circuits sol ch 01 circuit variables oct 2010

7/21/2019 Circuits Sol Ch 01 Circuit Variables Oct 2010

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October 2010

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بنت

ا

ـنع

ـ " !

#$ % ا

& ك

'( )*+( ةا

,&! -.-/& ت

3ا

ي

& ب

! 4

مو

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بVW Xد (YZ[ \]]]]2^[email protected]ة+ _ ` a &W d V*+( !eng-hs.com, eng-hs.netل

Chapter 1

Circuit Variables

ي + _ ة (N &W % ا& ك'

f د4 ا +(اد يV ا ت _ g! , % ة

f د4 ي dعى ا V*+(ا '(ك + ت Yة

h ـ نـ

ا

i ـ ـ

V!

4 j ـ ـ

dV ا

ZTk ع l' ي + mn ي

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October 2010

1Z1 :=ectr6cB= :;<6;eer6;<P I; OHerH6e• Electrical engineering is the profession concerned with sste!s that

produce" trans!it" and !easure electric signals#

J>e p6He ABqor c=B886p6cBt6o;8 op e=ectr6cB= 8E8teA8P1# 5oAA7;6cBt6o; 8E8teA8P $uch as ca!eras" trans!itters" recei%ers#&# 5oAM7ter 8E8teA8P $uch as personal co!puters#'# 5o;tro= 8E8teA8P $uch as control of te!peratures" pressures" and flow

rates in an oil refiner#(# oer 8E8teA8P $uch as generators#)# -6<;B=sMroce886;< 8E8teA8P $uch as i!age*processing sste!s#

I; e=ectr6c c6rc76tP+s a !athe!atical !odel that appro,i!ates the beha%ior of an actual electrical

sste!#

@ote t>BtP the ter! electric circuit is co!!onl used to refer to an actual

electrical sste! as well as to the !odel that represents it#

+n this te,t" when we tal- about an electric circuit" we alwas !ean a !odelZ

0234ا /.ة# 78ا 6 5 9;<= >?;< @?4 ة ABD م 9 ماF?< @?4 ة ;2A4 GH< @I 0234ا /.ة JK;24ة

هيادين

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'( )*+( ةا

,&! -.-/& ت

3ا

ي

& ب

! 4

مو





ل

*"! &N 3u

( ت

ا

Zv ـ

#( ا

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October 2010

1Z2 J>e Q;ter;Bt6o;B= -E8teA op w;6t8• Lhe $+ units are based on se%en basic units:

JILK: 1Z1 J>e 6;ter;Bt6o;B= -E8teA op w;6t8 x-Qy

z7B;t6tELB86c w;6t-EAbo=

Ke;<t>.eterA

.B88{6=o<rBA{<

J6Ae8eco;C8

:=ectr6c c7rre;tBAMereI

J>erAoCE;BA6c teAMerBt7reCe<ree {e=H6;|

IAo7;t op 87b8tB;ceAo=eAo=

K7A6;o78 6;te;86tEcB;Ce=BcC

+n addition" defined Muantities are co!bined to for! deri%ed units#

JILK: 1Z2 Der6HeC w;6t8 6; -Q

z7B;t6tEw;6t @BAe x-EAbo=ySorA7=B

Sre}7e;cE~ert• x~•y8s1

Sorce@eto; x@y|< Z A R 82

:;er<E or or{ Go7=e xGy@ ZA

oer€Btt x€yG R 8

:=ectr6c c>Br<e5o7=oAb x5yI Z 8

:=ectr6c Mote;t6B= xHo=tB<eyo=t xyG R 5

:=ectr6c re868tB;ceO>A x‚y R I

:=ectr6c co;C7ctB;ce-6eAe;8 x-yI R

:=ectr6c cBMBc6tB;ceSBrBC xSy5 R

.B<;et6c p=7ƒ€eber x€by Z 8

Q;C7ctB;ce~e;rE x~y€b R I

• +n !an cases" the $+ unit is either too s!all or too large to use con%enientl#

$tandard prefi,es corresponding to powers of 1N are applied to the basic unit#

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aNــت # ) ــل ــ ا !'„ كلZنع „+…a †4 ‡/W

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October 2010

• Oll of these prefi,es are correct" but engineers often use onl the ones for

powers di%isible b 'P thus centi" deci" de-a" and hecto are used rarel#

JILK: 1Z? -tB;CBrC6•eC rep6ƒe8 to -6<;6pE oer8 op 10rep6ƒ-EAbo=oer

BttoB10s1ˆ

peAtop 10s1‰

M6coM10s12

;B;o;10s\

A6croŠ10s^

A6==6A10s?

centic1N*&

decid1N*1

de-ada1N>ecto>102

{6=o{ 10?

Ae<B.10^

<6<B‹10\

terBJ1012

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Œ ( ع ىع X& $ا ةa( ا )Ž ZŒY

dل

ي

' ا

ت

ا

ة

!

‘(!

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October 2010

1Z^ o=tB<e B;C 57rre;t

8oAe 6AMortB;t c>BrBcter68t6c8 op e=ectr6c c>Br<eP

0234ا /.ة ;QRH4ا 5 .S./4ة T/ اU4

W Lhe charge is bipolar" !eaning that electrical effects are described in ter!s

of positi%e and negati%e charges#W Lhe electric charge e,ists in discrete Muantities" which are integral

!ultiples of the electronic charge" 1#XN&& Y 1N*1Z C#

• o=tB<e is the energ per unit charge created b the separation:

v=dw

dq

[here:

v=the voltage∈volts . w=the energy∈ joules. q=the charge∈coulombs.

• :=ectr6c c7rre;t is the rate of charge flow:i=

dq

dt

[here:i=the current ∈amperes .

q=the charge∈coulomb .

t =the time∈seconds.

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! +N (` اد ةg g_ ا ةa’ ن ( اZ

V’عى

د

* †4

Vn! †(*

• \uestion 1#Z:

Lhe current entering the upper ter!inal of ]ig# 1#) is:i=24cos 4000 t A

Ossu!e the charge at the upper ter!inal is ^ero at the instant the current is

passing through its !a,i!u! %alue# ]ind the e,pression for M_t`#

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October 2010

1Z‰ J>e QCeB= LB86c 56rc76t :=eAe;t /02 ا /.ة U; ع وA;4 4 U4ا 5

• On ideal basic circuit ele!ent has three attributes:

1# it has onl two ter!inals" which are points of connection to other circuit

co!ponents#&# it is described !athe!aticall in ter!s of current andor %oltage#'# it cannot be subdi%ided into other ele!ents#

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Solution:

i=dq

dt

∫q

o

q

dq=∫0

t

i dt

[ q ] qqo

=∫0

t

24cos 4000 t dt

¿ 24

4000[ sin4000 t ] t

0

q=6∗10−3

sin4000 t +qo

Current is !a,i!u! when cos 4000t is !a,i!u! @t =0

q=6∗10−3

sin 4000 (0 )+qo→ qo=0

q=6∗10−3

sin4000 t C

¿6sin 4000 t mC

_ـــ + ا ت )ـــ uا ـــ 'ي ي aـــ +Vدا `)ـــل ـ ـ ـ ـ ـ

m’ا

+_ ـــــب

U ـ ي ـ ـ ـ ـ

`

g

N “‘ ي

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October 2010

• passi%e sign con%ention:[hene%er the direction of current is sa!e as the direction of %oltage drop

across the ele!ent _as in ]ig#l#)`" use a positi%e sign in an e,pression

relateing %oltage with current# jtherwise" use a negati%e sign#

.?4ا #6 م k4 و4 0< ` @ _qع mة 4و K?S k4ع مة Tm ا;< ا< .?4ا I 6و ا< v ا

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! 4

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j بa! ) m ي d ا ل(V ا & ‘ اZ“ * &‘ ك4 + ه „'! ع udV ا

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October 2010

هيادين

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! 4

مو





• Ossess!ent proble! 1#':

The current at the terminals of the

element in Fig. 1.5 is

i=0 t <0

i=20e−5000t

A t ≥0

Calculate the total charge (in

microcoulombs) entering the element

at its upper terminal.

-o=7t6o;P

i=dq

dt

∫0

q

dq=∫0

∞

i dt

q=∫0

∞

20e−5000 t

dt

¿ 20

−5,000[ e

−5000t ] ∞

0

¿−4∗10−3

(e−∞

−e

0

) ¿−4∗10−3 (0−1 )

¿4∗10−3C

¿4 mC

¿4000 μC

ـ ـ ـ ـ– ـ uع )“ !—ــــ ل ـ ـ ـd نN 4 % د 4 ̃ا ™†

ـ ـ كـ

ـــ

( f ـــ

ـــة

, ت ـ ـ ـ

a’

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October 2010

1Z^ oer B;C :;er<E p=

dw

dt

where:

p x the power in watts" w x the energ in youles" t x the ti!e in

seconds#

Lhus 1[ is eMui%alent to 1 zs#

• Lhe power can be related with %oltage and current as follows:

p=dw

dt =( dw

dq )( dq

dt )so

p=vi

where:

p x the power in watts" v x the %oltage in %olts" i x the current in

a!peres#

• Olgebraic sign of power:+f the power is positi%e _that is" if p { N`" power is being deli%ered to the

circuit inside the bo,# +f the power is negati%e _that is" if p | N`" power is

being e,tracted fro! the circuit inside the bo,#

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ل

’ j ه

'N ي

4 T

VN T̃ا

™Z&œ ! /nN &# ا

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October 2010

هيادين

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• Ossess!ent proble! 1#):

Assume that a 20 V voltage rop occurs across an element

from terminal 2 to terminal 1 an that a current of ! A enters

terminal 2.a)"pecif# the values of v an i for the polarit# references

sho$n in Fig. 1.%(a)&().b)"tate $hether the circuit insie the bo' is absorbing or

elivering po$er.c` o$ much po$er is the circuit absorbing

+ ـــ ي 4 T ـــk( ا ـــ ــ

žg تــ

aNــ

Œ ــي

-o=7t6o;P

• $tart b drawing a picture of the circuit

described in the proble! state!ent:

a` }ow we ha%e to !atch the %oltage and current shown in the left figure with

the polarities shown in ]ig# 1#X#(a ) v=−20V i=−4 A (b ) v=−20V i=4 A

(c ) v=20V i=−4 A (d ) v=20V i=4 A

b"c` ~sing the reference sste! in ]ig# 1#X_d` and the passi%e sign

con%ention" p=vi=(20 ) (4 )=80! (absorbs)

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October 2010

هيادين

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! 4

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• Ossess!ent proble! 1#X:Ossu!e that the %oltage at the ter!inals of the ele!ent in ]ig# 1#) corresponding to the

current in Ossess!ent •roble! 1#' isv=0 t <0v=10e

−5000 t "V t ≥0

Calculate the total energ _in youles` deli%ered to the circuit ele!ent#

• Ossess!ent proble!s 1#':

The current at the terminals of the clement in Fig. 1.5 is

i=0 t <0

i=20e−5000t

A t ≥0

-o7=t6o;P p=vi=(10,000 e

−5000t ) (20e−5000 t )=200,000 e

−10,000t

¿2#105

e−10,000 t =

dw

dt

w=∫0

∞

p dt =∫0

∞

2∗105

e−10,000t

dt = 2∗10

5

−10,000[e−10,000 t ]∞

0

¿ 2∗10

5

−10,000 ( e−∞−e0 )= 2∗10

5

−10,000 (0−1 )=20$

4 ! 3 Tتd ة@o B6;8 6t>o7t B6;8Z

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October 2010

هيادين

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• \uestion 1#11:

jne Z V batter supplies 1NN !O to a ca!ping flashlight# €ow !uch energ

does the batter suppl in ) h

Solution:

p=vi

¿9∗(100∗10−3)

¿0.9!

p=dw

dt w=∫ p dt

¿∫0

5h

0.9dt

¿0.9 [t ]5h

0

¿0.9# (5#60#60 )

¿16,200 $

¿16.2"$

ـ ـ ـ ـ ـ ـ ـ ـ ـ ا ـــــــــ

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October 2010

هيادين

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• \uestion 1#1&:Lwo electric circuits" represented b bo,es O and ‚" are connected as shown in ]ig#

•1#1&#Lhe reference direction for the current i in the interconnection and the

reference polarit for the %oltage % across the interconnection are as shown in the

figure# ]or each of the following sets of nu!erical %alues" calculate the power in theinterconnection and state whether the power is flowing fro! O to ‚ or %ice %ersa#

a` i=5 A v=120V

b` i=−8 A v=250V

c` i=16 A v=−150V

d` i=−10 A v=−480V

ا

` اš †4 † نك ا ىع … W uN Tد ™

ع

V’ ا

ا

د

Ÿى

n#U

• -o7=t6o;P

$ince the current i flows into +ve ter!inal of %oltage v " we can use the

eMuation p=vi without % ve sign# if the power is +ve " so it flows fro! O

to ‚# if the power is −ve " so it flows fro! ‚ to O#

a¿ p=(120) (5)=600! 600! ¿ A ¿& b¿ p=(250 ) (−8 )=−2000! 2000! ¿& ¿ A

c ¿ p=(−150 ) (16 )=−2400! 2400! ¿& ¿ A d ¿ p=(−480 ) (−10 )=4800! 4800! ¿ A¿ &

a` p x _*(ƒN`_*1N` x (ƒNN [ (ƒNN [ fro! O to ‚

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October 2010

هيادين

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• \uestion 1#1):[hen a car has a dead batter" it can often be started b connecting the batter fro!

another car across its ter!inals# Lhe positi%e ter!inals are connected together as are the

negati%e ter!inals# Lhe connection is illustrated in ]ig# •1#1)# Ossu!e the current i

in ]ig# •1#1) is !easured and found to be 'N O#a` [hich car has the dead batter

b` +f this connection is !aintained for 1 !in" how !uch energ is transferred to the

dead batter

-o7=t6o;Pa` Current i=30 A flows fro! +ve %olt to % ve %olt in car O" $o car O is

absorbing power# Car O has the dead batter#

b¿ p=v i ¿12#30=360!

w=∫0

60

p dt =∫0

60

360dt =360 [t ]600

¿21,600 $ =21.6 "$

Œ ــW ــœ ا ب_ ــ, T ــN Œ ــW ا ا Ž _ــ & 3يت Ÿ Tـ &’4 ــ * –¡لZŒ ي

X&’+ ت

( ا

X

Y+ ا

ة

+

a ا

+k’

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October 2010

هيادين

كـل

عى

بنت

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• \uestion 1#1X:

Lhe !anufacturer of a Z V dr*cell flashlight batter sas that the batter will

deli%er &N !O for ƒN continuous hours# „uring that ti!e the %oltage will drop

fro! Z V to X V# Ossu!e the drop in %oltage is linear with ti!e# €ow !uchenerg does the batter deli%er in this ƒN h inter%al

-o7=t6o;P

p=vi

pt =0=9# (20#10−3 )=0.18!

pt =80h=6# (20#10−3

)=0.12!

w=∫0

80h

p dt = Area o' ( p−t )diagram

¿( 0.12+0.182 )# (80#60#60)

¿43,200 $ =43.2 "$

’ Xــ / ــa ا Xد ـ ـ* بــ—N ا Uــ Vبة +اـــ ـ ـ ـg! ة ـ ـ يـ¢& ا ) # تـــaN ل ـ ـ ـ ا [“£ ل

( ــت

,بــ

¤ Œ ــن

X

a ا

ا

4

n—N

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October 2010

هيادين

كـل

عى

بنت

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ـ " !

#$ % ا

& ك

'( )*+( ةا

,&! -.-/& ت

3ا

ي

& ب

! 4

مو





Lhe %oltage and current at the ter!inals of the circuit ele!ent in ]ig# 1#) are shown in

]ig# •1#1Z#

a` $-etch the power %ersus t" plot for N … t … )N s#

b` Calculate the energ deli%ered to the circuit ele!ent at t x (" 1&" 'X" and )N s#

• \uestion 1#1Z:

• "olution*

a` p=vi

t (sec)0−48−1620−3646−50

v (V )2.5 t 30−2.5 t t −3075−1.5 t

i ((A )1.0−1.00.4−0.6

p((! )2.5 t 2.5 t −300.4 t −120.9 t −45

N 3نـ[ ا ! 4ا Y [+ تى

ا % ـ ـ ـ ـ ’ ى ـ ـ ـ ـع ا ـ ـ ـ ـ¤

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October 2010

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مو





s ?Z^

2Z]

s]Z0

s10

10 10

• 5o;t6;7eC x}7e8t6o; 1Z1\yP

b)Calculate the area uner the curve from +ero up to 50 s*

w (4)=1

2(4 ) (10)=20 μ $

w (12 )=w (4 )+1

2(4 ) (−10 )=0 $

w (36 )=w (12)+1

2(4 ) (10 )+

1

2(10 ) (−4 )+

1

2(6 ) (2.4 )=7.2 μ $

w (50)=w (36)+ 12

(4 ) (−3.6 )=0 $

ـ ¥ـ

_ ب

N ــ

ــ

ــة

(_ ا

ــع

ـ ¥ ـ

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October 2010

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• \uestion 1#&&:

Lhe %oltage and current at the ter!inals of an auto*!obile batter during a charge ccle

are shown in ]ig# •1#&&#

a` Calculate the total charge transferred to the batter#

b` Calculate the total energ transferred to the batter#

•

-o7=t6o;Pa) q=areaunder (i−t ) plot

¿[ 12 (4 ) (5 )+ (4 ) (10 )+1

2(8 ) (4 )+(8 ) (6 )+

1

2(3 ) (6 )]#10

3

¿ [10+40+16+48+9 ]103=123,000C

ŒV ــŽ + ــ ي 4 ــ(( ا ــ ــ4% ـ ـ ـ ـ

ا

ــــ

n ن

ــــل

m’4 ــــ

_ ا

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October 2010

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5o;t6;7e x}7e8t6o; 1Z22yPb) w=∫ p dt =∫ vidt

v=0.2#10−3

t +9

• 0)t ) 4000 s

i=15−1.25#10−3

t

p=135−8.25#10−3 t −0.25#10−6 t 2

w1=∫

0

4000

(135−8.25#10−3

t −0.25#10−6

t 2 ) dt

¿ (540−66−5.3333 )103=468.667 "$

• 4,000) t )12,000

i=12−0.5#10−3

t

p=108−2.1#10−3

t −0.1#10−6

t 2

108−2.1#10−3

t −0.1

(¿#10−6

t 2)dt

w2= ∫4000

12,000

¿

¿ (864−134.4−55.467 )103=674.133"$

• 12,000)t )15,000

i=30−2#10−3

t

p=270−12#10−3−0.4#10

−6t 2

w3= ∫

12,000

15,000

(270−12#10−3

t −0.4#10−6

t 2 ) dt

¿ (810−486−219.6 )103=104.4 "$

w* =w1+w

2+w

3=468.667+674.133+104.4=1247.2 "$

Œ & $ X& œŸ V(W ‡/mNة 4“¦

ا

V

N § ( ع

n ن

N

4

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October 2010

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• \uestion 1#&':

Lhe %oltage and current at the ter!inals of the circuit ele!ent in ]ig# 1#) are ^ero for t |

N" and for t † N the are:

v=(16,000t +20) e−800t

V

i=(128 t +0.16 ) e−800 t

A

a` Ot what instant of ti!e is !a,i!u! power deli%ered to the ele!ent b` ]ind the !a,i!u! power in watts#c` ]ind the total energ deli%ered to the ele!ent in !illiyoules#

ŒV ــ

Ž

ـ +ـ

ي

4 ــ

(( ا

ـــ

ــ

4ا ــــ % ــــn ن ا _ــــ m’4 ــــل

• -o7=t6o;Pa` p=vi=[ (16,000 t +20 )e−8000 t ] [ (128 t +0.16 ) e

−800t ]¿2048#10

3t 2

e−1600 t +5120 t e

−1600 t +3.2e−1600 t

¿3.2e−1600 t [640,000 t

2+1600 t +1 ]

dp

dt =3.2 {e−1600 t [1280#10

3t +1600 ]−1600e

−1600 t [640,000 t 2+1600t +1 ] }

¿−3.2e−1600 t [128#10

4 (800 t 2+t ) ]

¿−409.6#104

e−1600t

t (800 t +1)

*here'ore dp

dt =0whent =0

so p ma+ occurs at t =0

b¿ pma+=3.2e−0 [0+0+1 ]=3.2!

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October 2010

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5o;t6;7eC x}7e8t6o; 1Z2?yP

c ¿w=∫0

t

p d+

640,000 t 2

e−1600 t

dt +¿∫0

t

1600 t e−1600t

dt +∫0

t

e−1600t

dt

w3.2

=∫0

t

¿

¿640,000 e

−1600 t

−4096#106 [256#10

4t 2+3200t +2 ] t

0+1600 e

−1600t

256#104

(−1600 t −1 ) t

0+[ e

−1600 t

−1600 ] t

0

!hent → ∞ all theupper limits evaluate ¿ ,ero hence

w3.2

= (640,000 )(2)4096#10

6 + 1600256#10

4+ 11600

w=10−3+2#10

−3+2#10−3=5m$

’ /` ن & ¤ … Wا ,ت ‘ (VN Tل ̃ا ™Z د V ا ‡ — ا ىع 3™ U_Nل

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October 2010

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• \uestion 1#&):

Lhe %oltage and current at the ter!inals of the circuit ele!ent in ]ig# 1#) are:

^ero for t | N" and for t † N the are:

v=(10,000t +5 ) e−400t

V t ≥0

i=(40 t +0.05) e−400t

A t ≥0

a` ]ind the ti!e _in !illiseconds` when the power deli%ered to the circuit ele!ent is

!a,i!u!# b` ]ind the !a,i!u! %alue of p in !illiwatts#c` ]ind the total energ deli%ered to the circuit ele!ent in !illiyoules#

• -o7=t6o;Pa` p=vi=400#10

3t 2

e−800 t +700 t e

−800t +0.25 e−800 t

¿e−800 t [400,000 t

2+700t +0.25 ]dp

dt ={e

−800 t [800#103

t +700 ]−800e−800 t [400,000 t

2+700 t +0.25 ] }

¿ [−3,200,000 t 2+2400t +5 ] 100e

−800 t

*here'ore dp

dt =0when3,200,000 t

2−2400 t −5=0

so p ma+ occurs at t =1.68ms

b` pma+=[400,000 (0.00168 )2+700 (0.00168 )+0.25 ] e−800(0.00168)

¿666m!

Œ & $ X& œŸ V(W ‡/mNة 4“¦

ا

V

N § ( ع

n ن

N

4

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October 2010

هيادين

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• 5o;t6;7eC x}7e8t6o; 1Z2‰yP

c` w=∫0

t

p dt

w=∫0

t

400,000 t 2

e−800t

dt +∫0

t

700 t e−800 t

dt +∫0

t

0.25e−800t

dt

¿ 400,000e

−800 t

−512#106 [64 #10

4t 2+1600t +2 ] t

0+700e

−800 t

64#104 [−800t −1 ] t

0+0.25 [ e

−800t

−800 ] t

0

whent =∞ all theupper limits evaluate¿ ,er hence

w=(400,000 )(2)

512#106 +

700

64#104+0.25

800 =2.97m$

¨

N&

N

Y © ع

ا

Y4 ª&VNهل

X د „“$ Œ˜ Œ ن( يPŒN

Y « ع

Ž

( ع ي

ن

, ا

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October 2010

هيادين

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بنت

ا

ـنع

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! 4

مو


• \uestion 1#&X:

The numerical values for the currents an voltages in the circuit in Fig.

,1.2% are given in Table ,1.2%.Fin the total po$er evelope in the

circuit.

• -o7=t6o;P

p=vi pa=vaia=(150) (0.6)=90m! pb=vbib=(150) (−1.4 )=−210m!

pc=−vc ic=−(100) (−0.8)=80m! pd=vd id=(250) (−0.8 )=−200m!

pe=−ve ie=−(300) (−2)=600m! p' =v ' i ' =(−300) (1.2 )=−360m!

• +f the power is positi%e" the circuit ele!ent is absorbing" otherwise de%eloping#

∑ pabsorbing=90+80+600=770m!

∑ pdeveloping=210+200+360=770m!

• Lhe power balances _j‡`

Z k نع ا+‘_! ا Œ˜ (Vل m’4ل &¬ –+Nي gة دي

circuits sol ch 01 circuit variables oct 2010

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