0.27 + 0.00259) Frms = 14659.0 = 0.3829
Compare this with the exact value of Frms = 6/1dttT2 1
0
2 =∫ = 0.4082
Chapter 17, Solution 28. This is half-wave symmetric since f(t − T/2) = −f(t).
ao = 0, T = 2, ωo = 2π/2 = π
an = ∫∫ π−=ω1
0
2/T
0 o dt)tncos()t22(24dt)tncos()t(f
T4
= 1
022 )tnsin(
nt)tncos(
n1)tnsin(
n14
π
π−π
π−π
π
= [4/(n2π2)][1 − cos(nπ)] = 8/(n2π2), n = odd
0, n = even
bn = 4 ∫ π−1
0dt)tnsin()t1(
= 1
022 )tncos(
nt)tnsin(
n1)tncos(
n14
π
π+π
π−π
π−
= 4/(nπ), n = odd
f(t) = ∑∞
=
π
π+π
π1k22 )tnsin(
n4)tncos(
n8 , n = 2k − 1
Chapter 17, Solution 29. This function is half-wave symmetric.
T = 2π, ωo = 2π/T = 1, f(t) = −t, 0 < t < π
For odd n, an = ∫π−
0dt)ntcos()t(
T2 = [ ]π+
π−
02 )ntsin(nt)ntcos(n2 = 4/(n2π)
bn = [ ]ππ−
π−=−
π ∫ 020)ntcos(nt)ntsin(
n2dt)ntsin()t(2 = −2/n
Thus,
f(t) = ∑∞
=
−
π1k2 )ntsin(
n1)ntcos(
n22 , n = 2k − 1
Chapter 17, Solution 30.
∫ ∫ ∫−
− −ω−
ω−ω==
2/T
2/T
2/T2/T
2/T2/T oo
tojnn tdtnsin)t(fjtdtncos)t(f
T1dte)t(f
T1c (1)
(a) The second term on the right hand side vanishes if f(t) is even. Hence
∫ ω=2/T
0on tdtncos)t(f
T2c
(b) The first term on the right hand side of (1) vanishes if f(t) is odd. Hence,
∫ ω−=2/T
0on tdtnsin)t(f
T2jc
Chapter 17, Solution 31.
If oo /T2
'T2'/T'T),t(f)t(h αω=
απ
=π
=ω→α=α=
∫∫ ωα=ω='T
0o
'T
0on tdt'ncos)t(f
'T2tdt'ncos)t(h
'T2'a
Let T'T,/dtd,,t =ααλ=λ=α
n
T
0on a/dncos)(f
T2'a =αλλωλα
= ∫
Similarly, nn b'b =
Chapter 17, Solution 32. When is = 1 (DC component)
i = 1/(1 + 2) = 1/3
For n 1, ω≥ n = 3n, Is = 1/n2∠0°
I = [1/(1 + 2 + jωn2)]Is = Is/(3 + j6n)
= )n2tan(n41n3
1)3/n6(tann413
0n1
2212
2−∠
+=
∠+
°∠
−
Thus,
i(t) = ∑∞
=
−−+
+1n
1
22))n2(tann3cos(
n41n3
131
Chapter 17, Solution 33.
For the DC case, the inductor acts like a short, Vo = 0. For the AC case, we obtain the following:
so
ooso
VVn5n5.2j1
04Vjn
n2jV
10VV
=
π−π+
=π
+π
+−
)5n5.2(jn4
n5n5.2j1
1n4A
n5n5.2j1
VV
22nn
so
−π+π=
π−π+π
=Θ∠
π−π+
=
π−π
−=Θ−π+π
= −n
5n5.2tan;)5n5.2(n
4A22
1n22222n
V)tnsin(A)t(v1n
nno ∑∞
=Θ+π=
Chapter 17, Solution 34. For any n, V = [10/n2]∠(nπ/4), ω = n. 1 H becomes jωnL = jn and 0.5 F becomes 1/(jωnC) = −j2/n
2 Ω jn
−j2/n
+ −
+
Vo
−
V Vo = −j(2/n)/[2 + jn − j(2/n)]V = −j2/[2n + j(n2 − 2)][(10/n2)∠(nπ/4)]
)]n2/)2n((tan)2/()4/n[(
4nn
20)n2/)2n((tan)2n(n4n
)2/)4/n((20
21
22
212222
−−π−π∠+
=
−∠−+
π−π∠=
−
−
vo(t) = ∑∞
=
−
−−
π−
π+
+1n
21
22 n22ntan
24nntcos
4nn20
Chapter 17, Solution 35.
If vs in the circuit of Fig. 17.72 is the same as function f2(t) in Fig. 17.57(b), determine the dc component and the first three nonzero harmonics of vo(t).
1 Ω 1 H
1 F 1 Ω
+ −
+
vo
−
vS
Figure 17.72 For Prob. 17.35
1
f2(t)
2
t
-2 -1 0 1 2 3 4 5
Figure 17.57(b) For Prob. 17.35
The signal is even, hence, bn = 0. In addition, T = 3, ωo = 2π/3. vs(t) = 1 for all 0 < t < 1 = 2 for all 1 < t < 1.5
ao = 34dt2dt1
32 5.1
1
1
0=
+ ∫∫
an =
π+π ∫∫
5.1
1
1
0dt)3/tn2cos(2dt)3/tn2cos(
34
= )3/n2sin(n2)3/tn2sin(
n26)3/tn2sin(
n23
34 5.1
1
1
0π
π−=
π
π+π
π
vs(t) = ∑∞
=
πππ
−1n
)3/tn2cos()3/n2sin(n12
34
Now consider this circuit,
j2nπ/3 1 Ω
-j3/(2nπ) 1 Ω
+ −
+
vo
−
vS
Let Z = [-j3/(2nπ)](1)/(1 – j3/(2nπ)) = -j3/(2nπ - j3) Therefore, vo = Zvs/(Z + 1 + j2nπ/3). Simplifying, we get
vo = )18n4(jn12
v9j22
s
−π+π−
For the dc case, n = 0 and vs = ¾ V and vo = vs/2 = 3/8 V. We can now solve for vo(t)
vo(t) = volts3
tn2cosA83
1nnn
Θ+
π+∑
∞
=
where
π−
π−=Θ
−
π+π
ππ= −
n23
3ntan90and
63
n4n16
)3/n2sin(n6
A 1on222
22
n
where we can further simplify An to this, 81n4n)3/n2sin(9
44n+ππ
A π=
Chapter 17, Solution 36.
vs(t) = ∑∞
==
θ−
oddn1n
nn )ntcos(A
where θn = tan−1[(3/(nπ))/(−1/(nπ))] = tan−1(−3) = 100.5°
An = 2
nsin9n1
2nsin
n1
n9 22
2222π
+π
=π
π+
π
ωn = n and 2 H becomes jωnL = j2n
Let Z = 1||j2n = j2n/(1 + j2n) If Vo is the voltage at the non-reference node or across the 2-H inductor. Vo = ZVs/(1 + Z) = [j2n/(1 + j2n)]Vs/1 + [j2n/(1 + j2n)] = j2nVs/(1 + j4n) But Vs = An∠−θn Vo = j2n An∠−θn/(1 + j4n)
Io = Vo/j = [2n An∠−θn]/ 2n161 + ∠tan−14n
= 2
2
n161
n22
nsin9n1
+
π+
π∠−100.5° − tan−14n
Since sin(nπ/2) = (−1)(n−1)/2 for n = odd, sin2(nπ/2) = 1
Io = 2
1
n161
n4tan5.100102
+
−°−∠π
−
io(t) = )n4tan5.100ntcos(n161
1102
oddn1n
1
2∑∞
==
−−°−+π
Chapter 17, Solution 37. From Example 15.1,
vs(t) = ∑∞
=
ππ
+1k
),tnsin(n1205 n = 2k − 1
For the DC component, the capacitor acts like an open circuit.
Vo = 5
For the nth harmonic, Vs = [20/(nπ)]∠0°
10 mF becomes 1/(jωnC) = −j/(nπx10x10−3) = −j100/(nπ)
vo = 22
1s
n25nn5tan90100
n20
100jn20100j
20n
100j
Vn
100j
π+ππ
+°−∠=
π−π−
=+
π−
π− −
vo(t) = ∑ π+°−π
π+π− )
n5tan90tnsin(
n25n
1100 1
22
Chapter 17, Solution 38.
∑∞
=+=π
π+=
1ks 1k2n,tnsin
n12
21)t(v
π=ωω+
ω= n,V
j1jV ns
n
no
For dc, 0V,5.0V,0 osn ===ω
For nth harmonic, os 90
n2V −∠π
=
22
1o
122
oo
n1
ntan290n2
ntann1
90nVπ+
π−∠=∠
π•
π∠π+
∠π=
−
−
1k2n),ntantncos(n1
2)t(v 1
1k 22o −=π−ππ+
= −∞
=∑
Chapter 17, Solution 39.
Comparing vs(t) with f(t) in Figure 15.1, vs is shifted by 2.5 and the magnitude is 5 times that of f(t). Hence
vs(t) = ∑∞
=
ππ
+1k
),tnsin(n1105 n = 2k − 1
T = 2, ωo = 2π//T = π, ωn = nωo = nπ
For the DC component, io = 5/(20 + 40) = 1/12 For the kth harmonic, Vs = (10/(nπ))∠0°
100 mH becomes jωnL = jnπx0.1 = j0.1nπ
50 mF becomes 1/(jωnC) = −j20/(nπ) 40 Ω
−j20/(nπ)
+ −
I Io
Z
20 Ω
VS j0.1nπ
Let Z = −j20/(nπ)||(40 + j0.1nπ) = π++
π−
π+π
−
n1.0j40n20j
)n1.0j40(n20j
= )20n1.0(jn40
800jn2n1.0jn4020j
n1.0j40(20j2222 −π+π
−π=
π+π+−π+−
Zin = 20 + Z = )20n1.0(jn40)1200n2(jn802
22
22
−π+π−π+π
I = )]1200n2(jn802[n
)200n(jn400ZV
22
22
in
s
−π+ππ−π+π
=
Io = )20n1.0(jn40
I20j
)n1.0j40(n20j
In20j
22 −π+π−
=π++
π−
π−
= )]1200n2(jn802[n
200j22 −π+ππ
−
= 2222
221
)1200n2()802(n)n802/()1200n2(tan90200
−π+π
π−π−°−∠ −
Thus
io(t) = ∑∞
=
θ−ππ
+1k
nn )tnsin(I200201 , n = 2k − 1
where π
−π+°=θ −
n8021200n2tan90
221
n
In = )1200n2()n804(n
1222 −π+π
Chapter 17, Solution 40. T = 2, ωo = 2π/T = π
ao = 2/12ttdt)t22(
21dt)t(v
T
1
0
21
0
T
0=
−=−= ∫∫
1
an = ∫∫ π−=π1
0
T
0dt)tncos()t1(2dt)tncos()t(v
T2
= 1
022 )tnsin(
nt)tncos(
n1)tnsin(
n12
π
π−π
π−π
π
= 2222
22
)1n2(4oddn,
n4
evenn,0)ncos1(
n2
−π==
π
==π−
π
bn = ∫∫ π−=π1
0
T
0dt)tnsin()t1(2dt)tnsin()t(v
T2
= π
=
π
π+π
π−π
π−
n2)tncos(
nt)tnsin(
n1)tncos(
n12
1
022
vs(t) = ∑ ϕ−π+ )tncos(A21
nn
where 4422n
21
n )1n2(16
n4A,
n2)1n2(
−π+
π=
−π−tan=φ
For the DC component, vs = 1/2. As shown in Figure (a), the capacitor acts like an open circuit.
+
Vx
−
i
Vo2Vx
+
Vo
−
Vx
− +
+ − 3 Ω
1 Ω 0.5V
(a)
Vo
(1/4)F
2Vx
+
Vo
−
Vx
− +
+ − 3 Ω
1 Ω VS (b)
Applying KVL to the circuit in Figure (a) gives –0.5 – 2Vx + 4i = 0 (1) But –0.5 + i + Vx = 0 or –1 + 2Vx + 2i = 0 (2) Adding (1) and (2), –1.5 + 6i = 0 or i = 0.25 Vo = 3i = 0.75 For the nth harmonic, we consider the circuit in Figure (b).
ωn = nπ, Vs = An∠–φ, 1/(jωnC) = –j4/(nπ) At the supernode, (Vs – Vx)/1 = –[nπ/(j4)]Vx + Vo/3 Vs = [1 + jnπ/4]Vx + Vo/3 (3) But –Vx – 2Vx + Vo = 0 or Vo = 3Vx Substituting this into (3), Vs = [1 + jnπ/4]Vx + Vx = [2 + jnπ/4]Vx = (1/3)[2 + jnπ/4]Vo = (1/12)[8 + jnπ]Vo
Vo = 12Vs/(8 + jnπ) = )8/n(tann64
A12122
n
π∠π+
φ−∠−
Vo = ))]n2/()1n2((tan)8/n([tan)1n2(
16n
4n64
12 11442222
−π−π∠−π
+ππ+
−−
Thus
vo(t) = ∑∞
=
θ+π+1n
nn )tncos(V43
where Vn = 442222 )1n2(
16n
4
n64
12−π
+ππ+
θn = tan–1(nπ/8) – tan–1(π(2n – 1)/(2n)) Chapter 17, Solution 41. For the full wave rectifier,
T = π, ωo = 2π/T = 2, ωn = nωo = 2n
Hence
vin(t) = ( )∑∞
= −π−
π 1n2 nt2cos
1n4142
For the DC component,
Vin = 2/π
The inductor acts like a short-circuit, while the capacitor acts like an open circuit.
Vo = Vin = 2/π
For the nth harmonic, Vin = [–4/(π(4n2 – 1))]∠0°
2 H becomes jωnL = j4n
0.1 F becomes 1/(jωnC) = –j5/n
Z = 10||(–j5/n) = –j10/(2n – j)
Vo = [Z/(Z + j4n)]Vin = –j10Vin/(4 + j(8n – 10))
=
−π°∠
−−+
−)1n4(
04)10n8(j4
10j2
= 22
1
)10n8(16)1n4()5.2n2(tan9040
−+−π
−−°∠ −
Hence vo(t) = ∑∞
=
θ++π 1n
nn )nt2cos(A2
where
An = 29n40n16)1n4(
2022 +−−π
θn = 90° – tan–1(2n – 2.5)
Chapter 17, Solution 42.
∑∞
==π
π+=
1ks 1-2kn ,tnsin
n1205v
π=ω=ωω
=→−ω=−
nn ,VRCjV)V0(Cj
R0V
onsn
oons
For n = 0 (dc component), Vo=0. For the nth harmonic,
22
5
9422o
oo
n210
10x40x10xn2090
n20
RCn901V
π=
π=−∠
ππ∠
=−
Hence,
∑∞
==π
π=
1k22
5o 1-2kn ,tncos
n1
210)t(v
Alternatively, we notice that this is an integrator so that
∑∫∞
==π
π=−=
1k22
5so 1-2kn ,tncos
n1
210dtv
RC1)t(v
Chapter 17, Solution 43.
(a) Vrms = )1020(2130)ba(
21a 222
1n
2n
2n
20 ++=++ ∑
∞
=
= 33.91 V
(b) Irms = )24(216 222 ++ = 6.782 A
(c) P = VdcIdc + ∑ Φ−Θ )cos(IV21
nnnn
= 30x6 + 0.5[20x4cos(45o-10o) – 10x2cos(-45o+60o)]
= 180 + 32.76 – 9.659 = 203.1 W Chapter 17, Solution 44.
(a) [ ] W73.300535.319.27045cos1025cos6021vip oo =++=++==
(b) The power spectrum is shown below.
p 27.19 3.535 0 1 2 3 ω Chapter 17, Solution 45. ωn = 1000n jωnL = j1000nx2x10–3 = j2n 1/(jωnC) = –j/(1000nx40x10–6) = –j25/n
Z = R + jωnL + 1/(jωnC) = 10 + j2n – j25/n
I = V/Z
For n = 1, V1 = 100, Z = 10 + j2 – j25 = 10 – j23 I1 = 100/(10 – j23) = 3.987∠73.89°
For n = 2, V2 = 50, Z = 10 + j4 – j12.5 = 10 – j8.5 I2 = 50/(10 – j8.5) = 3.81∠40.36°
For n = 3, V3 = 25, Z = 10 + j6 – j25/3 = 10 – j2.333 I3 = 25/(10 – j2.333) = 2.435∠13.13° Irms = 0.5 222 435.281.3987.3 ++ = 3.014 A
p = VDCIDC + ∑=
φ−θ3
1nnnnn )cos(IV
21
= 0 + 0.5[100x3.987cos(73.89°) + 50x3.81cos(40.36°) + 25x2.435cos(13.13°)]
= 0.5[110.632 + 145.16 + 59.28] = 157.54 watts Chapter 17, Solution 46. (a) This is an even function
Irms = ∫∫ =2/T
0
2T
0
2 dt)t(fT2dt)t(f
T1
f(t) = 2t1,01t0,t22
<<<<−
T = 4, ωo = 2π/T = π/2
Irms2 =
1
0
321
0
2 )3/ttt(2dt)t1(442
+−=−∫
= 2(1 – 1 + 1/3) = 2/3 or Irms = 0.8165 A (b) From Problem 16.14, an = [8/(n2π2)][1 – cos(nπ/2)], ao = 0.5
a1 = 8/π2, a2 = 4/π2, a3 = 8/(9π2), a4 = 0, a5 = 9/(25π2), a6 = 4/(9π2)
Irms = 66623.08116
62564
81641664
21
41A
21a 4
1n
2no =
++++
π+≅+ ∑
∞
=
Irms = 0.8162 A Chapter 17, Solution 47. Let I = IDC + I1 + I2
For the DC component
IDC = [5/(5 + 10)](3) = 1 A
Is
I
5 Ω
j8 10 Ω
For AC, ω = 100 jωL = j100x80x10–3 = j8
In = 5Is/(5 + 10 + j8)
For Is = 0.5∠–60° I1 = 10∠–60°/(15 + j8) or |I1| = 10/ 22 815 +
For Is = 0.5∠–120°
I2 = 2.5∠–120°/(15 + j8) or |I2| = 2.5/ 22 815 + p10 = (IDC
2 + |I1|2/2 + |I2|2/2)10 = (1 + [100/(2x289)] + [6.25/(2x289)])x10
p10 = 11.838 watts
Chapter 17, Solution 48. (a) For the DC component, i(t) = 20 mA. The capacitor acts like an open circuit so that v = Ri(t) = 2x103x20x10–3 = 40
For the AC component, ωn = 10n, n = 1,2
1/(jωnC) = –j/(10nx100x10–6) = (–j/n) kΩ Z = 2||(–j/n) = 2(–j/n)/(2 – j/n) = –j2/(2n – j)
V = ZI = [–j2/(2n – j)]I
For n = 1, V1 = [–j2/(2 – j)]16∠45° = 14.311∠–18.43° mV
For n = 2, V2 = [–j2/(4 – j)]12∠–60° = 5.821∠–135.96° mV
v(t) = 40 + 0.014311cos(10t – 18.43°) + 0.005821cos(20t – 135.96°) V
(b) p = VDCIDC + ∑∞
=
φ−θ1n
nnnn )cos(IV21
= 20x40 + 0.5x10x0.014311cos(45° + 18.43°) +0.5x12x0.005821cos(–60° + 135.96°)
= 800.1 mW Chapter 17, Solution 49.
(a) 5.2)5(21dt4dt1
21dt)t(z
T1Z
0
2T
0
2rms
2 =ππ
=
+
π== ∫ ∫∫
π π
π
581.1Zrms =
(b)
9193.2...251
161
91
411
181
41
n36
21
41)ba(
21aZ 2
1n 1n22n
2n
2o
2rms
2 =
+++++
π+=
π+=++= ∑ ∑
∞
=
∞
= 7086.1rms =Z
(c ) 071.8100x11.581
1.7086 %error =
−=
Chapter 17, Solution 50.
cn = ∫ π==ωω−T
0 ontj
1n2,dte)t(f
To
1
= ∫−π−1 tjn dtte
21
Using integration by parts,
u = t and du = dt dv = e–jnπtdt which leads to v = –[1/(2jnπ)]e–jnπt
cn = ∫−π−
−
π−
π+
π
1
1
tjn1
1
tjn dtejn21e
jn2t
−
= [ ]1
1
tjn222
tjnjn e)j(n2
1eenj
−
π−ππ−
−π++
π
= [j/(nπ)]cos(nπ) + [1/(2n2π2)](e–jnπ – ejnπ)
cn = π−
=ππ
+π−
n)1(j)nsin(
n2j2
n)1( n
22
nj
Thus
f(t) = ∑∞
−∞=
ω
n
tjnn
oec = ∑∞
−∞=
π
π−
n
tjnn enj)1(
Chapter 17, Solution 51.
π=π=ω= T/2,2T o
( ) 20
T
0
2222
03
tjntjn2tojn
n 2tjn2tn)jn(
e21dtet
21dte)t(f
T1c ∫ ∫ +π+π−
π−===
π−π−ω−
)jn1(n
2)n4jn4(n2j1c 22
2233n π+
π=π+π−
π=
∑∞
−∞=
ππ+π
=n
tjn22
e)jn1(n
2)t(f
Chapter 17, Solution 52.
cn = ∫ π==ωω−T
0 ontj
1n2,dte)t(f
To
1
= ∫−π−1 tjn dtte
21
Using integration by parts,
u = t and du = dt dv = e–jnπtdt which leads to v = –[1/(2jnπ)]e–jnπt
cn = ∫−π−
−
π−
π+
π
1
1
tjn1
1
tjn dtejn21e
jn2t
−
= [ ]1
1
tjn222
tjnjn e)j(n2
1eenj
−
π−ππ−
−π++
π
= [j/(nπ)]cos(nπ) + [1/(2n2π2)](e–jnπ – ejnπ)
cn = π−
=ππ
+π−
n)1(j)nsin(
n2j2
n)1( n
22
nj
Thus
f(t) = ∑∞
−∞=
ω
n
tjnn
oec = ∑∞
−∞=
π
π−
n
tjnn enj)1(
Chapter 17, Solution 53. ωo = 2π/T = 2π
cn = ∫ ∫ ω+−ω−− =1
0
t)jn1(T
0
tjnt dtedtee oo
= [ ]1en2j1
1en2j1
1 )n2j1(1
0
t)n2j1( −π+
−=
π+− π+−π+−
= [1/(j2nπ)][1 – e–1(cos(2πn) – jsin(2nπ))]
= (1 – e–1)/(1 + j2nπ) = 0.6321/(1 + j2nπ
f(t) = ∑∞
−∞=
π
π+n
tn2j
n2j1e6321.0
Chapter 17, Solution 54. T = 4, ωo = 2π/T = π/2
cn = ∫ ω−T
0
ntj dte)t(fT
o1
=
−+ ∫∫∫ π−π−π− 4
2
2/tjn2
1
2/tjn1
0
2/tjn dte1dte1dte241
= [ ]π−π−π−π−π− +−−+−π
jnn2j2/jnjn2/jn eeee2e2n2j
= [ ]π−π− +−π
jn2/jn e23e3n2j
f(t) = ∑∞
−∞=
ω
n
tjnn
oec
Chapter 17, Solution 55. T = 2π, ωo = 2π/T = 1
cn = ∫ ω−T
0
tjn dte)t(iT
o1
But i(t) = π<<π
π<<2t,0
t0),tsin(
cn = ∫∫π −−π π− −
π=
π 0
jntjtjt
0
tjn dte)ee(j2
121dte)tsin(
21
+
−+
−−
−=
+
+−π
=
+π−−π
π+−−
n11e
n11e
41
)n1(je
)n1(je
j41
)1n(j)n1(j
0
)n1(jt)n1(jt
[ ]nne1enne1e)1n(4
1 )n1(j)n1(j)n1(j)n1(j2 +−−+−+−−π
= +π−+π−−π−π
But ejπ = cos(π) + jsin(π) = –1 = e–jπ
cn = [ ])n1(2
e12neneee)1n(4
12
jnjnjnjnjn
2 −π+
=−+−−−−π
π−π−π−π−π−
Thus
i(t) = ∑∞
−∞=
ππ−
−π+
n
tjn2
jn
e)n1(2
e1
Chapter 17, Solution 56. co = ao = 10, ωo = π co = (an – jbn)/2 = (1 – jn)/[2(n2 + 1)]
f(t) = ∑∞
≠−∞=
π
+−
+
0nn
tjn2 e
)1n(2)jn1(10
Chapter 17, Solution 57. ao = (6/–2) = –3 = co cn = 0.5(an –jbn) = an/2 = 3/(n3 – 2)
f(t) = ∑∞
≠−∞= −
+
0nn
nt50j3 e
2n33−
Chapter 17, Solution 58. cn = (an – jbn)/2, (–1) = cos(nπ), ωo = 2π/T = 1
cn = [(cos(nπ) – 1)/(2πn2)] – j cos(nπ)/(2n)
Thus
f(t) = ∑+π
π
−π
−π jnt2 e
n2)ncos(j
n21)ncos(
4
Chapter 17, Solution 59. For f(t), T = 2π, ωo = 2π/T = 1. ao = DC component = (1xπ + 0)/2π = 0.5
For h(t), T = 2, ωo = 2π/T = π. ao = (3x1 – 2x1)/2 = 0.5
Thus by replacing ωo = 1 with ωo = π and multiplying the magnitude by five, we obtain
h(t) = ∑−1 ∞
≠−∞=
π+
π+0n
n
t)1n2(j
)1n2(e5j
2
Chapter 17, Solution 60. From Problem 16.17, ao = 0 = an, bn = [2/(nπ)][1 – 2 cos(nπ)], co = 0 cn = (an – jbn)/2 = [j/(nπ)][2 cos(nπ) – 1], n ≠ 0. Chapter 17, Solution 61.
(a) ωo = 1.
f(t) = ao + ∑ φ−ω )tncos(A non
= 6 + 4cos(t + 50°) + 2cos(2t + 35°)
+ cos(3t + 25°) + 0.5cos(4t + 20°) = 6 + 4cos(t)cos(50°) – 4sin(t)sin(50°) + 2cos(2t)cos(35°)
– 2sin(2t)sin(35°) + cos(3t)cos(25°) – sin(3t)sin(25°) + 0.5cos(4t)cos(20°) – 0.5sin(4t)sin(20°)
= 6 + 2.571cos(t) – 3.73sin(t) + 1.635cos(2t)
– 1.147sin(2t) + 0.906cos(3t) – 0.423sin(3t) + 0.47cos(4t) – 0.171sin(4t)
(b) frms = ∑∞
=
+1n
2n
2o A
21a
frms
2 = 62 + 0.5[42 + 22 + 12 + (0.5)2] = 46.625 frms = 6.828 Chapter 17, Solution 62.
(a) s3141.0202TT/220o =π
=→π==ω
(b) f ...)90t40cos(1.5)90t20cos(43)tncos(Aa)t( o
1n
onono +++++=φ+ω+= ∑
∞
=
....t80sin8.1t60sin7.2t40sin1.5t20sin43)t(f −−−−−= Chapter 17, Solution 63.
This is an even function.
T = 3, ωo = 2π/3, bn = 0.
f(t) = 5.1t1,2
1t0,1<<<<
ao =
+= ∫∫∫
5.1
1
1
0
2/T
0dt2dt1
32dt)t(f
T2 = (2/3)[1 + 1] = 4/3
an =
π+π=ω ∫∫∫
5.1
1
1
0
2/T
0 o dt)3/tn2cos(2dt)3/tn2cos(134dt)tncos()t(f
T4
=
π
π+
π
π
5.1
1
1
0 3tn2sin
n26
3tn2sin
n23
34
= [–2/(nπ)]sin(2nπ/3)
f2(t) = ∑∞
=
π
π
π−
1n 3tn2cos
3n3sin
n12
34
ao = 4/3 = 1.3333, ωo = 2π/3, an = –[2/(nπ)]sin(2nπt/3)
An =
π
π=+
3n2sin
n2b2
n2na
A1 = 0.5513, A2 = 0.2757, A3 = 0, A4 = 0.1375, A5 = 0.1103
The amplitude spectra are shown below. 1.333
0.1103 0.1378
0
0.275
0.551
4321n
5 0
An
Chapter 17, Solution 64.
The amplitude and phase spectra are shown below.
An 3.183 2.122 1.591 0.4244 0 2π 4π 6π ω nφ 0 2π 4π 6π ω
-180o
Chapter 17, Solution 65. an = 20/(n2π2), bn = –3/(nπ), ωn = 2n
An = 22442n n
9n400b
π+
π=+2
na
= 22n44.441
n3
π+
π, n = 1, 3, 5, 7, 9, etc.
n An 1 2.24 3 0.39 5 0.208 7 0.143 9 0.109
φn = tan–1(bn/an) = tan–1[–3/(nπ)][n2π2/20] = tan–1(–nx0.4712)
n φn 1 –25.23° 3 –54.73° 5 –67° 7 –73.14° 9 –76.74° ∞ –90°
0 10
φn
ωn
14 62 18
–25.23°
–54.73°–67°
–76.74° –73.14°
–60°
–30°
–90°
2.24
An
ωn
0.0.143 0.109 0.208
0.39
18 14 10 6 2 0
Chapter 17, Solution 66. The schematic is shown below. The waveform is inputted using the attributes of VPULSE. In the Transient dialog box, we enter Print Step = 0.05, Final Time = 12, Center Frequency = 0.5, Output Vars = V(1) and click enable Fourier. After simulation, the output plot is shown below. The output file includes the following Fourier components.
FOURIER COMPONENTS OF TRANSIENT RESPONSE V(1)
DC COMPONENT = 5.099510E+00 HARMONIC FREQUENCY FOURIER NORMALIZED PHASE NORMALIZED NO (HZ) COMPONENT COMPONENT (DEG) PHASE (DEG) 1 5.000E-01 3.184E+00 1.000E+00 1.782E+00 0.000E+00 2 1.000E+00 1.593E+00 5.002E-01 3.564E+00 1.782E+00 3 1.500E+00 1.063E+00 3.338E-01 5.347E+00 3.564E+00 4 2.000E+00 7.978E-01 2.506E-01 7.129E+00 5.347E+00 5 2.500E+00 6.392E-01 2.008E-01 8.911E+00 7.129E+00 6 3.000E+00 5.336E-01 1.676E-01 1.069E+01 8.911E+00 7 3.500E+00 4.583E-01 1.440E-01 1.248E+01 1.069E+01 8 4.000E+00 4.020E-01 1.263E-01 1.426E+01 1.248E+01 9 4.500E+00 3.583E-01 1.126E-01 1.604E+01 1.426E+01 TOTAL HARMONIC DISTORTION = 7.363360E+01 PERCENT
hapter 17, Solution 67.
he Schematic is shown below. In the Transient dialog box, we type “Print step = 0.01s, Final time = 36s, Center frequency = 0.1667, Output vars = v(1),” and click Enable Fourier. After simulation, the output file includes the following Fourier components,
C T
FOURIER COMPONENTS OF TRANSIENT RESPONSE V(1) DC COMPONENT = 2.000396E+00 HARMONIC FREQUENCY FOURIER NORMALIZED PHASE NORMALIZED
PHASE (DEG)
.000E+00 -8.996E+01 0.000E+00
NO (HZ) COMPONENT COMPONENT (DEG)
1 1.667E-01 2.432E+00 12 3.334E-01 6.576E-04 2.705E-04 -8.932E+01 6.467E-01 3 5.001E-01 5.403E-01 2.222E-01 9.011E+01 1.801E+02 4 6.668E-01 3.343E-04 1.375E-04 9.134E+01 1.813E+02 5 8.335E-01 9.716E-02 3.996E-02 -8.982E+01 1.433E-01 6 1.000E+00 7.481E-06 3.076E-06 -9.000E+01 -3.581E-02 7 1.167E+00 4.968E-02 2.043E-02 -8.975E+01 2.173E-01 8 1.334E+00 1.613E-04 6.634E-05 -8.722E+01 2.748E+00 9 1.500E+00 6.002E-02 2.468E-02 9.032E+01 1.803E+02
TAL HARMON DISTORT N = 2.280 E+01 PERC T TO IC IO 065 EN
Chapter 17, Solution 68. The schematic is shown below. We set the final time = 6T=12s and the center frequency = 1/T = 0.5. When the schematic is saved and run, we obtain the Fourier series from the output file as shown below.
FOURIER COMPONENTS OF TRANSIENT RESPONSE V(1) DC COMPONENT = 1.990000E+00 HARMONIC FREQUENCY FOURIER NORMALIZED PHASE NORMALIZED NO (HZ) COMPONENT COMPONENT (DEG) PHASE (DEG) 1 5.000E-01 1.273E+00 1.000E+00 9.000E-01 0.000E+00 2 1.000E+00 6.367E-01 5.001E-01 -1.782E+02 1.791E+02 3 1.500E+00 4.246E-01 3.334E-01 2.700E+00 1.800E+00 4 2.000E+00 3.185E-01 2.502E-01 -1.764E+02 -1.773E+02 5 2.500E+00 2.549E-01 2.002E-01 4.500E+00 3.600E+00 6 3.000E+00 2.125E-01 1.669E-01 -1.746E+0 -1.755E+02 7 3.500E+00 1.823E-01 1.431E-01 6.300E+00 5.400E+00 8 4.000E+00 1.596E-01 1.253E-01 -1.728E+02 -1.737E+02 9 4.500E+00 1.419E-01 1.115E-01 8.100E+00 7.200E+00 Chapter 17, Solution 69.
The schematic is shown below. In the Transient dialog box, set Print Step = 0.05 s, Final Time = 120, Center Frequency = 0.5, Output Vars = V(1) and click enable Fourier. After simulation, we obtain V(1) as shown below. We also obtain an output file which includes the following Fourier components.
FOURIER COMPONENTS OF TRANSIENT RESPONSE V(1) DC COMPONENT = 5.048510E-01 HARMONIC FREQUENCY FOURIER NORMALIZED PHASE NORMALIZED NO (HZ) COMPONENT COMPONENT (DEG) PHASE (DEG) 1 5.000E-01 4.056E-01 1.000E+00 -9.090E+01 0.000E+00 2 1.000E+00 2.977E-04 7.341E-04 -8.707E+01 3.833E+00 3 1.500E+00 4.531E-02 1.117E-01 -9.266E+01 -1.761E+00 4 2.000E+00 2.969E-04 7.320E-04 -8.414E+01 6.757E+00 5 2.500E+00 1.648E-02 4.064E-02 -9.432E+01 -3.417E+00 6 3.000E+00 2.955E-04 7.285E-04 -8.124E+01 9.659E+00 7 3.500E+00 8.535E-03 2.104E-02 -9.581E+01 -4.911E+00 8 4.000E+00 2.935E-04 7.238E-04 -7.836E+01 1.254E+01 9 4.500E+00 5.258E-03 1.296E-02 -9.710E+01 -6.197E+00 TOTAL HARMONIC DISTORTION = 1.214285E+01 PERCENT
Chapter 17, Solution 70. The schematic is shown below. In the Transient dialog box, we set Print Step = 0.02 s, Final Step = 12 s, Center Frequency = 0.5, Output Vars = V(1) and V(2), and click enable Fourier. After simulation, we compare the output and output waveforms as shown. The output includes the following Fourier components.
FOURIER COMPONENTS OF TRANSIENT RESPONSE V(1) DC COMPONENT = 7.658051E-01 HARMONIC FREQUENCY FOURIER NORMALIZED PHASE NORMALIZED NO (HZ) COMPONENT COMPONENT (DEG) PHASE (DEG) 1 5.000E-01 1.070E+00 1.000E+00 1.004E+01 0.000E+00 2 1.000E+00 3.758E-01 3.512E-01 -3.924E+01 -4.928E+01 3 1.500E+00 2.111E-01 1.973E-01 -3.985E+01 -4.990E+01 4 2.000E+00 1.247E-01 1.166E-01 -5.870E+01 -6.874E+01 5 2.500E+00 8.538E-02 7.980E-02 -5.680E+01 -6.685E+01 6 3.000E+00 6.139E-02 5.738E-02 -6.563E+01 -7.567E+01 7 3.500E+00 4.743E-02 4.433E-02 -6.520E+01 -7.524E+01 8 4.000E+00 3.711E-02 3.469E-02 -7.222E+01 -8.226E+01 9 4.500E+00 2.997E-02 2.802E-02 -7.088E+01 -8.092E+01 TOTAL HARMONIC DISTORTION = 4.352895E+01 PERCENT
Chapter 17, Solution 71. The schematic is shown below. We set Print Step = 0.05, Final Time = 12 s, Center Frequency = 0.5, Output Vars = I(1), and click enable Fourier in the Transient dialog box. After simulation, the output waveform is as shown. The output file includes the following Fourier components.
FOURIER COMPONENTS OF TRANSIENT RESPONSE I(L_L1) DC COMPONENT = 8.374999E-02 HARMONIC FREQUENCY FOURIER NORMALIZED PHASE NORMALIZED NO (HZ) COMPONENT COMPONENT (DEG) PHASE (DEG) 1 5.000E-01 2.287E-02 1.000E+00 -6.749E+01 0.000E+00 2 1.000E+00 1.891E-04 8.268E-03 8.174E+00 7.566E+01 3 1.500E+00 2.748E-03 1.201E-01 -8.770E+01 -2.021E+01 4 2.000E+00 9.583E-05 4.190E-03 -1.844E+00 6.565E+01 5 2.500E+00 1.017E-03 4.446E-02 -9.455E+01 -2.706E+01 6 3.000E+00 6.366E-05 2.783E-03 -7.308E+00 6.018E+01 7 3.500E+00 5.937E-04 2.596E-02 -9.572E+01 -2.823E+01 8 4.000E+00 6.059E-05 2.649E-03 -2.808E+01 3.941E+01 9 4.500E+00 2.113E-04 9.240E-03 -1.214E+02 -5.387E+01 TOTAL HARMONIC DISTORTION = 1.314238E+01 PERCENT Chapter 17, Solution 72. T = 5, ωo = 2π/T = 2π/5 f(t) is an odd function. ao = 0 = an
bn = ∫ ∫ π=ω2/T
0
10
0o dt)tn4.0sin(1054dt)tnsin()t(f
T4
= 1
0
)nt4.0cos(n2
5x8π
π− = )]n4.0cos(1[
n20
π−π
f(t) = ∑∞
=
ππ−π 1n
)tn4.0sin()]n4.0cos(1[n120
Chapter 17, Solution 73.
p = ∑+R
V21
RV 2
n2DC
= 0 + 0.5[(22 + 12 + 12)/10] = 300 mW
Chapter 17, Solution 74.
(a) An = 2n
2n ba + , φ = tan–1(bn/an)
A1 = 22 86 + = 10, φ1 = tan–1(6/8) = 36.87° A2 = 22 43 + = 5, φ2 = tan–1(3/4) = 36.87°
i(t) = 4 + 10cos(100πt – 36.87°) – 5cos(200πt – 36.87°) A
(b) p = ∑+ RI5.0RI 2n
2DC
= 2[42 +0.5(102 + 52)] = 157 W Chapter 17, Solution 75. The lowpass filter is shown below. R + + vs C vo - -
∑∞
=ω
πτ+
τ=
1nos tncos
Tnsin
n1
TA2
TAv
T/n2n,VRCj1
1V
Cj1R
Cj1
V onsn
s
n
no π=ω=ω
ω+=
ω+
ω=
For n=0, (dc component), TAVsoV τ
== (1)
For the nth harmonic,
o
n122
n2o 90
Tnsin
nTA2
RCtanCR1
1V −∠πτ
•ω∠ω+
=−
When n=1, 22
2o
CRT
41
1T
nsinTA2|V
π+
•πτ
=| (2)
From (1) and (2),
4222
222
10x09.39.30CRT
41
CRT
41
110
sinTA2x50
TA
=τ
=π
+→π
+
π=
τ
mF 59.2410x4
10x09.3x1010R2
TC10CRT
413
4251022
2=
π=
π=→=
π+
−
Chapter 17, Solution 76.
vs(t) is the same as f(t) in Figure 16.1 except that the magnitude is multiplied by 10. Hence
vo(t) = ∑∞
=
ππ
+1k
)tnsin(n1205 , n = 2k – 1
T = 2, ωo = 2π/T = 2π, ωn = nωo = 2nπ
jωnL = j2nπ; Z = R||10 = 10R/(10 + R) Vo = ZVs/(Z + j2nπ) = [10R/(10R + j2nπ(10 + R))]Vs
Vo = s2222
1
V)R10(n4R100
)R10)(R5/n(tanR+π+
+π−∠ −10
Vs = [20/(nπ)]∠0°
The source current Is is
Is = )R10(n2jR10
n20)R10(
n2jR10
R10V
n2jZV ss
+π+π
+=
π++
=π+
= 2222
1
)R10(n4R100
)R10)(3/n(tann20)R10(
+π+
+π−∠π
+ −
ps = VDCIDC + ∑ φ−θ )cos(IV21
nnsnsn
For the DC case, L acts like a short-circuit.
Is = R10
)R10(5
R10R
510
+=
+
, Vs = 5 = Vo
++π+
+
π+
π
+
+π+
+π
+
π
++
=
−
−
222
122
222
12
s
)R10(16R100
)R10(5
2tancos)R10(10
)R10(4R100
)R10(5
tancos)R10(20
21
R10)R10(25p
ps = ∑∞
=
+1n
onDC
RV
21
RV
=
+
+π++
+π+
+ 222222 )R10(10R100R100
)R10(4R100R100
21
R25
We want po = (70/100)ps = 0.7ps. Due to the complexity of the terms, we consider only the DC component as an approximation. In fact the DC component has the latgest share of the power for both input and output signals.
R10
)R10(25x107
R25 +
=
100 = 70 + 7R which leads to R = 30/7 = 4.286 Ω
Chapter 17, Solution 77.
(a) For the first two AC terms, the frequency ratio is 6/4 = 1.5 so that the highest common factor is 2. Hence ωo = 2.
T = 2π/ωo = 2π/2 = π
(b) The average value is the DC component = –2
(c) Vrms = ∑∞
=
++1n
2n
2no )ba(
21a
)136810(21)2( 2222222
rms +++++−=V = 121.5
Vrms = 11.02 V
Chapter 17, Solution 78.
(a) p = ∑ ∑+=+R
VR
VR
V21
RV 2
rms,n2DC
2n
2DC
= 0 + (402/5) + (202/5) + (102/5) = 420 W
(b) 5% increase = (5/100)420 = 21
pDC = 21 W = 105R21VtoleadswhichR
V 2DC
2DC ==
VDC = 10.25 V
Chapter 17, Solution 79. From Table 17.3, it is evident that an = 0,
bn = 4A/[π(2n – 1)], A = 10.
A Fortran program to calculate bn is shown below. The result is also shown.
C FOR PROBLEM 17.79 DIMENSION B(20) A = 10 PIE = 3.142 C = 4.*A/PIE DO 10 N = 1, 10 B(N) = C/(2.*FLOAT(N) – 1.) PRINT *, N, B(N) 10 CONTINUE STOP END
n bn 1 12.731 2 4.243 3 2.546 4 1.8187 5 1.414 6 1.1573 7 0.9793 8 0.8487 9 0.7498 10 0.67
Chapter 17, Solution 80. From Problem 17.55, cn = [1 + e–jnπ]/[2π(1 – n2)]
This is calculated using the Fortran program shown below. The results are also shown.
C FOR PROBLEM 17.80 COMPLEX X, C(0:20) PIE = 3.1415927
A = 2.0*PIE DO 10 N = 0, 10 IF(N.EQ.1) GO TO 10 X = CMPLX(0, PIE*FLOAT(N)) C(N) = (1.0 + CEXP(–X))/(A*(1 – FLOAT(N*N))) PRINT *, N, C(N)
10 CONTINUE STOP END
n cn 0 0.3188 + j0 1 0 2 –0.1061 + j0 3 0 4 –0.2121x10–1 + j0 5 0 6 –0.9095x10–2 + j0 7 0 8 –0.5052x10–2 + j0 9 0 10 –0.3215x10–2 + j0
Chapter 17, Solution 81. (a)
A
3T 2TT0
f(t) = ∑∞
=
ω−ππ 1n
o2 )tncos(1n4
1A4A−
2
The total average power is pavg = Frms
2R = Frms2 since R = 1 ohm.
Pavg = Frms2 = ∫
T
0
2 dt)t(fT1 = 0.5A2
(b) From the Fourier series above |co| = 2A/2, |cn| = 4A/[π(4n2 – 1)]
n ωo |cn| 2|cn|2 % power 0 0 2A/π 4A2/(π2) 81.1% 1 2ωo 2A/(3π) 8A2/(9π2) 18.01% 2 4ωo 2A/(15π) 2A2/(225π2) 0.72% 3 6ωo 2A/(35π) 8A2/(1225π2) 0.13% 4 8ωo 2A/(63π) 8A2/(3969π2) 0.04%
(c) 81.1% (d) 0.72% Chapter 17, Solution 82.
P = ∑∞
=1n
2n
2DC
RV
21
R+
V
Assuming V is an amplitude-phase form of Fourier series. But
|An| = 2|Cn|, co = ao |An|2 = 4|Cn|2
Hence,
P = ∑∞
=1n
2n
2o
Rc
R+ 2
c
Alternatively,
P = R
V
2rms
where
∑∑∑∞
−∞=
∞
=
∞
=
=+=+=n
2n
1n
2n
2o
1n
2n
2o
2rms cc2cA
21aV
= 102 + 2(8.52 + 4.22 + 2.12 + 0.52 + 0.22)
= 100 + 2x94.57 = 289.14
P = 289.14/4 = 72.3 W
Chapter 18, Solution 1. )2t()1t()1t()2t()t('f −δ+−δ−+δ−+δ=
2jjj2j eeee)(Fj ω−ω−ωω +−−=ωω ω−ω= cos22cos2
F(ω) = ω
ω−ωj
]cos2[cos2
Chapter 18, Solution 2.
<<=
otherwise,01t0,t
)t(f
-δ(t-1)
δ(t)
f ”(t)
t –δ’(t-1)
1
-δ(t-1)
0
1
f ‘(t)
t
f"(t) = δ(t) - δ(t - 1) - δ'(t - 1) Taking the Fourier transform gives
-ω2F(ω) = 1 - e-jω - jωe-jω
F(ω) = 2
j 1e)j1(ω
−ω+ ω
or F ∫ ω−=ω1
0
tj dtet)(
But ∫ +−= c)1ax(aedxex 2
axax
( )=−ω−
ω−=ω
ω−102
j
)1tj(j
e)(F ( )[ ]1ej11 j2 −ω+
ωω−
Chapter 18, Solution 3.
2t2,21)t('f,2t2,t
21)t(f <<−=<<−=
∫− −
ω−ω −ω−
ω−==ω
2
2
222
tjtj )1tj(
)j(2edtet
21)(F
[ ])12j(e)12j(e2
1 2j2j2 −ω−−ω−
ω−= ωω−
( )[ ]2j2j2j2j2 eeee2j
21 ω−ωωω −++ω−ω
−=
( )ω+ωω−ω
2sin2j2cos4j2
12−=
F(ω) = )2cos22(sinj2 ωω−ω
ω
Chapter 18, Solution 4.
1
–2δ(t–1)
2δ(t+1)
–1
g’ 2
0 –2
t
–2δ’(t–1)
–2δ(t+1)1
–2δ(t–1)
2δ’(t+1)
–1
g”
4δ(t)
0 –2
t
4sin4cos4ej2e24ej2e2)(G)j(
)1t(2)1t(2)t(4)1t(2)1t(2g
jjjj2
+ωω−ω−=ω−−+ω+−=ωω
−δ′−−δ−δ++δ′++δ−=′′
ω−ω−ωω
)1sin(cos4)(G 2 −ωω+ωω
=ω
Chapter 18, Solution 5.
1
0
h’(t)
–1
–2δ(t)
1 t
δ(t+1)
–δ(t–1)
1
0
h”(t)
–1
–2δ’(t)
1 t
ω−ω=ω−−=ωω
δ′−−δ−+δ=′′
ω−ω j2sinj2j2ee)(H)j(
)t(2)1t()1t()t(h
jj2
H(ω) = ωω
−ω
sinj2j22
Chapter 18, Solution 6.
dtetdte)1()(F tj0
1
1
0
tj ω−
−
ω−∫ ∫+−=ω
)1(cos1tsinttcos1tsin1
tdtcosttdtcos)(F Re
2102
01
0
1
1
0
−ωω
=
ω
ω+ω
ω+ω
ω−=
ω+ω−=ω
−
−∫ ∫
Chapter 18, Solution 7. (a) f1 is similar to the function f(t) in Fig. 17.6. )1t(f)t(f1 −=
Since ω−ω
=2F ω
j)1(cos)(
=ω=ω ω )(Fe)(F j1
ω−ωω−
j)1(cose2 j
Alternatively, )
)2t()1t(2)t()t(f '
1 −δ+−δ−δ= e2e(eee21)(Fj jjj2jj
1ωωω−ω−ω− +−=+−=ωω
)2cos2(e j −ω= ω−
F1(ω) = ω
−ωω−
j)1e j (cos2
(b) f2 is similar to f(t) in Fig. 17.14.
f2(t) = 2f(t)
F2(ω) = 2
)cos1(4ω
ω−
Chapter 18, Solution 8.
(a) 21
tj2
21
tj10
tj
2
1
tj1
0
tj
)1tj(e2ej4e
j2
dte)t24(dte2)(F
−ω−ω−
−ω−
+ω−
=
−+=ω
ω−ω−ω−
ω−ω− ∫∫
ω−ω−ω− ω+
ω−
ω−
ω+
ω+
ω=ω 2j
22jj
2 e)2j1(2ej4
j2e
j22)(F
(b) g(t) = 2[ u(t+2) – u(t-2) ] - [ u(t+1) – u(t-1) ]
ωω
−ω
ω=ω
sin22sin4)(G
Chapter 18, Solution 9.
(a) y(t) = u(t+2) – u(t-2) + 2[ u(t+1) – u(t-1) ]
ωω
+ωω
=ω sin42sin2)(Y
(b) )j1(e22)1tj(e2dte)t2()( 2
j
210
1
02
tjtj ω+
ω−
ω=−ω−
ω−
−=−=ω
ω−ω−ω−∫Z
Chapter 18, Solution 10.
(a) x(t) = e2tu(t) X(ω) = 1/(2 + jω)
(b)
<
>=
−−
0t,e0t,e
et
t)t(
∫ ∫ ∫− −
ω−−ωω +==ω1
1
0
1
1
0
tjttjttj dteedteedte)t(y)(Y
10
t)j1(0
1
t)j1(
)j1(e
j1e
ω+−+
ω−=
ω+−
−
ω−
ω+
ω−ω+
ω−ω+ω
−ω+
= −
j1sinjcos
j1sinjcose
12 1
2
Y(ω) = [ ])sin(cose11
2 12 ωω−ω−
ω+−
Chapter 18, Solution 11. f(t) = sin π t [u(t) - u(t - 2)]
( )∫ ∫ ω−π−πω− −=π=ω2
0
2
0
tjtjtjtj dteeej2
1dtetsin)(F
+∫ π+ω−π+ω−+2
0
t)(jt)(j dt)ee(j2
1=
π+ω−+
π−ω−
π+ω−π−ω− 2
0
t)(j20
t)(j
)(jee
)(j1
j21
=
ω+π
−+
ω−π− ω−ω− 2j2j e1e1
21
=
( )ω−π+πω−π
2j22 e22)(2
1=
F(ω) = ( )1e 2j22 −
π−ωπ ω−
Chapter 18, Solution 12. (a) F = dtedtee)(
0
2
0
t)j1(tjt∫ ∫∞ ω−ω− =ω
=ω−
= ω− 20
t)j1(ej1
1 ω−−ω−
j11e 2j2
(b) ∫ ∫−
ω−ω− −+=ω0
1
1
0
tjtj dte)1(dte)(H
( ) ( ) )cos22(j11e
j1e1
j1 jj ω+−
ω=−
ω+−
ωω−ω−=
=ωω−
=j
2/sin4 2 2
2/2/sinj
ωω
ω
Chapter 18, Solution 13. (a) We know that )]a()a([]at[cos +ωδ+−ωδπ=F .
Using the time shifting property,
)a(e)a(e)]a()a([e)]a3/t(a[cos 3/j3/ja3/j +ωδπ+−ωδπ=+ωδ+−ωδπ=π− ππ−ωπ−F
(b) sin tsinsintcoscostsin)1t( π−=ππ+ππ=+π g(t) = -u(t+1) sin (t+1)
Let x(t) = u(t)sin t, then 22 11
1)j(1)(X
ω−=
+ω=ω
Using the time shifting property,
1ee
11)(G 2
jj
2 −ω=
ω−−=ω
ωω
(c ) Let y(t) = 1 + Asin at, then Y )]a()a([Aj)(2)( −ωδ−+ωδπ+ωπδ=ω
h(t) = y(t) cos bt
Using the modulation property,
)]b(Y)b(Y[21)(H −ω++ω=ω
[ ] [ ])ba()ba()ba()ba(2Aj)b()b()(H −−ωδ−−+ωδ++−ωδ−++ωδπ
+−ωδ++ωδπ=ω
(d) )14j(ej
e1)1tj(ej
edte)t1()(2
4j4j
2402
tj4
0
tjtj +ω
ω−
ω−
ω=−ω−
ω−−
ω−=−=ω
ω−ω−ω−ω−ω−∫I
Chapter 18, Solution 14.
(a) )t3cos()0(t3sin)1(t3cossint3sincost3cos)t3cos( −=−−=π−π=π+ (f )t(ut3cose)t t−−=
F(ω) = ( )( ) 9j1
j12 +ω+ω+−
(b)
[ ])1t(u)1t(utcos)t('g −−−ππ=
g(t)
t 1 -1
-1
1
-π
-1 1
g’(t)
t
π
)1t()1t()t(g)t("g 2 −πδ++πδ−π−= ω−ω π+π−ωπ−=ωω− jj22 ee)(G)(G
( ) ωπ−=−π−=ωω−π ω−ω sinj2)ee()(G jj22
G(ω) = 22
sinj2π−ωωπ
Alternatively, we compare this with Prob. 17.7 f(t) = g(t - 1) F(ω) = G(ω)e-jω
( ) ( )ωω−ω −π−ω
π=ω=ω jj
22j eee)(FG
22
sin2jπ−ωωπ−
=
G(ω) = 22
sinj2ω−πωπ
(c) tcos)0(tsin)1(tcossintsincostcos)1t(cos π−=π+−π=ππ+ππ=−π Let ex = )t(he)1t(u)1t(cos)t( 2)1t(2 −=−−π−−
and )t(u)tcos(e)t(y t2 π= −
22)j2(j2)(Y
π+ω+ω+
=ω
)1t(x)t(y −= ω−ω=ω je)(X)(Y
( )( ) 22
j
j2ej2)(Xπ+ω+
ω+=ω
ω
)(He)(X 2 ω−=ω )(Xe)(H 2 ω−=ω −
= ( )( ) 22
2j
j2ej2π+ω+
ω+− −ω
(d) Let x )t(y)t(u)t4sin(e)t( t2 −=−−= −
)t(x)t(p −=where )t(ut4sine)t(y t2=
( ) 22 4j2
j2)(Y+ω+ω+
=ω
( ) 16j2
j2)(Y)(X 2 +ω−ω−
=ω−=ω
=ω−=ω )(X)(p( ) 162j
2j2 +−ω
−ω
(e) 2j2j ej1)(23e
j8)(Q ω−ω−
ω
+ωπδ−+ω
=ω
Q(ω) = 2j2j e)(23ej6 ω−ω ωπδ−+ω
Chapter 18, Solution 15. (a) F =−=ω ω−ω 3j3j ee)( ω3sinj2
(b) Let g ω−=ω−δ= je2)(G),1t(2)t(
=ω)(F F
∫ ∞−
tdt)t(g
)()0(Fj
)(Gωδπ+
ωω
=
)()1(2je2 j
ωδ−πδ+ω
=ω−
= ω
ω−
je j2
(c) F [ ] 121)t2( ⋅=δ
=ω−⋅=ω j211
31)(F
2j
31 ω−
Chapter 18, Solution 16.
(a) Using duality properly
2
2tω−
→
ωπ→− 2t
22
or ωπ−→ 4t42
F(ω) = F =
2t4
ωπ4−
(b) tae− 22aa2ω+
22 taa2+
ω−π ae2
22 ta8+
ω−π 2e4
G(ω) = F =
+ 2t48
ω−π 2e4
Chapter 18, Solution 17.
(a) Since H(ω) = F ( ) ( ) ([ ]000 FF21)t(ftcos ω−ω+ω+ω=ω )
where F(ω) = F ( )[ ] ( ) 2,j1tu 0 =ωω
+ωπδ=
( ) ( ) ( ) ( ) ( )
−ω
+−ωπδ++ω
++ωπδ=ω2j
122j
1221H
( ) ( )[ ] ( )( )
−ω+ω−ω++ω
−−ωδ++ωδπ
=2222
2j22
2
H(ω) = ( ) ( )[ ]4
j222 2 −ω
ω−−ωδ++ωδ
π
(b) G(ω) = F [ ] ( ) ([ ]000 FF2j)t(ftsin ω−ω−ω+ω=ω )
where F(ω) = F ( )[ ] ( )ω
+ωπδ=j1tu
( ) ( ) ( ) ( ) ( )
−ω
−−ωπδ−+ω
++ωπδ=ω10j
11010j
1102jG
( ) ( )[ ]
+ω−
−ω+−ωδ−+ωδ
π=
10j
10j
2j1010
2j
= ( ) ( )[ ]100
1010102 2
j−ω
−−ωδ−+ωδπ
Chapter 18, Solution 18.
Let f ( ) ( )tuet t−= ( )ω+
=ωjj
1F
( ) tcostf ( ) ([ ]1F1F21
+ω+−ω )
Hence ( ) ( ) ( )
+ω+
+−ω+
=ω1j1
11j1
121Y
( )[ ] ( )[ ]
+ω+−ω+−ω+++ω+
=1j11j1jj1jj1
21
1jjjj1
j12 +ω−−ω++ω+
ω+=
= 2j2
j12 +ω−ωω+
Chapter 18, Solution 19.
( ) ( )∫ ∫∞
∞−
ω−π−πω +==ω dteee21dte)t(fF tj1
0
t2jt2jtj
( ) ( ) ( )[ ]∫ π−ω−π+ω− +=ω1
0
t2jt2j dtee21F
( )( )
( )( )
1
0
t2jt2j e2j
1e2j
121
π−ω−
+π+ω−
= π−ω−π+ω−
( )
( )( )
( )
π−ω−
+π+ω−
−=π−ω−π+ω−
2j1e
2j1e
21 2j2j
But π−π ==π+π= 2j2j e12sinj2cose
( )
π−ω+
π+ω
−−=ω
ω−
21
21
j1e
21F
j
= ( )1e4
j j22 −
π−ωω ω−
Chapter 18, Solution 20.
(a) F (cn) = cnδ(ω) F ( ) ( )on
tjnn ncec o ω−ωδω =
F =
∑∞
−∞=
ω
n
tjnn
oec ( )∑∞
−∞=
ω−ωδn
on nc
(b) π= 2T 1T2
o =π
=ω
( )∫ ∫
+⋅
π==
π −ω−T
0 0
jnttjnn 0dte1
21dtetf
T1c o
( )1en2
jejn1
21 jn
0jnt −
π=
−
π= π−π
But e njn )1(ncosnsinjncos −=π=π+π=π−
( )[ ]
=−−
π= =
≠=π−
evenn,0
0n,oddn,n
jn
n 11n2jc
for n = 0
∫π
=π
=0n 2
1dt121c
Hence
∑∞
=≠−∞= π
−=
oddn0n
n
jntenj
21)t(f
F(ω) = ( )∑∞
=≠−∞=
−ωδπ
−δω
oddn0n
nn
nj
21
Chapter 18, Solution 21.
Using Parseval’s theorem,
∫∫∞
∞−
∞
∞−ωω
π= d|)(F|
21dt)t(f 22
If f(t) = u(t+a) – u(t+a), then
∫∫ ∫∞
∞−
∞
∞− −ω
ωω
π=== d
aasina4
21a2dt)1(dt)t(f
22a
a22
or
aa4a4d
aasin
2
2 π=
π=ω
ωω
∫∞
∞− as required.
Chapter 18, Solution 22.
F [ ] ( )∫∞
∞−
ω−ω−ω −
=ω dtej2ee)t(ftsin)t(f tj
tjtj
o
oo
( ) ( )
−= ∫ ∫
∞
∞−
∞
∞−
ω+ω−ω−ω− dtedte)t(fj2
1 tjtj oo
= ( ) ([ ]oo FFj2
1ω+ω−ω−ω )
Chapter 18, Solution 23.
(a) f(3t) leads to ( )( ) ( )( )ω+ω+=
ω+ω+⋅
j15j630
3/j53/j210
31
F [ ]( ) =− t3f ( )( )ω−ω− j15j630
(b) f(2t) ( )( ) ( )( )ω+ω+=
ω+ω+⋅
j10j420
2/j152/j210
21
f(2t-1) = f [2(t-1/2)] ( )( )ω+ω+
ω−
j10j4e20 2/j
(c) f(t) cos 2t ( ) ( )2F212F
21
+ω++ω
= [ ]( ) ( )[ ] ( ) ( )[ ][ ]2j52j25
2j52j2 −ω+−ω++
+ω++ω+5
(d) F [ ]( ) ( ) ( )( )ω+ω+ω
=ωω=j5j2
10jFjt'f
(e) ( )∫ f ∞−
tdtt ( )
( ) ( ) ( )ωδπ+ωω 0F
jF
( )( ) ( )5x2
10xj5j2j
ωπδ+ω+ω+ω
10=
= ( )( ) ( )ωπδ+ω+ω+ω j5j2j
10
Chapter 18, Solution 24. (a) ( ) ( )ω=ω FX + F [3]
= ( ) ( )1e j
ω+ωπδ ω−j6 −
(b) ( ) ( )2tfty −=
( ) ( ) =ω=ω ω− FeY 2j ( )1eje j2j
−ω
ω−ω−
(c) If h(t) = f '(t)
( ) ( ) ( ) =−ω
ω=ωω=ω ω− 1ejjFjH j ω−− je1
(d) ( )
ω+
ω=ω
=
53F
53x10
23F
23x4)(G,t
35f
3ft +
10t24g
( ) ( )1e
53
j61e
23
j6 5/3j2/3j −ω
+−ω
⋅= ω−ω−
= ( ) ( )1e10j1e4j 5/3j2/3j −ω
+−ω
ω−ω−
Chapter 18, Solution 25.
(a) ( ) ( ) ω=+
+=+
= js,2s
Bs2ss
s A10F
52
10B,52
10A −=−
===
( )2j
5j5F
+ω−
ω=ω
f(t) = ( ) ( )tue5t2
t2−−sgn5
(b) ( ) ( )( ) 2jB
1jA
2j1j4jF
+ω+
+ω=
+ω+ω−ω
=ω
( ) ( )( ) ω=+
++
=++
−= js,
2sB
1sA
2s1s4ssF
A = 5, B = 6
( )ω+
+ω+
−=ω
j26
j15F
f(t) = ( ) ( )tue6e5 t2t −− +−
Chapter 18, Solution 26.
(a) )t(ue)t( )2t( −−=f (b) )t(ute)t( t4−=h
(c) If ωω
=ω→−−+=sin2)(X)1t(u)1t(u)t(x
By using duality property,
ttsin2)t(g)1(u2)1(u2)(G
π=→−ω−+ω=ω
Chapter 18, Solution 27.
(a) Let ( ) ( ) ω=+
+=+
= js,10s
Bs10ss
s A100F
1010
100B,1010100A −=
−===
( )10j
10j10F
+ω−
ω=ω
f(t) = ( ) ( )tue10tsgn t10−−5
(b) ( ) ( )( ) ω=+
+−
=+−
= js,3s
Bs2
As3s2
s10sG
6530B,4
520A −=
−===
( )3j
62j
4G+ω
−+ω−=
=ω
g(t) = ( ) ( )tue6tue t3t2 −−−4
(c) ( )( ) ( ) 90020j
60130040jj
60H 22 ++ω=
+ω+ω=ω
h(t) = ( )tu)t30sin(e t20−2
( ) ( )( )( )∫
∞
∞−
ω
=⋅π=+ωω+ωωδ
π=
21
21
1jj2de
21ty
tj
π41
Chapter 18, Solution 28.
(a) ∫∫∞
∞−
ω∞
∞−
ω ωω+ω+
ωπδπ
=ωωπ
= d)j2)(j5(
e)(21de)(F
21)t(f
tjtj
===201
)2)(5(1
21 0.05
(b) ∫∞
∞−
−ω
+−−π=ω
+ωω+ωδ
π=
)12j)(2j(e
210de
)1j(j)2(10
21)t(f
t2jtj
=−π
=−
2j1e
25j t2j
π
+− −
2e)j2( t2j
(c) ∫∞
∞−
ω
++π=ω
ω+ω+−ωδ
π=
)j3)(j2(e
220d
)53)(j2(e)1(20
21)t(f
jttj
=+π
=)j55(2
e20 jt
π− jte)j1(
(d) Let )(F)(F)j5(j
5)j5()(5)( 21 ω+ω=
ω+ω+
ω+F ωπδ
=ω
∫∞
∞−
ω =⋅ππ
=ωω+ωπδ
π= 5.0
51
25de
j5)(5
21)t(f tj
1
1B,1A,5s
BsA
)s5(s5)s(2 −==
++=
+=F
5j
1j1)(F2 +ω−
ω=ω
t5t52 e)t(u
21e)tsgn(
21)t( −+−=−= −f
=+= )t(f)t(f)t(f 21 u t5e)t( −−
Chapter 18, Solution 29.
(a) f(t) = F -1[ +ωδ )]( F -1 [ ])3(4)3(4 −ωδ++ωδ
π
+π
=t3cos4
21 = ( )t3cos81
21
+π
(b) If )2t(u)2t(u)t(h −−+=
ωω
=ω2sin2)(H
)(H4)(G ω=ω t
t2sin821)t( ⋅π
=g
g(t) = t
t2sin4π
(c) Since
cos(at) ↔ )a()a( −ωπδ++ωπδ Using the reversal property,
)2t()2t(2cos2 −πδ++πδ↔ωπ or F -1 [ ] =ω2cos6 )2t(3)2t(3 −δ++δ
Chapter 18, Solution 30.
(a) ω+
=ωω
=ω→=ja
1)(X,j2)(Y)tsgn()t(y
)]t(u)t(u[a)t(2)t(hja22
j)ja(2
)(X)(Y)(H −−+δ=→
ω+=
ωω+
=ωω
=ω
(b) ω+
=ωω+
=ωj2
1)(Y,j1
1)(X
)t(ue)t()t(hj2
11j2j1)(H t2−−δ=→
ω+−=
ω+ω+
=ω
(c ) In this case, by definition, )t(u btsine)t(y)t(h at−==
Chapter 18, Solution 31.
(a) ω+
=ωω+
=ωja
1)(H,)ja(
1)(Y 2
)t(ue)t(xja
1)(H)(Y)(X at−=→
ω+=
ωω
=ω
(b) By definition, )1t(u)1t(u )t(y)t(x −−+==
(c ) ω
=ωω+
=ωj2)(H,
)ja(1)(Y
)t(ue2a)t(
21)t(x
)ja(2a
21
)ja(2j
)(H)(Y)(X at−−δ=→
ω+−=
ω+ω
=ωω
=ω
Chapter 18, Solution 32.
(a) Since 1j
j
+ω
ω−e ( ) )1t(ue 1t −−−
and F f(-t) ( )ω−
( )1j
e j
1 +ω−=ω
ω
F ( ) ( ) ( )1tuetf 1t1 −−= −−−
f1(t) = e ( ) ( )1tu1t −−+
(b) From Section 17.3,
1t
22 +
ω−πe2
If ( ) ,e2F2
ω−=ω then
f2(t) = ( )1t2
2 +π
(b) By partial fractions
( )( ) ( ) ( ) ( ) ( ) 1j
41
1j41
1j41
1j41
1j1j1F 22223 −ω
−−ω
++ω
++ω
=−ω+ω
=ω
Hence ( ) ( ) ( )tueteete41t tttt
3 −++= −−f
= ( ) ( ) ( ) (tue1t41tue1t
4tt −++ − )1
(d) ( ) ( ) ( )∫ ∫∞
∞−
∞
∞−
ωω
ω+ωδ
π=ωω
π=
2j1e
21deF
21tf
tjtj
14 = π21
Chapter 18, Solution 33. (a) Let ( ) ( ) ( )[ ]1tu1tutsin2tx −−+π=
From Problem 17.9(b),
( ) 22
sinj4Xω−πωπ
=ω
Applying duality property,
( ) ( ) ( )22 ttsinj2tX
21tf
−π−
=−π
=
f(t) = 22ttsinj2
π−
(b) ( ) ( ) ( )ω−ωω
−ω−ωω
=ω sinjcosj2sinj2cosjF
( )ω
−ω
=−ω
ωω−ω−ω
je
jeeej 2jj
j2j=
( ) ( ) ( )2tsgn211tsgn
21tf −−−=
But sgn( 1)t(u2)t −=
( ) ( ) ( )212tu
211tutf +−−−−=
= ( )u ( 2tu1t −−− ) Chapter 18, Solution 34. First, we find G(ω) for g(t) shown below.
( ) ( ) ( )[ ] ( ) ( )[ ]1tu1tu102tu2tu10tg −−++−−+= ( ) ( ) ( )[ ] ( ) ( )[ ]1t1t102t2t10t'g −δ−+δ+−δ−+δ= The Fourier transform of each term gives
20
t
g(t)
10
0
–2 –1 1 2
10δ(t+2) 10δ(t+1)
–1 –2 1 2t
g ‘(t)
0
–10δ(t-2) –10δ(t-1)
( ) ( ) ( )ω−ωω−ω −+−=ωω jj2j2j ee10ee10Gj ω+ω= sinj202sinj20
( ) =ω
ω+
ωω
=ωsin202sin20G 40 sinc(2ω) + 20 sinc(ω)
Note that G(ω) = G(-ω).
( ) ( )ω−π=ω G2F
( ) ( )tG21tfπ
=
= (20/π)sinc(2t) + (10/π)sinc(t) Chapter 18, Solution 35. (a) x(t) = f[3(t-1/3)]. Using the scaling and time shifting properties,’’
)j6(ee
3/j21
31)(X
3/j3/j
ω+=
ω+=ω
ω−ω−
(b) Using the modulation property,
−ω
++ω
=
−ω+
++ω+
=−ω++ω=ω3j
17j
121
)5(j21
)5(j21
21)]5(F)5(F[
21)(Y
(c ) ω+
ω=ωω=ω
j2j)(Fj)(Z
(d) 2)j2(1)(F)(F)(Hω+
=ωω=ω
(e) 22 )j2(1
)j2()j0(j)(F
ddj)(I
ω+=
ω+
−=ω
ω=ω
Chapter 18, Solution 36.
( ) ( )( )ωω
=ωi
o
VV
H
( ) ( ) ( ) ( )ω+ω
=ωω=ωj2
V10VHV iio
(a) vi = 4δ(t) Vi(ω) = 4
( )ω+
=ωj2
40Vo
( )tue40)t(v t2
o−=
vo(2) = 40e–4 = 0.7326 V
(b) ( )tue6v ti
−= ( )ω+
=ωj1
6Vi
( ) ( )( )ω+ω+=ω
j1j260Vo
( ) ( )( ) ω=+
++
=++
= js,2s
B1s
A1s2s
60sVo
601
60B,60160A −=
−===
( )ω+
−ω+
=ωj2
60j1
60Vo
[ ] )t(uee60)t(v t2t
o−− −=
[ ] ( )01831.013533.060ee60)2(v 42o −=−= −−
= 7.021 V (c) vi(t) = 3 cos 2t
Vi(ω) = π[δ(ω + 2) + δ(ω- 2)]
( ) ( )[ ]ω+
−ωδ++ωδπ=
j22210Vo
( )∫∞
∞−
ω ωωπ
= deV21)t(v tj
oo
( ) ( )∫∫∞
∞−
ω∞
∞−
ω ωω+
−ωδ+ω
ω++ωδ
= dj2
e25dej225
tjtj
( ) ([ ]°−°−−+−
+=+
+−
= 45t2j45t2jt2jt2j
ee22
52j2
e52j2
e5 )
( )°−= 45t2cos2
5
( ) ( ) ( )°−°=°−= 4518.229cos2
5454cos2
52vo
vo(2) = –3.526 V Chapter 18, Solution 37.
ω+ω
=ωj2
2jj2
By current division,
( ) ( )( ) ω++ω
ω=
ω+ω
+
ω+ω
=ωω
=ω4j82j
2j
j22j4
j22j
II
Hs
o
H(ω) = ω+
ω3j4
j
Chapter 18, Solution 38.
( ) ( )ω
+ωπδ=ωj1Vi
( ) ( ) ( )
ω
+ωπδω+
=ωω+
=ωj1
j55V
2j1010V io
Let ( ) ( ) ( ) ( )( )ω+ω
+ω+ωπδ
=ω+ω=ωj5j
5j5
521o VVV
( ) ( ) 5sB
sA
5ss5V2 +
+=+
=ω A = 1, B = -1, s = jω
( )ω+
−ω
=ωj5
1j1V2 ( ) t5
2 e)tsgn(21t −−=v
( )ω+ωπδ
=j5
5V1 ( ) ( )∫∞
∞−
ω ωω+ωπδ
π= de
j55
21tv tj
1
v1(t) 5.051
25
=⋅ππ
=
( ) ( ) ( ) ( ) t5
210 etsgn5.05.0tvtvtv −−+=+= But ( ) ( )tu21tsgn +−=
( ) ( ) ( ) =−+−+= − tuetu5.05.0t t5
ov ( ) ( )tuetu t5−− Chapter 18, Solution 39.
ω−ω−∞
∞− ω−
ω+
ω=−=ω ∫ j
22tj
s e11j1dte)t1()(V
ω−
ω+
ωω+=
ω+
ω=ω ω−
−j
226
3
33s e11
j1
j1010
10xj10
)(V)(I
Chapter 18, Solution 40. )2t()1t(2)t()t(v −δ+−δ−δ= 2jj2 ee21)(V ωω− +−=ωω−
( ) 2
2jj ee21Vω−+−
=ωω−ω−
Now ( )ωω+
=ω
+=ωj
2j1j12Z
( )( ) ω+
ω⋅
ω−−
=ωω
=ωω
2j1j1ee2
ZVI 2
2jj
( ) ( )ω−ω− −+ω+ω
j2j ee5.05.0j5.0j
=1
But 5.0s
BsA
)5.0s(s1
++=
+ A = 2, B = -2
( ) ( ) ( )ω−ω−ω−ω −+ω+
−−+ω
=ω j2jj2j ee5.05.0j5.0
2ee5.05.0j2I
i(t) = )1t(ue2)2t(ue)t(ue)1tsgn()2tsgn(21)tsgn(
21 )1t(5.0)2t(5.0t5.0 −−−−−−−−+ −−−−−
Chapter 18, Solution 41. 2
ω+ j21
+ − 1/s 0.5s
+
V
−
2
( )ω+
ω+ω−=
ω+ω
+ω=+ω−ω
=−ω
+ω+ω+
−
j29j4
j2j4jV42j
02jV2Vj
2j2
1V
22
V(ω) = )j24)(j2(
)2j5.4(j22 ω+ω−ω+
ω+ω
Chapter 18, Solution 42.
By current division, ( )ω⋅ω+
= Ij2
2Io
(a) For i(t) = 5 sgn (t),
( )
)j2(j20
j10
j22I
j10I
o ω+ω=
ω⋅
ω+=
ω=ω
Let 10B,10A,2s
BsA
)2s(s20
o −==+
+=+
I =
( )ω+
−ω
=ωj2
10j10
oI
io(t) = 5 A)t(ue10)tsgn( t2−− (b)
)1t(4)t(4)t('i −δ−δ=
4 i(t)
t 1
–4δ(t–1)
4δ(t) i’(t)
t 1
( ) ω−−=ωω je44Ij
( ) ( )ω−
=ωω−
je14I
j
( ) ( )ω−
ω−
−
ω+
−ω
=ω+ω
−= j
j
o e1j2
1j14
)j2(je18I
ω+
+ω
−ω+
−ω
=ω−ω−
j2e4
je4
j24
j4 jj
io(t) = ( )A1tue4)t(ue4)1tsgn(2)tsgn(2 )1t(2t2 −+−−− −−− Chapter 18, Solution 43.
ω+=→=
ω=
ω=
ω→ −
− j51Ie5i,
j50
10x20j1
Cj1 mF 20 s
ts3
ω=++
=ω
•
ω+
= js,)5s)(25.1s(
50j50I
j5040
40V so
+ω
−+ω
=+
++
=5j
125.1j
1340
5sB
25.1sAVo
)t(u)ee(340)t(v t5t25.1
o−− −=
Chapter 18, Solution 44. 1H jω
We transform the voltage source to a current source as shown in Fig. (a) and then combine the two parallel 2Ω resistors, as shown in Fig. (b).
(a)
+
Vo
−
1 Vs/2
2
Io
+
Vo
−jω2
Vs/2
(b)
Io
jω
,122 Ω= 2
Vj1
1 so ⋅
ω+=I
)j1(2Vj
IjV soo ω+
ω=ω=
( ) )2t(10t10)t(vs −δ−δ= ( ) ω−−=ωω 2j
s e1010Vj
( ) =ωsV ( )ω− ω−
je110 2j
Hence ( ) ω−ω−
ω+−
ω+=
ω+−
= 2j2j
o ej1
5j1
5j1
e15V
)2t(ue5)t(ue5)t(v )2t(t
o −−= −−−
=−−= − 01e5)1(v 1o 1.839 V
Chapter 18, Solution 45.
)t(ute2)t(v)1j(
2)2(
j1j2
j1
V to2o
−=→+ω
=
ω+ω+
ω=
Chapter 18, Solution 46.
F41
ω−
=ω
4j
41j
1
2H jω2 3 )t(3δ
)t(ue t−
ω+ j11
The circuit in the frequency domain is shown below: 2 Ω Vo
Io(ω)
+ − 3
+ − 1/(1+jω)
–j4/ω
j2ω
At node Vo, KCL gives
ω
=
ω−−
+−
ω+2jV
4jV3
2
Vj1
1oo
o
ω
−=ω−ω+−ω+
ooo
V2jVj3jV2j1
2
ω−ω+
ω+ω+= 2jj2
3jj1
2
Vo
( )
ω−ω+ω
ω+ω−ω+
=ω
=ω2jj22j
j133j2
2jV
I
2
oo
Io(ω) = )28(j64
3j32
22
ω−ω+ω−ω−ω+2
Chapter 18, Solution 47.
Transferring the current source to a voltage source gives the circuit below:
1/(jω)2 Ω
+ −Vo
+ − 1 Ω
8 V jω/2
Let ω+ω+
=ω
+
ω
+=ω
+=j23j4
2j1
2j
22j12inZ
By voltage division,
( ) ( )ω+ω+ω
+=
ω+=⋅
+ω
ω=ω
j23j4j1
8Zj1
88Z
j1
j1
Vin
in
o
= ( )234jj2
j2ω−ω+ω+
8 ω+
= ( )235j2
j2ω−ω+
8 ω+
Chapter 18, Solution 48.
0.2F ω
−=ω
5jCj
1
As an integrator,
4.010x20x10x20RC 63 == −
∫−=t
o io dtvRC1v
( )
ωδπ+
ω−= )0(V
jV
RC1V i
io
( ) ( )
ωπδ+
ω+ω−=
j2j2
4.01
( ) ( )
ωδπ+
ω+ω−==
j2j2125.0mA
20VI o
o
( )ωπδ−ω+
+ω
−= 125.0j2
125.0j125.0
( ) ( )∫ ω− ωπδπ
−+−= dte2125.0tue125.0)tsgn(125.0)t(i tjt2
o
2125.0)t(ue125.0)t(u25.0125.0 t2 −++= −
io(t) = mA)t(ue125.0)t(u25.0625.0 t2−+−
Chapter 18, Solution 49.
Consider the circuit shown below:
2 Ω
jω
i2i1
+ −
j2ω
jω
1 Ω
+
vo
−
VS
( ) ([ ]21Vs −ωδ+ )+ωδπ= For mesh 1, ( ) 0IjI2I2j2V 221s =ω−−ω++− ( ) ( ) 21s Ij2Ij12V ω+−ω+= (1) For mesh 2, ( ) 112 IjI2Ij30 ω−−ω+=
( )( )ω+
ω+=
2I3
I 21 (2)
Substituting (2) into (1) gives
( )( ) ( ) 22
s Ij2j2
Ij3j122V ω+−ω+
ω+ω+=
( ) ( ) ( )[ ] 2
22s I4j44j322V ω−ω+−ω−ω+=ω+
( )22 4j2I ω−ω+=
( )
ω=++
+= js,
2s4sV2s
I 2s
2
== 2o IV ( ) ( ) ( )[ ]( ) 24jj
112j2 +ω+ω
−ωδ++ωδπ+ω
( )∫∞
∞−
ω ωωπ
= dev21)t(v tj
oo
( ) ( )
( )
( ) ( )
( )∫∞
∞−
ωω
+ω+ω
ω−ωδ+ω+
+ω+ω
ω+ωδ+ω=
24jj
d1e2j21
24jj
d1e2j21
2
tj
2
tj
( ) ( )
24j1
e2j21
24j1
e2j21 jtjt
++−
++
+−−
+−=
( )( ) ( )( )17
e4j1j221
e17
4j1j221
)t(vjt
jto
−−+
+−=
( ) ( ) jtjt e7j6341e7j6
341
−++=
( ) ( )°−°−− += 64.13tj64.13tj e271.0e271.0
vo(t) = 0 V)64.13tcos(542. °− Chapter 18, Solution 50.
Consider the circuit shown below:
jωi2i1
+ −
+
vo
− jω
1 Ω
1 Ωj0.5ω
VS
For loop 1, ( ) 0I5.0jIj12 21 =ω+ω++− (1) For loop 2, ( ) 0I5.0jIj1 12 =ω+ω+ (2) From (2),
( ) ( )
ωω+
−=ω−
ω+=
jIj12
5.0jIj1I 22
1
Substituting this into (1),
( )
22 I
2j
jIj12
2 ω+
ωω+−
=
22 I
234j4j2
ω−ω+−=ω
22 5.14j4j2I
ω−ω+ω
=
( )22o j5.14j4
j2IVω+ω+
ω−==
( )2
o
j3
8j38
j34
Vω+
ω+
ω=
2222
38j
34
316
38j
34
j344
+
ω+
+
+
ω+
ω+−
=
=)t(Vo V)t(ut38sine657.5)t(ut
38cose4 3/t43/t4
+
− −−
Chapter 18, Solution 51.
ω+=
ω+
ω=ω
=j1
1
j11
j1
j1//1Z
ω=++
=
ω+ω+=
ω+∗
ω++
ω+=
+=
js,)5.1s)(1s(
1
j12
j231
j12
j112
j11
V2Z
ZV oo
)t(u)ee(2)t(v5.1s
21s
25.1s
B1s
AV t5.1too
−− −=→+
−+
=+
++
=
∞∞ −−−−−−
∞
∞−
∞−−
∫
∫ ∫
−+
−=+−=
−==
00
t3t5.2t2t3t5.2t2
0
2t5.1t2
3e
5.2e2
2e4dt)ee2e(4
dt)ee(4dt)t(fW
J 1332.0)31
5.22
21(4W =+−=
Chapter 18, Solution 52.
J = ∫∫∞∞
ωωπ
=0
2
0
2 d)(F1dt)t(f2
= 23
1)3/(tan31d
911
0
1
0 22
ππ
=ωπ
=ωω+π
∞−∞
∫ = (1/6)
Chapter 18, Solution 53.
J = ∫∫∞
∞−
∞π=ωω dt)t(f2d)(F 2
0
2
f(t) = 0t,e0t,e
2
t2
>
<− t
J =
−+π=
+π
∞−
∞−
∞ −
∞− ∫∫0
t40t4
0
t40 t4
4e
4e2dtedte2 = 2π[(1/4) + (1/4)] = π
Chapter 18, Solution 54.
W1Ω =
∞−∞ −∞
∞−−== ∫∫ 0
t2
0
t22 e8dte16dt)t(f = 8 J
Chapter 18, Solution 55.
f(t) = 5e2e–tu(t) F(ω) = 5e2/(1 + jω), |F(ω)|2 = 25e4/(1 + ω2)
W1Ω = ∞
−∞∞ω
π=ω
ω+π=ωω
π ∫∫0
14
0 2
4
0
2 )(tane25d1
1e25d)(F1
= 12.5e4 = 682.5 J
or W1Ω = 12.5e== ∫∫∞ −∞
∞− 0
t242 dtee25dt)t(f 4 = 682.5 J
Chapter 18, Solution 56.
W1Ω = ∫ ∫∞ −∞
∞−=
0
2t22 dt)t2(sinedt)t(f
But, sin2(A) = 0.5(1 – cos(2A))
W1Ω = ∞−∞−∞ − +−
+−
−=−
0
t2
0
t2
0
t2 )]t4sin(4)t4cos(2[164
e2
e21dt)]t4cos(1[5.0e ∫
= (1/4) + (1/20)(–2) = 0.15 J
Chapter 18, Solution 57.
W1Ω = 0t20 t22 e2dte4dt)t(i∞−∞−
∞
∞−== ∫∫ = 2 J
or I(ω) = 2/(1 – jω), |I(ω)|2 = 4/(1 + ω2)
W1Ω = 2
4)(tan4d)1(
124d)(I
21
0
12
2 ππ
=ωπ
=ωω+π
=ωωπ
∞−∞
∞−
∞
∞− ∫∫ = 2 J
In the frequency range, –5 < ω < 5,
W = )373.1(4)5(tan4tan4 15
0
1
π=
π=ω
π−− = 1.7487
W/ W1Ω = 1.7487/2 = 0.8743 or 87.43% Chapter 18, Solution 58.
ωm = 200π = 2πfm which leads to fm = 100 Hz (a) ωc = πx104 = 2πfc which leads to fc = 104/2 = 5 kHz (b) lsb = fc – fm = 5,000 – 100 = 4,900 Hz (c) usb = fc + fm = 5,000 + 100 = 5,100 Hz
Chapter 18, Solution 59.
ω+−
ω+=ω+
−ω+=
ωω
=ωj4
3j2
52
j46
j210
)(V)(V)(H
i
o
ω=++
−++
=
ω+
ω+
−ω+
=ωω=ω
js,)4s)(1s(
12)2s)(1s(
20j1
4j4
3j2
5)(V)(H)(V io
Using partial fraction,
ω++
ω+−
ω+=
++
++
++
+=ω
j44
j220
j116
4sD
1sC
2sB
1sA)(Vo
Thus, ( ) V)t(ue4e20e16)t(v t4t2t
o−−− +−=
Chapter 18, Solution 60. 2
1/jω jω +
V
−
Is
ω+ω−
ω=
ω++ω
ωω=
2j1
Ij
j2j1
j1
IjV 2s
s
°−∠=+−
°∠=
π+π−
°∠π= 92.712418.1
566.12j48.389027.50
4j418902V
.componentDCnoisthere,inductortheacrossappearsvoltagetheSince
21
°−∠=+−
°∠=
π+π−
°∠π= 9.803954.0
13.25j91.1569083.62
8j1615904V 22
mV)1.129t4cos(3954.0)92.41t2cos(2418.1)t(v °+π+°−π=
Chapter 18, Solution 61.
lsb = 8,000,000 – 5,000 = 7,995,000 Hz usb = 8,000,000 + 5,000 = 8,005,000 Hz Chapter 18, Solution 62.
For the lower sideband, the frequencies range from 10,000 – 3,500 Hz = 6,500 Hz to 10,000 – 400 Hz = 9,600 Hz
For the upper sideband, the frequencies range from 10,000 + 400 Hz = 10,400 Hz to 10,000 + 3,500 Hz = 13,500 Hz Chapter 18, Solution 63.
Since fn = 5 kHz, 2fn = 10 kHz
i.e. the stations must be spaced 10 kHz apart to avoid interference. ∆f = 1600 – 540 = 1060 kHz The number of stations = ∆f /10 kHz = 106 stations Chapter 18, Solution 64.
∆f = 108 – 88 MHz = 20 MHz The number of stations = 20 MHz/0.2 MHz = 100 stations Chapter 18, Solution 65.
ω = 3.4 kHz fs = 2ω = 6.8 kHz
Chapter 18, Solution 66.
ω = 4.5 MHz fc = 2ω = 9 MHz Ts = 1/fc = 1/(9x106) = 1.11x10–7 = 111 ns Chapter 18, Solution 67.
We first find the Fourier transform of g(t). We use the results of Example 17.2 in conjunction with the duality property. Let Arect(t) be a rectangular pulse of height A and width T as shown below.
Arect(t) transforms to Atsinc(ω2/2)
–ωm/2
G(ω)
ω
ωm/2
F(ω)
ω
–T/2
A f(t)
t
T/2
According to the duality property, Aτsinc(τt/2) becomes 2πArect(τ) g(t) = sinc(200πt) becomes 2πArect(τ) where Aτ = 1 and τ/2 = 200π or T = 400π i.e. the upper frequency ωu = 400π = 2πfu or fu = 200 Hz
The Nyquist rate = fs = 200 Hz
The Nyquist interval = 1/fs = 1/200 = 5 ms
Chapter 18, Solution 68.
The total energy is
WT = ∫ ∞
∞−dt)t(v2
Since v(t) is an even function,
WT = ∞−∞ −
−=
0
t4
0
t4
4e5000dte2500∫ = 1250 J
V(ω) = 50x4/(4 + ω2)
W = ∫ ωω+π
=ωωπ
5
1 22
25
1
2 d)4(
)200(21d|)(V
2 ∫ |1
But C)a/x(tana1
axx
a21dx
)xa(1 1
222222 +
+
+=
+−∫
W = 5
1
12
4
)2/(tan21
48110x
ω+
ω+ω
π−2
= (2500/π)[(5/29) + 0.5tan-1(5/2) – (1/5) – 0.5tan–1(1/2) = 267.19
W/WT = 267.19/1250 = 0.2137 or 21.37%
Chapter 18, Solution 69.
The total energy is
WT = ∫∫∞
∞−
∞
∞−ω
ω+π=ωω
πd
4400
21d)(F
2 2221
= [ ]2
100)4/(tan)4/1(4000
1 ππ
=ωπ
∞− = 50
W = [ ]2
0
12
0
2 )4/(tan)4/1(2400d)(F
21
ωπ
=ωωπ
−∫
= [100/(2π)]tan–1(2) = (50/π)(1.107) = 17.6187
W/WT = 17.6187/50 = 0.3524 or 35.24%
Chapter 19, Solution 1. To get z and , consider the circuit in Fig. (a). 11 21z
1 Ω 4 Ω I2 = 0
+
V2
−
Io+
V1
−
2 Ω6 ΩI1
(a)
Ω=++== 4)24(||611
111 I
Vz
1o 21
II = , 1o2 2 IIV ==
Ω== 11
221 I
Vz
To get z and , consider the circuit in Fig. (b). 22 12z
1 Ω 4 ΩI1 = 0
Io' +
V2
−
+
V1
−
2 Ω6 Ω I2
(b)
Ω=+== 667.1)64(||22
222 I
Vz
22'
o 61
1022
III =+
= , 2o1 '6 IIV ==
Ω== 12
112 I
Vz
Hence,
=][z Ω
667.1114
Chapter 19, Solution 2.
Consider the circuit in Fig. (a) to get and . 11z 21z
+
V2
−
I2 = 0
1 Ω
1 Ω
+
V1
−
1 Ω
Io
1 Ω
1 Ω
Io'1 Ω 1 Ω
I1
1 Ω 1 Ω 1 Ω 1 Ω (a)
])12(||12[||121
111 +++==
IV
z
733.21511
24111)411)(1(
243
2||1211 =+=+
+=
++=z
'
o'
oo 41
311
III =+
=
11'
o 154
41111
III =+
=
11o 151
154
41
III =⋅=
1o2 151
IIV ==
06667.0151
121
221 ==== z
IV
z
To get z , consider the circuit in Fig. (b). 22
1 Ω 1 Ω 1 Ω1 ΩI1 = 0
+
V2
−
+
V1
−
1 Ω1 Ω1 Ω I2
1 Ω 1 Ω 1 Ω 1 Ω(b)
733.2)3||12(||12 112
222 ==++== z
IV
z
Thus,
=][z Ω
733.206667.0
06667.0733.2
Chapter 19, Solution 3.
(a) To find and , consider the circuit in Fig. (a). 11z 21z
Io +
V2
−
+
V1
−
1 Ωj Ω
I2 = 0-j Ω
I1
(a)
j1j1j)j1(j
)j1(||j1
111 +=
−+−
=−==IV
z
By current division,
11o jj1j
jIII =
−+=
1o2 jIIV ==
j1
221 ==
IV
z
To get z and , consider the circuit in Fig. (b). 22 12z
I1 = 0
+
V2
−
+
V1
−
1 Ω
-j Ω
j Ω I2
(b)
0)jj(||12
222 =−==
IV
z
21 jIV =
j2
112 ==
IV
z
Thus,
=][z Ω
+0jjj1
(b) To find and , consider the circuit in Fig. (c). 11z 21z
+ V1 −
-j Ω 1 Ω
+ V2 −
1 Ω
I2 = 0j Ω -j Ω
I1
(c)
5.0j5.1j1
j-j1-j)(||11j
1
111 +=
−++=++==
IV
z
12 )5.0j5.1( IV −=
5.0j5.11
221 −==
IV
z
To get z and , consider the circuit in Fig. (d). 22 12z
+ V2 −
-j Ω 1 Ω
+ V1 −
1 Ω
I1 = 0 j Ω -j Ω
I2
(d)
j1.5-1.5(-j)||11-j2
222 =++==
IV
z
21 )5.0j5.1( IV −=
5.0j5.12
112 −==
IV
z
Thus,
=][z Ω
−−−+
5.1j5.15.0j5.15.0j5.15.0j5.1
Chapter 19, Solution 4. Transform the Π network to a T network.
Z1 Z3
Z2
5j12120j
5j10j12)10j)(12(
1 +=
−+=Z
j512j60-
2 +=Z
5j1250
3 +=Z
The z parameters are
j4.26--1.77525144
j5)--j60)(12(22112 =
+=== Zzz
26.4j775.1169
)5j12)(120j(1212111 +=+
−=+= zzZz
739.5j7758.1169
)5j12)(50(2121322 −=+
−=+= zzZz
Thus,
=][z Ω
−−−+
739.5j775.126.4j775.1-26.4j775.1-26.4j775.1
Chapter 19, Solution 5. Consider the circuit in Fig. (a).
s 1 I2 = 0
Io
1/s1/s +
V2
−
+
V1
−
1 I1
(a)
s1
s11s
1s1
s11s
1
s1
s1||
s1
1
s1
s1
s1||s1
||111
+++
+
++
+
=
+++
=
++=z
1s3s2s1ss
23
2
11 +++++
=z
12
11o
1ss1s
s1s
s
s1
s11s
11s
1
s1
s1s1
||1
s1
||1IIII
++++
+=+++
+
+=+++
=
123o 1s3s2ss II
+++=
1s3s2ss1
231
o2 +++==
IIV
1s3s2s123
1
221 +++
==IVz
Consider the circuit in Fig. (b).
s 1 I1 = 0
1/s1/s 1 +
V1
−
+
V2
−
I2
(b)
+++=
++==1s
1s1||
s1
s1
||1s1||s1
2
222 I
Vz
1ss
ss1
1s1
s1
1s1
s1s1
1s1
s1s1
222
++++
+++
=
++++
+++
=z
1s3s2s2s2s
23
2
22 +++++
=z
2112 zz =
Hence,
=][z
+++++
+++
++++++++
1s3s2s2s2s
1s3s2s1
1s3s2s1
1s3s2s1ss
23
2
23
2323
2
Chapter 19, Solution 6.
To find and , connect a voltage source to the input and leave the output open as in Fig. (a).
11z 21z 1V
Vo
+
V2
−
20 ΩI1
30 Ω0.5 V2+ −
10 Ω
V1
(a)
505.0
10o
2o1 V
VVV
+=−
, where oo2 53
302030
VV =+
V =
oo
oo1 2.455
35 VVVVV =+
+=
ooo1
1 32.010
2.310
VVVV
I ==−
=
Ω=== 125.1332.02.4
o
o
1
111 V
VIV
z
Ω=== 875.132.06.0
o
o
1
221 V
VIV
z
To obtain and , use the circuit in Fig. (b). 22z 12z
I2
0.5 V2 + −
10 Ω
+
V1
−
30 Ω
20 Ω
V2
(b)
22
22 5333.030
5.0 VV
VI =+=
Ω=== 875.15333.01
2
222 I
Vz
2221 -9)5.0)(20( VVVV =−=
Ω=== -16.8755333.0
9-
2
2
2
112 V
VIV
z
Thus,
=][z Ω
1.8751.87516.875-125.13
Chapter 19, Solution 7. To get z11 and z21, we consider the circuit below. I2=0 I1 20 100Ω Ω + + + vx 50Ω 60 Ω V1
- V2 - -
12vx - +
1xxxxx1 V
12140V
160V12V
50V
20VV
=→+
+=−
88.29IVz)
20V(
12181
20VVI
1
111
1x11 ==→=
−=
37.70IVzI37.70
I81
121x20)12140(
857V)
12140(
857V
857V12)
160V13(60V
1
2211
11xxx
2
−==→−=
−=−=−=−=
To get z12 and z22, we consider the circuit below. I2 I1=0 20 100Ω Ω + + + vx 50Ω 60 Ω V1
- V2 - -
12vx - +
2x22
222x V09.060
V12V150VI,V
31V
5010050V =
++==
+=
11.1109.0/1IVz
2
222 ===
704.3IVzI704.3I
311.11V
31VV
2
112222x1 ==→====
Thus,
Ω
−
=11.1137.70
704.388.29]z[
Chapter 19, Solution 8. To get z11 and z21, consider the circuit below.
j4Ω I1 -j2Ω 5 Ω I2 =0 •• + j6Ω j8Ω + V2 V1 10Ω - -
4j10IVzI)6j2j10(V
1
11111 +==→+−=
)4j10(IVzI4jI10V
1
221112 +−==→−−=
To get z22 and z12, consider the circuit below.
j4Ω I1=0 -j2Ω 5 Ω I2 •• + j6Ω j8Ω + V2 V1 10Ω - -
8j15IVzI)8j105(V
2
22222 +==→++=
)4j10(IVzI)4j10(V
2
11221 +−==→+−=
Thus,
Ω
++−+−+
=)8j15()4j10()4j10()4j10(
]z[
Chapter 19, Solution 9.
It is evident from Fig. 19.5 that a T network is appropriate for realizing the z parameters.
R2 R1 2 Ω 6 Ω
R3 4 Ω
=−=−= 410R 12111 zz Ω6 =−=−= 46R 12222 zz Ω2
=== 21123R zz Ω4 Chapter 19, Solution 10.
(a) This is a non-reciprocal circuit so that the two-port looks like the one shown in Figs. (a) and (b).
+ −
+
V1
−
I1
+ −
+
V2
−
z21 I1 z12 I2
z22 I2 z11
(a)
(b) This is a reciprocal network and the two-port look like the one shown in
Figs. (c) and (d).
+
V1
−
I2I1
+
V2
−
z12
z22 – z12z11 – z12
(c)
+ −
+
V1
−
I2 I1
+ −
+
V2
−
5 I1 20 I2
10 Ω25 Ω
(b)
s5.01
1s2
11211 +=+=− zz
s21222 =− zz
s1
12 =z
1 F +
V1
−
I2I1
+
V2
−
0.5 F 2 H 1 Ω
(d)
Chapter 19, Solution 11. This is a reciprocal network, as shown below. 1+j5 3+j 1Ω j5Ω 3 Ω 5-j2 5Ω -j2Ω Chapter 19, Solution 12. This is a reciprocal two-port so that it can be represented by the circuit in Figs. (a) and (b).
Ω
2 Ω
2 Ω8 ΩI1
+
V1
−
+
V2
−
I2
4 Ω
+
V1
−
I2I1
+
V2
−
(a)
z12
z22 – z12z11 – z12
j1
Io
2 Ω
(b) From Fig. (b),
111 10)4||48( IIV =+=
By current division,
1o 21
II = , 1o2 2 IIV ==
==1
1
1
2
10II
VV
1.0
Chapter 19, Solution 13. This is a reciprocal two-port so that the circuit can be represented by the circuit below.
40 Ω 50 Ω 20 Ω
10 I2 30 I1 + −
+ − 120∠0° V
rms I1 I2 +
− 100 Ω
We apply mesh analysis. For mesh 1,
2121 91201090120- IIII +=→=++ (1) For mesh 2,
2121 -4012030 IIII =→=+ (2) Substituting (2) into (1),
3512-
-35-3612 2222 =→=+= IIII
=
== )100(3512
21
R21
P2
2
2I W877.5
Chapter 19, Solution 14. To find , consider the circuit in Fig. (a). ThZ
I2I1
+
V1
−
+ −
ZS Vo = 1
(a)
2121111 IzIzV += (1)
2221212 IzIzV += (2) But
12 =V , 1s1 - IZV =
Hence, 2s11
1212121s11
-)(0 I
Zzz
IIzIZz+
=→++=
222s11
1221-1 Iz
Zzzz
+
+=
===22
2Th
1II
VZ
s11
122122 Zz
zzz
+−
To find , consider the circuit in Fig. (b). ThV
+ −
I2 = 0ZS
+
V2 = VTh
−
I1
+
V1
−
VS
(b)
02 =I , V s1s1 ZIV −=
Substituting these into (1) and (2),
s11
s1111s1s Zz
VIIzZIV
+=→=−
s11
s211212 Zz
VzIzV
+==
== 2Th VVs11
s21
ZzVz+
Chapter 19, Solution 15.
(a) From Prob. 18.12,
24104060x80120
ZzzzzZ
s11
211222Th =
+−=
+−=
Ω== 24ZZ ThL
(b) 192)120(1040
80VZz
zs
s11
21Th =
+=
+=V
W19224x8
192R8
VP2
Th
Th2
max ===
Chapter 19, Solution 16. As a reciprocal two-port, the given circuit can be represented as shown in Fig. (a).
+ − j6 Ω
b
a
(a)
15∠0° V
5 Ω 4 – j6 Ω10 – j6 Ω
j4 Ω
At terminals a-b, )6j105(||6j)6j4(Th −++−=Z
6j4.26j415
)6j15(6j6j4Th ++−=
−+−=Z
=ThZ Ω4.6
==°∠−++
= 6j)015(6j1056j
6jThV V906 °∠
The Thevenin equivalent circuit is shown in Fig. (b).
6.4 Ω
+
Vo
−
+ − 6∠90° V j4 Ω
(b)From this,
°∠=+
= 14818.3)6j(4j4.6
4joV
=)t(vo V)148t2cos(18.3 °+
Chapter 19, Solution 17. To obtain z and , consider the circuit in Fig. (a). 11 21z
4 Ω Io I2 = 0
Io' +
V2
−
+
V1
−
2 Ω
8 ΩI1
(a)6 Ω
In this case, the 4-Ω and 8-Ω resistors are in series, since the same current, , passes through them. Similarly, the 2-Ω and 6-Ω resistors are in series, since the same current,
, passes through them.
oI
'oI
Ω===++== 8.420
)8)(12(8||12)62(||)84(
1
111 I
Vz
11o 52
1288
III =+
= 1'
o 53
II =
But 024- '
oo2 =+− IIV
111'
oo2 52-
56
58-
2-4 IIIIIV =+=+=
Ω=== -0.452-
1
221 I
Vz
To get z and , consider the circuit in Fig. (b). 22 12z
+
V1
−
I1 = 0
8 Ω
2 Ω +
V2
−
4 Ω
I2
(b)6 Ω
Ω===++== 2.420
)14)(6(14||6)68(||)24(2
222I
Vz
Ω== -0.42112 zz
Thus,
=][z Ω
4.20.4-0.4-8.4
We may take advantage of Table 18.1 to get [y] from [z].
20)4.0()2.4)(8.4( 2z =−=∆
21.020
2.4
z
2211 ==
∆=
zy 02.0
204.0-
z
1212 ==
∆=
zy
02.020
4.0-
z
2121 ==
∆=
zy 24.0
208.4
z
1122 ==
∆=
zy
Thus,
=][y S24.002.002.021.0
Chapter 19, Solution 18. To get y and , consider the circuit in Fig.(a). 11 21y
I2I1
+ −
+
V2 = 0
−
(a)
6 Ω
3 Ω
3 Ω
6 ΩV1
111 8)3||66( IIV =+=
81
1
111 ==
VI
y
12-
832-
366- 11
12
VVII ==
+=
121-
1
221 ==
VI
y
To get y and , consider the circuit in Fig.(b). 22 12y
6 Ω 3 ΩIoI1 I2
+ −
+
V1 = 0
−
3 Ω6 Ω V2
(b)
21
6||31
)6||63(||31
2
222 ==
+==
VI
y
2- o
1
II = , 22o 3
163
3III =
+=
12-
21
61-
6- 2
22
1
VV
II =
==
212
112 12
1-y
VI
y ===
Thus,
=][y S
21
121-
121-
81
Chapter 19, Solution 19. Consider the circuit in Fig.(a) for calculating and y . 11y 21
1/s
I2I1
+ −
+
V2 = 0
−
s 1
1
V1
(a)
1111 1s22
)s1(2s2
2||s1
IIIV+
=+
=
=
5.0s2
1s2
1
111 +=
+==
VI
y
2-
1s2-
2)s1()s1-( 11
12
VIII =
+=
+=
-0.51
221 ==
VI
y
To get y and , refer to the circuit in Fig.(b). 22 12y
1 I1 I2
1/s s 1
+ −
+
V1 = 0
−
V2
(b)
222 2ss2
)2||s( IIV+
==
s1
5.0s22s
2
222 +=
+==
VI
y
2-
s22s
2ss-
2ss- 2
221
VVII =
+⋅
+=
+=
-0.52
112 ==
VI
y
Thus,
=][y Ss10.50.5-
0.5-5.0s
+
+
Chapter 19, Solution 20. To get y11 and y21, consider the circuit below.
3ix I1 2Ω I2 + ix + I1 V1 4 6Ω Ω V2 =0
- -
Since 6-ohm resistor is short-circuited, ix = 0
75.0VIyI
68)2//4(IV
1
111111 ==→==
5.0VIyV
21)V
86(
32I
244I
1
2211112 −==→−=−=
+−=
To get y22 and y12, consider the circuit below.
3ix I1 2Ω + ix + V1=0 4 6Ω Ω V2 I2
-
-
1667.061
VIy
6V
2Vi3iI,
6Vi
2
222
22xx2
2x ===→=+−==
0VIy0
2Vi3I
2
112
2x1 ==→=−=
Thus,
S1667.05.0075.0
]y[
−
=
Chapter 19, Solution 21. To get and , refer to Fig. (a). 11y 21y
At node 1,
10 Ω
I2I1
+ −
+
V2 = 0
−
(a)
5 Ω
0.2 V1V1
V1
4.04.02.05 1
11111
11 ==→=+=
VI
yVVV
I
-0.2-0.21
22112 ==→=
VI
yV
and 12y , refer to the circuit in Fig. (b).
I
o get
ince , the dependent current source can be replaced with an open circuit.
22y
0.2 V1
T
+
V1 0 +
I1 I2
V2 5 Ω
(b)
10 Ω
V1
− =
−
01 =VS
1.0101
102
22222 ===→=
VI
yIV
02
112 ==
VI
y
Thus,
onsequently, the y parameter equivalent circuit is shown in Fig. (c)
=][y S1.02.0-
04.0
C .
hapter 19, Solution 22.
and refer to the circuit in Fig. (a).
At node 1,
I I1 2
0.1 S 0.2 V1 + +
(c)
0.4 S V1
−
V2
−
C
(a) To get y11 21y
Ω Ω2
1 Ω
(a)
+
V2 0 V1 =
−
+
I1 I2 V1
+
Vx
−
3
Vx/2
Vo
−
o11o11
1 5.05.121
VVIVVV
−=→−
+ (1)
At node 2,
o1o1o 2.1
322VV
VVV=→=+ (2)
ubstituting (2) into (1) gives,
I =
1V −
S
9.09.06.05.11
1111111 ==→=−=
VI
yVVVI
-0.4-0.43
-
1
2211
o2 ==→==
VI
yVV
I
o get and refer to the circuit in Fig. (b).
so that the dependent current source can be replaced by an
open circuit.
22y 12y
2 Ω 3 Ω
T
1 Ω
(b)
+
V1 0
I1 I2V1
+
x Vx/2 + V2 − V
−
=
−
01x == V
V
2.051
5)023(2
222222 ===→=++=
VI
yIIV
-0.2-0.2-2
112221 ==→==
VI
yVII
Thus,
=][y S0.20.4-0.2-9.0
(b) o get and refer to Fig. (c). 11y 21y
j Ω
T
)5.0j5.1(||jj1
1||j-j)||11(||jin −= −
+=+=Z j-
-j Ω
1 Ω1 Ω
V1
(c)
+
V2 0
I1 I2
+
Io
Io'
Io''
Zin
− =
−
8.0j6.05.0j5.1
)5.0j5.1(j+=
+−
=
8.0j6.08.0j6.0
11
in1
1111in1 −=
+===→=
ZVI
yIZV
1o 5.0j5.1j
II+
= , 1'
o 5.0j5.15.0j5.1
II+−
=
1I
j2)5.0j5.1)(j1(j1j- 1
o''
o −=
+−=
−=
III
111
2''
o'
o2 )4.0j2.1(5
j25.2
)5.0j5.1(- IIIIII −=
++
−=+=
112 )2.1j4.0()8.0j6.0)(4.0j2.1(- VVI −=−−=
121
221 j1.2-0.4 y
VI
y =+==
To get refer to the circuit in Fig.(d).
22y
8.0j.0)j-||11(||jout =+=Z
-j Ω
1 Ω1 Ω
(d)
+
V1 = 0
I1 I2
Zout
+ −
V2
j Ω
−
6 +
8.0j6.01
out22 −==
Zy
Thus,
=][y S8.0j6.0j1.20.4-
j1.20.4-8.0j6.0
−++−
Chapter 19, Solution 23. (a)
1s1y
1s1
s1//1y 1212 +
−=→+
==−
1s2s
1s11y1y1yy 12111211 +
+=
++=−=→=+
1s
1ss1s
1sysysyy2
12221222 +++
=+
+=−=→=+
+++
+−
+−
++
=
1s1ss
1s1
1s1
1s2s
]y[ 2
(b) Consider the network below. I1 I2 1 + + + Vs V1 V2 2 - - -
[y]
11s VIV += (1)
22 I2V −= (2)
2121111 VyVyI += (3)
2221212 VyVyI += (4) From (1) and (3)
212111s2121111s VyV)y1(VVyVyVV ++=→+=− (5) From (2) and (4),
22221
12221212 V)y5.0(y1VVyVyV5.0 +−=→+=− (6)
Substituting (6) into (5),
++−
=→=
+++
−=
)y5.0)(y1(y1y
s/2Vs2
VyVy
)y5.0)(y1(V
221121
12
2
212221
2211s
)5.3s5.7s6s2(s)1s(2
1s1ss
21
1s3s2)1s(
1s1
s/2V 2322+++
+=
+++
+
++
+++
−
=
Chapter 19, Solution 24. Since this is a reciprocal network, a Π network is appropriate, as shown below.
Y2
4 Ω
8 Ω 4 Ω
1/4 S
1/8 S 1/4 S
(a)
Y3 Y1
(b) (c)
S41
41
21
12111 =−=+= yyY , =1Z Ω4
S41
- 122 == yY , =2Z Ω4
S81
41
83
21223 =−=+= yyY , =3Z Ω8
Chapter 19, Solution 25. This is a reciprocal network and is shown below. 0.5 S 0.5S 1S Chapter 19, Solution 26. To get y and , consider the circuit in Fig. (a). 11 21y
4 Ω
+
Vx
−
2 Ω
2 Vx
1 I2
+ −
+
V2 = 0
−
1 Ω
2
V1
(a)
At node 1,
x1xx
xx1 -2
412
2VV
VVV
VV=→+=+
− (1)
But 5.15.122
2 1
1111
11x11 ==→=
+=
−=
VI
yVVVVV
I
Also, 1x2xx
2 -3.575.124
VVIVV
I ==→=+
-3.51
221 ==
VI
y
To get y and , consider the circuit in Fig.(b). 22 12y
4 Ω
+ −
+
Vx
−
2 Ω
2 Vx
1 I2
1 Ω
2
I1 V2
(b) At node 2,
42 x2
x2
VVVI
−+= (2)
At node 1,
x2xxxx2
x -23
1242 VVV
VVVVV =→=+=
−+ (3)
Substituting (3) into (2) gives
2xxx2 -1.55.121
2 VVVVI ==−=
-1.52
222 ==
VI
y
5.022
-
2
112
2x1 ==→==
VI
yVV
I
Thus,
=][y S5.1-5.3-5.05.1
Chapter 19, Solution 27. Consider the circuit in Fig. (a).
+
V2 = 0
−
(a)
20 I1 10 Ω 0.1 V2 − +
I1
+ −
4 Ω I2
V1
25.04
41
1
1
11111 ===→=
II
VI
yIV
55201
221112 ==→==
VI
yVII
Consider the circuit in Fig. (b).
4 Ω I1 I2
+
V1 = 0
−
− +
+ −
20 I1 10 Ω0.1 V2 V2
(b)
025.041.0
1.042
11221 ===→=
VI
yVI
6.06.01.05.010
202
222222
212 ==→=+=+=
VI
yVVVV
II
Thus,
=][y S6.05
025.025.0
Alternatively, from the given circuit,
211 1.04 VIV −=
212 1.020 VII += Comparing these with the equations for the h parameters show that
411 =h , -0.112 =h , 2021 =h , 1.022 =h Using Table 18.1,
25.0411
1111 ===
hy , 025.0
41.0-
11
1212 ===
hh
y
54
20
11
2121 ===
hh
y , 6.04
24.0
11
h22 =
+=
∆=
hy
as above. Chapter 19, Solution 28. We obtain and by considering the circuit in Fig.(a). 11y 21y
I2
+
V2 = 0
−
(a)
1 Ω
2 Ω
+
V1
−
4 Ω
6 ΩI1
4.34||61in =+=Z
2941.01
in1
111 ===
ZVI
y
11
12 346-
4.3106-
106-
VV
II =
==
-0.176534
6-
1
221 ===
VI
y
To get y and , consider the circuit in Fig. (b). 22 12y
1 Ω 4 ΩI1 Io
+
V2
−
+
V1 = 0
−
2 Ω6 Ω I2
(b)
2
2
22 IV
2434
)734(2)734)(2(
764||2)1||64(||21
==+
=
+=+=
y
7059.03424
22 ==y
o1 76-
II = 222o 347
4814
)734(22
VIII ==+
=
-0.176534
6-34
6-
2
11221 ===→=
VI
yVI
Thus,
=][y S0.70590.1765-0.1765-2941.0
The equivalent circuit is shown in Fig. (c). After transforming the current source to a voltage source, we have the circuit in Fig. (d).
6/34 S
18/34 S 4/34 S 1 A 2 Ω
(c)
8.5 Ω 5.667 Ω
1.889 Ω+ −
(d)
+
V
−
8.5 V 2 Ω
5454.0167.149714.0
)5.8)(9714.0(667.55.8889.1||2)5.8)(889.1||2(
V =+
=++
=
===2
)5454.0(R
VP
22
W1487.0
Chapter 19, Solution 29.
(a) Transforming the ∆ subnetwork to Y gives the circuit in Fig. (a).
+
V2
−
+
V1
−
Vo1 Ω 1 Ω
2 Ω10 A -4 A
(a) It is easy to get the z parameters
22112 == zz , 32111 =+=z , 322 =z
54921122211z =−=−=∆ zzzz
22z
2211 5
3y
zy ==
∆= ,
52--
z
122112 =
∆==
zyy
Thus, the equivalent circuit is as shown in Fig. (b).
2/5 S I2I1
1/5 S+
V2
−
+
V1
−
1/5 S 10 A -4 A
(b)
21211 235052
53
10 VVVVI −=→−== (1)
21212 3-220-53
52-
-4 VVVVI +=→+==
2121 5.1105.110 VVVV +=→−= (2) Substituting (2) into (1),
=→−+= 222 25.43050 VVV V8
=+= 21 5.110 VV V22
(b) For direct circuit analysis, consider the circuit in Fig. (a). For the main non-reference node,
122
410 oo =→=− V
V
=+=→−
= o1o1 10
110 VV
VVV22
=−=→−
= 41
4- o2o2 VV
VVV8
Chapter 19, Solution 30.
(a) Convert to z parameters; then, convert to h parameters using Table 18.1. Ω=== 60211211 zzz , Ω= 10022z
24003600600021122211z =−=−=∆ zzzz
241002400
22
z11 ==
∆=
zh , 6.0
10060
22
1212 ===
zz
h
-0.6-
22
2121 ==
zz
h , 01.01
2222 ==
zh
Thus,
=][h
ΩS0.010.6-
6.024
(b) Similarly,
Ω= 3011z Ω=== 20222112 zzz
200400600z =−=∆
1020200
11 ==h 12020
12 ==h
-121 =h 05.0201
22 ==h
Thus,
=][h
ΩS0.051-
110
Chapter 19, Solution 31. We get h and by considering the circuit in Fig. (a). 11 21h
2 Ω 4 I1
1 Ω1 Ω
+
V1
−
V3 V4
(a)
2 Ω I2
I1
At node 1,
431433
1 2222
VVIVVV
I −=→−
+= (1)
At node 2,
14
24
143 V
IVV
=+−
431431 6-2163-8 VVIVVI +=→+= (2)
Adding (1) and (2),
1441 6.3518 IVVI =→=
1143 8.283 IIVV =−=
1131 8.3 IIVV =+=
Ω== 8.31
111 I
Vh
-3.6-3.61
-
1
2211
42 ==→==
II
hIV
I
To get h and h , refer to the circuit in Fig. (b). The dependent current source can be replaced by an open circuit since
22 12
04 1 =I .
1 Ω1 Ω 2 ΩI1 I2
+ −
+
V1
−
4 I1 = 0 2 Ω V2
(b)
4.052
1222
2
112221 ==→=
++=
VV
hVVV
S2.051
5122 2
222
222 ===→=
++=
VI
hVV
I
Thus,
=][h
ΩS0.23.6-
4.038
Chapter 19, Solution 32.
(a) We obtain and by referring to the circuit in Fig. (a). 11h 21h
1 s s I2
+
V1
−
+
V2 = 0
−
1/s I1
(a)
1211 1ss
s1s1
||ss1 IIV
+++=
++=
1ss
1s 21
111 +
++==IV
h
By current division,
1s1-
1s-
s1ss1-
21
221
112 +
==→+
=+
=II
hI
II
To get h and h , refer to Fig. (b). 22 12
1 s s I1 = 0 I2
+ −
+
V1
−
1/s V2
(b)
1s1
1ss1ss1
22
1122
221 +
==→+
=+
=VV
hV
VV
1ss
s1s1
s1
s 22
22222 +
=+
==→
+=
VI
hIV
Thus,
=][h
++
++++
1ss
1s1-
1s1
1ss
1s
22
22
(b) To get g and g , refer to Fig. (c). 11 21
1 s s I1 I2 = 0
+ −
+
V2
−
1/s V1
(c)
1sss
s1s11
s1
s1 21
11111 ++
=++
==→
++=VI
gIV
1ss1
1sss1s1s1
21
2212
112 ++
==→++
=++
=VV
gV
VV
To get g and g , refer to Fig. (d). 22 12
I1
+
V2
−
+
V1 = 0
−
I2s 1
1/s
s
I2
(d)
222 s1s1s)1s(
s)1s(||s1
s IIV
+++
+=
++=
1ss1s
s 22
222 ++
++==
IV
g
1ss1-
1ss-
s1s1s1-
22
1122
221 ++
==→++
=++
=II
gI
II
Thus,
=][g
+++
+++
++++
1ss1s
s1ss
11ss
1-1ss
s
22
22
Chapter 19, Solution 33. To get h11 and h21, consider the circuit below.
4 j6Ω Ω I2 + -j3Ω + I1 V1 5 VΩ 2=0 - -
2821.1j0769.3IVh
6j9I)6j4(5I)6j4//(5V
1
111
111 +==
++
=+=
Also, 2564.0j3846.0IIhI
6j95I
1
22112 +−==→
+−=
To get h22 and h12, consider the circuit below. 4 j6Ω Ω I2 I1 + + -j3Ω + V1 5 VΩ 2
- -
2564.0j3846.06j9
5VVhV
6j95V
2
11221 −=
+==→
+=
2821.0j0769.0)6j9(3j
3j9)6j9//(3j
1VIh
I)6j9//(3jV 2
222
22+=
+−+
=+−
==→+−=
Thus,
++−−+
=2821.0j0769.02564.0j3846.02564.0j3846.02821.1j0769.3
]h[
Chapter 19, Solution 34. Refer to Fig. (a) to get h and . 11 21h
300 Ω
100 Ω
10 Ω 1
+
Vx
−
+
V1
−
(a)
10 Vx
50 Ω 2
+
V2 = 0
−
− +
I2
I1
At node 1,
x1xx
1 4300300
0100
VIVV
I =→−
+= (1)
11x 754
300IIV ==
But Ω==→=+= 8585101
1111x11 I
VhIVIV
At node 2,
111xxxx
2 75.1430075
575
300530050100
IIIVVVV
I =−=−=−+
=
75.141
221 ==
II
h
To get h and h , refer to Fig. (b). 22 12
300 Ω
I2
+ −
I1 = 0 2 50 Ω
10 Vx − +
+
V1
−
+
Vx
−
1 10 Ω
100 Ω V2
(b)
At node 2,
x22x22
2 8094005010
400VVI
VVVI +=→
++=
But 4400
100 22x
VVV ==
Hence, 2222 29209400 VVVI =+=
S0725.040029
2
222 ===
VI
h
25.041
4 2
112
2x1 ===→==
VV
hV
VV
=][h
ΩS0725.075.14
25.085
To get g and g , refer to Fig. (c). 11 21
300 Ω
I1 I2 = 0
+ −
2 50 Ω
10 Vx − +
+
V2
−
+
Vx
−
1 10 Ω
100 ΩV1
(c) At node 1,
x1xxx
1 5.14350350
10100
VIVVV
I =→+
+= (2)
But x11x1
1 1010
VVIVV
I −=→−
=
or (3) 11x 10IVV −=
Substituting (3) into (2) gives
11111 5.144951455.14350 VIIVI =→−=
S02929.0495
5.14
1
111 ===
VI
g
At node 2,
xxx2 -8.42861035011
)50( VVVV =−
=
1111 4955.14)286.84(-8.4286286.84-8.4286 VVIV
+=+=
-5.96-5.961
22112 ==→=
VV
gVV
To get g and g , refer to Fig. (d). 22 12
300 Ω
I1
+
V1 = 0
−
Io Io
+
V2
−
50 Ω
10 Vx − +
+
Vx
−
10 Ω
100 Ω I2
(d)
091.9100||10 =
091.93005010 2x2
2 ++
+=
VVVI
x22 818.611818.7091.309 VVI += (4)
But 22x 02941.0091.309
091.9VVV == (5)
Substituting (5) into (4) gives
22 9091.309 VI =
Ω== 34.342
222 I
Vg
091.30934.34
091.30922
o
IVI ==
)091.309)(1.1(34.34-
110100- 2
o1
III ==
-0.1012
112 ==
II
g
Thus,
=][g
Ω34.345.96-
0.101-S02929.0
Chapter 19, Solution 35. To get h and h consider the circuit in Fig. (a). 11 21
I2
+
V2 = 0
−
4 Ω1 Ω
(a)
+
V1
−
1 : 2
I1
144
n4
Z 2R ===
Ω==→=+= 22)11(1
111111 I
VhIIV
-0.521-
2-NN-
1
221
1
2
2
1 ===→==II
hII
To get h and h , refer to Fig. (b). 22 12
− +
1 : 2
+
V1
−
I21 Ω 4 ΩI1 = 0
V2
(b) Since , . 01 =I 02 =IHence, . 022 =h At the terminals of the transformer, we have and which are related as 1V 2V
5.021
2nNN
2
112
1
2
1
2 ===→===VV
hVV
Thus,
=][h
Ω05.0-5.02
Chapter 19, Solution 36. We replace the two-port by its equivalent circuit as shown below.
+
V2
−
100 Ω
2 I1
-2 I1 + −
+ −
+
V1
−
I1 4 Ω
3 V2
16 Ω I2
10 V 25 Ω
Ω= 2025||100
112 40)2)(20( IIV == (1)
032010- 21 =++ VI
111 140)40)(3(2010 III =+=
141
1 =I , 1440
2 =V
14136
316 211 =+= VIV
708-)2(
125100
12 =
= II
(a) ==13640
1
2
VV
2941.0
(b) =1
2
II
1.6-
(c) ==136
1
1
1
VI
S10353.7 -3×
(d) ==140
1
2
IV
Ω40
Chapter 19, Solution 37.
(a) We first obtain the h parameters. To get h and h refer to Fig. (a). 11 21
6 Ω 3 Ω I2
+
V2 = 0
−
+
V1
−
3 Ω6 ΩI1
(a)
26||3 =
Ω==→=+= 88)26(1
111111 I
VhIIV
32-
32-
636-
1
221112 ==→=
+=
II
hIII
To get h and h , refer to the circuit in Fig. (b). 22 12
6 Ω 3 ΩI1 = 0 I2
+ −
+
V1
−
3 Ω6 Ω V2
(b)
49
9||3 =
94
49
2
22222 ==→=
VI
hIV
32
32
366
2
112221 ==→=
+=
VV
hVVV
=][h
Ω
S94
32-
32
8
The equivalent circuit of the given circuit is shown in Fig. (c).
8 Ω I1 I2
-2/3 I1 +
V2
−
2/3 V2 9/4 Ω + −
+ − 10 V 5 Ω
(c)
1032
8 21 =+ VI (1)
1112 2930
2945
32
49
||532
IIIV =
=
=
21 3029
VI = (2)
Substituting (2) into (1),
1032
3029
)8( 22 =+
VV
==252300
2V V19.1
(b) By direct analysis, refer to Fig.(d).
6 Ω 3 Ω
3 Ω
+
V2
−
+ − 6 Ω10 V 5 Ω
(d)
Transform the 10-V voltage source to a 6
10-A current source. Since
, we combine the two 6-Ω resistors in parallel and transform
the current source back to
Ω= 36||6
V536
10=× voltage source shown in Fig. (e).
3 Ω 3 Ω
+
V2
−
+ −
5 V 3 || 5 Ω
(e)
815
8)5)(3(
5||3 ==
==+
=6375
)5(8156
8152V V19.1
Chapter 19, Solution 38. We replace the two-port by its equivalent circuit as shown below.
50 I1 +
V2
−
200 kΩ + −
+ −
800 Ω
10-4 V2
+
V1
−
I1 200 Ω I2
10 V 50 kΩ
1
sin I
VZ = , Ω= k4050||200
1
6312 )10-2()1040(-50 IIV ×=×=
For the left loop,
12
-4s
100010
IVV
=−
11
6-4s 1000)10-2(10 IIV =×−
111s 8002001000 IIIV =−=
==1
sin I
VZ Ω800
Alternatively,
L22
L211211sin 1 Zh
ZhhhZZ
+−+=
=××+
×−+=
)1050)(105.0(1)1050)(50)(10(
800200 35-
3-4
inZ Ω800
Chapter 19, Solution 39. To get g11 and g21, consider the circuit below which is partly obtained by converting the delta to wye subnetwork. I1 R1 R2 I2 + + R3 V2 V1 10Ω - -
2.320
8x8R,R6.1488
8x4R 321 ====++
=
8919.0VVgV8919.0V
6.12.132.13V
1
221112 ==→=
+=
06757.08.14
1VIgI8.14)102.36.1(IV1
111111 ===→=++=
To get g22 and g12, consider the circuit below. 1.6Ω 1.6Ω I1 + V2 V1=0 I2 13.2Ω -
8919.0IIgI8919.0I
6.12.132.13I
2
112221 −==→−=
+−=
027.3IVgI027.3)6.1//2.136.1(IV
2
222222 ==→=+=
−=
027.38919.08919.006757.0
]g[
Chapter 19, Solution 40. To get g and g , consider the circuit in Fig. (a). 11 21
I2 = 0j10 Ω
12 Ω
-j6 ΩI1
+ −
+
V2
−
(a)
V1
S0333.0j0667.06j12
1)6j12(
1
11111 +=
−==→−=
VI
gIV
4.0j8.0j2
2)6j12(
12
1
1
1
221 +=
−=
−==
II
VV
g
To get g and g , consider the circuit in Fig. (b). 12 22
I1 -j6 Ω
12 Ω
j10 Ω
+
V1 = 0
−
I2
I2
(b)
4.0j8.0--j6-12
12-j6-12
12-21
2
11221 −====→= g
II
gII
22 -j6)||1210j( IV +=
Ω+=+== 2.5j4.2j6-12-j6))(12(
10j2
222 I
Vg
=][g
Ω++
−+2.5j4.24.0j8.0
4.0j8.0-S0333.0j0667.0
Chapter 19, Solution 41.
For the g parameters 2121111 IgVgI += (1)
2221212 IgVgV += (2) But and s1s1 ZIVV −=
222121L22 - IgVgZIV +==
2L22121 )(0 IZgVg ++=
or 221
L221
)(-I
gZg
V+
=
Substituting this into (1),
221
122111L11221 -
)(I
ggggZgg
I−+
=
or gL11
21
1
2
Zgg-
II
∆+=
Also, 222s1s212 )( IgZIVgV +−=
2221s21s21 IgIZgVg +−= 2222gL11ss21 )( IgIZgZVg +∆++=
But L
22
-ZV
I =
+∆+−=
L
222sgLs11s212 ][
ZV
gZZZgVgV
s21L
22sgLs11L2 ][Vg
ZgZZZgZV
=+∆++
22sgLs11L
L21
s
2
gZZZgZZg
VV
+∆++=
22s1221s2211Ls11L
L21
s
2
gZggZggZZgZZg
VV
+−++=
=s
2
VV
s2112L22s11
L21
Zgg)Zg)(Zg1(Zg
−++
Chapter 19, Solution 42.
(a) The network is shown in Fig. (a).
20 ΩI1 I2
100 Ω 0.5 I1 +
V2
−
+
V1
−
-0.5 I2+ −
(a)
(b) The network is shown in Fig. (b).
s 2 ΩI1 I2
10 Ω 12 V1 +
V2
−
+
V1
−
+ −
(b)
Chapter 19, Solution 43.
(a) To find and , consider the network in Fig. (a). A C
+
V2
−
I2I1
+ −
Z
V1
(a)
12
121 ==→=
VV
AVV
002
11 ==→=
VI
CI
To get B and , consider the circuit in Fig. (b). D
+
V2 = 0
−
I2I1
+ −
Z
V1
(b)
11 IZV = , 12 -II =
ZIIZ
IV
B ===1
1
2
1
---
1-
2
1 ==II
D
Hence,
=][T
10Z1
(b) To find and , consider the circuit in Fig. (c). A C
I1 I2
Z +
V2
−
+ −
V1
(c)
12
121 ==→=
VV
AVV
YZV
ICVIZV ===→==
1
2
1211
To get B and , refer to the circuit in Fig.(d). D
I2
+
V1
−
Y +
V2 = 0
−
I1
(d)
021 == VV 12 -II =
0-
2
1 ==IV
B , 1-
2
1 ==II
D
Thus,
=][T
1Y01
Chapter 19, Solution 44. To determine and , consider the circuit in Fig.(a). A C
j15 Ω
I1
+
V2
−
(a)
+ − 20 Ω
-j20 Ω
Io'
I2 = 0
Io
-j10 Ω
Io
V1
11 ])20j15j(||)10j-(20[ IV −+=
111 310
j20j15-
-j10)(-j5)(20 IIV
−=
+=
1'
o II =
11o 32
5j10j-10j- III
=
−
=
'20-j20)( oo2 IIV += = 11 I20I340j +− = 1I
340j20
−
=
−
−==
1
1
2
1
I340j20
)310j20( IVVA 0.7692 + j0.3461
=−
==
340j20
1
2
1
VIC 0.03461 + j0.023
To find and , consider the circuit in Fig. (b). B D
j15 Ω
+
V2 = 0
−
I2I1 -j10 Ω
+ −
-j20 Ω
20 ΩV1
(b) We may transform the ∆ subnetwork to a T as shown in Fig. (c).
10j20j10j15j
-j10))(15j(1 =
−−=Z
340
-jj15-
)-j10)(-j20(2 ==Z
20jj15--j20))(15j(
3 ==Z
j10 Ω j20 ΩI1 I2
+
V2 = 0
−
+ − 20 – j40/3 ΩV1
(c)
112 j32j3
20j340j20340j20
- III+−
=+−
−=
6923.0j5385.02j3j3-
2
1 +=−+
==II
D
11 20j340j20)340j20)(20j(
10j IV
+−−
+=
)18j24(j])7j9(210j[ 111 −=++= IIV
j55)-15(136
j3j2)-(3-
)18j24(j--
1
1
2
1 +=
+
−==
I
IIV
B
Ω+= j25.385-6.923B
=][T
++
Ω++j0.69230.5385Sj0.0230.03461
j25.3856.923-3461.0j7692.0
Chapter 19, Solution 45. To obtain A and C, consider the circuit below. I1 sL 1/sC I2 =0 + + V1 R1 V2 - - R2
1
21
2
11
21
12 R
sLRRVVAV
sLRRRV ++
==→++
=
12
1112 R
1VICRIV ==→=
To obtain B and D, consider the circuit below.
I1 sL 1/sC I2 + + V1 R1 V2=0 - - R2
CsRCsR1
IIDI
CsR1CsRI
sC1R
RI1
1
2
11
1
11
1
12
+=−=→
+−=
+−=
[ ]2
C1
1
1
1211
1
1
21 IsR
)CsR1(CsR1
R)sLR)(CsR1(I
sC1R
sCR
sLRV ++
+++−=
+++=
[ ])sLR)(CsR1(RCsR
1IVB 211
12
1 +++=−=
Chapter 19, Solution 46. To get and C , refer to the circuit in Fig.(a). A
+
V2
−
I1
+ −
1 Ω
2 Ω 4 Ix
1 Ω
Ix
2 I2 = 01
+
Vo
−
V1
(a)At node 1,
2o12oo
1 23212
VVIVVV
I −=→−
+= (1)
At node 2,
2ooo
x2o -2
24
41
VVVV
IVV
=→===−
(2)
From (1) and (2),
S-2.525-
-522
121 ===→=
VI
CVI
But 21o1
1 1VV
VVI +=
−=
21212 -3.52.5- VVVVV =→+=
-3.52
1 ==VV
A
To get B and , consider the circuit in Fig. (b). D
+
V2 = 0
−
+
Vo
−
Ix
+ −
1 Ω1 2 I2
4 Ix 2 Ω
1 Ω I1
V1 (b) At node 1,
o1oo
1 3212
VIVV
I =→+= (3)
At node 2,
041 x
o2 =++ I
VI
– I (4) o2oo2 -302 VIVV =→=+=
Adding (3) and (4),
2121 -0.502 IIII =→=+ (5)
5.0-
2
1 ==II
D
But o11o1
1 1VIV
VVI +=→
−= (6)
Substituting (5) and (4) into (6),
2221 65-
31-
21-
IIIV =+=
Ω=== 8333.065-
2
1
IV
B
Thus,
=][T
Ω0.5-S2.5-
0.83333.5-
Chapter 19, Solution 47. To get A and C, consider the circuit below. 6Ω I1 1Ω 4Ω I2=0 + + + + Vx 2Ω 5Vx V2
V1 - - - -
x1xxxx1 V1.1V
10V5V
2V
1VV
=→−
+=−
3235.04.3/1.1VVAV4.3V5)V4.0(4V
2
1xxx2 ===→=+−=
02941.04.3/1.0VICV1.0VV1.1
1VVI
2
1xxx
x11 ===→=−=
−=
Chapter 19, Solution 48.
(a) Refer to the circuit below.
I2I1
+ −
[ T ]+
V2
−
V1 ZL
221 304 IVV −= (1)
221 1.0 IVI −= (2) When the output terminals are shorted, 02 =V . So, (1) and (2) become
21 -30IV = and 21 -II = Hence,
==1
1in I
VZ Ω30
(b) When the output terminals are open-circuited, 02 =I .
So, (1) and (2) become 21 4 VV =
21 1.0 VI = or 12 10IV =
11 40IV =
==1
1in I
VZ Ω40
(c) When the output port is terminated by a 10-Ω load, . 22 I-10V =
So, (1) and (2) become 2221 -7030-40 IIIV =−=
2221 -2- IIII =−=
11 35IV =
==1
1in I
VZ Ω35
Alternatively, we may use DZCBZA
+Z
+=
L
Lin
Chapter 19, Solution 49. To get and C , refer to the circuit in Fig.(a). A
1/s
+ − 1 Ω 1/s
(a)
1 Ω 1/s +
V2
−
I1 I2 = 0
V1
1s1
s11s1
s1
||1+
=+
=
12 s1||1s1s1||1
VV+
=
1s2s
1s1
s1
1s1
1
2
+=
++
+==VV
A
++
+
=
++
+
=)1s(s
1s2||
1s1
1s1
s1
||1s
1111 IIV
)1s3)(1s(1s2
)1s(s1s2
1s1
)1s(s1s2
1s1
1
1
+++
=
++
++
++
⋅
+
=IV
But s
1s221
+⋅= VV
Hence, )1s3)(1s(
1s2s
1s2
1
2
+++
=+
⋅IV
s)1s3)(1s(
1
2 ++==
IV
C
To get B and , consider the circuit in Fig. (b). D
+
V2 = 0
−
I2
1 Ω 1/s
I1
+ −
1/s
1/s 1 Ω V1
(b)
1s2s21
||1s1
||s1
||1 1111 +
=
=
=I
IIV
1
1
2 12ss-
s1
1s1
1s1-
II
I+
=+
+
+=
s1
2s
1s2-
2
1 +=+
==II
D
s1-
s-s-1s2
1s21
2
1221 ==→=
+
+=
IV
BI
IV
Thus,
=][T
+++
+
s1
2s
)1s3)(1s(s1
1s22
Chapter 19, Solution 50. To get a and c, consider the circuit below.
I1=0 2 s I2
+ + V1 4/s V2
- -
21
22221 s25.01VVaV
4s4V
s/4ss/4V +==→
+=
+=
4ss25.0s
VIc
s/4sV)s25.01(
s/4sVI
or I)s/4s(V
2
3
1
212
22
22
+
+==→
++
=+
=
+=
To get b and d, consider the circuit below. I1 2 s I2
+ +
V1=0 4/s V2
- -
s5.01IId
2sI2I
s/42s/4I
1
2221 +=−=→
+−=
+−
=
2ss5.0IVbI
2)2s
2s)(4s2s(
I2s
)4s2s(I)s4//2s(V
2
1
21
2
22
22
++=−=→+
+++
−=
+++
=+=
++
++++
=1s5.0
4sss25.0
2ss5.01s25.0]t[
2
222
Chapter 19, Solution 51. To get a and c , consider the circuit in Fig. (a).
j2 Ω j Ω
I1 = 0 I2
+
V1
−
(a)
+ −
j Ω1 Ω -j3 Ω
V2
222 -j2)3jj( IIV =−=
21 -jIV =
2j-
j2-
2
2
1
2 ===II
VV
a
jj-
1
1
2 ===VI
c
To get b and d , consider the circuit in Fig. (b).
+
V1 = 0
−
I1
j2 Ω j Ω
I2
+ −
1 Ω -j3 Ωj Ω
V2
(b) For mesh 1,
21 j)2j1(0 II −+=
or j2j
2j1
1
2 −=+
=II
j-2-
1
2 +==II
d
For mesh 2, 122 j)3jj( IIV −−=
1112 )5j-2(j)2j-)(j2( IIIV −=−−=
5j2-
1
2 +==IV
b
Thus,
=][t
++
j2-j5j22
Chapter 19, Solution 52. It is easy to find the z parameters and then transform these to h parameters and T parameters.
+
+=
322
221
RRRRRR
][z
223221z R)RR)(RR( −++=∆
133221 RRRRRR ++=
(a)
++
++++
=
∆
=
3232
2
32
2
32
133221
2222
21
22
12
22
z
RR1
RRR-
RRR
RRRRRRRR
1-][
zzz
zz
zh
Thus,
32
32111 RR
RRRh
++= , 21
32
212 h-
RRR
=+
=h , 32
22 RR1+
=h
as required.
(b)
+
+++
=
∆
=
2
32
2
2
133221
2
21
21
22
21
21
z
21
11
RRR
R1
RRRRRRR
RRR
1][
zz
z
zzz
T
Hence,
2
1
RR
1A += , )RR(RR
RB 322
13 ++= ,
2R1
C = , 2
3
RR
1D +=
as required. Chapter 19, Solution 53.
For the z parameters, 2121111 IzIzV += (1)
2221122 IzIzV += (2) For ABCD parameters,
221 IBVAV −= (3)
221 IDVCI −= (4) From (4),
21
2 ICD
CI
V += (5)
Comparing (2) and (5),
Cz
121 = ,
CD
z =22
Substituting (5) into (3),
211 IBC
ADI
CA
V
−+=
21 IC
BCADICA −
+= (6)
Comparing (6) and (1),
CA
z =11 CC
BCADz T
12
∆=
−=
Thus,
[Z] =
∆
CD
C1
CCA T
Chapter 19, Solution 54.
For the y parameters 2121111 VyVyI += (1)
2221212 VyVyI += (2) From (2),
221
22
21
21 V
yy
yI
V −=
or 221
212
221
1-I
yV
yy
V += (3)
Substituting (3) into (1) gives
221
112122
21
22111
-I
yy
VyVy
yyI ++=
or 221
112
21
y1
-I
yy
Vy
I +∆
= (4)
Comparing (3) and (4) with the following equations
221 IBVAV −=
221 IDVCI −= clearly shows that
21
22
yy-
A = , 21y1-
B = , 21
y
y-
C∆
= , 21
11
yy-
D =
as required. Chapter 19, Solution 55.
For the z parameters 2121111 IzIzV += (1)
2221212 IzIzV += (2) From (1),
211
121
111
1I
zz
Vz
I −= (3)
Substituting (3) into (2) gives
211
1221221
11
212 I
zzz
zVzz
V
−+=
or 211
z1
11
212 I
zV
zz
V∆
+= (4)
Comparing (3) and (4) with the following equations
2121111 IgVgI +=
2221212 IgVgV += indicates that
1111 z
1g = ,
11
1212 z
z-=g ,
11
2121 z
z=g ,
11
z22 z
∆=g
as required. Chapter 19, Solution 56.
(a) 5j74j)j3)(j2(y +=+−+=∆
Ω
−−−−−
=
∆∆−∆−∆
=0405.0j2568.00676.0j0946.03784.0j2703.02973.0j2162.0
/y/y/y/y
]z[y11y21
y12y22
(b)
++−−−−
=
∆−
=6.0j8.32.0j4.06.1j8.02.0j4.0
y/y/yy/yy/1
]h11y1121
111211[
(c )
++−
+−=
−∆−−−
=75.0j25.075.1j25.1
25.0j5.0j25.0y/yy/
y/1y/y]t
122212y
121211[
Chapter 19, Solution 57.
1)1)(20()7)(3(T =−=∆
=
∆
=
CD
C
CCA
z 1][T
Ω
7113
=
∆
=
BA
B
BBD
y 1-
-
][T
S
203
201-
201-
207
=
∆
=
DC
D
DDB
h 1-][T
Ω
S71
71-
71
720
=
∆
=
AB
A
AAC
g 1
-
][T
Ω3
2031
31-
S31
=
∆∆
∆∆=
TT
TT][ AC
BD
t
Ω3S1
207
Chapter 19, Solution 58.
The given set of equations is for the h parameters.
Ω=
S0.42-21
][h 4.4-2))(2()4.0)(1(h =−=∆
(a) =
∆=
11
h
11
21
11
12
11
-1
][
hhh
hh
hy S
4.42-2-1
(b) =
∆
=
2121
22
21
11
21
h
1--
--
]
hhh
hh
hT[
Ω5.0S2.0
5.02.2
Chapter 19, Solution 59.
2.00.080.12-0.4)(2)()2)(06.0(g =+=−=∆
(a) =
∆=
11
g
11
21
11
12
11
-1
[
ggg
gg
gz] Ω
333.3333.3667.6667.16
(b) =
∆
=
2222
21
22
12
22
g
1-][
ggg
gg
gy S0.50.1-0.2-1.0
(c) =
∆∆
∆∆=
g
11
g
21
g
12
g
22
-
-
][ gg
gg
h
ΩS0.31-
210
(d) =
∆=
21
g
21
11
21
22
21
1
][
ggg
gg
gT
Ω1S3.0
105
Chapter 19, Solution 60.
28.002.03.021122211y =−=−=∆ yyyy
(a) =
∆∆
∆∆=
y
11
y
21
y
12
y
22
-
-
][ yy
yy
z Ω
143.23571.0
7143.0786.1
(b) =
∆=
11
y
11
21
11
12
11
-1
][
yyy
yy
yh
ΩS0.46670.1667-
3333.0667.1
(c) =
∆=
12
22
12
y
1212
11
--
1--
][
yy
y
yyy
t
Ω5.2S4.1
53
Chapter 19, Solution 61.
(a) To obtain and , consider the circuit in Fig. (a). 11z 21z
1 ΩIo
+
V1
−
1 Ω 1 Ω I2 = 0
+
V2
−
1 ΩI1
(a)
1111 35
32
1])11(||11[ IIIV =
+=++=
35
1
111 ==
IV
z
11o 31
211
III =+
=
0- 1o2 =++ IIV
1112 34
31
IIIV =+=
34
1
221 ==
IV
z
To obtain and , consider the circuit in Fig. (b). 22z 12z
1 Ω
+
V2
−
I1
+
V1
−
1 Ω 1 Ω
1 Ω I2
(b) Due to symmetry, this is similar to the circuit in Fig. (a).
35
1122 == zz , 34
1221 == zz
=][z Ω
35
34
34
35
(b) =
∆
=
2222
21
22
12
22
z
1-][
zzz
zz
zh
Ω
S53
54-
54
53
(c) =
∆
=
21
22
21
21
z
21
11
1][
zz
z
zzz
T
Ω
45
S43
43
45
Chapter 19, Solution 62. Consider the circuit shown below.
20 kΩ
a 40 kΩ10 kΩ
Ib
b
I1
50 kΩ
30 kΩ
+
V1
−
+−
+
V2
−
I2
Since no current enters the input terminals of the op amp,
13
1 10)3010( IV ×+= (1)
But 11ba 43
4030
VVVV ===
133b
b 10803
1020V
VI
×=
×=
which is the same current that flows through the 50-kΩ resistor. Thus, b
32
32 10)2050(1040 IIV ×++×=
133
23
2 10803
10701040 VIV×
⋅×+×=
23
12 1040821
IVV ×+=
2
31
32 104010105 IIV ×+×= (2)
From (1) and (2),
=][z Ω
k
40105040
8
21122211z 1016×=−=∆ zzzz
=
∆
=
=
21
22
21
21
z
21
11
1][
zz
z
zzz
DCBA
T
µ
Ω381.0S52.9
k24.15381.0
Chapter 19, Solution 63. To get z11 and z21, consider the circuit below. I1 1:3 I2=0 •• + + + + V’1 V’2 9Ω V2 V1 4 - - - -
Ω
3n,1n9Z 2R ===
8.0IVzI
54I)Z//4(V
1
11111R1 ==→==
4.2IVzI)5/4(3nV'nV'VV
1
22111122 ==→====
To get z21 and z22, consider the circuit below. I1=0 1:3 I2 •• + + + + V’1 V’2 9Ω V2 V1 4 - - - -
Ω
3n,36)4(n'Z 2R ===
2.7IVzI
4536x9I)'Z//9(V
2
22222R2 ==→==
4.2IVzI4.2
3V
nVV
2
1212
221 ==→===
Thus,
Ω
=
2.74.24.28.0
]z[
Chapter 19, Solution 64.
Ω==ω
→µ kj-)10)(10(
j-Cj
1F 6-31
Consider the op amp circuit below.
40 kΩ
+
V1
−
1 +
V2
−
I2 I1
2
Vx20 kΩ 10 kΩ
-j kΩ
−+
At node 1,
100
j-20xxx1 −
+=− VVVV
x1 )20j3( VV += (1)
At node 2,
2x2x
41-
400
100
VVVV
=→−
=−
(2)
But 3x1
1 1020×−
=VV
I (3)
Substituting (2) into (3) gives
26-
16-
321
1 105.121050102025.0
VVVV
I ×+×=×
+= (4)
Substituting (2) into (1) yields
21 )20j3(41-
VV +=
or (5) 21 )5j75.0(0 VV ++= Comparing (4) and (5) with the following equations
2121111 VyVyI +=
2221212 VyVyI += indicates that and that 02 =I
=][y S5j75.01
105.121050 -6-6
+××
-6-6
y 10)250j65(10)5.12.25j5.77( ×+=×−+=∆
=
∆=
11
y
11
21
11
12
11
-1
][
yyy
yy
yh
+×Ω×
S5j3.11020.25-102
4
4
Chapter 19, Solution 65. The network consists of two two-ports in series. It is better to work with z parameters and then convert to y parameters.
For aN ,
=
2224
][ az
For bN ,
=
1112
][ bz
=+=
3336
][][][ ba zzz
9918z =−=∆
=
∆∆
∆∆=
z
11
z
21
z
12
z
22
-
-
][ zz
zz
y S
32
31-
31-
31
Chapter 19, Solution 66. Since we have two two-ports in series, it is better to convert the given y parameters to z parameters.
-6-3-321122211y 10200)1010)(102( ×=−××=−=∆ yyyy
Ω
Ω=
∆∆
∆∆=
10000500
-
-
][
y
11
y
21
y
12
y
22
a yy
yy
z
=
+
=
200100100600
100100100100
10000500
][z
i.e. 2121111 IzIzV +=
2221212 IzIzV += or (1) 211 100600 IIV +=
212 200100 IIV += (2) But, at the input port,
11s 60IVV += (3) and at the output port,
2o2 -300IVV == (4) From (2) and (4),
221 -300200100 III =+
21 -5II = (5) Substituting (1) and (5) into (3),
121s 60100600 IIIV ++=
22 100-5))(660( II += = (6) 2-3200I
From (4) and (6),
==2
2
2
o
3200-300-
II
VV
09375.0
Chapter 19, Solution 67. The y parameters for the upper network is
=
21-1-2
][y , 314y =−=∆
=
∆∆
∆∆=
32
31
31
32
-
-
][
y
11
y
21
y
12
y
22
a yy
yy
z
=
1111
][ bz
=+=
35343435
][][][ ba zzz
19
16925
z =−=∆
=
∆
=
21
22
21
21
z
21
11
1][
zz
z
zzz
T
Ω25.1S75.0
75.025.1
Chapter 19, Solution 68.
For the upper network , [ aN
=
42-2-4
]ay
and for the lower network , [ bN
=
211-2
]by
For the overall network,
=+=
63-3-6
][][][ ba yyy
27936y =−=∆
=
∆=
11
y
11
21
11
12
11
-1
][
yyy
yy
yh
Ω
S29
21-
21-
61
Chapter 19, Solution 69. We first determine the y parameters for the upper network . aNTo get y and , consider the circuit in Fig. (a). 11 21y
21
n = , s4
ns12R ==Z
111R1 s4s2
s4
2)2( IIIZV
+
=
+=+=
)2s(2s
1
111 +
==VI
y
2ss-
-2n
- 11
12 +
===V
II
I
2ss-
1
221 +
==VI
y
To get y and , consider the circuit in Fig. (b). 22 12y
2 : 1
+
V1 =0
−
+
V2
−
I1 2 Ω 1/s I2
I2
(b)
21
)2(41
)2)(n( 2'R =
==Z
222'
R2 s22s
21
s1
s1
IIIZV
+
=
+=
+=
2ss2
2
222 +
==VI
y
2221 2ss-
2ss2
21-
n- VVII
+
=
+
==
2ss-
2
112 +
==VI
y
++
++=
2ss2
2ss-
2ss-
)2s(2s
][ ay
For the lower network , we obtain y and by referring to the network in Fig. (c). bN 11 21y
2 I1 I2
+
V2 = 0
−
+ −
s V1
(c)
21
21
11111 ==→=
VI
yIV
21-
2-
-1
221
112 ==→==
VI
yV
II
To get y and , refer to the circuit in Fig. (d). 22 12y
2 I1 I2
+
V2
−
+
V1 = 0
−
s I2
(d)
s22s
2ss2
)2||s(2
222222
+==→
+==
VI
yIIV
2-
s22s
2ss-
2ss-
- 2221
VVII =
+
+
=+
⋅=
21-
2
112 ==
VI
y
+
=s2)2s(21-
21-21][ by
=+= ][][][ ba yyy
+++
++
++
++
)2s(s24s4s5
2)(s22)(3s-
2)(s22)(3s-
2s1s
2
Chapter 19, Solution 70.
We may obtain the g parameters from the given z parameters.
=
1052025
][ az , 150100250az =−=∆
=
30252550
][ bz , 8756251500bz =−=∆
∆=
11
z
11
21
11
12
11
zzz
zz-
z1
][g
=
60.20.8-04.0
][ ag ,
=
17.50.50.5-02.0
][ bg
=+= ][][][ ba ggg
Ω23.50.7
1.3-S06.0
Chapter 19, Solution 71. This is a parallel-series connection of two two-ports. We need to add their g parameters together and obtain z parameters from there. For the transformer,
2121 I2I,V21V −==
Comparing this with
221221 DICVI,BIAVV −=−=
shows that
=
2005.0
]T[ 1b
To get A and C for Tb2 , consider the circuit below. I1 4 IΩ 2 =0 + + 5Ω V1 V2 - 2Ω -
1211 I5V,I9V ==
2.05/1VIC,8.15/9
VVA
2
1
2
1 ======
Chapter 19, Solution 72. Consider the network shown below.
+
V1
−
Ia1 Ia2
+
V2
−
I2
+ Va2
+ Vb2
Ib2
+ Va1
Nb
Na
Ib1
+ Vb1
I1
2a1a1a 425 VIV += (1)
2aa12a 4- VII += (2)
2b1b1b 16 VIV += (3)
2bb12b 5.0- VII += (4)
1b1a1 VVV +=
2b2a2 VVV ==
2b2a2 III +=
1a1 II = Now, rewrite (1) to (4) in terms of and 1I 2V
211a 425 VIV += (5)
212a 4 - VII += (6)
21b1b 16 VIV += (7)
2b12b 5.0- VII += (8) Adding (5) and (7),
21b11 51625 VIIV ++= (9) Adding (6) and (8),
21b12 5.14- VIII +−= (10)
11a1b III == (11) Because the two networks and are independent, aN bN
212 5.15- VII += or (12) 212 6667.0333.3 IIV += Substituting (11) and (12) into (9),
2111 5.15
5.125
41 IIIV ++=
211 333.367.57 IIV += (13)
Comparing (12) and (13) with the following equations
2121111 IzIzV +=
2221212 IzIzV += indicates that
=][z Ω
6667.0333.3333.367.57
Alternatively,
=
14-425
][ ah ,
=
0.51-116
][ bh
=+=
1.55-541
][][][ ba hhh 5.86255.61h =+=∆
=
∆
=
2222
21
22
12
22
h
1-][
hhh
hh
hz Ω
6667.0333.3333.367.57
as obtained previously.
Chapter 19, Solution 73. From Example 18.14 and the cascade two-ports,
==
2132
][][ ba TT
=
==
2132
2132
]][[][ ba TTT
Ω7S4
127
When the output is short-circuited, 02 =V and by definition
21 - IBV = , 21 - IDI = Hence,
===DB
IV
Z1
1in Ω
712
Chapter 19, Solution 74. From Prob. 18.35, the transmission parameters for the circuit in Figs. (a) and (b) are
=
101
][ a
ZT ,
=
1101
][ b ZT
Z
We partition the given circuit into six subcircuits similar to those in Figs. (a) and (b) as shown in Fig. (c) and obtain [ for each. ]T
(b)
Z
(a)
s
T5 T6 T3 T4T1 T2
1 1/s 1 1/s
s
=
1101
][ 1T , ,
=
10s1
][ 2T
=
1s01
][ 3T
][][ 24 TT = , ][][ 15 TT = , ][][ 36 TT =
==
1s01
1101
][][][][][][][][][][][ 4321654321 TTTTTTTTTTT
=
+
=
+ 11s
0110s1
][][][11s01
][][][][ 3214321 TTTTTTT
=
+++
11ss1ss
1s01
][][2
21 TT
=
++++++
1s1s2ss
s1ss10s1
][ 223
2
1T
=
+++++++++
1s1s2sss2s1s2s3ss
1101
223
3234
=][T
++++++++++++
1s2ss2s4s4s2ss2s1s2s3ss
23234
3234
Note that as expected. 1=−CDAB Chapter 19, Solution 75. (a) We convert [za] and [zb] to T-parameters. For Na, 162440 =−=∆ z .
=
∆=
25.125.042
z/zz/1z/z/z
]T[212221
21z2111a
For Nb, . 88880y =+=∆
−−−−
=
−∆−−−
=4445.05
y/yy/y/1y/y
]T[211121y
212122b
−−−−
==125.525.5617186
]T][T[]T[ ba
We convert this to y-parameters. .3BCADT −=−=∆
−=
−
∆−=
94.100588.01765.03015.0
B/AB/1B/B/D
]y[ T
(b) The equivalent z-parameters are
−
=
∆=
0911.00178.00533.03067.3
C/DC/1C/C/A
]z[ T
Consider the equivalent circuit below. I1 z11 z22 I2 + + + + ZL Vi z12 I2 z21 I1 Vo - - - -
212111i IzIzV += (1)
222121o IzIzV += (2)
But (3) Lo2L2o Z/VIZIV −=→−= From (2) and (3) ,
+=→−=
21L
22
21o1
L
o22121o zZ
zz1VI
ZVzIzV (4)
Substituting (3) and (4) into (1) gives
0051.0VV3.194
Zz
Zzzz
zz
VV
i
.o
L
12
L21
2211
21
11
o
i −=→−=−
+=
Chapter 19, Solution 76. To get z11 and z21, we open circuit the output port and let I1 = 1A so that
21
2211
1
111 V
IVz,V
IVz ====
The schematic is shown below. After it is saved and run, we obtain
122.1Vz,849.3Vz 221111 ==== Similarly, to get z22 and z12, we open circuit the input port and let I2 = 1A so that
22
2221
2
112 V
IVz,V
IVz ====
The schematic is shown below. After it is saved and run, we obtain
849.3Vz,122.1Vz 222112 ==== Thus,
Ω
=
849.3122.1122.1949.3
]z[
Chapter 19, Solution 77. We follow Example 19.15 except that this is an AC circuit. (a) We set V2 = 0 and I1 = 1 A. The schematic is shown below. In the AC Sweep Box, set Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592. After simulation, the output file includes
FREQ IM(V_PRINT2) IP(V_PRINT2) 1.592 E–01 3.163 E–.01 –1.616 E+02 FREQ VM($N_0001) VP($N_0001) 1.592 E–01 9.488 E–01 –1.616 E+02
From this we obtain
h11 = V1/1 = 0.9488∠–161.6°
h21 = I2/1 = 0.3163∠–161.6°.
(b) In this case, we set I1 = 0 and V2 = 1V. The schematic is shown below. In the AC Sweep box, we set Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592. After simulation, we obtain an output file which includes
FREQ VM($N_0001) VP($N_0001) 1.592 E–01 3.163 E–.01 1.842 E+01 FREQ IM(V_PRINT2) IP(V_PRINT2) 1.592 E–01 9.488 E–01 –1.616 E+02
From this,
h12 = V1/1 = 0.3163∠18.42°
h21 = I2/1 = 0.9488∠–161.6°.
Thus, [h] =
°−∠°−∠°∠°−∠6.1619488.06.1613163.0
42.183163.06.1619488.0
Chapter 19, Solution 78 For h11 and h21, short-circuit the output port and let I1 = 1A. 6366.02/ == πωf . The schematic is shown below. When it is saved and run, the output file contains the following: FREQ IM(V_PRINT1)IP(V_PRINT1) 6.366E-01 1.202E+00 1.463E+02 FREQ VM($N_0003) VP($N_0003) 6.366E-01 3.771E+00 -1.350E+02 From the output file, we obtain
o1
o2 135771.3V,3.146202.1I −∠=∠=
so that o2
21o1
11 3.146202.11Ih,135771.3
1Vh ∠==−∠==
For h12 and h22, open-circuit the input port and let V2 = 1V. The schematic is shown below. When it is saved and run, the output file includes: FREQ VM($N_0003) VP($N_0003) 6.366E-01 1.202E+00 -3.369E+01 FREQ IM(V_PRINT1)IP(V_PRINT1) 6.366E-01 3.727E-01 -1.534E+02 From the output file, we obtain
o1
o2 69.33202.1V,4.1533727.0I −∠=−∠=
so that o2
22o1
12 4.1533727.01Ih,69.33202.1
1Vh −∠==−∠==
Thus,
−∠∠−∠−∠= o
oo
4.1533727.03.146202.169.33202.1135771.3]h[
Chapter 19, Solution 79 We follow Example 19.16. (a) We set I1 = 1 A and open-circuit the output-port so that I2 = 0. The schematic is shown below with two VPRINT1s to measure V1 and V2. In the AC Sweep box, we enter Total Pts = 1, Start Freq = 0.3183, and End Freq = 0.3183. After simulation, the output file includes
FREQ VM(1) VP(1) 3.183 E–01 4.669 E+00 –1.367 E+02 FREQ VM(4) VP(4) 3.183 E–01 2.530 E+00 –1.084 E+02
From this, z11 = V1/I1 = 4.669∠–136.7°/1 = 4.669∠–136.7° z21 = V2/I1 = 2.53∠–108.4°/1 = 2.53∠–108.4°.
(b) In this case, we let I2 = 1 A and open-circuit the input port. The schematic is shown below. In the AC Sweep box, we type Total Pts = 1, Start Freq = 0.3183, and End Freq = 0.3183. After simulation, the output file includes
FREQ VM(1) VP(1) 3.183 E–01 2.530 E+00 –1.084 E+02 FREQ VM(2) VP(2) 3.183 E–01 1.789 E+00 –1.534 E+02
From this, z12 = V1/I2 = 2.53∠–108.4°/1 = 2.53∠–108..4° z22 = V2/I2 = 1.789∠–153.4°/1 = 1.789∠–153.4°.
Thus,
[z] = 2
°−∠°−∠°−∠°−∠
4.153789.14.10853.4.10853.27.136669.4
Chapter 19, Solution 80 To get z11 and z21, we open circuit the output port and let I1 = 1A so that
21
2211
1
111 V
IVz,V
IVz ====
The schematic is shown below. After it is saved and run, we obtain
37.70Vz,88.29Vz 221111 −==== Similarly, to get z22 and z12, we open circuit the input port and let I2 = 1A so that
22
2221
2
112 V
IVz,V
IVz ====
The schematic is shown below. After it is saved and run, we obtain
11.11Vz,704.3Vz 222112 ==== Thus,
Ω
−
=11.1137.70
704.388.29]z[
Chapter 19, Solution 81 (a) We set V1 = 1 and short circuit the output port. The schematic is shown below. After simulation we obtain
y11 = I1 = 1.5, y21 = I2 = 3.5
(b) We set V2 = 1 and short-circuit the input port. The schematic is shown below. Upon simulating the circuit, we obtain
y12 = I1 = –0.5, y22 = I2 = 1.5
[Y] =
−5.15.35.05.1
Chapter 19, Solution 82 We follow Example 19.15. (a) Set V2 = 0 and I1 = 1A. The schematic is shown below. After simulation, we obtain
h11 = V1/1 = 3.8, h21 = I2/1 = 3.6
(b) Set V1 = 1 V and I1 = 0. The schematic is shown below. After simulation, we obtain
h12 = V1/1 = 0.4, h22 = I2/1 = 0.25
Hence, [h] =
25.06.34.08.3
Chapter 19, Solution 83
To get A and C, we open-circuit the output and let I1 = 1A. The schematic is shown below. When the circuit is saved and simulated, we obtain V1 = 11 and V2 = 34.
02941.0341
VIC,3235.0
VVA
2
1
2
1 =====
Similarly, to get B and D, we open-circuit the output and let I1 = 1A. The schematic is shown below. When the circuit is saved and simulated, we obtain V1 = 2.5 and I2 = -2.125.
4706.0125.21
IID,1765.1
125.25.2
IVB
2
1
2
1 ==−===−=
Thus,
=
4706.002941.01765.13235.0
]T[
Chapter 19, Solution 84
(a) Since A = 0I2
1
2VV
=
and C = 0I2
1
2VI
=
, we open-circuit the output port and let V1
= 1 V. The schematic is as shown below. After simulation, we obtain A = 1/V2 = 1/0.7143 = 1.4 C = I2/V2 = 1.0/0.7143 = 1.4
(b) To get B and D, we short-circuit the output port and let V1 = 1. The schematic is shown below. After simulating the circuit, we obtain B = –V1/I2 = –1/1.25 = –0.8
8
Thus =
D = –I1/I2 = –2.25/1.25 = –1.
DCBA
−−
8.14.18.04.1
lution 85
(a) Since A =
Chapter 19, So
0I2
1
2VV
=
and C = 0I2
1
2VI
=
, we let V1 = 1 V and open-
circuit the output port. The schematic is shown below. In the AC Sweep box, we set n
FREQ IM(V_PRINT1) IP(V_PRINT1) 01
REQ VM($N_0002) VP($N_0002)
From this, we obtain
A
Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592. After simulation, we obtaian output file which includes
1.592 E– 6.325 E–01 1.843 E+01 F1.592 E–01 6.325 E–01 –7.159 E+01
= °∠=°−∠
= 59.71581.159.716325.0
11 V2
C = °∠=°−∠°∠
= 90159.716325.0
43.186325.0I1
V2
= j
(b) Similarly, since B =
0V2
1
2IV
= 0V2
1
2II
=
− , we let V1 = 1 V and short-
circuit the output port. The schematic is shown below. Again, we set Total Pts = 1, Start
FREQ IM(V_PRINT1) IP(V_PRINT1) 01
REQ IM(V_PRINT3) IP(V_PRINT3) 01
From this,
B
Freq = 0.1592, and End Freq = 0.1592 in the AC Sweep box. After simulation, we get an output file which includes the following results:
and D =
1.592 E– 5.661 E–04 8.997 E+01 F1.592 E– 9.997 E–01 –9.003 E+01
= j901909997.0
11
2
−=°∠−=°−∠
−=− I
D =°−∠°∠
−=−−
909997.097.8910x661.5I 4
2
1
I = 5.561x10–4
=
DCBA
−°∠−410x661.5j
j59.71581.1
Chapter 19, Solution 86
(a) By definition, g11 = 0I1
1
2VI
=
, g21 = 0I2
1
2VV
=
.
We let V1 = 1 V and open-circuit the output port. The schematic is shown below. After simulation, we obtain g11 = I1 = 2.7 g21 = V2 = 0.0
(b) Similarly,
g12 = I
, g22 = 0V2
1
1
I
= 0V2
2
1IV
=
g12 = I1 = 0
We let I2 = 1 A and short-circuit the input port. The schematic is shown below. After simulation,
g22 = V2 = 0
Thus [g] =
000S727.2
hapter 19, Solution 87
a =
C
(a) Since 0I1
2
1VV
=
and c = 0I1
2
1VI
=
,
we open-circuit the input port and let V2 = 1 V. The schematic is shown below. In the C Sweep box, set Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592. After
IP(V_PRINT2) 1.592 E–01 5.000 E–01 1.800 E+02
1) ) .592 E–01 5.664 E–04 8.997 E+01
From this,
Asimulation, we obtain an output file which includes
FREQ IM(V_PRINT2)
FREQ VM($N_000 VP($N_00011
a = °= 97.891 −∠°∠− 1765
97.8910x664.5 4
c = °−∠−=°∠
°∠− 97.8928.882
97.8910x664.51805.04
(b) Similarly,
b = 0V1
2
1I
=
V− and d =
0V1
2
1II
=
−
We short-circuit the input port and let V2 = 1 V. The schematic is shown below. After simulation, we obtain an output file which includes
FREQ IM(V_PRINT2) IP(V_PRINT2) 1.592 E–01 5.000 E–01 1.800 E+02
FREQ IM(V_PRINT3) IP(V_PRINT3) 1.592 E–01 5.664 E–04 –9.010 E+01
From this, we get
b = °−∠
− = –j1765 − 1.9010x664.514
d = °−∠
°∠− = j888.28 − 1.9010x664.5
1805.04
Thus [t] =
−−2.888j2.888j
1765j1765j
Chapter 19, Solution 88 To get Z , consider the network in Fig. (a). in
I2I1
+ −
(a)Zin
+
V1
−
Vs Two-Port+
V2
−
RL
Rs
2121111 VyVyI += (1)
2221212 VyVyI += (2)
But 222121L
22 R
-VyVy
VI +==
L22
1212 R1
-+
=y
VyV (3)
Substituting (3) into (1) yields
+⋅+=
L22
121121111 R1
-y
VyyVyI ,
LL R
1=Y
1L22
L11y1 V
YyYy
I
+
+∆= , 21122211y yyyy −=∆
or ==1
1inZ
IV
L11y
L22
YyYy
+∆+
+
+=
+==
L22
121
1
22in21
1
222121
1
2i
-ZA
YyVy
Iyy
IVyVy
II
+
−
+∆+
=+
−L22
212221
L11y
L22
L22
in2122in21
ZZ
Yyyyy
YyYy
Yyyyy=
=iAL11y
L21
YyYy
+∆
From (3),
==1
2vA
VV
L22
21
Yyy-+
To get Z , consider the circuit in Fig. (b). out
Zout
I2I1
(b)
+ −
+
V1
−
Rs Two-Port V2
222121
2
2
2outZ
VyVyV
IV
+== (4)
But 1s1 R- IV = Substituting this into (1) yields
2121s111 R- VyIyI +=
2121s11 )R1( VyIy =+
s
1
s11
2121 R
-R1
Vy
VyI =
+=
or s11
s12
2
1
R1R-
yy
VV
+=
Substituting this into (4) gives
s11
s211222
out
R1R
1Z
yyy
y+
−=
s2221s221122
s11
RRR1
yyyyyy
−++
=
=outZs22y
s11
YyYy
+∆+
Chapter 19, Solution 89
Lfereoeieie
Lfev R)hhhh(h
Rh-A
−+=
54-6-
5
v 10)72106.210162640(26401072-
A⋅××−××+
⋅=
=+⋅
=182426401072-
A5
v 1613-
=== )1613(log20Alog20gain dc v 15.64
Chapter 19, Solution 90
(a) Loe
Lfereiein Rh1
RhhhZ
+−=
L6-
L-4
R10201R12010
20001500×+×
−=
L5-
-3
R10211012
500×+×
=
L
-3L
-2 R1012R10500 ×=+
L2 R2.010500 =×
=LR Ωk250
(b) Lfereoeieie
Lfev R)hhhh(h
Rh-A
−+=
34-6-
3
v 10250)1012010202000(200010250120-
A×××−××+
××=
=×+×
×= 33
6
v 1071021030-
A 3333-
=×××+
=+
= 36-Loe
fei 1025010201
120Rh1
hA 20
120101020)2000600(2000600
hhh)hR(hR
Z 4-6-fereoeies
iesout ×−××+
+=
−++
=
=Ω= k40
2600Zout Ωk65
(c) =××==→== 3-svc
s
c
b
cv 104-3333AA VV
VV
VV
V13.33-
Chapter 19, Solution 91 Ω= k2.1R s , Ω= k4R L
(a) Lfereoeieie
Lfev R)hhhh(h
Rh-A
−+=
34-6-
3
v 104)80105.110201200(120010480-
A××××−××+
××=
==124832000-
A v 25.64-
(b) =×××+
=+
= 36-Loe
fei 10410201
80Rh1
hA 074.74
(c) ireiein AhhZ −=
≅××−= 074.74105.11200Z -4in Ωk2.1
(d) fereoeies
iesout hhh)hR(
hRZ
−++
=
==××−××
+=
0468.02400
80105.11020240012001200
Z 4-6-out Ωk282.51
Chapter 19, Solution 92 Due to the resistor , we cannot use the formulas in section 18.9.1. We will need to derive our own. Consider the circuit in Fig. (a).
Ω= 240R E
hoe hfe Ib
hie
+ −
hre Vc
Zin
IE
+ −
Ib Ic
+
Vc
−RE
(a)
+
Vb
−
Rs
RL Vs
cbE III += (1)
Ecbcrebieb R)(hh IIVIV +++= (2)
oeE
cbfec
h1R
h+
+=VII (3)
But (4) Lcc R-IV = Substituting (4) into (3),
c
oeE
Lbfec
h1R
Rh III
+−=
or Loe
oeEfe
b
ci R(h1
)hR1(hA
++
==II
(5)
)240000,4(10301)10x30x2401(100A 6-
6
i +×++
=−
=iA 79.18
From (3) and (5),
oeE
cbfeb
ELoe
oeEfec
h1R
h)RR(h1
h)R1(h+
+=++
+=
VIII (6)
Substituting (4) and (6) into (2),
EccrebEieb Rh)Rh( IVIV +++=
EL
ccre
feELoe
oeEfe
oeE
Eiecb R
Rh
h)RR(h1
)hR1(hh1R
)Rh( VVVV −+
−
+++
+
+=
L
Ere
feELoe
oeEfe
oeE
Eie
c
b
v RRh
h)RR(h1
)hR1(hh1R
)Rh(A1
−+
−
+++
+
+==
VV
(7)
400024010
100424010301
)10x30x2401(10010x301240
)2404000(A1 4-
6-
6
6v
−+
−
××++
+
+=
−
−
-0.06606.01010x06.6A1 4-3
v
=−+−= −
=vA –15.15
From (5),
bLoe
fec Rh1
hII
+=
We substitute this with (4) into (2) to get
cLreEbEieb )RhR()Rh( IIV −++=
++
+−++= b
ELoe
oeEfeLreEbEieb )RR(h1
)hR1(h)RhR()Rh( IIV
)RR(h1)hR1)(RhR(h
RhZELoe
oeELreEfeEie
b
bin ++
+−++==
IV
(8)
424010301)10x30x2401)(10410240)(100(2404000Z 6-
63-4
in ××++×××
++=−
=inZ 12.818 kΩ
To obtain , which is the same as the Thevenin impedance at the output, we introduce a 1-V source as shown in Fig. (b).
outZ
hie Rs IcIb
(b)
+
Vb
−RE
+
Vc
−
+ −
IE
Zout
hre Vc
+ −
hfe Ib hoe
1 V
From the input loop, 0)(Rh)hR( cbEcreiesb =++++ IIVI
But 1c =VSo,
0Rh)RhR( cEreEiesb =++++ II (9) From the output loop,
bfeoeE
oebfe
oeE
cc h
1hRh
h
h1R
IIVI ++
=++
=
or oeE
fe
oe
fe
cb hR1
hh
h +−=
II (10)
Substituting (10) into (9) gives
0hR1
hh)hRR(
Rhh
)hRR(oeE
fe
oeieEs
cErefe
cieEs =
+
++
−++
++ II
refe
oe
oeE
ieEscEc
fe
ieEs hhh
hR1hRR
Rh
hRR−
+
++=+
++ II
feieEsE
reoeE
ieEsfeoe
c h)hRR(R
hhR1
hRR)hh(
+++
−
+
++
=I
fereoeoeE
ieEs
ieEsfeE
cout
hhhhR1
hRRhRRhR1Z
−
+
+++++
==I
10010103010x30x240140002401200
)40002401200(100240Z4-6-
6
out
×−××
+++
+++×=
−
=+
=152.0
544024000Zout 193.7 kΩ
Chapter 19, Solution 93
We apply the same formulas derived in the previous problem.
L
Ere
feELoe
oeEfe
oeE
Eie
v RRh
h)RR(h1
)hR1(hh1R
)Rh(A1
−+
−
+++
+
+=
3800200105.2
15004.01
)002.01(150)10200(
)2002000(A1 4-
5v
−×+
−
++
+
+=
-0.0563805263.0105.2004.0A1 4-
v
=−×+−=
=vA –17.74
=+×+
+=
+++
=−
)3800200(101)10x2001(150
)RR(h1)hR1(h
A 5-
5
ELoe
oeEfei 5.144
)RR(h1)hR1)(RhR(h
RhZELoe
oeELreEfeEiein ++
+−++=
04.1)002.1)(108.3105.2200)(150(2002000Z
3-4
in×××−
++=
289662200Zin +=
=inZ 31.17 kΩ
fereoeoeE
ieEs
ieEsfeEout
hhhhR1
hRRhRRhR
Z−
+
+++++
=
0.0055-33200
150105.2002.1
10320020002001000150200Z
4-5-out =
××−
×+++×
=
=outZ –6.148 MΩ
Chapter 19, Solution 94 We first obtain the ABCD parameters.
Given ,
= 6-10100
0200][h -4
21122211h 102×=−=∆ hhhh
×=
∆
= 2-8-
6-
2121
22
21
11
21
h
10-10-2-102-
1--
-
][
hhh
hh
hT
The overall ABCD parameters for the amplifier are
××≅
×
×= 4-10-
-2-8
2-8-
-6
2-8-
-6
1010102102
10-10-2-102-
10-10-2-102-
][T
0102102 -12-12
T =×−×=∆
=
∆
= 6-4-
T
1010-0200
1-][
DC
D
DDB
h
Thus, , h200h ie = 0re = , h , h -4fe -10= -6
oe 10=
=××−×+
×= 34-
34
v 104)0102(200)104)(10(
A 5102×
=−=+
−= 0200Rh1Rhh
hZLoe
Lfereiein Ω200
Chapter 19, Solution 95
Let s5s
8s10s13
24
22A +
++==
yZ
Using long division,
B13
2
A Lss5s8s5
s ZZ +=++
+=
i.e. and H1L1 = s5s8s5
3
2
B ++
=Z
as shown in Fig (a).
y22 = 1/ZA
L1
ZB
(a)
8s5s5s1
2
3
BB +
+==
ZY
Using long division,
C22B sC8s5
s4.3s2.0 YY +=
++=
where and F2.0C2 = 8s5s4.3
2C +=Y
as shown in Fig. (b).
43
2
CC Cs
1Ls
s4.38
4.3s5
s4.38s51
+=+=+
==Y
Z
i.e. an inductor in series with a capacitor
H471.14.3
5L3 == and F425.0
84.3
C4 ==
Thus, the LC network is shown in Fig. (c).
Chapter 19, Solution 96 This is a fourth order network which can be realized with the network shown in Fig. (a).
1.471 H0.425 F
L1
1 Ω
L3
C4C2
(c)
0.2 F
1 H
C2
(b)
Yc = 1/ZC
L1
(a))s613.2s613.2()1s414.3s()s( 324 ++++=∆
s613.2s613.21s414.3s
1
s613.2s613.21
)s(H
3
24
3
+++
+
+=
which indicates that
s613.22.613s1-
321 +=y
s613.2s613.21s414.3s
3
4
22 +++
=y
We seek to realize . By long division, 22y
A43
2
22 Css613.2s613.2
1s414.2s383.0 Yy +=
++
+=
i.e. and F383.0C4 = s613.2s613.21s414.2
3
2
A ++
=Y
as shown in Fig. (b).
L1
C4C2
L3
YA
y22(b)
1s414.2s613.2s613.21
2
3
AA +
+==
YZ
By long division,
B32A Ls1s414.2
s531.1s082.1 ZZ +=
++=
i.e. and H082.1L3 = 1s414.2s531.1
2B +=Z
as shown in Fig.(c).
L1
C4C2
L3
ZB
(c)
12
BB Ls
1Cs
s531.11
s577.11
+=+==Z
Y
i.e. and F577.1C2 = H531.1L1 = Thus, the network is shown in Fig. (d).
1.531 H 1.082 H
1.577 F 0.383 F 1 Ω
(d) Chapter 19, Solution 97
s12s24s6
1
s12ss
)24s6()s12s(s
)s(H
3
2
3
3
23
3
++
+
+=+++
=
Hence,
A3
3
2
22 Cs1
s12s24s6
Zy +=++
= (1)
where is shown in the figure below. AZ
C1 C3
L2
ZA y22 We now obtain C and using partial fraction expansion. 3 AZ
Let 12sCBs
sA
)12s(s24s6
22
2
++
+=++
CsBs)12s(A24s6 222 +++=+
Equating coefficients : 2 0s : AA1224 =→= : 1s C0 = 2s : 4BBA6 =→+=Thus,
12ss4
s2
)12s(s24s6
22
2
++=
++
(2)
Comparing (1) and (2),
F21
A1
C3 ==
s3
s41
s412s1 2
A+=
+=
Z (3)
But 2
1A Ls
1sC
1+=
Z (4)
Comparing (3) and (4),
F41
C1 = and H31
2 =L
Therefore, =1C F25.0 , =2L H3333.0 , =3C F5.0
Chapter 19, Solution 98
2.08.01h =−=∆
−−−−
=
−−−∆−
== − 005.010x5.210001.0
h/1h/hh/hh/
]T[]T[ 6212122
211121hba
== −−
−
58
5ba
10x510x5.106.010x6.2]T][T[]T[
We now convert this to z-parameters
=
∆= 37
3T
10x33.310x667.60267.010x733.1
C/DC/1C/C/A
]z[
1000 I1 z11 z22 I2 + + + + ZL Vs z12 I2 z21 I1 Vo - - - -
212111s IzI)z1000(V ++= (1)
121222o IzIzV += (2)
But (3) Lo2L2o Z/VIZIV −=→−= Substituting (3) into (2) gives
+=
L21
22
21o1 Zz
zz1VI (4)
We substitute (3) and (4) into (1)
V74410x136.210x653.7
VZzV
Zzz
z1)z1000(V
54
oL
12o
L21
22
1111s
µ=−=
−
++=
−−
Chapter 19, Solution 99
)(|| abc31ab ZZZZZZ +=+=
cba
bac31
)(ZZZ
ZZZZZ
+++
=+ (1)
)(|| cba32cd ZZZZZZ +=+=
cba
cba32
)(ZZZ
ZZZZZ
+++
=+ (2)
)(|| cab21ac ZZZZZZ +=+=
cba
cab21
)(ZZZ
ZZZZZ
+++
=+ (3)
Subtracting (2) from (1),
cba
acb21
)(ZZZ
ZZZZZ
++−
=− (4)
Adding (3) and (4),
=1Zcba
cb
ZZZZZ++
(5)
Subtracting (5) from (3),
=2Zcba
ba
ZZZZZ++
(6)
Subtracting (5) from (1),
=3Zcba
ac
ZZZZZ++
(7)
Using (5) to (7)
2cba
cbacba133221 )(
)(ZZZ
ZZZZZZZZZZZZ
++++
=++
cba
cba133221 ZZZ
ZZZZZZZZZ
++=++ (8)
Dividing (8) by each of (5), (6), and (7),
=aZ1
133221
ZZZZZZZ ++
=bZ3
133221
ZZZZZZZ ++
=cZ2
133221
ZZZZZZZ ++
as required. Note that the formulas above are not exactly the same as those in Chapter 9 because the locations of and are interchanged in Fig. 18.122. bZ cZ