Download - Solution Chap6
Irwin, Basic Engineering Circuit Analysis, 9/E
6.1 An u1charged 100-~-tF cap cit 1' i chru·g Cl by a constant c1.1.rrent of I mA. Hnd tb v.o]tag aero s tb capacitm· after 4 s.
SOLUTION:
V(t) --'
V(-1) --
V(:t) --
V(r)
v
T
I f J_ (f) dt (
()
lf
I J I m cit (00)-J..
0
I ['m(4)-0] (00 _).)
Lfov ()9-{_
Lfov
Chapter 6: Capacitance and Inductance Problem 6.1
Irwin, Basic Engineering Circuit Analysis, 9/E
6.10 Th voltage acmss a 25-!J..F capacitor j hown in Fig. P6JO. Det rmin the cur1'ent w v f nn.
v(t) (v)
Figure P6.1o
SOLUTION:
t 1 - 0 ·2. m~
_±2 = 0 -y 1)105
.13= 0·8-rn.S
-*u =- I 'YYl s. Js =- 1-2rns
•
0 ) /) __ ,._, ,· = 0
) V= ~...-~..r LA \A..
O~.:i<.J-, ) v= ro5 * v
) V=20V J l=-0
) v= 6u- to5 .::tv
Chapter 6: Capacitance and Inductance Problem 6.1 0
2 Irwin, Basic Engineering Circuit Analysis, 9/E
Problem 6.1 0
:k3 ~-± <1-t., , V=O J i:=- 0
.~ v= -12ox ros x
J.C :t) = 0
2·5
0 -2'5 0
2·5
0
2··5A
;t<O
() ~ J < o·2ml
0·.2m.s s :t < o·y ms O·l.frm( ~ :t < a·· c9m s
0· gms. ~ J:: < 1-rr.s
)"rfi.s ~ 1: < 1·2mS
J: 7/ /·2 rns
Chapter 6: Capacitance and Inductance
Irwin, Basic Engineering Circuit Analysis, 9/E
6.11 Th voUage aero sa 2-F capacitor is given by tb wav -onn jn Flg. P6.ll . Fi.nd tb wav tlorm for th cun· nt in
tlle capacit r.
vc{t) v)
- t2
Figure P6.11
SOLUTION:
-t<o ; v=O J .i= 0
0 ( :t < 205 )
J- cdv 2(0·6) ct:1:
J = /·2A
2o s < f < 3o~ J v=6o-2·YfV
J= 2G2·LtJ J..- -::: - Lf· ~ A
3 OS < i < 50-S ) V= - 3o -t 0·6;t V
J=2[o·6]- 1·2A
J 7 sos ) v:::::o )
Chapter 6: Capacitance and Inductance Problem 6.11
2 Irwin, Basic Engineering Circuit Analysis, 9/E
J (.t) - 0 ct <o -1·2A 0 <: J < 'los -y·~A 2o.s <..:t < ~OS
1·2A 3o~ < ;t < :ru.s 0 ~ 7 SDs
Problem 6.11 Chapter 6: Capacitance and Inductance
Irwin, Basic Engineering Circuit Analysis, 9/E
6 .12 The voltage .across a 2-jiiF c:apadt r j given by the waveform in Ag. P6.12. C mput th c11..1uent wavefonn.
v(t v)
2 3 6
- 11.2
Figure P6 .12
SOLUTION:
t<o ) v-o
o < J: <.. 2 ms
2m.s <' x< 3ms
3 ms ( :t < 6 m.s }
t ~ms)
) v : -6 000 ;t v l = C dV
ell
}
J = 2 u ( -6ooo) J-= -12rnA
v= -I2V
._1 = 0
v = - 2Y-t LrooqJ V
. J V=::O 1 vl=o
J( j:) - ()
-12rnA
0 gmP\
0 Chapter 6: Capacitance and Inductance
:t<O 0<1<1Y'Y'S '2m~ ( .1.. < 3rns ~Ms < .t < Gms
t 7 6ms Problem 6.12
Irwin, Basic Engineering Circuit Analysis, 9/E
6.13 Dmw the warv form fort 1e cun· nt in a 24-~J,.F capacitor wh nth capacitor voltage is as described in Fig. P6.13.
v t (v
Figure P6.13
SOLUTION:
-t<o ) J_==O
v= lo~ ;t v
L=c%'f
~ :t < IOOJ..,tJ
u.= (2'-1}...{) (-2·SXro 5)
vl:::. -6A
I oo U S $ .:t < J 6 o u ~ ) V = -5 2 + I 0 6 t V
3 I)
J - 24u [ ',~'] ~ =- /· G A
Chapter 6: Capacitance and Inductance Problem 6.13
2 Irwin, Basic Engineering Circuit Analysis, 9/E
s // 160JJS ) v=o ) i==O
J.{ i) -:::- 0 ;t<O
2.·Y~ 0~ J < 6 o)-JS
-2·L.fA 60J.JS ~ t < IOCJ-.i~
J·6 A /OO).J. S ~ ~ <. J6o)..1S
0 J- //(bOLLS
Problem 6.13 Chapter 6: Capacitance and Inductance
o·o~
....-.... <C '-' ........ 0 ~
.......... -~
-o·o<-f
Irwin, Basic Engineering Circuit Analysis, 9/E
6.14 Th.e vo tag acms a 10-p.F capacit r i give.n by the wav form in Fg. P6.14l. Plot th wav l!onn f 1' the capac· tor Clm' nt.
SOLUTION:
v(t v
Figure P6.14
V(.t) := 12 _2inw.f
w==2TI I
2TL I om
W::: 2 OO)[ 1"'-c:J./s itt) = (\oM) (__\4.) (_w) Los. t.Ot
j__ U.) '; 7 S: • '-1 (_o.t w-t "''V' A
t L-t) vs. t -=:-----------·· · -····----··-·---------~
; ·
\ \
I
\ ~..--
o·OIS \ 0 ·005 0·01 ., \
0'01 \ \
Chapter 6: Capacitance and Inductance Problem 6.14
Irwin, Basic Engineering Circuit Analysis, 9/E
6.15 The warvefonn for the current ]o a 50-~J..F capacitor is howu in Fig. P6.15. D tennine the wavefonn for the
capadtOl' voUag .
i(t) (rnA)
110
0 10 20 30 40 t ( ms)
Figure P6.15
SOLUTION:
t. < o } i{ x) = o J V(:t)-= 0
) 1 ( :t ) :::: 0· 2 5 .t
V(.:t) - d f J.(.t)dJ: -tVa
V{;:) f 0·25 .±d..t
\lc:t) - (~)
± >/ 40mJ ) lC:1) = o
'v(t) - Joct..t ~ 1soo{ Yoh"'J'-
50 ,u.
V( 1.) == YV
Chapter 6: Capacitance and Inductance Problem 6.15
2
V(t )=
Problem 6.15
0 '25DO.t 2 V
l-JV
Irwin, Basic Engineering Circuit Analysis, 9/E
::t<o o ~· .:t < ltam s
.f 4 40tn~
Chapter 6: Capacitance and Inductance
Irwin, Basic Engineering Circuit Analysis, 9/E
6.16 The waveti rm f r th cmrent in a 50-p.F injfally unch rged capac"tor · shown in Hg. P6.16. D termine th wav form or th capacit r 's voltag .
i(t) mA) .
10
0
0
-1 0
Figure P6.16
SOLUTION:
10 20 30 40 50 t (ms)
VcCtl - -t S Jc C:t J cU i V0
l(f) - /Omf\
Vcct)-=_1_ (lomrutv0
50 ).A J Vc(t) - 2oo_t + Vo
J.( 1) -lornA
Vc(t) -SD
vc{U = - 2oo t -+ Vo
~o~ t <o Vc t.t-) = 0
Vc(t) 20o~V
Chapter 6: Capacitance and Inductance Problem 6.16
2 Irwin, Basic Engineering Circuit Analysis, 9/E
'-{-o .?\. I am s ~ :1: ~ 'l o Y'Y'-S
Problem 6.16
Vc Lt) = -200.t t Y V
~o"- 1 oms ~ ..:t S. 3 o fY!.S
Vc (t) - 2 OO;t-- L( V
~ f ~ lforru
- 200 .f: -t g v
4-0h- 4 OyYl~ ·~ J: S,. 5-0(Y).S
vcll) = 1oo~ - ~ v
~91- t 7 5ams \; c c ;t) = ov
0
2 OO.t V
"-!- L- OO;i V -Lf+ 100± v ? - L_oot..V
- ~t L_OotV
0
-t<a O~ f~ lcYn"l~
' /DmJ $ :J $.. '2 OfYJS
20m5 $ 1 £ ~oYY's
3orn5' ~ j { lfc:::.ms
yo ms ~ ):: ~ rom~
roms ~ ± ~ Gums
Chapter 6: Capacitance and Inductance
Irwin, Basic Engineering Circuit Analysis, 9/E
6.17 The waveform for the cmrent flowing through th ! 0-J.l.F capaci t r in Fig. P6. l7a is bown in Fig. P6.17b. If Vc (t = 0) = 1 V, d tem1ine vc(t) at t = 1 ms, rns , 4 rns, and 5 rns.
SOLUTION:
i t) ( mA)
15 +------.....,
3 5 + I I I
I I I
10~F 1 2 4
- 10
(a) Cb)
Figure P6.17
VcCo ) lv -);
Vc r :t) - Vc (to J -t ~ f .i(r)d:t c tb
hn
Vc C o) -t 7 ) (lr;m) ctJ: 0 I vn
I"-+ J ( t5m)clt I OJ-A
0
Vc ( lrns) -
Vc ( I rn s ) = ~ +- I '5o o ( 1 m )
Vc .( lms) - 2 ·5 V
I t (m ) I
6
Chapter 6: Capacitance and Inductance Problem 6.17
2
Problem 6.17
Irwin, Basic Engineering Circuit Analysis, 9/E
t
Vc(Jrns) = Vc. ( lms) -t- _.l_ f 15 md.:t lo.u
lrn t
VcC3"'rllS) =-2·5 -ttooXIo3 (1S'XIa3) CtJ] lro
Vc C 1 ms) -= 2 · 5 + t 5bo [ t - I m J
Vc. ( 3 Y'r1 ~) - 5 · 5 V J:.
Vc ( C{ms) = \Jc C sm) +_I_ f -1 o 1"nd..:t I 0 )..J.
)n"
J ~ Vc ( lf me; ) =- 5 · 5" X ( o o )( I 0 (-16 X I o)) [ ..t ]
3 vn
VcCLtrn~)= 5·5- looo[.:t-3m]
Vc ( '-1ms) ::: 5·5- Joo o [ Ym- 3W~) ~
;t
Vc. (5ms) = Vc(~m) "t_/ J -lomcU lop_
C,m .x Vc C 5 ~5) == Lt · 5 + 1 oo x , o3 ( -1 ox 1 c?) [ _t J
1-ft"n
Chapter 6: Capacitance and Inductance
Irwin, Basic Engineering Circuit Analysis, 9/E 3
Vc C 5ms) L-r·5 -tooo [.x-LrmJ Y·5- tooo [5m- Ltm]
Chapter 6: Capacitance and Inductance Problem 6.17
Irwin, Basic Engineering Circuit Analysis, 9/E
6.18 The waveform for the voltage aero 1 00-~ capacitor shown Fig. 6. l8a is given in Fig. 6.18b. Detemline the following quantities: (a) the energy stored in the capacitor at t = 2.5 ms, (b) the energy stored in the capacitor at t = 5.5 ms, (c) ic(t) at t = 1.5 m , (d) ic(t ) at t = 4.75 ms, and (e) ic( t ) at t = 7.5 ms.
v (t) 100 11F 0~(1)
(a)
Figure P6.t8
SOLUTION:
v(t) (V)
(b)
c foour=-
(O-J UJ( t) =- _, c v1-c t) 2.
-- I
_L ( Joo.u) C 15 fl-2
(b) W(t) - ( V2 [t) 2..
LJJ ( 5"5niS)
we s·s )'y)s)
Chapter 6: Capacitance and Inductance
_)_ ( 100.4) (- 5 }L 2..
1·2.o mJ
Problem 6.18
2
Problem 6.18
(C) vc~)
/YI::: 15-5 "LX I 0- 3 - I X I 0-]
)'Y) = 10 J()-00
V(-J:)= /OOOO.t f g
lo = 5~( lm)+ g 13= -~
Irwin, Basic Engineering Circuit Analysis, 9/E
\J(;t) -= IDOoot - 5 V '1o9-t ~ ,·~:tcrvoJ.. et irt+e~t
V( t} =- I 0/)00 X t. - 5 VoJ.::t!
i, (i) -=. c ~(t) Jc (t )= fOOD., [loooo]
~c {lj = I A
( cLJ VC :t) =- -5 V '{ott -Hvz. 1 hfe1 voJ. ef-lnte~:t
= C (i.v Ctl eXt-
Chapter 6: Capacitance and Inductance
Irwin, Basic Engineering Circuit Analysis, 9/E 3
icC:t} = oA
s-- 0 -=moo
Vc.t J = 5ooo :t + B
5 ==- 50oo( ~ rn) t B B= -35
vc f) ~ 5 ooo.f - 35 V ~o7c ~ , nf-fovuJ
Ef ,ntc>-~t
ic C;t)
~( c .t)
( C!VC.t) clt
I oo 1J [ Sooo]
lc ( -=/·5 ms) = o·s A
Chapter 6: Capacitance and Inductance Problem 6. 18
I
Irwin, Basic Engineering Circuit Analysis, 9/E
6.19 If vc(t = 2 s) = 10 V in the ci.rcu.it in Fig. P6.I9, find the n:ergy tore in the cap.adtor and th pow r uppli. d
by til s me at t = 6 s.
3fl: 60.
Figure P6.19
SOLUTION:
6
V(J h) - L s 2. oLt i I 0
)._..
2-6V
Chapter 6: Capacitance and Inductance
1
Problem 6.19
2
Problem 6.19
Irwin, Basic Engineering Circuit Analysis, 9/E
We u,J ::: fC [ "c f;hJ J 2
WcC .t1J =- f ( -t) [ 2(;] 2
Wc(1 1 ) = /bq]
VR ( ct1) = J(-tt) [~5 ~)J -L [ ~f J ·VR (1L) = ~V
Vs C tl) = Vc c tl) t vR.. C ;tL)
\Js ( .t 2- ) = 2_ ~-+ Y
v ~ c 1-'l) ::: 3 0 v
P5
C:t2-) = \JsC:tl-) is C.t~.._)
p ~ ( ;t") = ~ 0 c 2)
?s(-1,_)=- ~ow
Chapter 6: Capacitance and Inductance
Irwin, Basic Engineering Circuit Analysis, 9/E
6.2 A I2-jiiP capacitor has an accl n1Uiat d charg of 480 ~J~.C . Determin tb voltage acmss the capac"tor.
SOLUTION:
c Q
v
v= en_ c
v= L( ~0~_
I 2 JJ
V= lfov
Chapter 6: Capacitance and Inductance Problem 6.2
Irwin, Basic Engineering Circuit Analysis, 9/E
6.20 Th cun nt in au inductor chnngeCI fmm 0 to 200 mAin 4tns and induce a voltage of 100 mV. What is the value of the inductor?
SOLUTION:
Vc:tJ L cU.' c .t) Ckt
V==L AT
n..t
[J.t -
L=
L=
L==
Chapter 6: Capacitance and Inductance
I oom ( Lfm -) 2oom
2.rnH
Problem 6.20
Irwin, Basic Engineering Circuit Analysis, 9/E
6.21 Th curr nt in a 100'-mH inductor is i( t) = 2 sin 377t A. Rnd (a) tb voUage acm s tb indllctor and (b) th ex:p.ress 'on fm· th . enet·gy sl:o1' dl io the ]em nt
SOLUTION:
(Q) V{t) = L di c J) d..t .
V( t ) -:: 0 'I ( l ) ( ] 7 7 ) COS 3 ll ±
Vet> -
(b) w( :t) L . 2(.1-) k'
X_ (O· I) [ 2 ~in 3 ll.* j 2-
wt
Chapter 6: Capacitance and Inductance Problem 6.21
Irwin, Basic Engineering Circuit Analysis, 9/E
6.22 If the Clll'r ilt i(t) = 1l.5! A flows th.rough a 2-H inductor, fi.nd th energy stored at t = 2s.
SOLUTION:
wt 2
LuC 2) = iC l) (_ t· ~ ( 1.)) 2-
'2
C~apter 6: Capacitance and Inductance Problem 6.22
Irwin, Basic Engineering Circuit Analysis, 9/E
6.23 Tihe current in a 25 ~mH inducto1~ is gi.ven by the expres i.on
i(t)=O t < O
i(t) = 10{1 - e- 1t) rnA t > 0
Hnd (a) the vo]tag acm the inductor and b) the XJUe sion f 1' the e:r~:ergy tor d in "t
~~/): L(t) - IO(l-e-t) SOLUTION:
(Q) V(J..) L di CtJ eli
[,~e-* t<O
)
) k /0
V(:t J ~ 0 ) j: < 0 -..t
JJV I f_ /0 25D€
\b) wCtJ _L Li 1 (:t) 2
w(:t) - L o5 , t <o -~ 1. t 1·25 0-e ) IJ'J" ) 70
Chapter 6: Capacitance and Inductance Problem 6.23
Irwin, Basic Engineering Circuit Analysis , 9/E
6.24 Giv 11 d1e data in the previ.on pr blem, wd the voltage acros til indJJct rand the n rgy tored ·nit after 1 s.
SOLUTION:
V ( t) ) :t 70
V( I)
w( f)
W(l)
Chapter 6: Capacitance and Inductance
-1:) )_ 1·25 ( 1--€'
o ·5 IJ-I
UJ , £70
Problem 6.24
Irwin, Basic Engineering Circuit Analysis, 9/E
6.2.5 Th curr 11t in -o-mH i11duct r is specified a foUows .
i(t) = 0 t < 0
i(t) = 2te - 41 A t > 0
F".nd (a) the V· ao acm induct r, ( ) the thne t which th current i.s a maximum, and (c) d1 tim at which th volt ge i a minimum.
SOLUTION:
i.(t) -=-oA , -4tA
i(l)-= 2.te
t <o J t 70
\j L (t ) := CJ V );t<O
(b) r Ju t1 ~ wh.o.n i( 1:)
~UJ"(J.. ~
d; c 1) 0
oil
Chapter 6: Capacitance and Inductance
) -;t 70 ~
Problem 6.25
-
Irwin, Basic Engineering Circuit Analysis, 9/E
6.26 The vohag ac a 2-H inductor i g:iv 11 by th wave-f rm hown in Fig. P6.26. Find tb wav form for the ct tTent in the ]mluct r.
v t v
Figure P6.26
SOLUTION:
i (t) := _I_J V(:f) ill L
V(.t) V-1 a. ~on.t o~, eac.J.1 -tr~ {pan, otncl :tlv- c~ con btL. ex_p~ed.. cu _.-
i{_t) =- jL * -t Io L
t <o
0 ::{_ t s:: 2.5
J(t-)=2-t 2._·5t- A 2
i C ;t) = Of 2 · 5 ( 1)
Chapter 6: Capacitance and Inductance Problem 6.26
S
S
S
S
ss
Irwin, Basic Engineering Circuit Analysis, 9/E
6.27 Th voltage across a 4-H induct r i giv n by the waveform h wn in Fig. P6.27. Find the wavef rm forth cun ut in the induct r. v(t) = 0, t < 0.
v t
0 10 20 30 40 50 t(ms)
Figure P6.2]
SOLUTION:
i.{_i) = + J \1 ( J::) cU:
V(t) ~ WN;l:Qr\t Qc.J>-OSS ~ f:IY)"..Q, -~~Q~, 0 ret i._ ( ±-) Celt"~ brt w~n 0:
l( i) == :y_t + 'Io L
bo~ -J: < 0 ) l ( t) = o
i (t) = 2·Lfm tY
L{ t) - Goot JJ-A
ton roms s f ~ 1oms
J(t) :::: o-f(_~ o o)( fOm)
i(:t) =
Chapter 6: Capacitance and Inductance Problem 6.27
t
2
Problem 6.27
Irwin, Basic Engineering Circuit Analysis, 9/E
- b -= 6 oo :l - 6 M f\
J(t) = 12-)J.A
vi( t) = - 12-f6oot )).A
~O)L. t ?/ 5Drf\~
vt(.t) = fg ).JA
i(~) - 0 600t J)A
b)-(1\
- 6 f6oot .UA 12M A
_,'L-+ 6 ool )JA
ltMA
t <a 0$ -J L... /0YY>'5.
1 o mJ ~ .t ~ 20'fY1J
20'MS ~ t ~ ~~J
] o m s .$ :): ~ l.f Oyy-.J'
L( on1 r- S f ~ r-orrd'
r 7/ 5bms
Chapter 6: Capacitance and Inductance
Irwin, Basic Engineering Circuit Analysis, 9/E
6.28 The vo tag aero a 10-mH ]nductor · sh w.n in Fg. P6.28. D renuine th wavelonn ~ r the inducto Cli.UTent
v(t) (mv
10-
0 11
Fl gu re P6 .28
SOLUTION:
._jl (;t)
Chapter 6: Capacitance and Inductance
.2 t (ms)
IO,k"Yn V
t s VL!t)c{i;
/00 J {Oj Oi
500 1 1 "m A
1 oo Jt-Jot t 2o) d..:t
-'-5o at L t '20°0t m A
Problem 6.28
2 Irwin, Basic Engineering Circuit Analysis, 9/E
t ( l'fr1S)
Problem 6.28 Chapter 6: Capacitance and Inductance
Irwin, Basic Engineering Circuit Analysis, 9/E
6.29 The current in a 10-mH inductor is shown in Flg. P6.29. D tenuin.e th wav form for the voUag aero t 1
inductm·.
i(t) ~mA) .
012 3 4 56
Figure P6.29
SOLUTION:
'\1'-+-) _ L di.Ctl cU:-
) V(t)= 0
V(l)
0 -somv
6omV ()
Chapter 6: Capacitance and Inductance
-3omV
6omv
:t < 0
0( ~~Lime:;.
~ Y'(\..5 <. t ~ 6YY" 5
;t 7 6ms
Problem 6.29
Irwin, Basic Engineering Circuit Analysis, 9/E
6.3 A capaci.tor h.as an accumulated charge of 600 ~C with 5 V acros H. What is th vah] of capacUanc ?
SOLUTION:
c (Q_
v
c= 6 QO J-1. 5
c -· 110 uF
Chapter 6: Capacitance and Inductance Problem 6.3
Irwin, Basic Engineering Circuit Analysis, 9/E
6.30 Th cnl'ent in a 50-mH induct r · given in Fig. P6.30. S!<;, lch the indu.ct r vol.tage.
i(t (mA
1 00 - - - - - - - - - - - - - - -
0
-1 00
Figure P6.3o
SOLUTION:
J-0"-'
~Ofv
V(.J) = L di (1) elf-
0 < .:f ~ 1.nts
vc ;t):::: 0
LYY1 s <( t ~ lf))"LS
vc .1-) = SOm [-5b]
V( :t) = -2·5 v
Ltm.s <t.~ ~rns
V(f) JOI"tl [50]
€ms < t ~ /OYY's
'v(l) = :>om[- 50]
Chapter 6: Capacitance and Inductance Problem 6.30
2 Irwin, Basic Engineering Circuit Analysis, 9/E
V( t)::.- 2 · 5V
\X:t)(V)
'" 2'5 ~
. y le ~
6 8 11. ~c
2·5
Problem 6.30 Chapter 6: Capacitance and Inductance
Irwin, Basic Engineering Circuit Analysis, 9/E
6.31 The lUTent in a -o-mH induct r i shown in Fig. P6.3l. Fi11d the voltage .acros the inductor.
i(t ~mA)
+10
0
Figure P6.31
SOLUTION:
V(:J)
{-o"-
-- L eLl l .:t) cU
-t<O V(i: )= 0 v
0 < J: ~ '20yyt<;
V(t) = 50TY1 [-I J 'v( t) = - 50/"() v
2-om S. S :t £. YOYYI ~
V ( -J:) = 50)"() [I· 5] 'v(:t-) = l 5 m V
Ltoms < .t:_ .::: 6om ~ v(:t) = Ov
6orns z t ~ lorns Vl:t)= 50m[-1]
V(:J.)=
Chapter 6: Capacitance and Inductance Problem 6.31
2
V(tJ
Problem 6.31
± /lams v( :1:) = ov
0\1
-somv l5mV 0\J
-50)')'")\J
ov
Irwin, Basic Engineering Circuit Analysis, 9/E
-L ..( 0 0 <. !- ~ 20ri'-1
1 OY"il s ~ j. S: ltorns
YoYY1s .S t ~ 6o)'Y'IJ
6orru ~;t ~ (om')
t /lOY'f\.C
Chapter 6: Capacitance and Inductance
Irwin, Basic Engineering Circuit Analysis, 9/E
6.32 Dr, w the wavef rm. for the voUage acros a 24-mH induc tor when the inductor cun· ntis given by lit wav -
rm t wn in Fig. P6.3 .
i(t) (A) 8
4
-2
Figure P6.32
SOLUTION:
V(f) L cit (f) d.t
t~O ) V(:JJ=-=0
V[-):) 2Ym [ 4~J
V(t) 3lO')f)V
~9t. 0· 3 s < .J: $ 0 '6 s
V(:J-) 24m [ - 2 o]
\J(:t)
o·6s < :Jc~ o·qs
V{J) ::: 0
Chapter 6: Capacitance and Inductance Problem 6.32
2
Problem 6.32
Irwin, Basic Engineering Circuit Analysis, 9/E
~o~ O·q :S < .k $ I· IS
V(:1) =- 2Lrm [sc] V(:t)= 12oom\J
~.91. :t 71· IS V(;t)-= ()
V( t) = 0 32orn\J -ygomV
0
12oornV
0
t_.{ 0
0 (_ t ~ 0· 3 s o·3s < :ts 0·6 s 0 · 6 s < :k ~ o· q s o·qs< :t~ l·ls
u / I· Is
Chapter 6: Capacitance and Inductance
3 Irwin, Basic Engineering Circuit Analysis, 9/E
VCt) vs. ;t
? .....:14Jo:;
·--J2o-
-loao '
-8oo -·
- boO ...... -~-
-4oo ----
--200 -------------- ·----~--· ..
,... ' l ., .._ .... _ -f\i lo O•i. 0·4 0 ~ .,. 0·~ 1·0 I~ 2
-vo·l;' --
-~-· -·-
Problem 6.32.. Chapter 6: Capacitance and Inductance
Irwin, Basic Engineering Circuit Analysis, 9/E
6.33 The current in a 24l~mH jnductol' is given by tb wav form in Hg. P6.33. Hnd the wavef nn for the voltage acmss the in.ductol.'.
it (A
Figure P6.33
SOLUTION:
V(t) = L di(:t) cit
{o9t i ~ 0
V(t-) = L.~m [- 12 "X lo3]
- 2~3'V V[tJ =
2m~ L t ~ 5h"S
\.(t) =- 2Lfl11 [ 11- 1<163]
V( -t) == L~ 3V
5m.s. < f ~ qm.s V(t) ::::0\1
Cims ~ t~Jims v(t) -= 2-Ym [ -IL.XI o3]
VLt) = -2 '9'tV
I h··nS' L .t ~ 12 )"Y)S
'v( t) = 2 Lfln [ I 2. X I a'~ J V(i) = 2'? S V
Chapter 6: Capacitance and Inductance Problem 6.33
2 Irwin, Basic Engineering Circuit Analysis, 9/E
--f>oJL t / /2rY"S
V(t) ::: ov
ov t~o
v(t) -2~ ~ 'V 0 ~ t: s. 2h'L.S
~~rgv 2YY1~ z J: ~ 5 rf\3
()V 5 rns < :i ..c ~Yf\ s
- 2-\?~ v CjyY!_s ( :t ~ II 'YYtS
2~~v IIYn5 <. t ~ !'Lrll8
ov t ?l2r0'S..
Problem 6.33 Chapter 6: Capacitance and Inductance
Irwin, Basic Engineering Circuit Analysis, 9/E
6.34 Th Cltnnt "11 a 4-mH .indlllctor i given by l:be wave~orm in Fig. P6 .34. PI t the v Hage acros the indu t r.
i(t (mA) .
0.12
1·0 t (ms)
Figure P6.34
SOLUTION:
ll;t) = -120 ,gin wJ: J._.,tA.
w= L.TI T
T= lrnS
w=== 200011 9tQd/~
..l ( -t) = -120 A i Y'l 2.000 TCt
v c t) L cii C ;t:J cU
~A
V( J) y )'Y) [- 1'2 0 1--1 ( 2000 Jl) Co.& 2000 JT:): J
v(f:) - 3·()2.. co.s 2oooJlt rnv
Chapter 6: Capacitance and Inductance Problem 6.34
2 Irwin, Basic Engineering Circuit Analysis, 9/E
V(t) vs.t
~ :)
1.. ,........,
> ~ 0 ~ --.. ......
-I 0•0015 '1
-2
-s -y ------------- --~------------
±(~)
Problem 6.34 Chapter 6: Capacitance and Inductance
Irwin, Basic Engineering Circuit Analysis, 9/E
6.35 The waveform forth current in th 2-H inductor shown Fig. P6.35a is given in Fig. P6.35b. D termine the following quantities (a) the energy stored in the inductor at t = 1.5 ms, b the energy tored in the inductor at t = 7.5 ms, (c vL(t) at t = 1.5 m , (d) vL(t) at t = 6.25 ms, and vL(t) at t = 2.75 ms.
i(t) (rnA)
(a) (b)
Figure P6.35
SOLUTION:
w(±)
w(t·Sms)
w(I ·Sms)=
(b) w( t) 11_ L i 2 ( t)
lo( 1·5 Y'A8) !{_(2) (l·S~Io-s )L.
(C)
.t(t)=-
Chapter 6: Capacitance and Inductance
L &i Ct) ru
3om A
Problem 6.35
2
Problem 6.35
vL(t)= 2[o] VL(t) = OV
VL( 1·5r0s)= OV
(d) VL( ;t) - Lch(1) clt
£(1) ::: (Ct -f 60
VL(;t) = 20V
vL(b·25ms) =-2ov
-tot-do-~- ~cy'\to-~
~ X I a-s - 2 X I a-s
Irwin, Basic Engineering Circuit Analysis, 9/E
A 1n ~ tnKtvol
0 b I ntt sc...st
= -2(J
Chapter 6: Capacitance and Inductance
Irwin, Basic Engineering Circuit Analysis, 9/E
i.Lt):: -2ok te
.((t) -
3 D X I 0- 3 ::: - 2 0 ( 2 !'t ICf 3) f g &= CrOi
- 2at t Q· 0 7 A tn ~ irtfi'r-val e(- I )'\j-(~
Chapter 6: Capacitance and Inductance
3
Problem 6.35
Irwin, Basic Engineering Circuit Analysis, 9/E
6.36 F ind the pos ible capacitanc range of the foUowing capacitOl's.
(a) 0.0068 1jUJ.F with a ~olerance of 10%.
(b) 120 pF with a tolerance of 20o/ .
(c) 39 ~tF wi.th a to] rnnce of 20o/ .
SOLUTION:
(O.) C = Q.(X)6 8' )..J. F wi~
(b)
nanr--: G I· 2 r, F ~ C s 7Y' 8' n F
(;·('L ~ ~ c ~ 7· ~&''h
c-=- 12opr: w~-rl> Rnnr:j- ;
([) C -::: ] q ).J F w i :H-> 2 0 X tolrl -ya ~ Q
Range,·.
Chapter 6: Capacitance and Inductance Problem 6.36
Irwin, Basic Engineering Circuit Analysis, 9/E
6.37 Fnd the po sibl inductanc range of the foUowiog indnc~ors.
(a) ll O mH wHh a toleranc of W% .
(b) 2.0 nH wHh a toleranc of -%.
(c) 68 IJ-H with a to I ranee of 1 Oo/ .
SOLUTION:
(Ct) L- {OmH wit--h to~ fC)-tnuna qY¥1H ~ L ~ /1-mH
(b) L-=- 1 n H (J._)j th 5 % f&-lo1 G Ylt9..
l·q Y'IH ~ L ~ '2· IY)H
(C) L== 6 8' .L1 H w i~ J 0 ~ ;t-o.Q_t r O..Y1 U.
6 ) · '2 u H ~ L ~ a- Y • 8.J.J t-f
Chapter 6: Capacitance and Inductance Problem 6.37
2.24
Irwin, Basic Engineering Circuit Analysis, 9/E
·6.38 Tt1e capndtor in Rg. P6.38a is 51 nF with a to]erance of 10%. Given tb voltage wavefmfll in Rg. 6.38b, graph the cmnnt i ( 1) or the nuniml tn and maxin l tn capadt r va u s.
60
40
20 S' ..._.. 0 <:" ..._..
o;:, - 20
-40
~60
SOLUTION:
c
(a)
t-
r- 1\ t- v \ t- \ J
' v 0 1 2 3 4 5 6 7
Time (ms)
{b)
i[t) =- c d v (~) clt
{a""' 0 (. .t ~ In"~ j_ l:t) = OA
I l'YU < f ~ 2W'S.
-== t I·I{'Jln) {Lf')(I0'-1)-=-
Chapter 6: Capacitance and Inductance Problem 6.38
2 Irwin, Basic Engineering Circuit Analysis, 9/E
pO~ 2tYlS $ X~ "3 ms 1 YYl~ ( t) =- I· J (51n) (~g)< /OY) =y·U i rnA
i"rninlt) = 0·9 ( SJn) (-'&'X IO"'~)= 3·(lmA
~o7L 3rns-<. t .( Lfms i-Ct) =at\
bO"t. LtY'ns < t ~ 5 ms imo.x Lt) -= \• \ ( 5 h,) (~X IO '1) = 2-l Y "rYYI\
im;YlL;t):: O·Cf( 51YI) {Lt1-.fo'1)::: 1·8LfmA
..JOh 5n1~ <X~ 6 mJ
l(.t) =- 0 A
i(x) Cm") 1!\
3 ' 2,ly ..
'
1·8'1 ------ -----
..... /
I 2 3 4 5 6 1 t
1·61.- I· I c. ----
4·41 t ----- o·ctc.
Problem 6.38 Chapter 6: Capacitance and Inductance
,
Irwin, Basic Engineering Circuit Analysis, 9/E
6.39 Giv 11 Ul capacitors in Fig. 6.39 ;u· C 1 = 2.0 t.tF wi.th a o]erance of 2o/ and C2 = 2.0 p.F with a to]erance of
20%, find the foHowing.
(a) T 1e nonnnal value of Ceq·
(b) T 1 minimum and maximum po sib]e va tu of Ceq .
(c) T 1e p rcent enor of th minimum and maximum vah es.
SOLUTION:
(9.)
(b)
o------i 1----1
cl Figure P6.39
Cett =
c "YYC.J{
C 1.-'ldo-6 ) C 2- K Ia-' J 2 X ta-6 -t 2 X 10_,
(LD L1 X to-<:. ) ( l·L(X /o-6 J 204,>\ I o-6 + 1·Y ~~o-6
Chapter 6: Capacitance and Inductance Problem 6.39
I
2 Irwin, Basic Engineering Circuit Analysis, 9/E
(C}
Problem 6.39
cnYM - ,,, .Ll r
CMin =- C1rnin C1..N1iYl
c, min cl-YY">;n
CM;n :- U·96X la-6 ) ( l-b Xto-6 )
( I·Y 6 X I o-<;) + 1·6 i-lo-6
-r o /o e«a---b - Ch1a.x- c~ cecv
T fo eor-o'"'~ -· , .. , 'XlQ_, - ;f ~ 10-&
--+ % erro1
-% (!YYDV
IX lo-6
lo /o
-· Cm;Yl- C e_q,
CectJ
0·8 Kl)( Jo-6 - I '/do-6
))(/0-6
-I f -q :;;
Chapter 6: Capacitance and Inductance
Irwin, Basic Engineering Circuit Analysis, 9/E
·6.4 A ..:5-ptF capacitor laitiaUy charged to - 10 V i charged by a constaRt curr nt o 2.5 IJI.A. Find til voltage across the capacit •.i after 2t min.
SOLUTION:
V(t-)
V(t)
V(J)
V( -t)
t-
+f J- C :J-) ct:.t -t VC o)
0 150
_j_ f 2·5 jJ d1 -10 25 }..J
0
2· 5 JJ- [I 50] - 10 25 }v{
5V
Chapter 6: Capacitance and Inductance Problem 6.4
,
Irwin, Basic Engineering Circuit Analysis, 9/E
6.40 T"he inductor in Fi.g. P6AOa i.s 4. 7 ~J-H w· h a tolel'anc of 0%. G"ven lli CliH' nt waveform ·n Fig. 6.40b, gntrpll the v · tage v(t) for the III.inimum and n aximwn inductorv,h s.
(a)
f-
.--. 5
1 0
[7 \ J ~ \ I
\ I ~ I
- 115 0 10 .20 30 40 50 60 70 80
Tlrne ~ms)
(b)
Figure P6.-4o
SOLUTION:
Vrrp_t L ;t) :::: ( 1· 2-) ( '-{' 7~) {- 1·5) =- -';·l.f6 1..1 V
Vm.-Y1Lt) -=-(b~S)( Lf·l,u.)(-1·5)= -5·46uV,
Chapter 6: Capacitance and Inductance Problem 6.40
2
,
Problem 6.40
Irwin, Basic Engineering Circuit Analysis, 9/E
~oh ~Om~ S:. t: ~ Ltul'Y'&
V'(Y)Qt. { t):::: O·lJ ( ~·l..u) (o·5) =: -2 · 8 2 J.-t V
V m i Yl ( .f) -::: (6 · ~) ( Lr 1LI) (- G · 5 )= - I . ~ 8 1-A V
~ Dh- YuYYlJ ~ -;t ~ 5 () YY"S
Vm~ (t) = 0·2.) ( Lf·l.u) (f)= 5 •6'-f, ~v
V m in ( f. ) ={ o. K ) l Y · 1t.-t) ( I) = ~ · l ~ .U v
50rrL~ 5. t < 6o~s V(t) = av
v Ct) ::: ov
Chapter 6: Capacitance and Inductance
/
Irwin, Basic Engineering Circuit Analysis, 9/E 3
v< '" /
-~
-~ ----.... 1------~-
2-f2
~-- - --- ._,.... --
so l(o
lo 20 So 6o 10
-1·88 t(ms}
~- ~- -----2. ~2. - '""
- S·6Y -... ~----
'
'· 2(
----- o~s>c.
-1-· C= ~-1.4 F
Chapter 6: Capacitance and Inductance Problem 6.40
Irwin, Basic Engineering Circuit Analysis , 9/E
6.41 If the t tal energy ~ored in t re circuit in Fg. P6.41 is 80 mJ what i.s l:h valll · L?
200{} 80 p.F
Figure P641
SOLUTION:
.l c ::: c. dv, ~
• v~.,.= L oUL.
~
1<2.. d '"J CoJJ :tPu_
IPI 2oo..Q
Chapter 6: Capacitance and Inductance
l
;-80.JJ - v ~
-::: o ) v (_
50{}
L ..t._
----·-
A..h CA..
=o, iL ..lj) a
' [_ 't r-(.U...L:t"
-t Vc.
Q...b:
conAJof\i
CD('-) -lo r.J
Problem 6.41
c
2 Irwin, Basic Engineering Circuit Analysis, 9/E
Reey= ? oo 11 SD - 2 oQ{ 5-o)
Rea;::. yo.J2
c I)( ~0) _L c v 1. l.
20a+50
lAJc = k_ ( J'O J-.l) ( LID),_
We: GYmJ
,( L - ( LOO ) ( I) 2oo+5o
L L = D ·z A
WL=- '~t_ LiL'l. L=- liiVL
.i_L )...
WtotcU = WL i We
WL =. gem- 6 '-f YY1
WL =- /6mH
L= 2..[ Ibm) (p ·g )'1-
L::::. 5ornH
Problem 6.41 Chapter 6: Capacitance and Inductance
Irwin, Basic Engineering Circuit Analysis, 9/E
6.42 Find 1e val.ue of C i.f tb . energy s~ored in th ca adtor .in Hg. P6A2 eqt a. s lhe energy stored in the indl1ctor.
c
0.1 H
Figure P6 .42
SOLUTION: -t Vc-
c
l 00-fl..
12-V
•
1 oo.tl-
j_,_
0 .1 H
Ac. (_ d. Vc = 0 J Vc i.A 0 Co'N}:t1(\i cJ.;I.
cliL =-o , leU
Chapter 6: Capacitance and Inductance
Ci ~cu.i.:J -t Vc----. I 00..>\-
• JL
Problem 6.42
2
Problem 6.42
12 20oftoo
i L = Lt C.Ym A
~cv1..= 2.-
Irwin, Basic Engineering Circuit Analysis, 9/E
c= L(t5) (:: o ·I ( C l.f~ to->,f)
Chapter 6: Capacitance and Inductance
Irwin, Basic Engineering Circuit Analysis, 9/E
6.43 Given th network in Fig. P6.43, find fu power diss]pat di in the 3-0. re i.stot' and th energy stored j u th
capacitor.
3fi
i 2 v sn 2 F
Figure P6-43
SOLUTION:
. VL L d.JL=O ) L L .JJ'J (gJ~f"\t. -
, .A. c.
12. v
otA
c_ d V c. = o J v c ..J...A coru:taAt Olt
GJL
Chapter 6: Capacitance and Inductance Problem 6.43
2 Irwin, Basic Engineering Circuit Analysis, 9/E
' .AR. /2-3'-
3f6 • _g A J.R -
]
We= !{ C Vc 'l = k_(2.)(12-)"'L
PR - iR2 c ~) -= (- g/~) l..( 3)
pR.-=- 21·33 w
Problem 6.43 Chapter 6: Capacitance and Inductance
Irwin, Basic Engineering Circuit Analysis, 9/E
6.44 What va]{l s of capaci.tanc can b obtain d by interconnecting a 4-p.F capacitor, a 6-IJLF capac"tor, and a 12-IJI.F capacitor?
SOLUTION:
Cet• l 1 l --!i>
I c., Jc~ Jc1
Ce1..1 -= C, + cL +c 3 -= Lf).{ t 6 M t 12 L-1 = 2l...U F
1 Ce1}-Ic, Cetl----7 1 CL Ce.z~. Tc!
Ce22
Chapter 6: Capacitance and Inductance
--
_L -L_L -f J_ cl cl-- cs
_L +j_+_L Lf..u I:,V. ll..U
2..uF
CC2. -tc3) C c, ) c,-t C1- t-C1
Problem 6.44
2
l l~ Cez~ T ~ ---}. T c.
l h Cets
~
c1- 1 Jc'
Problem 6.44
Irwin, Basic Engineering Circuit Analysis, 9/E
Cezt.t - c3 t c, c}_
c,-1- cl.
Ce2r., I \..t . '--1 J.J F
Ce1s- - (C,tcJ ( C ~) -C14cl tc!
Cez~ - 5· '15 ...u F -
Ce7.~:: (CI+c 3)Cc,_) e~+cL f LJ
Chapter 6: Capacitance and Inductance
Irwin, Basic Engineering Circuit Analysis, 9/E
c~ Ce21
) c1-
I I
1 Cect") - Ct(CLfCl) -C,-f c.J_ 1CJ
c, c~1 = p )..A F
Ceq, 8 :: c 1 c c I + c J )
C1-t c1_ + c3
C4.f'oci1a.t1~ v~ po-1sib&.<
Ce~1 2 2 ..u r Ce12 :: '2uf (e~ 3 = · 3 , 2- 7 )--(. f
c e.z"' ::. '~'l.f ,u F C e1s :::. '5 · 4 5 LJ F
. (~, ~ L1<1t:,.UF C et7 -= g }....t F
Cei~ ~ qu F
3
Chapter 6: Capacitance and Inductance Problem 6.44
Irwin, Basic Engineering Circuit Analysis, 9/E
6.45 Given •our 2-~F cap< itors, lind th maximl m va1lle aud minim111.m value tb,at can be obtain.ed by iuterconnecth g th c. pac"to.r i.t ries/paraUe] combinations.
SOLUTION:
Clv\;Y\ ?
l l ' c1 cl.
C'1 c3 I I I
-' -t..Ltj_fJ_ c, c1.- c1 cy
0·5)-.tF
~
u.ih-tn CU1 C.Onhll~ 111 pu~· ·
Chapter 6: Capacitance and Inductance Problem 6.45
2 Irwin, Basic Engineering Circuit Analysis, 9/E
C ma_x. 8 .u. F
Problem 6.45 Chapter 6: Capacitance and lnd'uctance
Irwin, Basic Engineering Circuit Analysis, 9/E
6.4·6 Giv n a l -, 3-, and 4-~J.F capaci or, can. bey be inl:erooonected to obtain an equival nt 2-11F capaci.~or?
SOLUTION:
I ,.u_ F
Ceq_,- ( I X I o-6 + ~ X I a 6) ( Y X I o -t;") IX{t>~6 +~ )( lo-6 -t Y Xlo-'
Chapter 6: Capacitance and Inductance Problem 6.46
Irwin, Basic Engineering Circuit Analysis, 9/E
·6.47 D tenu"ne the v iue of inductance that can be obl'lined by intet'COIUl ct".ng a 4-mH lnduc~or, a 6-mH inductor, and a 1--mH induct r.
SOLUTION:
I
Le.t2. ---::::>
let,3 ' -
----:)
L.t
\
j_+ ... L.:+ j_ L, Lt L~
2mH
~1-::: Ll.
Leq,2 = L1
L-, Le.z_3
Let_3
Chapter 6: Capacitance and Inductance
L~
L, + L1.... tL3
2 )....-n H
L l. L! -t Ll
L}__ + Lj
- ~mH -
Problem 6.47
2
1
Let~ l ~ L \ I "">
~ )
L~
L2.
I I
L~
LL
L~t' ---7
L, Ls
Problem 6.47
Irwin, Basic Engineering Circuit Analysis, 9/E
Lecv'-f = l L,+ L,J I J
Lez_Lt - ( L, f L2-) ( L3) -
L, -f LL -t LJ
~:: 5· L-f5 mH
Le.!s ::. L, Ll_ -t Ll L 1-t L 2
Le.!5 = IY'YY'nH
Ll
Le..t~ ::. ( L, ll L ~ ) + LL
le26 == L, LJ + L2.
l,+ L~
le.t' =qrn H
Chapter 6: Capacitance and Inductance
Irwin, Basic Engineering Circuit Analysis, 9/E
L~a ---;;>
L~1 = (LL f L~) L,
L1t L,+ L3
L, Ll.
L3
Lect~ =- ( L1 + L3) C L2J . L1 -t L l--t L3
Chapter 6: Capacitance and Inductance
3
leq_? -= I '-1· y "r() H
~6-= qmH
le't7 == s· L l m H Leq,t -= Lt · 3 ~ m H
Problem 6.47
Irwin, Basic Engineering Circuit Analysis, 9/E
6.4.8 Giv; o f m 4--mH jnductors, det rmin the max.inmm and tninimum vallues of inductan:c that c<1n b obtained by illteroonnectil]g th.e inductor ill erieslparaU l combinations.
SOLUTION:
L, L~
Chapter 6: Capacitance and Inductance Problem 6~48
Irwin, Basic Engineering Circuit Analysis, 9/E
6.49 Given <1 6- , 9- and 18-mH lndlnctor, can th y "ntet·omulecred to obtaill an equi.val nt n -mH indll.ctor?
SOLUTION:
:: \ 1 mH
Chapter 6: Capacitance and Inductance Problem 6.49
Irwin, Basic Engineering Circuit Analysis, 9/E
6 .. 5 Til energy that i.s stored in a 25 -~F capacitor i.s w t) = 12 si112 377t. Find the current in til: capacUor.
SOLUTION:
w(t) _L c V 2(f) 2
Lu( j_-) - 12 Sl.n2 3ll ;;t
V 1 C±) - c /2_ s;i;?31l t) (2)
25 X \o-6
V( :t) -+ ql Cf · S' Sin 377 :t
C c1 V( :t-)
d.:t
v
J-(1:) ::: ( 25X to-t:) [ T q 1q. ~ (311) cos31l.:t J j(f)
Chapter 6: Capacitance and Inductance Problem 6.5
Irwin, Basic Engineering Circuit Analysis, 9/E
6.50 Find th . tot, I. c pacitance Cr of tb iletw rk itl Fig P6.50.
1 Cr --- 1,2~·
2 fl.F 3 IIJ>F
Figure P6.so
SOLUTION:
1 /..u_F
\----It-~---t--------J
')uf' \---~ . ....__ _ ___.. ,-
j,uF" L...--~ }'-----'
C, = ( Lu-+ 2).(t 3u )( 4v)
c, =-
l.t.t -t l..Ll-1'1 ,Lt t 4/J
6 ).J ( '-f-U)
6u-t ltM.
C, + /1M
Chapter 6: Capacitance and Inductance Problem 6.50
Irwin, Basic Engineering Circuit Analysis, 9/E
6 •. 51 Find the t tal. capac"tanc Cy of th network in Fig. P6.- 1.
t
G· T ~F ~~f-F-....J Figure P6.51
SOLUTION:
~----------r----t c1
-c .___a...-c-'1-1 ...... _____.
C1 = 1ul=' .)cL:::GuF )C~-= 3MF;
Ct..t ::: Gt...U F ) C5 = 6 U. F
C'=-6.u. +c1
c' y.u. i 6M. r1.uJ Gu. +34
C'
Chapter 6: Capacitance and Inductance Problem 6.51
2 Irwin, Basic Engineering Circuit Analysis, 9/E
c..' T JbuF
CT = C/(6.u) -t-c, c'-t 64
CT - §ML 6Lt) T3J..-t -Gu+ 64
Cr GuF
Problem 6.51 Chapter 6: Capacitance and Inductance
Irwin, Basic Engineering Circuit Analysis, 9/E
6.52 Hml the t tal capacitanc Cr shown in tb netwo1·k in Hg. P6.52 .
• ~F T~... ____ c_ T_ ....... r _____ ____JT 2 ~F Figure P6.52
SOLUTION: ..
c, :: G M F ) c 1 = ~ JJ.. F 1
c3 ;; ~ .u. F)
1
Cy =- J 1-M F ) Cs-:::- 4 ,u r:- ) C6 = 2 JJ FJ
QY'\Cl ( 1 ~ 2J..LF
C' =- Ccs + c,) C C. to, 2. -t- c 3
CL-i+Cs+(&
C 1 (Lf_u+l2.J-t)(12_L..f) i-i,U., ll.u_ t y..U. + 2).{
C' -
Chapter 6: Capacitance and Inductance Problem 6.52
2 Irwin, Basic Engineering Circuit Analysis, 9/E
Rad~!
I lc, c., I
cl. 1 C1~) c1
T
t_Lt-_L (/' c.., cl. c._ I
-' I t +_L_ c'' 6J.A C,f,..u ll.u
c" Cll l..U. F
Problem 6.52 Chapter 6: Capacitance and Inductance
Irwin, Basic Engineering Circuit Analysis, 9/E
·6.53 Cmnpnt the qu]va] nt caprrdtance of Ute n tw rk ]u Fig. P6. if a 1 the cap it are 4 ~J.F.
'-'
;;: ::;::' ::::::::::
If :::::: :: 1\
::: :::
,..,
Figure P6.53
SOLUTION:
c ' = c 2- +c J = '? J.J ,::-
I I lc5 Cecv I c,
~c~ J ~ · c_l
I C"=- c c 1
I 2·61J.J F
cr+C1
Chapter 6: Capacitance-and Inductance Problem 6.53
2 Irwin, Basic Engineering Circuit Analysis, 9/E
Problem 6.'53 Chapter 6: Capacitance and Inductance
Irwin, Basic Engineering Circuit Analysis, 9/E
6.54 Find CT ·n fu network in Fig. P6.54 if (a) th. switch is op n nd (b) h swi.tch is do d.
SOLUTION:
(a)
1
3,f!!-F
Figure P6.54
Swi.J-01 op.en :
Cr :)
C'=
~p. I Tct 6J.J.- ~c2
b (3)
6-tJ
Chapter 6: Capacitance and Inductance
6 ,f!!-F
sMl c~
I 6u Jc~
..
l c"
J Problem 6.54
2 Irwin, Basic Engineering Circuit Analysis, 9/E
( b) S'wi-JC.h dO:\E'cL
Problem 6.54
C 1 ~.. c.l3 .. ~ '}~ J1 ~
-
---.a....-· ___ c._~j__, bM C 6..U. I ~I
lc, I c''
T Cr :::::- c'cu
c.'+ C''
Chapter 6: Capacitance and Inductance
Irwin, Basic Engineering Circuit Analysis, 9/E
6.55 In th 11 l:w rk in Fig. P6.5- md the capacHance CT if (a) th switch i.s p nand (b ) th switch is do ed.
112 j,~~.F 6 f1F
Cr -
3 f.i>F 112 !JI.F
Figure P6 ·55
SOLUTION:
c,
C6
c.= 61-fF Jcl= l2_ur) C3=6..t.~~
Chapter 6: Capacitance and Inductance Problem 6.55
2 Irwin, Basic Engineering Circuit Analysis, 9/E
(G.J Cr = c,( C6 2 -t C1-(Cy) -I- C3CCs-) c,+c6 (2---tC..y C~tCs-
(T:: Gu(I?M) + ,, J.J._ ( 6.U ) + b't-tf?u} Gu-t-12..u 12),{ -t6u 6/...t-f 1U
(6) SwitCh cto.recL ·.
Cs
c'= C c 2. tc 3 ) C c'-1 +C5 ) = C 12)..{ + 6 ~-t)(6.u r3J...tJ cl t (3 t cl.f+Cs IIJ..t +6u + 6U +J.M
Problem 6.55 Chapter 6: Capacitance and Inductance
Irwin, Basic Engineering Circuit Analysis, 9/E 3
Chapter 6: Capacitance and Inductance Problem 6.55
Irwin, Basic Engineering Circuit Analysis, 9/E
6.56 Select tile value of C to p.roduce the d sired t tal capac-Hanc of Cr = 10 ~J..F in the c '_rcu't in Hg. P6.56.
o------
l e Cr = 10 f.LF
0~--8-~-F ~][~------~li 1 6~F Figure P6.s6
SOLUTION:
Cr-= ( C1 + Cz._) (c)
c,+c)..+c.
Chapter 6: Capacitance and Inductance Problem 6.56
2 Irwin, Basic Engineering Circuit Analysis, 9/E
Problem 6.56
C = _CrC C1+ ck)
C1+C2--cr
C== IOJ...t ( 8..u t I bu)
F u -t I l;.Lt - I 0 -4
Chapter 6: Capacitance and Inductance
Irwin, Basic Engineering Circuit Analysis, 9/E
6.57 S .ec t the vah of C to pmduc b , e ired tota] capacitance of CT = 1 p.:F in the c"rcu"t in Fi.g. P6.57.
0
Figure P6.57
SOLUTION:
"t I Cr ---4- he
c, T 1 Ct_
Cx -= ( C+ C J) ( C 4) C+ cJ+ c'1
C'-1 ~ (Cr+Ct) ( () c tc, -1-c l.
Cr :::- ~+Cy
J c~
Cy::: (C+C3 )CC42 t CCrtCl-)(C) ct c,-t c~
Chapter 6: Capacitance and Inductance Problem 6.57
2 Irwin, Basic Engineering Circuit Analysis, 9/E
lu= (CttJ..t)(Ju) -+ (tu-+2,u)Cc) C t I Lt -t (l.J C -t I Ll + 2 LA.
Ct2u
I= Cfl + 3c C+2 Ct-l
c~i.n u~
C t I + 3C ( C t l) Ct-1
(Ct2.) (Ct3)=(C-ti)('Ct3)+~c ( C-+2..)
3C1--t- 5C. -3 ==-o
c = -5 :t ,)25-lt( st 3)~ L.( J)
c = -5 ::t 1-·gl
' c-= L(65 nF
Problem 6.57 Chapter 6: Capacitance and Inductance
Irwin, Basic Engineering Circuit Analysis, 9/E
6.58 Th tw capacHm·s ]n Fi.g. P6.58 were charg d and l:h 11
connected as hown. D tennine the eqniva1 nt ca acitance, the "nitia1 voltage at the t rminals, and the t tal
n gy t .red in the 11etwork.
SOLUTION:
+
~4~F
Figure P6.s8
C,: /2JAF Ct ~ C1-- l1..Lt F
V-t 6 ::: 2 v = - yv
Chapter 6: Capacitance and Inductance
j
Problem 6.58
2 Irwin, Basic Engineering Circuit Analysis, 9/E
Problem 6.58 Chapter 6: Capacitance and Inductance
Irwin, Basic Engineering Circuit Analysis, 9/E
6 •. 59 Th two capacitor shown in Fig. P6.59 hav been connect d for som tim aud hav .reached their p.resent values. Fnd VQ.
SOLUTION:
+
~4~F
Figu re P6.59
SQT'I\.a
CQ = cv Co \Jo =- C,\1,
V 0 = y)..L( t bj ll.u
V0 = 5·33 V
Chapter 6: Capacitance and Inductance Problem 6.59
Irwin, Basic Engineering Circuit Analysis, 9/E
6.6 A capacitor i charg d by a constant curren.t of 2 mA and r suUs in a voltage increas of n V in a 1 0- interval \\rhat i.s th vatu of th c paci.tanc ?
SOLUTION:
12 = l_( ID) c
C 2 X 10~ ( 1o)
12.
c 1·67 mF
Chapter 6: Capacitance and Inductance Problem 6.6
Irwin, Basic Engineering Circuit Analysis, 9/E
6.60 The th.ree capadtor lmwo in Fig. P6.60 have e.n co.IUJ!ected for orne Ume and have l' ached their present vah s. Find Vi and v2.
+ vl
+ y.
Figure P6.6o
SOLUTION:
V,Ct)
'1 (t)
B,IJ!IF
+
4l ,pJF
12V
11.2V
v. --v.
2..
-t 12V
...L c, I -cl.
v I -
cl --\J,_ c.,
Chapter 6: Capacitance and Inductance Problem 6.60
2
Problem 6.60
v,:: CL VL c,
c2- vL-tVl== 12 c,
Irwin, Basic Engineering Circuit Analysis, 9/E
v2 -= ~v
v,+g= 12. V 1 ::: 4V
Chapter 6: Capacitance and Inductance
Irwin, Basic Engineering Circuit Analysis, 9/E
6.61 D term.in the inductanc at terminals A-B in th netw rk in Fi.g. 6.61.
1 mHI 1 mH
B o------~
Figure P6.61
2mH
2 mH
SOLUTION:
A
lr l~
L I = L] = L6 = I YY\ H L2..= I'L m~
Ly = G mH J Lr- = L(mt-1
L l:: Lg =- lq ::: 2mH
Chapter 6: Capacitance and Inductance Problem 6.61
2 Irwin, Basic Engineering Circuit Analysis, 9/E
~tLq::: LtmH
Le - ~. L b~_Ldl_ L1 i-Lb-+ Ltt
Le = 3mH
Problem 6.61 Chapter 6: Capacitance and Inductance
B
A
Irwin, Basic Engineering Circuit Analysis, 9/E
·6.62 Detenui.ne tb indue nc at te.rmi.1uds A-Bin the network in Hg. P6 .62.
1 mi-t A
.2 mH
4ml-l
B
Figure P6 .62
SOLUTION:
L 1 =1mti
LAg ::: ~ yY) f ~ yY")
~FI~ =- b mH
Chapter 6: Capacitance and Inductance Problem 6.62
Irwin, Basic Engineering Circuit Analysis, 9/E
6.63 Find the total inductance at th terminals of tt1 n rwol'k in Hg. P6.63.
SOLUTION:
Figure P6.63
Rod~ CicrcuA! :
0 l'1
Ll. '('
Sho9t..h.cl.·
Lr == [c L'-11 L~) -n., J /I L . .,
L r = [ ( I 2 ~ ) I G m) + l-m J I/ Lt m
lr ::: [ 11 m(f>m) + 'l~ J II Lf M
l2m-f6m
Chapter 6: Capacitance and Inductance Problem 6.63
2 Irwin, Basic Engineering Circuit Analysis, 9/E
Lr
Problem 6.63 Chapter 6: Capacitance and Inductance
Irwin, Basic Engineering Circuit Analysis, 9/E
6.64 Comput th equiv lent indm::ta11c of th network i.n Fig. P6.64 if an inductm's ar 4 mH.
SOLUTION:
~
Figure P6.64
lt•
CL
b
c
La :::_L_, L_t._ L,-f LL
L1 LL.f
Chapter 6: Capacitance and Inductance
Lf
b
c
2mH
Red oC.U..U i -rut d-
c
Problem 6.64
2 Irwin, Basic Engineering Circuit Analysis, 9/E
Le~ - [~IIY)+2-]~H
L - b { y) -r 2 -e.ay 2-
- ~·~ "'(Y\.'t-1 -
Problem 6.64 Chapter 6: Capacitance and Inductance
Irwin, Basic Engineering Circuit Analysis, 9/E
·6.·65 Find Lr ·n tll n rwork in Fig. P6.65 {a) wi.th the w"tch op u and (b) with the switch c o eel AU induct rs ar 12mH.
LT -.
Figure P6.65
SOLUTION:
Lr ---'}
L8
LLt
L1
leA. (L 1-t Lg ) I J L6 24yy, J I 12m
Lo. ~mH
Lb (LL II L~ )
12m I) I2W)
Lb 6mH
Chapter 6: Capacitance and Inductance Problem 6.65
2 Irwin, Basic Engineering Circuit Analysis, 9/E
L a.
Lr - [ ( LytLQ.) II C Lb + Ls) J + L1
Lr= [ct2m-tJ'rn) II a>'YltiLm)]+l2YYl
Ly=- (2o-m /IIZm) +12Yl-l
(b) ~witch CJos~cL:
L,
Lr Lb Lt.,
- ~
Ls L~
Problem 6.65 Chapter 6: Capacitance and Inductance
Irwin, Basic Engineering Circuit Analysis, 9/E
Lr = ( La II Ls ) + ( L6 I I Ly ) i L 1
Lr ::. ( 8m ll I 2.. m ) -t ( 6 m J J 12 rn)-+- 12 YYI
Lr- 2a·~'Y'nH
3
Chapter 6: Capacitance and Inductance Problem 6.65
Irwin, Basic Engineering Circuit Analysis, 9/E
6.66 Given. th 11 twork shown i.n fig. P6.6 , find , a ll equ.ivai 11 indnctanc at terminals A -B w "tb t rminai C-D bon circait d, and (b) the equival nt inductance at termina]s C-D with tenninals A-B p n cll'CilitOO.
20 mHI
6 mH
Figure P6.66
SOLUTION:
(a) C-D s:ho~
A
lLf
L 1 - 20 rnH ) L2. = 5mH LJ =- )2mH
a rd.- LL.f = 6mrl L,
L~
L'-1
Chapter 6: Capacitance and Inductance
D
Problem 6.66
2 Irwin, Basic Engineering Circuit Analysis, 9/E
l.J1R = (_ ll II LL) t ( L3 II Ly)
lAg= ( 1o r-n 115m) ·-t ( 11 m II 6 Y"() J
LAJ] = g'mH
(b) wd--l-1 A--e of-0n:
c
L~
ly
b
Lc 0 = c L I + l L ) 11 ( LJ + ~'-f)
Leo= (2omt5m) I/ (1'lrnT6mJ
Leo = 25'rn ) I IFrn
LeD - la·l.ilmH
Problem 6.66 Chapter 6: Capacitance and Inductance
Irwin, Basic Engineering Circuit Analysis, 9/E
·6. 67 Find the value of Lin l:be network in F'g. P6.6i so that the total indudance L, wi.U b · mH.
Ly -
Figure P6.67
SOLUTION:
'lm:::
1m=
2 mHI
6 mHI
A ... L
[c L II L 3 ) -t l2. J /I L 1
+ '-rn J /I 4rf1
[ ( 6m)L
Ltm Lt~m
Y'Y"n -r 6m ( L)
Li6m + 2W'\
L.m l6m+ 6m(L) J==-ym[6m(L) +lm] L L+ 6m L+ 6m
Chapter 6: Capacitance and Inductance Problem 6.67
2
(12 rn) ( L) + y m Lf6m
~1 Y)'l)( L) -(6m) ( L) Lt6m
Irwin, Basic Engineering Circuit Analysis, 9/E
6m -t-(GYYI)L L t Grn
2m
(bm) ( L) - 2m ( L-+ 6Y¥') -
(Ym) ( L) - }2 )A -
L= 12J.A Lrm
L= ~1'nH
Problem 6.67 Chapter 6: Capacitance and Inductance
Irwin, Basic Engineering Circuit Analysis, 9/E
6.68 .Ftnd the valu of Lin th 11 tw rk in Fig. P6 .68 · that
th valu of Lr will b · mH.
2 mHI
Figure P6.68
SOLUTION:
2
L, s ( ~ L .
L-r \LT 4L ) II L y-t·L
L-r ( L-+ YL) L i...1tL
Lt 2-* l.fL
y+L
LT-= 2L -+ L( L 1...
y-fL
l( \....f+L) -f- 2( 4+L) -t L.(L
LriL
Chapter 6: Capacitance and Inductance Problem 6.68
2 Irwin, Basic Engineering Circuit Analysis, 9/E
Problem 6.68
LT = 2l(LftL) + YL~ YtL
~Lt LL +S?t'LL t yL
YtL
l T = ~L + '2LL -t Y L ')..
?L + L Lt~ -t'l..L
LT::: GL,_ + 2L
L2 tiDL1.?
L T = 6 L 'L t~ L
L 1 i IOL -tf
2L'l. -t 20l +I b
L( L 't - I'LL -I b = 0
L'l- sL-Y-=0
( L-LI) ( L -t I ) = 0
) L===-1
Chapter 6: Capacitance and Inductance
Irwin, Basic Engineering Circuit Analysis, 9/E
6.69 A 20-mH indtuctot· and 1 -mH imJuctor are com1 . cted
in sedes with l -A curent urc . Find (a) th equival.e.nt 'ndl ctan.c and b) th motaJ 11 rgy stored.
SOLUTION:
(~)
JA
(b) \N t4>-\:a..l
W.t~fu·t
w1+ w,_ Yt_L1I'l. +){L2-IJ..
I; ('LOm)(1)2--t ,k'(ilm)(1)1. /'L 1-
I G mJ
Chapter 6: Capacitance and Inductance Problem 6.69
Irwin, Basic Engineering Circuit Analysis, 9/E
6. 7 Th CUIIl'ent j n 1 00-!i-F capaci t r is sh wn · n Fi.g. P6. 7. D t rm.ine th \V v f rm for th v ltage acr s th apa i -or jf it j inillaUy 1.1.ncharged.
i(t) (mA)
10P-----.
0 2 t (ms)
Figure P6.7
SOLUTION:
1 1 = 2mS
* <O 1 i=O J Qhd V=O
) l = Ia 'YYlA ) o.rd
L_W\
s torn c:J..L Joo.u. 0
2N~
V= lcrm [ t] lOOt V
IOO)J. D
v = 0·2V
Chapter 6: Capacitance and Inductance Problem 6.7
2 Irwin, Basic Engineering Circuit Analysis, 9/E
V(t) ov J :t<.o
lOOt V ,
0·2 v
Problem 6.7 Chapter 6: Capacitance and Inductance
Irwin, Basic Engineering Circuit Analysis, 9/E
6.]0 For tb network i.ll Hg. P6.70, Vs( t) = no c 377t v. Find Vo(t) .
1 kO
+
Figure P6.7o
SOLUTION:
' c _l.,
C= l,t.Jf="" a.rcl R = I k_Q_
Vs ( -t) ::: 12 0 C06 3 ll-t V
C ctVs eLf
( irucu op- etm_p)
- Vo -~
V0 =- - Q c clVs_ cit
Vo= -(I K) C /.Lt) Gl2o( j 77) cos ~lli]
Chapter 6: Capacitance and Inductance Problem 6. 70
2 Irwin, Basic Engineering Circuit Analysis, 9/E
Vo (t)
Problem 6.70 Chapter 6: Capacitance and Inductance
Irwin, Basic Engineering Circuit Analysis, 9/E
·6.71 For tbe n twodc in Fig. P6. 71 cbo · C l ch that
v0 = -1l0 J v5 dt .
Source mode'! C
Fl gu re P6.71
SOLUTION:
+
V0 = - I o S VscU:
Vs-O
Recv
R s t 7 0 K. = f OK + l 0 1<.
Recy =- ~ 0 K.Jl.
-CdVo cU
V0 I fVsct.t f<ettc
Chapter 6: Capacitance and Inductance Problem 6.71
2
c-=
Problem 6. 71
I 10
1·2Sur
Irwin, Basic Engineering Circuit Analysis, 9/E
Chapter 6: Capacitance and Inductance
3
ttt
2 Irwin, Basic Engineering Circuit Analysis, 9/E
Vo(t) -21·11 Stn37lt V
Problem 6. 72 Chapter 6: Capacitance and Inductance
Irwin, Basic Engineering Circuit Analysis, 9/E
6.73 Th circuit shown in Fig. P6 .73 i known as a "D bo ' int grat r.
(a) Express the utput vol. age in t rms o. tb ' npu voUag and circuit p ram ters.
(b) How j d1 D bo int grn~or 's performance differe11 t' m hat of a standard integrator?
(c) \Vhat kind of application w uld j u ti y d1 use of tbi device?
-y R~
... Figure P6.73
SOLUTION:
+ R ..
l's
+
Vo
-i cr R,., ·' ~-
CO.)
V5-v,... C. JVA t VA-Vo R CiT R3
Vs. cdVA + VA-Vo. -tVA --R dl ~ . R
Chapter 6: Capacitance and Inductance Problem 6. 73
2 Irwin, Basic Engineering Circuit Analysis, 9/E
l~J Vs = ;z, R c_ d.Vo -t 0 R1 R. - R ) Vo (( ,-t 12.,_ C}.J;: (R,+RJ ~~ (<~
-t ( R1 ) v R ,+ r<,_ o
'vs =- R,f?-.C dVo t [~I ( Rif<d - (!_ ]Vo R,-t-R.l.. cl.:t R3 ( R.,-t£.z..) R~
(b) Co-n~IJ.e.y the. c. ~se. w h.e. Y\ ;
Problem 6. 73 Chapter 6: Capacitance and Inductance
Irwin, Basic Engineering Circuit Analysis, 9/E 3
t
M n!l.tu prui Ji ve .
(c)
Chapter 6: Capacitance and Inductance Problem 6. 73
Irwin, Basic Engineering Circuit Analysis, 9/E
6.7 4 An integrator is required that ha the oiiowing perf rman e
v0 ( t) = lcfl J v~ dt
w her b capaci.tor value must b great r than 10 nF and the r si t r values mu t b gr at r than
10k.fl.
(a) Design tb int gratol.'.
(b) If ::!: 1 0-V uppUe are used, what are the maxi.mum
and minimum vah1.e of v0?
(c) S 1ppose v~ = 1 V. \\'hat .i the rate of d1, l'lge of v0?
SOLUTION: c
v. = rJ.. ~ .,...., J ~<.o cU:
20flF )
(<.1. -;:: 2... O~_!L
rz ~::: ~ Mft-
--
Chapter 6: Capacitance and Inductance Problem 6.74
2 Irwin, Basic Engineering Circuit Analysis, 9/E
(b) -tt0\1
(C) \Is= IV
Problem 6. 7 4 Chapter 6: Capacitance and Inductance
Irwin, Basic Engineering Circuit Analysis, 9/E
6.75 A driverless au~om ·bile .i under deve]opm nt On critical i sue i braking, patticubdy at red lights. It · d cided that th braking ffort should d p nd on distance to the tight (i f you're c] , you bett r top now) and peed if you're going fast, you'll. need more brake ). Th
resulting design q~ atio n i
braking eff rt = Kt [ d~~t)] + K2x(t)
where x, the distance from th vebi.cle o th .i11te1·s cdon, i m asured by a ns ..r who output is prop rtionai to X, Vsen.se = CiX. U SUp .!pOSition to l1 W that th circuit in Fig. P6.75 can produc the brak"ng effort signal.
Ra
V euse
Figure P6.75
SOLUTION:
c
Chapter 6: Capacitance and Inductance
~.
Problem 6.75
3
Irwin, Basic Engineering Circuit Analysis, 9/E
6.8 Th voltage across a 50-!J..F capacitor i hown io Fig. P6.8. D tennine the current wav fomi.
v t v
Figure P6.8
SOLUTION:
i(t)
iC t) 50 ).A [ )000]
25omA
2 ms <: t- S:.. Llms
OA
~Oh- LpYIS < :t~ ?ms
i(t) 5o ~.A- [- s-ooo] i(t-) -L5b ~nA
'-bo~ gms ~ t~ IOm.s
j(t-) = OA
Chapter 6: Capacitance and Inductance Problem 6.8
2 Irwin, Basic ,Engineering Circuit Analysis, 9/E
loms ~j: ~ /2mS
i(r) = 50..u. [ 5Qoo]
~ah l(;t} = 25DmA
l(f) --(tn mA)
Problem 6.8
:t 7 /2m.s
iCt)-= oA
250
0 -250
0 250
0
0 ~ :t :S 2ms 2ms ~ :t ~ Wrns
Lrms' *~ ~ms
~ms.~ ~ ~ I a'rf"'6
/O'YnS ~ :J: ~ 12 YYIS
t- 7 I 2... m s
Chapter 6: Capacitance and Inductance
Irwin, Basic Engineering Circuit Analysis, 9/E
6.9 Draw the wavef rm fi .1!' the m'fellt in a 1 -j.LF capacitor wh nth c paci.tor voltag i as described in Fig. P6.9.
v(t) (v)
12
t (fl-S) - 8
Figu re P6.9
SOLUTION:
l(t)
i(:t) = 2LJA.
6 u.S <.. t: ~ IOJ.A.S
~(;t) 12.U [ - 5 'X/06 J
i (:t) -6oA
~Jt, lop-s ~ :t .$. 16J.JS
i(:t) 12j.J-[1 · 3~~/D6] j(:J.:) -= 16A
~o9\. r / 16 J.A~
i(J:.)=O
Chapter 6: Capacitance and Inductance Problem 6.9
2
'30
2..0 r------- -- ----
IO
0
-lo ,........ 4: -20 ..._, r--+I. -3,() ........... ·-.(_
-40
-s-o t----------- --
-60 -:to
Problem 6.9
Irwin, Basic Engineering Circuit Analysis, 9/E
2YA.
-6oA
16 A
OA
i(:t) vs·t=
. ----------------------·
:t(u~)
D ~ t ~ 6 I-tS
6w ~ s :t ~togs
lOLLS~ t~ !6J.AS t>16us
·-·-·----~-------- ·-·---·--
20
Chapter 6: Capacitance and Inductance
Irwin, Basic Engineering Circuit Analysis, 9/E 1
Chapter 6: Capacitance and Inductance Problem 6.FE-1
SOLUTION:
The correct answer is a. Yes. The capacitors should be connected as shown.
F2 F4
F6
eqC
FCeq
366
)6(6
Irwin, Basic Engineering Circuit Analysis, 9/E 1
Chapter 6: Capacitance and Inductance Problem 6.FE-2
SOLUTION:
The correct answer is c.
dttitq )()(
stC
stCt
tC
tq
1,6
10,6
0,0
)(
C
tqtv
)()(
stV
stVtx
tV
tv
1,6
10,106
0,0
)( 6
)(2
1)( 2 tvCtw
stJ
stJtx
tJ
tw
1,18
10,1018
0,0
)( 26
Irwin, Basic Engineering Circuit Analysis, 9/E 1
Chapter 6: Capacitance and Inductance Problem 6.FE-3
SOLUTION:
The correct answer is b.
The voltage across the unknown capacitor Cx is (using KVL):
xV 824
VVx 16vCq
The capacitors are connected in series and the charge is the same.Cq 480)8(60
Fv
qCx
3016
480
Irwin, Basic Engineering Circuit Analysis, 9/E 1
Chapter 6: Capacitance and Inductance Problem 6.FE-4
SOLUTION:
mH6
eqL
mH12
mH3
mH3 mH9
mH2
The correct answer is d. mmmmmmLeq 2]36]12)93[([
mmmmmLeq 2]36]12)12[([
mmmmLeq 2]366[
mmmLeq 2]33[
mmLeq 25.1
mHLeq 5.3
Irwin, Basic Engineering Circuit Analysis, 9/E 1
Chapter 6: Capacitance and Inductance Problem 6.FE-5
SOLUTION:
The correct answer is a.
dt
tdiLtv
)()(
tttt teeetedt
tdi 2222 402020)2(20)(
tt teemtv 22 402010)(
0,4.02.0
0,0)(
22 tVtee
tVtv
tt