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


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


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 )



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


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


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



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


o·o~

....-.... <C '-' ........ 0 ~

.......... -~

-o·o<-f


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 \ \



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


2

V(t )=

Problem 6.15

0 '25DO.t 2 V

l-JV


::t<o o ~· .:t < ltam s

.f 4 40tn~



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



'-{-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



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


2

Problem 6.17


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


Irwin, Basic Engineering Circuit Analysis, 9/E 3

Vc C 5ms) L-r·5 -tooo [.x-LrmJ Y·5- tooo [5m- Ltm]



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)


_)_ ( 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= -~


\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-



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


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


1

Problem 6.19

2

Problem 6.19


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



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



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==


I oom ( Lfm -) 2oom

2.rnH

Problem 6.20


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



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


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


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)


-1:) )_ 1·25 ( 1--€'

o ·5 IJ-I

UJ , £70

Problem 6.24


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


) -;t 70 ~

Problem 6.25

-


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)


S

S

S

S

ss


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) =


t

2

Problem 6.27


- 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



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)


.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


t ( l'fr1S)



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 ()


-3omV

6omv

:t < 0

0( ~~Lime:;.

~ Y'(\..5 <. t ~ 6YY" 5

;t 7 6ms

Problem 6.29


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



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]



V( t)::.- 2 · 5V

\X:t)(V)

'" 2'5 ~

. y le ~

6 8 11. ~c

2·5



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.)=


2

V(tJ

Problem 6.31

± /lams v( :1:) = ov

0\1

-somv l5mV 0\J

-50)')'")\J

ov


-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



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


2

Problem 6.32


~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



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


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



--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..



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



V(t) vs.t

~ :)

1.. ,........,

> ~ 0 ~ --.. ......

-I 0•0015 '1

-2

-s -y ------------- --~------------

±(~)



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)=-


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


A 1n ~ tnKtvol

0 b I ntt sc...st

= -2(J



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-(~


3

Problem 6.35


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,·.



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


2.24


·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)-=-



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.


,


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


I


(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 :;;



·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


,


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,


2

,

Problem 6.40


~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


/


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


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


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


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



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


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.-


c= L(t5) (:: o ·I ( C l.f~ to->,f)



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



' .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



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


--

_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


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



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



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~· ·



C ma_x. 8 .u. F

Problem 6.45 Chapter 6: Capacitance and lnd'uctance


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-'



·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


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


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



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


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


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


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



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)



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



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'



c..' T JbuF

CT = C/(6.u) -t-c, c'-t 64

CT - §ML 6Lt) T3J..-t -Gu+ 64

Cr GuF



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' -



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



·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


Problem 6.'53 Chapter 6: Capacitance and Inductance


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


6 ,f!!-F

sMl c~

I 6u Jc~

..

l c"

J Problem 6.54


( 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''



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.~~



(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





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.



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



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~



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



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


j

Problem 6.58




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



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



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.,


2

Problem 6.60

v,:: CL VL c,

c2- vL-tVl== 12 c,


v2 -= ~v

v,+g= 12. V 1 ::: 4V



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



~tLq::: LtmH

Le - ~. L b~_Ldl_ L1 i-Lb-+ Ltt

Le = 3mH


B

A


·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



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



Lr



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


Lf

b

c

2mH

Red oC.U..U i -rut d-

c

Problem 6.64


Le~ - [~IIY)+2-]~H

L - b { y) -r 2 -e.ay 2-

- ~·~ "'(Y\.'t-1 -



·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



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~



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



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


D

Problem 6.66


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



·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


2

(12 rn) ( L) + y m Lf6m

~1 Y)'l)( L) -(6m) ( L) Lt6m


6m -t-(GYYI)L L t Grn

2m

(bm) ( L) - 2m ( L-+ 6Y¥') -

(Ym) ( L) - }2 )A -

L= 12J.A Lrm

L= ~1'nH



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



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



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



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



V(t) ov J :t<.o

lOOt V ,

0·2 v



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]



Vo (t)



·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


2

c-=

Problem 6. 71

I 10

1·2Sur



3

ttt


Vo(t) -21·11 Stn37lt V

Problem 6. 72 Chapter 6: Capacitance and Inductance


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



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


t

M n!l.tu prui Ji ve .

(c)



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-

--



(b) -tt0\1

(C) \Is= IV

Problem 6. 7 4 Chapter 6: Capacitance and Inductance


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


~.

Problem 6.75

3


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


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



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


2

'30

2..0 r------- -- ----

IO

0

-lo ,........ 4: -20 ..._, r--+I. -3,() ........... ·-.(_

-40

-s-o t----------- --

-60 -:to

Problem 6.9


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 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



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



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



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



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

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Solution Chap6