Solutions Chapter 02, 03, 04 Network Analysis 3rd Edition - M. E. v. Valkenburg

8/12/2019 Solutions Chapter 02, 03, 04 Network Analysis 3rd Edition - M. E. v. Valkenburg

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Muhammad Irfan Yousuf

Dedicated to: Prof. Dr. Sohail Aftab Qureshi

Conventions for Describing Networks

2-1. For the controlled (monitored) source shown in the figure, re!re ! lotsimil!r to th!t given in Fig. 2-"(b).

v2

v1# $b

$b

v1# $!

$!

i2 Fig. 2-" (b)

%olution&

'en our book see the figure (*+)

t is volt!ge controlled current source.

i2

/$e !0is

v2-$e !0is

gv1

i2

gv1 /

v2current source

-


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2-2. ee!t *rob. 2-1 for the controlled source given in the !ccom!ning figure.

%olution&


t is current controlled volt!ge source.

v2ri1

i2

2-. 3he network of the !ccom!ning figure is ! model for ! b!tter of oen-circuit

termin!l volt!ge $ !nd intern!l resist!nce b. For this network, lot i !s ! function

v. dentif fe!tures of the lot such !s sloes, intercets, !nd so on.

%olution&


3ermin!l volt!ge

v # $ - ibib #$ - vi#($ - v)+b4hen v # 5

i#($ - v)+bi#($ - 5)+bi#$+b !m

4hen v # $

i#($ - $)+bi#(5)+bi#5 !m

v # 5 i # $+

v # $ i # 5

i

$+b

$ v

%loe&


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# m0 / c

(01, 1) # (5, $+b)

(02, 2) # ($, 5)

m # (26 1)+(026 01) # (5 6 $+b)+($ - 5) # (-$+b)+$ # (-$+b)(1+$) # -1+b-intercet # $+b0-intercet # $

%loe -intercet 0-intercet

-1+b $+b $

2-. 3he m!gnetic sstem shown in the figure h!s three windings m!rked 1-17, 2-27,

!nd -7. 8sing three different forms of dots, est!blish ol!rit m!rkings for thesewindings.

%olution&


9ets !ssume current in coil 1-17 h!s direction u !t 1 (incre!sing). t roduces flu0

(incre!sing) in th!t core in clockwise direction.

1 17 2 27 7

:ccording to the 9en;7s l!w current roduced in coil 2-27 is in such ! direction th!t

it ooses the incre!sing flu0 . %o direction of current in 2-27 is down !t 27.


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1 17 2 27 7

(!)

1 17 2 27 7

(b)2-. 3he figure shows four windings on ! m!gnetic flu0-conducting core. 8sing

different sh!ed dots, est!blish ol!rit m!rkings for the windings.

%olution&

'en our book see the figure (*+>)

i1 i

i

1

(Follow Fleming7s right h!nd rule)

2->. 3he !ccom!ning schem!tic shows the e?uiv!lent circuit of ! sstem with

ol!rit m!rks on the three-couled coils. Dr!w ! tr!nsformer with ! core simil!r to

th!t shown for *rob. 2- !nd l!ce windings on the legs of the core in such ! w! !s

2

i2

Coil 1

Coil

Coil 2

Coil


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to be e?uiv!lent to the schem!tic. %how connections between the elements in the

s!me dr!wing.

%olution&


9

1 2

2-". 3he !ccom!ning schem!tics e!ch show two inductors with couling but with

different dot m!rkings. For e!ch of the two sstems, determine the e?uiv!lent

induct!nce of the sstem !t termin!ls 1-17 b combining induct!nces.

%olution&


9et ! b!tter be connected !cross it to c!use ! current i to flow. 3his is the c!se of

!dditive flu0.

M

91 92

$

i

(!)

$ # self induced e.m.f. (1) / self induced e.m.f. (2) / mutu!ll induced e.m.f. (1) /

mutu!ll induced e.m.f. (2)

$ # 91di+dt / 92di+dt / @ di+dt / @ di+dt

9et 9e?be the e?uiv!lent induct!nce then $ # 9e?di+dt

9e?di+dt # (91/ 92/ @ / @) di+dt

2

i2

91

92


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9e? # 91/ 92/ @ / @

9e? # 91/ 92/ 2@

@

91 92

i

$

(b)

3his is the c!se of subtr!ctive flu0.

$ # 91di+dt / 92di+dt - @ di+dt - @ di+dt

9et 9e?be the e?uiv!lent induct!nce then $ # 9e?di+dt9e?di+dt # (91/ 92- @ - @) di+dt

9e? # 91/ 92- @ - @

9e? # 91/ 92- 2@

2-A. : tr!nsformer h!s 155 turns on the rim!r (termin!ls 1-17) !nd 255 turns on

the second!r (termin!ls 2-27). : current in the rim!r c!uses ! m!gnetic flu0,

which links !ll turns of both the rim!r !nd the second!r. 3he flu0 decre!ses

!ccording to the l!w # e-t 4eber, when t 5. Find& (!) the flu0 link!ges of the

rim!r !nd second!r, (b) the volt!ge induced in the second!r.

%olution&N1# 155

N2# 255

# e-t (t 5)

*rim!r flu0 link!ge 1# N1# 155 e-t

%econd!r flu0 link!ge 2# N2# 255 e-t

@!gnitude of volt!ge induced in second!r v2# d2+dt # d+dt(255 e-t)

v2# -255 e-t


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b

2

1

! c

=

d

d c

=

R1

b 2 !

2-11. 3hree gr!hs !re shown in figure. Cl!ssif e!ch of the gr!hs !s l!n!r or

nonl!n!r.

%olution&

'en our book see the figure (*+")

:ll !re l!n!r.

n th!t the m! be dr!wn on ! sheet of !er without crossing lines.


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2-12. For the gr!h of figure, cl!ssif !s l!n!r or nonl!n!r, !nd determine the

?u!ntities secified in e?u!tions 2-1 2-1.

%olution&


Cl!ssific!tion&

Nonl!n!rNumber of br!nches in tree # number of nodes 6 1 # = 6 1 #

Number of chords # br!nches 6 nodes / 1 # 15 6 = / 1 # 15 6 #

Chord me!ns B: str!ight line connecting two oints on ! curve7.

2-1. n (!) !nd (b) of the figure for *rob. 2-11 !re shown two gr!hs, which m! be

e?uiv!lent. f the !re e?uiv!lent, wh!t must be the identific!tion of nodes !, b, c, d

in terms of nodes 1, 2, , if ! is identic!l with 1

%olution&


(b)

! is identic!l with 1b is identic!l with

c is identic!l with 2

d is identic!l with

2-1. 3he figure shows ! network with elements !rr!nged !long the edges of ! cube.

(!) Determine the number of nodes !nd br!nches in the network. (b) C!n the gr!h

of this network be dr!wn !s ! l!n!r gr!h

%olution&


Number of nodes # "

Number of br!nches # 11

(b) es it c!n be dr!wn.

2-1=. 3he figure shows ! gr!h of si0 nodes !nd connecting br!nches. ou !re to !dd

non!r!llel br!nches to this b!sic structure in order to !ccomlish the following

different obEectives& (!) wh!t is the minimum number of br!nches th!t m! be

!dded to m!ke the resulting structure nonl!n!r (b) 4h!t is the m!0imum

number of br!nches ou m! !dd before the resulting structure becomes

nonl!n!r

%olution&

'en our book see the figure (*+A)

@!ke the structure nonl!n!r

@inimum number of br!nches #

@!0imum number of br!nches # >

2-1. Disl! five different trees for the gr!h shown in the figure. %how br!nches

with solid lines !nd chords with dotted lines. (b) ee!t (!) for the gr!h of (c) in

*rob. 2-11.

%olution&



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1) )

2) )

=)

b)&

1) 2)


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) )

=)

2-1>. Determine !ll trees of the gr!hs shown in (!) of *rob. 2-11 !nd (b) of *rob. 2-

15. 8se solid lines for tree br!nches !nd dotted lines for chords.

%olution&


:ll trees&

1) 2) ) )

=) ) >) ")

A) 15) 11) 12)


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1) 1) 1=) 1)

1>) 1") 1A) 25)

21) 22) 2) 2)

2=) 2) 2>) 2")


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2A) 5)

:ll trees of

%olution&

1)

2)

)

)

efore solving e0ercise following terms should be ket in mind&

1. Node

2. r!nch

. 3ree

. 3r!nsformer theor

=. %loe

. %tr!ight line e?u!tion

>. ntercet

". %elf induction

A. @utu!l induction

15. Current controlled volt!ge source


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11. $olt!ge controlled current source

12. Coordin!te sstem

Network e?u!tions

-1. 4h!t must be the rel!tionshi between Ce?!nd C1!nd C2in (!) of the figure of

the networks if (!) !nd (c) !re e?uiv!lent ee!t for the network shown in (b).

%olution&

'en our book see the figure (*+">)

/ - / -

+

C1 Cv(t) i

!

kirchhoff7s volt!ge l!w&

A""A# M$#AMMAD %P.&.$.#'


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v(t) # 1+C1i dt / 1+C2i dt

v(t) # (1+C1/ 1+C2)i dt

n second c!se

/ -

v(t) Ce?i

v(t) # 1+Ce?i dt

f (!) (c) !re e?uiv!lent

1+Ce?i dt # (1+C1/ 1+C2)i dt

1+Ce?# (1+C1/ 1+C2)

(b) / - ! / -

i C1 / i2 C

- C2i1

b

i # i1/ i2

i # C2dv!+dt / Cdv!+dt when v!is volt!ge !cross !b.3he e?uiv!lent c!!cit!nce between ! b be Ce?7

3hen i # Ce?7dv!+dt

Ce?7dv!+dt # C2dv!+dt / Cdv!+dt

Ce?7 # C2/ C

Di!gr!m (b) reduces to

/ -

/

C1 /

v

Ce?7

-

-

From result obt!ined b (!)

1+Ce?# (1+C1/ 1+Ce?7)


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1+Ce?# (1+C1/ 1+C2 / C)

-2. 4h!t must be the rel!tionshi between 9e?!nd 91, 92!nd @ for the networks

of (!) !nd of (b) to be e?uiv!lent to th!t of (c)

%olution&

'en our book see the figure (*+">)n network (!) !ling G$9

v # 91di+dt / 92di+dt / @di+dt / @di+dt

v # (91/ 92/ @ / @)di+dt

v # (91/ 92/ 2@)di+dt

n network (c)

v # 9e?di+dt

f (!) (c) !re e?uiv!lent

(91/ 92/ 2@)di+dt # 9e?di+dt

(91/ 92/ 2@) # 9e?

n network (b) !ling G$9

v # 91di+dt / 92di+dt - @di+dt - @di+dt

v # (91/ 92- @ 6 @)di+dt

v # (91/ 92- 2@)di+dt

n network (c)

v # 9e?di+dt

f (b) (c) !re e?uiv!lent

(91/ 92- 2@)di+dt # 9e?di+dt

(91/ 92- 2@) # 9e?

-. ee!t *rob. -2 for the three networks shown in the !ccom!ning figure.%olution&

'en our book see the figure (*+">)

/

@v

i1 91 92

loo 1 loo 2 i2

-


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:ling G$9 in loo 1

v # 91d(i16 i2)+dt / @di2+dt

v # 91di1+dt - 91di2+dt / @di2+dt

v # 91di1+dt / @di2+dt - 91di2+dt

v # 91di1+dt / (@ - 91)di2+dt

:ling G$9 in loo 2

5 # 92di2+dt / 91d(i26 i1)+dt / H-@di2+dtI / H-@d(i26 i1)+dtI

5 # 92di2+dt / 91di2+dt - 91di1+dt - @di2+dt - @d(i26 i1)+dt

5 # 92di2+dt / 91di2+dt - 91di1+dt - @di2+dt - @di2+dt / @di1+dt

5 # 92di2+dt / 91di2+dt - 91di1+dt - 2@di2+dt / @di1+dt

5 # (@ 6 91) di1+dt / (91/ 926 2@) di2+dt

4riting in m!tri0 form

91 @ 6 91 di1+dt v

#

@ 6 91 91/ 926 2@ di2+dt 5

v @ 6 91

5 91/ 926 2@

di1+dt #

91 @ 6 91

@ 6 91 91/ 926 2@

v @ 6 91

5 91/ 926 2@

# (v)( 91/ 926 2@) 6 5 # (v)(91/ 926 2@)

91 @ 6 91

@ 6 91 91/ 926 2@

# (91)(91/ 926 2@) 6 (@ 6 91)(@ 6 91)

# (91)(91/ 926 2@) 6 (@ 6 91)2

# (912/ 91926 291@) 6 @

26 912 / 2@91


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# 912/ 91926 291@ 6 @

26 912 / 2@91

# 91926 @2

di1+dt # (v)(91/ 926 2@)+91926 @2

di1+dt H(91926 @2)+(91/ 926 2@)I # v

n network (c)

v i1 9e?

v # 9e?di1+dtFor (!) (c) to be e?u!l

di1+dt H(91926 @2)+(91/ 926 2@)I # 9e?di1+dt

(91926 @2)+(91/ 926 2@) # 9e?

(b)

/

@

vi1 91 i2 92

-

:ling G$9 in loo 1

v # 91d(i16 i2)+dt - @di2+dt

v # 91di1+dt - 91di2+dt - @di2+dt

v # 91di1+dt / @di2+dt - 91di2+dt

v # 91di1+dt - (91 / @)di2+dt

:ling G$9 in loo 2

5 # 92di2+dt / 91d(i26 i1)+dt / @di2+dt / @d(i26 i1)+dt

5 # 92di2+dt / 91di2+dt - 91di1+dt / @di2+dt / @d(i26 i1)+dt

5 # 92di2+dt / 91di2+dt - 91di1+dt / @di2+dt / @di2+dt - @di1+dt

5 # 92di2+dt / 91di2+dt - 91di1+dt / 2@di2+dt - @di1+dt

5 # - (91 / @) di1+dt / (91/ 92/ 2@) di2+dt


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4riting in m!tri0 form

91 - (91 / @) di1+dt v

#

- (91 / @) 91/ 92/ 2@ di2+dt 5

v - (91 / @)

5 91/ 92/ 2@

di1+dt #

91 - (91 / @)

- (91 / @) 91/ 92/ 2@

v - (91 / @)

5 91/ 92/ 2@

# (v)( 91/ 92/ 2@) 6 5 # (v)(91/ 92/ 2@)

91 - (91 / @)

- (91 / @) 91/ 92/ 2@

# (91)(91/ 92/ 2@) - (91 / @)(91 / @)

# (91)(91/ 92/ 2@) - (91 / @)2

# (912/ 9192/ 291@) - @

2- 912 - 2@91

# 912/ 9192/ 291@ - @

2- 912 - 2@91

# 91926 @2

di1+dt # (v)(91/ 92/ 2@)+91926 @2

di1+dt H(91926 @2)+(91/ 92/ 2@)I # v

n network (c)

v i1 9e?


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v # 9e?di1+dt

For (!) (c) to be e?u!l

di1+dt H(91926 @2)+(91/ 92/ 2@)I # 9e?di1+dt

(91926 @2)+(91/ 92/ 2@) # 9e?

-. 3he network of inductors shown in the figure is comosed of ! 1-< inductor on

e!ch edge of ! cube with the inductors connected to the vertices of the cube !s

shown. %how th!t, with resect to vertices ! !nd b, the network is e?uiv!lent to th!t

in (b) of the figure when 9e? # =+


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1+-< 1+-< 1+-

1 2

3

4

5

1

0

1

2

3

4

5

6

7

8

9

time

Series2

Series3

Series2 0 0 5 8 7

Series3 1

1 2 3 4 5

-1>. For e!ch of the four networks shown in the figure, determine the number of

indeendent loo currents, !nd the number of indeendent node-to-node volt!ges

th!t m! be used in writing e?uilibrium e?u!tions using the kirchhoff l!ws.

%olution&

'en our book see (*+A5)

(!) Number of indeendent loos # 2


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1i2/ 92d(i26 i)+dt / 1+c2(i26 i)dt # 5

i&

91d(i6 i1)+dt / 92d(i6 i2)+dt / (i6 i) # v(t)

i&

2i/ (i6 i) / 1+c2(i6 i2)dt # 5

-22. @!ke use of the G$9 to write e?u!tions on the loo b!sis for the fournetworks of *rob. -1".

%olution&

'en our book see (*+A1).

(!)

i1&

1i1 / 1+c(i16 i2)dt # - v(t)

i2&

1+c(i26 i1)dt / 1(i26 i) # 5

i&

1+c1idt / 1(i6 i2) / (i6 i) # 5

i&

1+c(i6 i=)dt / 2(i6 i) # 5

i=&

2i= / 1+c(i=6 i)dt / 1+c2i=dt # - v(t)

i&

2(i6 i=)/ (i6 i>) # - v(t)

i>&

1+c=i>dt / (i>6 i) # 5

(b)

i1&

92di1+dt / 1+c1

(i16 i2)dt / 1+c

(i16 i)dt / 9d(i16 i)+dt # v(t) i2&

91di2+dt / 1+c2(i26 i)dt / 1+c1(i26 i1)dt # 5

i&

9di+dt / 1+c2(i6 i2)dt / 1+c(i6 i1)dt / 9d(i6 i1)+dt / i# 5

(c)

i1&

1+c(i16 i)dt / 1(i16 i2) # v(t)

i2&

1+c(i26 i)dt / 1(i26 i1) / 9(i26 i) # 5

i&

i/ (i6 i) / 1+c

(i6 i2)dt / 1+c

(i6 i1)dt # 5

i&

9(i6 i2) / (i6 i) / 1+c1idt # 5

(d)

i1&

1+c!(i16 i2)dt / 291di1+dt / 9bd(i16 i)+dt / 1+cb(i16 i)dt # v(t)

i2&


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9!d(i26 i)+dt / 1+c!(i26 i1)dt # 5

i&

292d(i6 i)+dt / (i6 i) / 1+c!(i6 i=)dt / 9bd(i6 i1)+dt / 1+cb(i6 i1)dt # 5

i&

9!d(i6 i2)+dt / 9bdi+dt / 1+cbidt / 292d(i6 i)+dt / (i6 i) # 5

(!)

/

-


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i1

i2

i

i

i=

i

i>

(b)

i2

/

-


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i1i

(c)

i

i

/

-

/

-


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i1 i2

(d)

i2 i


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i1

i

i=

-2.

4rite ! set of e?uilibrium e?u!tions on the loo b!sis to describe the network in the

!ccom!ning figure. Note th!t the network cont!ins one controlled source. Collect

terms in our formul!tion so th!t our e?u!tions h!ve the gener!l form of O?s. (-

>).

i2

/

-

/

-

!

+


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i1i

i1&

i1/ (i16 i2) / (i16 i) / 1(i16 i)dt # v1(t)

i2&

1(i26 i1) / 9di2+dt # 5

i&

i/ (i6 i1) / 1(i6 i1)dt 6 k1i1# 5

-2. For the couled network of the figure, write loo e?u!tions using the G$9. n

our formul!tion, use the three loo currents, which !re identified.

%olution&

'en our book see (*+A2).i1&

1i1/ (91/ 92)di1+dt / @di2+dt # v1i2&

9di2+dt / @di1+dt / 1+c(i26 i)dt # v2

i&

2i/ 1+c(i6 i2)dt # 5

-2=. 8sing the secified currents, write the G$9 e?u!tions for this network.

%olution&

'en our book see (*+A2).

i1&

1(i1 / i2/ i) / 91di1+dt / @12di2+dt / 2i1 -@1di2+dt # v1(t)i2&

1(i1 / i2/ i) / 92di2+dt / @12di1+dt /@2di2+dt # v1(t)

i&

1(i1 / i2/ i) / 9di+dt - @1di1+dt /@2di2+dt / 1+cidt# v1(t)

-2. : network with m!gnetic couling is shown in figure. For the network, @12# 5.

Formul!te the loo e?u!tions for this network using the G$9.

i1&

1i1 / 91di1+dt / @1d(i16 i2)+dt /9d(i16 i2)+dt / @2d(-i2)+dt / @1di1+dt / 2(i16 i2) #

v1(t)

i2&

i2 / 92di2+dt / @2d(i26 i1)+dt /9d(i26 i1)+dt / @2d(i2)+dt / @1d(-i1)+dt / 2(i26 i1)

# 5

-2>. 4rite the loo-b!sis volt!ge e?u!tions for the m!gnetic!ll couled network

with k closed.

%olution&

%!me !s .2.

-2". 4rite e?u!tions using the GC9 in terms of node-to-d!tum volt!ge v!ri!bles

for the four networks of *rob. -1>.


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(!)

1 2v1

v2v

9

C

v(t)

2

v1 v2

9

1C

v(t)+1

/

-


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Node-v1:ccording to GC9

%um of currents entering into the Eunction # %um of currents le!ving the

Eunction

v(t)+1 #v1+1 / (v16 v2)+2 / cd(v16 v2)+dt

v(t)+1 #v1+1 / v1+26 v2+2/ cdv1+dt 6 cdv2+dt

v(t)+1 #v1+1 / v1+2 / cdv1+dt 6 v2+26 cdv2+dt

v(t)+1 #v1(1+1 / 1+2 / cd+dt) / (6 1+26 cd+dt)v2

v(t)+1 #v1(P1 / P2 / cd+dt) / (6 P26 cd+dt)v2 ec!use P # 1+


%um of currents entering into the Eunction # %um of currents le!ving theEunction

5#(v26 v1)+2 / cd(v26 v1)+dt / 1+9v2dt / v2+

5#(v26 v1)+2 / cd(v26 v1)+dt / Qv2dt / v2+

5#v2+26 v1+2 / cdv2+dt 6 cdv1+dt / Qv2dt / v2+

5#v2+2 / cdv2+dt / v2+/ Qv2dt 6 v1+26 cdv1+dt

5#v2(1+2 / cd+dt / 1+/ Qdt) / v1(6 1+26 cd+dt)

5#v2(P2 / cd+dt / P/ Qdt) / v1(6 P26 cd+dt) ec!use P # 1+, Q # 1+9

(b)

1

9 C

v(t)

2

/

-


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v1

v(t)+1 9 C

1

2



Eunction

v(t)+1 #v1+1 / v1+2 / cdv1+dt / 1+9v1dt

v(t)+1 #v1+1 / v1+2 / cdv1+dt / Qv1dt Hec!use 1+9 # QI

v(t)+1 #v1(1+1 / 1+2 / cd+dt / Qdt)

(c)

9

C

v(t)

v1

/

-


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9

C

1+9v(t)dt



Eunction1+9v1dt / v1+/ cdv1+dt # 1+9v(t)dt

v1(1+9dt / 1+/ cd+dt) # 1+9v(t)dt

(d)

v(t)

C1 91

92

-

/


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1C2

2

-

/

-

/


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1+91

v(t)dt

91

c1dv(t)+dt

C1

v1v

v2


74/163



Node-v1&c1dv(t)+dt / 1+91v(t)dt # c1dv1+dt / 1+91v1dt / 1+92(v16 v)dt / (v16 v2)+1

(c1d+dt / 1+91dt)v(t) # c1dv1+dt / 1+91v1dt / 1+92v1dt - 1+92vdt / v1+16 v2+1

(c1d+dt / 1+91dt)v(t) # c1dv1+dt / 1+91v1dt / 1+92v1dt / v1+16 v2+1 - 1+92vdt

(c1d+dt / 1+91dt)v(t) # (c1d+dt / 1+91dt / 1+92dt / 1+1)v16 v2+1 - 1+92vdt

Node-v2&

c2d(v26 v)+dt / (v26 v1)+1 / v2+2 # 5

c2dv2+dt - c2dv+dt / v2+16 v1+1/ v2+2 # 5

6 v1+1/ v2+2 /c2dv2+dt / v2+1 - c2dv+dt # 5

6 v1+1/ (1+2 /c2d+dt / 1+1)v2 - c2dv+dt # 5

Node-v&

1+92(v6 v1)dt / c2d(v6 v2)+dt / v+ # 5

1+92vdt - 1+92v1dt / c2dv+dt - c2dv2+dt / v+ # 5

- 1+92v1dt - c2dv2+dt / 1+92vdt / v+ / c2dv+dt # 5

- 1+92v1dt - c2dv2+dt / (1+92dt / 1+ / c2d+dt)v# 5

-2A. @!king use of the GC9, write e?u!tions on the node b!sis for the fournetworks of *rob. -1".

(!)


75/163



/

-

/

-


76/163



v(t)+1

v2

v


77/163



vv1(t)+2

:ccording to GC9


Eunction

Node-v2&

v2+1/ v2+1 / cdv2+dt / c1d(v26 v)+dt # v(t)+1v2+1/ v2+1 / cdv2+dt / c1dv2+dt6 c1dv+dt # v(t)+1

v2(1+1/ 1+1 / cd+dt / c1d+dt)6 c1dv+dt # v(t)+1

Node-v&

$+2 / cdv+dt / c1d(v6 v2)+dt / c2d(v6 v)+dt # 5

$+2 / cdv+dt / c1dv+dt - c1dv2+dt / c2dv+dt - c2dv+dt # 5

- c1dv2+dt / $+2 / cdv+dt / c1dv+dt / c2dv+dt - c2dv+dt # 5

- c1dv2+dt / $(1+2 / cd+dt / c1d+dt / c2d+dt) - c2dv+dt # 5

Node-v&

v+ / c2d(v6 v)+dt / c=dv+dt / v+2 # v1(t)+2v+ / c2dv +dt - c2dv +dt / c=dv+dt / v+2 # v1(t)+2

- c2dv +dt / (c=d+dt / 1+2 /1+ / c2d+dt)v# v1(t)+2

(b)

/

-


78/163



91

v1 v2 v

9C1 C2

C 92

" 1+92v(t)dt

:ccording to GC9%um of currents entering into the Eunction # %um of currents le!ving the

Eunction

Node-v1&1+92v(t)dt # 1+92v1d t / 1+91(v16 v)d t / c1d(v16 v2)+dt

1+92v(t)dt # 1+92v1d t / 1+91v1d t - 1+91vd t / c1dv1+dt 6 c1dv2+dt

1+92v(t)dt # 1+92v1d t / 1+91v1d t / c1dv1+dt 6 c1dv2+dt- 1+91vd t


79/163



1+92v(t)dt # v1(1+92d t / 1+91d t / c1d+dt) 6 c1dv2+dt- 1+91vd t

:ccording to GC9


Eunction

Node-v2&c1d(v26 v1)+dt / c1d(v26 v)+dt / cdv2+dt / 1+92v2dt # 5c1dv2+dt - c1dv1+dt / c1dv2+dt - c1dv+dt / cdv2+dt / 1+92v2dt # 5

- c1dv1+dt / c1dv2+dt / c1dv2+dt / cdv2+dt / 1+92v2dt - c1dv+dt # 5

- c1dv1+dt / v2(c1d+dt / c1d+dt / cd+dt / 1+92dt) - c1dv+dt # 5

:ccording to GC9


Eunction

Node-v&1+91(v6 v1)d t / c1d(v6 v2)+dt / 1+9vd t / v+ # 5

1+91

v d t - 1+91

v1d t / c1dv+dt - c1dv2+dt / 1+9

vd t / v+ # 5

- 1+91v1d t - c1dv2+dt / c1dv+dt / 1+9vd t / v+ /1+91v d t # 5

- 1+91v1d t - c1dv2+dt / v(c1d+dt / 1+9d t / 1+ /1+91d t) # 5

(c)

/

-


80/163



*1

/

-

/

-


81/163



v2 v

C

1 C1

9C

Cdv(t)+dt

:ccording to GC9


Eunction

Node-v1&v(t)+ # v1+ / (v16 v)+ / c1dv1+dt

v(t)+ # v1+ / v1+ 6 v+/ c1dv1+dt

v(t)+ # v1(1+ / 1+ / c1d+dt) - v+

:ccording to GC9


Eunction

Node-v2&cdv(t)+dt # cdv2+dt / v2+1 / cd(v26 v)+dt

cdv(t)+dt # cdv2+dt / v2+1 / cdv2+dt - cdv+dt

cdv(t)+dt # v2(cd+dt / 1+1 / cd+dt) - cdv+dt

:ccording to GC9


Eunction

Node-v&5 # v+9/ (v6 v1)+ / cd(v6 v2)+dt

5 # v+9/ v+ 6 v1+ / cdv+dt - cdv2+dt

5 # 6 v1+ - cdv2+dt / cdv+dt / v+9/ v+

5 # 6 v1+ - cdv2+dt / v(cd+dt / 1+9/ 1+)

(d)

%t',R


82/163



/

-


83/163



/

-


84/163



291

/

-


85/163



v1 v2

291

1+291v(t)dt

v v

:ccording to GC9


Eunction

Node-v1&1+291v(t)dt# 1+291(v16 v)dt/ c!d(v16 v2)+dt / 1+9!(v16 v2)dt / 1+9b(v16 v)dt / cbd(v16v)+dt

:ccording to GC9


Eunction

Node-v2&


86/163



c!d(v26 v1)+dt / cbd(v26 v)+dt / 1+9b(v26 v)dt/ 1+29(v26 v)dt / (v26 v) # 5

:ccording to GC9


Eunction

Node-v&c!d(v6 v)+dt / 1+9b(v6 v2)dt/ cd(v6 v2)+dt / 1+291(v6 v1)dt # 5

Node-v&

1+9!(v6 v)dt / c!d(v6 v)+dt / 1+292(v6 v2)dt / 1+9b(v6 v1)dt / cbd(v6 v1)+dt # 5

-5. For the given network, write the node-b!sis e?u!tions using the node-to-d!tum

volt!ges !s v!ri!bles.

2

1 =

v2

v1 v v=

v:ccording to GC9


Eunction

Node-v1&

(v16 v2)+(1+2) / (1+2)d(v16 v)+dt / (v16 v)+(1+2) # 5

(v16 v2)+(2) / (2)d(v16 v)+dt / (v16 v)+(2) # 5

Node-v2&

i2# (v26 v1)+(1+2) / (v26 5)+(1+2)

i2# (v26 v1)+2 / v2+2

Node-v&

i2# (1+2)d(v6 v)+dt / (1+2)d(v6 v1)+dt / (1+2)d(v6 5)+dt


87/163



i2# (2)d(v6 v)+dt / (2)d(v6 v1)+dt / (2)dv +dt

Node-v&

5 # (1+2)d(v6 v)+dt / (v6 5)+(1+2) / (v6 v1)+(1+2)

5 # (2)d(v6 v)+dt / (v)+(2) / (v6 v1)+(2)

-1. 3he network in the figure cont!ins one indeendent volt!ge source !nd two

controlled sources. 8sing the GC9, write node-b!sis e?u!tions.


88/163



*1

1

*-

/

:ccording to GC9

-

/

+!


89/163




Eunction

Node-$1&

($16 v1)+1/ C1d$1+dt / $1+2# 5

Node-$2 vk&

vk6 $2# (v1- vk)

Node-$2&

($6 $2)+ / $+ / 1+9vdt / ($6 $)+= # 5

Node-$&

($6 $)+= / $+# i2 Hwhere i2# $+I

-2. 3he network of the figure is ! model suit!ble for Rmidb!ndS oer!tion of the

Rc!scode-connectedS @'% tr!nsistor !mlifier.

%olution&

'en our book see (*+A).

%imlified di!gr!m&

-gm$

i$

$2

rd

rd 9

gm$1 i1i2

9oo-b!sis&

i2# -gm$1i# gm$i1& (i16 i)rd / i19 - $# 5


90/163



(i16 i)rd / i19 # $

i1rd6 ird / i19 # $

i1 i2 i

(rd/ 9) 5 -rd i1 $5 -gm$1 5 i2 # 5

5 5 gm$ i 5

Node-b!sis&

Node-$&

gm$1 # $+rd-gm$ / ($6 $2)+rdT gm$1 / gm$ # $+rd /$+rd -$2+rdNode-$2&

- gm$ # $2+9/ ($26 $)+rdT- gm$ # $2+9/ $2+rd6 $+rd

$2 $ $2 - gm$ (1+9/ 1+rd) -1+rd -1+rd 2+rd $ gm$1 / gm$

-. n the network of the figure, e!ch br!nch cont!ins ! 1-ohm resistor !nd four

br!nches cont!in ! 1-$ volt!ge source. :n!l;e the network on the loo b!sis.

%olution&/

-

/

-

/

-

+!


91/163



1 >

0

2 3

O?. $olt!ge i1 i2 i i i= i i> i" iA


92/163


93/163



1

2 =

Coefficients of

O?. $olt!ge di1+dt di2+dt di+dt di+dt

/

-


94/163



1 5 -1 -1 5

2 1 -1 5 -1

5 -1 5 -1

5 5 -1 -1

/

-

/

-

/

-

/

-


95/163




96/163



2h 1h

/

-

/

-

/

-

/

-


97/163


98/163



h

1h

$2

$

1+2dt 1+dt

h 1h

2h

Node-$1

1+2

dt # 1+2

$1dt /

($16 $)dt / 1+2

($16 $2)dt

1+2dt # 1+2$1dt / $1dt - $dt / 1+2$1dt - 1+2$2dt

Node-$2

5 # 1+2$2dt / ($26 $)dt / 1+2($26 $1)dt

5 # 1+2$2dt / $2dt - $dt / 1+2$2dt - 1+2$1dt

1+dt # 1+$dt / $dt / ($6 $1)dt / ($6 $2)dt

1+

dt # 1+

$dt /

$dt /

$dt -

$1dt /

$dt -

$2dt

O?. Current $1 $2 $


99/163



1 Jdt 2dt -1+2dt -dt

2 5 -1+2

dt 2

dt -

dt

1+dt -dt -dt dt

$2


100/163


101/163



O?. Current $1 $2 $ $

1 5 1.= -5.2= 5 -1

2 5 -5.2= 1.= -5.2= -1

5 5 -5.2= 1.= -1

-1 -1 -1

9oo-b!sis&


102/163



O?. $olt!ge $1 $2 $ $

1 5 -1 -1 5

2 5 -1 5 -1

+

!


103/163



- -1 5 -1

5 -1 -1

-. For the network shown in the figure, determine the numeric!l v!lue of the

br!nch current i1. :ll sources in the network !re time inv!ri!nt.

$1 $2 2$


104/163



($16 2)+(1+2) / ($16 $2)+1 / ($16 2)+2 # 5

($16 2)2 / ($16 $2) / ($16 2)+2 # 5

2$16 / $16 $2/ $1+2 - 1 # 5

.=$16 $2- = # 5 (i)

$2+(1+2) / ($26 $1)+1 / ($26 2)+1 # 12$2/ $26 $1/ $26 2 # 1

$26 $1#

$1# - / $2 (ii)

*ut $1in (i)

.=$16 $2- = # 5

.=(- / $2) 6 $2- = # 5

-15.= / 1$26 $26 = # 5

1$26 1=.= # 5

1$2# 1=.=

$2# 1=.=+1 # 1.1A2$olts

*ut v!lue of $2in (ii)

$1# - / $2 (ii)

$1# - / (1.1A2)$1# - / (1.1A2)

$1# 1.>"$olts

i1# ($16 $2)+1 # $16 $2 #1.>" $olts -1.1A2 $olts # 5.=> !meres.

->. n the network of the figure, !ll sources !re time inv!ri!nt. Determine the

numeric!l v!lue of i2.

/

-

/

-


105/163



1

:ccording to GC9


Eunction

Node-v1&

+ 1 4 1,1 + %15 ',1 + %15 ',1

# $1/ ($16 $2) / ($16 $)

# $1/ $16 $2/ $16 $

# $16 $26 $ (i)

Node-v2&

($26 $1)+1 / ($26 $)+2 / ($2- 5)+1 # 1

$26 $1/ $2+2 6 $+2 / $2# 1

$26 $1/ $2+2 6 $+2 / $2# 1

2.=$26 $16 5.=$# 1 (ii)

Node-v&

($6 $1)+1 / ($6 $2)+2 / $+1 # 1

$6 $1/ $+2 6 $2+2 / $# 1

2.=$6 $16 $2+2 # 1 (iii)


106/163



# $16 $26 $$1# / $2/ $$1# ( / $2/ $)+

2.=$6 $16 $2+2 # 1

2.=$6 (( / $2/ $)+) 6 $2+2 # 1

2.=$6 (+ / $2+ / $+) 6 $2+2 # 12.=$6 1 - $2+ - $+ 6 $2+2 # 1

2.=$- $2+ - $+ 6 $2+2 # 2

2.=$6 5.$2- 5.$6 5.=$2# 2

2.1$6 5."$2# 2

%ubtr!cting (ii) (iii)

2.=$26 $16 5.=$# 1 (ii)

2.=$6 $16 $2+2 # 1 (iii)

2.=$26 2.=$6 5.=$/ $2+2 # 5

$26 $ # 5

$2# $

$2# $

2.1$6 5."$2# 2

utting $2# $2.1$6 5."$# 2

1.2$# 2

$# 2+1.2 # 1.=51 $

$# 1.=51 $

i2# (2 6 $)+2 # (2 6 1.=51)+2 # 5.2A= !meres.

i2# 5.2A= !meres.

-". n the given network, !ll sources !re time inv!ri!nt. Determine the br!nch

current in the 2 ohm resistor.

+!

+

!


107/163



+

!

+

!

+

!

+

!


108/163



i1

+

!

+

!

+!


109/163


110/163



# M(=+2)(2) / 2 # >

-1 2

i2# >+>.>= # 5.A5 !meres. :ns.

-A. %olve for the four node-to-d!tum volt!ges.

$2$1


111/163


112/163



i

i2

i1

1 17

e e?uiv!lent with resect to the !ir of termin!ls


113/163



1 1

(!)

(b)

1 1

O?u!ting (!) (b)


114/163


115/163



1-=. %olution&

v # $5sin

tC # C5(1 - cost)

U # t

U # C$

i # d(?)+dt # d(Cv)+dt # Cdv+dt / vdC+dt

i # Cdv+dt / vdC+dt

i # C5(1 - cost)d($5sint)+dt / $5sintdC5(1 - cost)+dt

i # C5(1 - cost) $5cost / $5sintHC5sintI

1-15.

t

w # vi dt

-

For !n inductor

v9# 9di+dt

utting the v!lue of volt!ge

t

w #

vi dt

-

t

w # (9di+dt)i dt

-

t

w # 9idi

-

t

w # 9i2+2

-

w # 9Mi2(t)+2 - i2(-)+2

w # 9Mi2(t)+2 6 (i(-))2+2

w # 9Mi2(t)+2 6 (5)2+2

w # 9Mi2(t)+2 Hec!use i(-) # 5 for !n inductorI


116/163



:s we know

# 9i

2# 92i2

2+9 # 9i2

w # 9Mi2(t)+2

w # 9i2+2 utting the v!lue of 9i2

w # (2+9)+2

w # 2+29 Hwhere # flu0 link!geI

1-11.

t

w # vi dt

-

For ! c!!citor

i # Cdv+dt

utting the v!lue of current

t

w # vi dt

-

t

w # (Cdv+dt)v dt

-

tw # Cvdv

-

t

w # Cv2+2

-

w # CMv2(t)+2 - v2(-)+2

w # CMv2(t)+2 6 (v(-))2+2

w # CMv2(t)+2 6 (5)2+2

w # CMv2(t)+2 Hec!use v(-

) # 5 for !n inductorI

:s we know

U # C$

$ # U+C

w # CMv2(t)+2

w # CM(?+C)2+2

w # CM?2+2C2


117/163


118/163



w # >2W

1-1".

vc# 255 $

C # 1

Fm!ss # 155 lb # =. kg

work done # Fd # mgd # (=.)(A.")d

work done # energ # (1+2)C(vc)2 #(1+2)(115-)(255)2 # 5.52 Eoule

work done # (=.)(A.")d

5.52 # (=.)(A.")d

d # 5.52+(=.)(A.") # 5.52+.A

d # .=5=15-=m :ns.

1-1A.

%olution& $m

v

5 1 2 time

-$m

for 5

t

1

(01, 1) # (1, $m)

(05, 5) # (5, 5)

m # (16 5)+(016 05) # ($m6 5)+(1 6 5)

m # $m

# m0 / c # $m(t) / 5 # $mt

%loe # m


119/163



for 1t

%tr!ight-line e?u!tion

-intercet # c # 5


120/163



(01, 1) # (1, $m)

m # (6 1)+(06 01) # (-$m6 $m)+( - 1) # -2$m+2 # -$m

m # -$m (0, ) # (, -$m)

# m0 / c # -$m(t) / 2$m# -$mt / 2$m

for t

(0, ) # (, 5)

(0, ) # (, -$m)

m # (6 )+(06 0) # (5 6 (-$m))+( 6 )

m # $m

# m0 / c # $m(t) - $m# $mt 6 $m

9et c!!cit!nce be C

%loe # m


-intercet # c # 2$m

%loe # m


-intercet # c # -$m


121/163

i # C$m



for 5t1

i # Cdv+dt # Cd($mt)+dt # C$m

i # C$m

for 1

t

i # Cdv+dt # Cd(-$mt / 2$m)+dt # -C$m

i # -C$m

for t

i # Cdv+dt # Cd($mt 6 $m)+dt # C$m

C$m

-C$mfor 5t1

? # C$

? # C$mt

for 1t

? # C$


122/163



? # C$m(2 6 t)

for t

? # C$

? # C$m(t - )

for 5t1 ? # C$mt t # 5, ? # 5 t # 1, ? # C$m

for 1t ? # C$m(2 6 t) t # , ? # -C$m

for t ? # C$m(t - ) t # , ? # 5

Ch!rge w!veform s!me !s volt!ge w!veform.

(b)

i(t)


123/163



5 1 2 time

for 5

t

1

(01, 1) # (1, m)


124/163



(05, 5) # (5, 5)

m # (16 5)+(016 05) # (m6 5)+(1 6 5)

m # m

# m0 / c # m(t) / 5 # mt

for 1t

(01, 1) # (1, m)

m # (6 1)+(06 01) # (-m6 m)+( - 1) # -2m+2 # -m

m # -m (0, ) # (, -m)

# m0 / c # -m(t) / 2m# -mt / 2m

for t

(0, ) # (, 5)

%loe # m


-intercet # c # 5

%loe # m


-intercet # c # 2m


125/163



(0, ) # (, -m)

m # (6 )+(06 0) # (5 6 (-m))+( 6 )

m # m

# m0 / c # m(t) - m# mt 6 m

for 5t1

t

v(t) # (1+C)id(t) / v(t1)

t1 t

v(t) # (1+C)mtd(t) / 5

5

t

v(t) # (1+C)mtd(t)

5

t

v(t) # (1+C)mt2+2

5

v(t) # (1+C)mM(t2+2) - ((5)2+2)

v(t) # (1+C)m(t2+2)

v(1) # (1+C)m((1)2+2) # (1+C)m(1+2) # m+2C

for 1

t

t

v(t) # (1+C)id(t) / v(t1)

t1 t

v(t) # (1+C)m(2 6 t)d(t) / m+2C

1

%loe # m


-intercet # c # -m


126/163

v() # (1+C)M(2() 6 ()2+2) - (+2) / m+2C



t

v(t) # (1+C)2t 6 t2+2/ m+2C

1

v(t) # (1+C)M(2t 6 t2+2) - (2(1) 6 12+2) / m+2C

v(t) # (1+C)M(2t 6 t2

+2) - (2 6 1+2) / m+2C

v(t) # (1+C)M(2t 6 t2+2) - (+2) / m+2C

!t time t #

v() # (1+C)M 6 .= 6 1.= / m+2C

v() # m+2C

for

t

t

v(t) # (1+C)id(t) / v()

t1 t

v(t) # (1+C)m(t - )d(t) / m+2C

1

t

v(t) # (1+C)mt2+2 6 t/ m+2C

1

v(t) # (1+C)mM(t

2+2 6 t) 6 (1+2 - ) / m+2C

v(t) # (1+C)mM(t2+2 6 t) / 2.= / m+2C

!t time t #

v() # (1+C)mM(()2+2 6 ()) / 2.= / m+2C

v() # (1+C)mM1+2 6 12 / 2.= / m+2C

v() # (1+C)mM" 6 12 / 2.= / m+2C

v() # (1+C)mM61.= / m+2C

v() # -m+C

v(5) 5

v(1) m+2C # 5.=(m+C)

v(2) m2+C # 2(m+C)

v() m+2C # 5.=(m+C)

v() -m+C # -1(m+C)


127/163



v(t) # (1+C)m(t2+2)

!t time t # 2

v(2) # (1+C)m((2)2+2) # m(2)+C

1

2

3

4

5

1

2

3

4

5

-2

-1

0

1

2

3

4

5

time

voltage

Series2

Series1

Series2 0 0.5 2 0.5 -1

Series1 0 1 2 3 4

1 2 3 4 5

for 5t1

? # C$

? # C(mt2+2C) # mt

2+2

for 1t

? # C$

? # Cm(t 6 t2- 2)+2C


128/163



for t

? # C$

? # C(1+C)MmM(t2+2 6 t) / 2.= / m+2C # mM(t

2+2 6 t) / 2.= / m+2C

:t time t # 5

? # C(mt2+2C) # mt

2+2 # m(5)2+2 # 5 C

:t time t # 1

? # C(mt2+2C) # mt

2+2 # m12+2 # m+2 C

:t time t # 2

? # C(mt2+2C) # mt

2+2 # m22+2 # 2mC

:t time t #

? # Cm(t 6 t2- 2)+2C # Cm(() 6

2- 2)+2C # (5.=m+C) C

:t time t4 /

? # C(1+C)MmM(t2+2 6 t) / 2.= / m+2C # mM(t

2+2 6 t) / 2.= / m+2C

? # mM(2+2 6 ()) / 2.= / m+2C # -m+C

Ch!rge w!veform s!me !s volt!ge w!veform.

1-25.

%olution&

# 1:

9 # J $

(e)

w # (1+2)9i2 #(1+2)(1)(1 6 e-t)2

w # (1+2)(1 6 e-t)2

w # (1+2)(1 / 2e-2t6 2e-t)

w(t5) # (1+2)(1 / 2e-2t56 2e-t5)

t5# 1 sec.

w(1) # (1+2)(1 / 2e-2(1)6 2e-1)

w(1) # (1+2)(1 / 2e-26 2e-1)

w(1) # (1+2)(1 / 2(1+e2)6 2(1+e)) H1+e # 5.>T 1+e2# 5.1=I

w(1) # (1+2)(1 / 2(5.1=)6 2(5.>))

w(1) # (1+2)(1 / 5.2>6 5.>)

w(1) # 5.2= Woule

(f)

v#

v # i #(1 6 e-t)(1) # (1 6 e-t)

v # i # (1 6 e-t)

v(t5)# (1 6 e-t5)

!t time t5# 1 sec.

v(1)# (1 6 e-1) # 5. $

(g)

w # (1+2)(1 / 2e-2t6 2e-t)

dw+dt # d((1+2)(1 / 2e-2t6 2e-t))+dt

dw+dt # (1+2)d(1 / 2e-2t6 2e-t)+dt

dw+dt # (1+2)Hd(1)+dt / d(2e-2t)+dt - d(2e-t)+dtI

dw+dt # (1+2)H5 / 2e-2t)(-2) - 2e-t)(-1)I

dw+dt # (1+2)H-e-2t/ 2e-t)I

dw(t5)+dt # (1+2)H-e-2t5/ 2e-t5)I

dw(1)+dt # (1+2)H-e-2/ 2e-1)I

dw(1)+dt # (1+2)H-(1+e2) / 2(1+e)I

dw(1)+dt # (1+2)H-(5.1=) / 2(5.>)IH1+e # 5.>T 1+e2# 5.1=I

dw(1)+dt # (1+2)H-5.= / 5.>I # 5.1 w!tts

(h)

*# i2 # (1 / e-2t6 2e-t)(1)

*# i2 # (1 / e-2t6 2e-t)

*(t5) # i2 # (1 / e-2t56 2e-t5)


132/163



:t time t5# 1 sec.

*(1) # i2 # (1 / e-2(1)6 2e-(1))

*(1) # i2 # (1 / e-26 2e-1)

*(1) # i2 # (1 / 1+e26 2(1+e))

*(1) # i2 # (1 / 5.1= 6 2(5.>))

*(1) # i2

# (1 / 5.1= 6 5.>)

*(1) # i2 # (5.A=) w!tts

(i)

*tot!l# vi # (1)(1 6 e-t) # (1 6 e-t)

:t time t5# 1 sec.

*tot!l(t5) # vi # (1)(1 6 e-t) # (1 6 e-t5)

*tot!l(1) # (1 6 e-1)

*tot!l(1) # (1 6 e-1) # 5. w!tts.

1-2=.$olt!ge !cross the c!!citor !t time t # 5

vc(5) # 1 $olt

k is closed !t t # 5

i(t) # e-t, tL5

i(t5) # 5.> :

5.> # e-t5

3!king log!rithm of both the sides

log5.> # loge-t5

-t5loge # -5.2

t5(5.) # 5.2 Hec!use e # 2.>1"I

t5 # 1 sec.

(!) dvc(t5)+dt #

8sing loo e?u!tion

vc(t) # i # e-t(1) # e-t$olts

dvc(t)+dt # -e-t$olts

dvc(t5)+dt # -e-t5$olts

t5# 1 sec.

dvc(t5)+dt # -e-1$olts

dvc(t5)+dt # -5.> $+sec

(b)

Ch!rge on the c!!citor # ? # Cv # (1)(e -t) # e-t coulomb

Ch!rge on the c!!citor # ?(t5) # Cv # (1)(e-t) # e-t5 coulomb

t5# 1 sec.


133/163



Ch!rge on the c!!citor # ?(1) # Cv # (1)(e -t) # e-1 coulomb # 5.> coulomb

(c) d(Cv)+dt # Cdv+dt # Cde-t+dt # -Ce-t

d(Cv(t5))+dt # -Ce-t5

t5# 1 sec.

d(Cv(t5))+dt # -Ce-1

:s C # 1F

d(Cv(t5))+dt # -e-1 # -5.> coulomb+sec.

(d) vc(t) # e-t

t5# 1 sec.

vc(t5) # e-t5

vc(1) # e-1 # 5.> $olt

(e) wc# wc# (1+2)Cv

2 # (1+2)(1)(e-t)2# (1+2)e-2t

wc(t5) # (1+2)Cv2 # (1+2)(1)(e-t5)2# (1+2)e-2t5

t5# 1 sec.

wc(1) # (1+2)e-2(1)

wc(1) # (1+2)e-2

wc(1) # (1+2)(1+e2) H1+e2# 5.1=I

wc(1) # (1+2)(5.1=)

wc(1) # (1+2)(5.1=) # 5.5> Woules

(f)v(t) # i # e

-t(1) # e-t $olts

v(t5) # i # e-t(1) # e-t5 $olts

t5# 1 sec.

v(1) # i # e-t(1) # e-1 $olts # 5.> $olts

(g) dwc+dt #

wc# (1+2)e-2t

dwc+dt # d(1+2)e-2t+dt

dwc+dt # (1+2)e-2t(-2) # -e-2t

dwc(t5)+dt # (1+2)e-2t(-2) # -e-2t5

t5# 1 sec.dwc(1)+dt # (1+2)e-2t(-2) # -e-2(1)

dwc(1)+dt # (1+2)e-2t(-2) # -e-2

dwc(1)+dt # (1+2)e-2t(-2) # -e-2 # - 5.1= w!tts.

(h) * # i2 # (e-t)2(1) # e-2t

*(t5) # i2 # (e-t)2(1) # e-2t5


134/163


135/163



5 + +2 =+

for vc-5.=$olt

C # (-1.5 / 5.=)+(-1.= / 5.=) # -5.=+-1 # 5.= F

for 65.= vc5.=

C # (5.= / 5.=)+(5.= / 5.=) # 1+1 # 1F

for 5.= vc1.=

C # (1.5 - 5.=)+(1.= - 5.=) # 5.=+1 # 5.=F

for 5 vc5.= for 5 t +

for 5.= vc 1 for + t =+

for 5.= vc5 for =+ t

ic(t) # d(Cv)+dt # Cdv+dt / vdC+dt

ic(t) # Cdv+dt / vdC+dt

for 5

t

+C # 1F

$ # sint

ic(t) # (1)dsint+dt / sintd(1)+dt

ic(t) # cost



for + t =+

C # 5.=F

v # sintic(t) # (5.=)dsint+dt / sintd(5.=)+dt

ic(t) # (5.=)cost



for =+ t


136/163



C # 1F

v # sint

ic(t) # (1)dsint+dt / sintd(1)+dt

ic(t) # cost

vc(t)

5 + +2 =+

time

ic(t)


137/163



5 + +2 =+ time

1-5.


138/163



5 1 2 time

5.2= 5.2= 5.2= 5.2= 5.2= 5.2= 5.2= 5.2=

2 2

vc(t) interv!l

2t for 5 t 1

-2t / for 1 t 2

2t 6 for 52t

-2t / " for t

5 for t

vc(t) interv!l C!!citor(v!lue)

2t for 5 t 5.2= 1F

2t for 5.2= t 1 5.=F

-2t / for 1 t 1.>= 5.=F

-2t / for 1.>= t 2 1F

2t 6 for 2 t 2.2= 1F


139/163



2t 6 for 2.2= t 5.=F

-2t / " for t .>= 5.=F

-2t / " for .>= t 1F

5 for t 1F

For the rem!ining !rt see 1-A for reference.

1-2> 6 1-". (%ee ch!terV for reference)

efore solving ch!terV1 following oints should be ket in mind&

1. $olt!ge !cross !n inductor

2. Current through the c!!citor

. Pr!hic!l !n!lsis

. *ower dissi!tion

-1.

%olution&

1 *osition B17

/ switch

$

-

9


140/163

n the ste!d st!te

inductor beh!ves like

! short circuit



: ste!d st!te current h!ving reviousl been est!blished in the 9 circuit.

1

$

i

%hort circuit

i(5-) # $+1 (current in 9 circuit before switch Bk7 is closed)

t me!ns th!t i(5-) # i(5/) # $+1G is moved from osition 1 to osition 2 !t t # 5.

1

9

2

4h!t does th!t me!n

n !n inductor i(5-) # i(5/) # 5


141/163



for t 5

:ccording to kirchhoffs volt!ge l!w

%um of volt!ge rise # sum of volt!ge dro

Circuit simlific!tion&

(2/ 1)

9

(!) (b)

9di+dt / (1/ 2)i # 5

9di+dt # -(1/ 2)i

di+dt # -(1/ 2)i+9

di+i # -(1/ 2)dt+9

ntegr!ting both the sides,

di+i #

-(1/ 2)dt+9

di+i # -(1/ 2)+9dt

lni # -(1/ 2)t+9 / C

i # e-(1 / 2)t+9 / C

i # e-(1 / 2)t+9eC

i # ke-(1 / 2)t+9

:ling initi!l condition

i(5/) # $+1i(5/) # ke-(1 / 2)(5)+9

i(5/) # ke5

i(5/) # k

i(5/) # $+1i(5/) # k

O?u!ting

G # $+1


142/163



i # ke-(1 / 2)t+9

i # ($+1)e-(1 / 2)t+9 is the !rticul!r solution.

-2.

%olution&

%witch is closed to b !t t # 5

niti!l conditions v2(5/) # 5 i(5/) # $5+1for t L 5

(1+C1)idt / (1+C2)idt / 1i # 5

Differenti!ting both sides with resect to Bt7

(1+C1)i / (1+C2)i / 1di+dt # 5

(1+C1/ 1+C2)i / 1di+dt # 5

i+Ce?/ 1di+dt # 5

i+Ce?# -1di+dt

di+i # (-1+Ce?1)dt

ntegr!ting both the sidesdi+i # (-1+Ce?1)dt

di+i # (-1+Ce?1)dt

lni # (-1+Ce?1)t / k1i # e(-1+Ce?1)t /k1

i # e(-1+Ce?1)tek1

i # ke(-1+Ce?1)t


i(5/) # ke(-1+Ce?1)(5)

i(5/) # ke5

i(5/) # k(1)

i(5/) # ki(5/) # $5+1O?u!ting

G # $5+1

3herefore

i # ke(-1+Ce?1)t

i # ($5+1)e(-1+Ce?1)t

t

v2(t) # (1+C2)

idt

-

5 t

v2(t) # (1+C2)idt / (1+C2)idt

- 5

t

# v2(5/) / (1+C2)($5+1)e(-1+Ce?1)tdt


143/163



5

t

# 5 / (1+C2) ($5+1)(-Ce?1)e(-1+Ce?1)t

5

t

# (1+C2) ($5)(-Ce?)

e(-1+Ce?1)t

5

# (1+C2) ($5)(-Ce?)Me(-1+Ce?1)t- e(-1+Ce?1)(5)

# (1+C2) ($5)(-Ce?)Me(-1+Ce?1)t6 e5

# (1+C2) ($5)(-Ce?)Me(-1+Ce?1)t6 1

v2(t) # (1+C2) ($5)(Ce?)M1 - e(-1+Ce?1)t

t

v1(t) # (1+C1)idt

-

5 tv1(t) # (1+C1)idt / (1+C1)idt

- 5

t

# v1(5/) / (1+C1)($5+1)e(-1+Ce?1)tdt

5

t

# -$5/ (1+C1) ($5+1)(-Ce?1)e(-1+Ce?1)t

5

t

# (1+C1) ($5)(-Ce?)e(-1+Ce?1)t

5

# (1+C1) ($5)(-Ce?)Me(-1+Ce?1)t- e(-1+Ce?1)(5)

# (1+C1) ($5)(-Ce?)Me(-1+Ce?1)t6 e5

# (1+C1) ($5)(-Ce?)Me(-1+Ce?1)t6 1

v1(t) # (1+C1) ($5)(Ce?)M1 - e(-1+Ce?1)t

$(t) # i1# (($5+1)e(-1+Ce?1)t)1

$(t) # i1# ($5)e(-1+Ce?1)t

! b

/ /

G 1 $5


144/163

C!!citor is ch!rging

$olt!ge !cross the

c!!citor # $5

C!!citor is disch!rging

$olt!ge !cross the c!!citor # - $5



- v2

C1 C2

-

! b

1C2

//////// ------------

--------------- /////// C1-.

%olution&

k is closed !t t # 5

niti!l condition i(5/) # ($16 $2)+1for t L 5

(1+C1)idt / (1+C2)idt / i # 5


(1+C1)i / (1+C2)i / di+dt # 5

(1+C1/ 1+C2)i / di+dt # 5

i+Ce?/ di+dt # 5

i+Ce?# -di+dt

di+i # (-1+Ce?)dt

ntegr!ting both the sides

di+i # (-1+Ce?)dt

di+i # (-1+Ce?)dt

lni # (-1+Ce?)t / k1i # e(-1+Ce?)t /k1

i # e(-1+Ce?)tek1

i # ke(-1+Ce?)t


i(5/) # ke(-1+Ce?)(5)

i(5/) # ke5

i(5/) # k(1)

i(5/) # k

i(5/) # ($16 $2)+


145/163



O?u!ting

G # ($16 $2)+

3herefore

i # ke(-1+Ce?1)t

i # ($16 $2)+)e(-1+Ce?)t

t

v2(t) # (1+C2)idt

-

5 t

v2(t) # (1+C2)idt / (1+C2)idt

- 5

t

# v2(5/) / (1+C2)

(( $16 $2)+)e(-1+Ce?)tdt 5

t

# $2/ (1+C2) (($16 $2)+)(-Ce?)e(-1+Ce?)t

5

t

# (1+C2) ($16 $2))(-Ce?)e(-1+Ce?)t

/ $2

5

# (1+C2) ($16 $2)(-Ce?)Me(-1+Ce?)t- e(-1+Ce?)(5) / $2

# (1+C2) ($16 $2)(-Ce?)Me(-1+Ce?)t6 e5 / $2

# (1+C2) ($16 $2)(-Ce?)Me(-1+Ce?)t6 1

v2(t) # (1+C2) ($16 $2)(Ce?)M1 - e(-1+Ce?)t / $2 (i)

t

v1(t) # (1+C1)idt

-

5 t

v1(t) # (1+C1)idt / (1+C1)idt

- 5

t

# v1(5/) / (1+C1)

(($16 $2)+)e(-1+Ce?)t

dt 5

t

# -$1/ (1+C1) (($16 $2)+)(-Ce?)e(-1+Ce?)t

5

t


146/163



# (1+C1) ($16 $2)(-Ce?)e(-1+Ce?)t

- $1

5

# (1+C1) ($16 $2)(-Ce?)Me(-1+Ce?)t- e(-1+Ce?)(5) 6 $1

# (1+C1) ($16 $2)(-Ce?)Me(-1+Ce?)t6 e5 6 $1

# (1+C1) ($16 $2)(-Ce?)Me(-1+Ce?)t6 1 6 $1

v1(t) # (1+C1) ($16 $2)(Ce?)M1 - e(-1+Ce?)t 6 $1 (ii)

from (i)

v2(t) # (1+C2) ($16 $2)(Ce?)M1 - e(-1+Ce?)t / $2

v2() # (1+C2) ($16 $2)(Ce?)M1 - e(-1+Ce?)() / $2

v2() # (1+C2) ($16 $2)(Ce?)M1 - e-() / $2

v2() # (1+C2) ($16 $2)(Ce?)M1 - 5 / $2

v2() # (1+C2) ($16 $2)(Ce?) / $2

v2() # (1+C2) ($16 $2)(Ce?) / $2

from (ii)

v1(t) # (1+C1) ($16 $2)(Ce?)M1 - e(-1+Ce?)t 6 $1

v1() # (1+C1) ($16 $2)(Ce?)M1 - e(-1+Ce?)() 6 $1

v1() # (1+C1) ($16 $2)(Ce?)M1 - e-() 6 $1

v1() # (1+C1) ($16 $2)(Ce?)M1 - 5 6 $1 He-()# 5I

v1() # (1+C1) ($16 $2)(Ce?) 6 $1

v1() # (1+C1) ($16 $2)(Ce?) 6 $1


147/163


148/163



0

1

2

1

1.634

1.666

0

0.5

1

1.5

2

2.5

time

Seri es1

Seri es2

Series1 0 1 2

Series2 1 1.634 1.666

1 2 3

:t t # 5 switch is moved to osition b.

niti!l condition i91(5-) # i91(5/) # $+ # 1+1 # 1:.

$2(5/) # (-1)(1+2) # -5.= volts

for t 5, GC9

(1+1)v2dt / v2+(1+2) / (1+2)v2dt # 5


149/163

n c!se of D.C.


! short circuit



(1 / 1+2)v2dt / 2v2# 5

(+2)v2dt / 2v2# 5


(+2)v2/ 2dv2+dt # 5

Dividing both the sides b 2

H(+2)+2Iv2/ (2+2)dv2+dt # 5(+)v2/ dv2+dt # 5

%olving b method of integr!ting f!ctor

* # X, U # 5

v2(t) # e-*te*t.Udt / ke-*t

v2(t) # e-*te*t.Udt / ke-*t

v2(t) # e-(+)t

e(+)t.(5)dt / ke-(+)t

v2(t) # ke-(+)t


v2(5/) # ke-(+)(5/)

v2(5/) # ke5

v2(5/) # k(1)

v2(5/) # k

-5.= # k

v2(t) # -5.=e-(+)t

efore switching

%hort circuit

-

! b


150/163



1 e0!nding

+

!

O?uiv!lent network !t t # 5/

Coll!sing

-

/

-=.

%olution&

%witch is closed !t t # 5

niti!l condition&-

i(5-) # i(5/) # (25 / 15)+(5 / 25) # 5+=5 # += :

for t 5,

:ccording to G$9

%um of volt!ge rise # sum of volt!ge dro

25i / (1+2)di+dt # 15

@ultiling both the sides b B27

2(25i) / 2(1+2)di+dt # 15(2)

5i / di+dt # 25

di+dt / 5i # 25

%olving b the method of integr!ting f!ctor

* # 5 U # 25

i(t) # e-*te*t.Udt / ke-*t


151/163


152/163


153/163


154/163

n c!se of D.C.

inductor beh!ves like! short circuit



1+C # 155

C # 1+155 # 5.51 secs.

5.

5.=

- 5.1

efore switching&

%te!d

st!te

lustr!nsien

t

%te!d

st!te

lus

tr!nsien

t


155/163



:fter switching&

->.

%olution&

niti!l condition vc(5-) # vc(5/) # v2(5/) # 5

for t 5

(v26 v1)+1/ Cdv2+dt / v2+2# 5

v2+16 v1+1/ Cdv2+dt / v2+2# 5

v2+1/ Cdv2+dt / v2+2# v1+1v2+1/ v2+2/ Cdv2+dt # v1+1v2(1+1/ 1+2) / Cdv2+dt # v1+1Dividing both the sides b BC7

v2(1+1/ 1+2)+C / Cdv2+Cdt # v1+C1v2(1+1/ 1+2)+C / dv2+dt # v1+C1

C # (1+25) F 1# 15-ohm 2# 25-ohm

n c!se of D.C.


! short circuit


156/163



v2(1+15 / 1+25)+(1+25) / dv2+dt # e-t+H(1+25)(15)I

v2(5.1 / 5.5=)+(5.5=) / dv2+dt # e-t+H5.=I

v2(5.1=)+(5.5=) / dv2+dt # e-t+H5.=I

v2/ dv2+dt # 2e-t


157/163


158/163



import java.io.*;public class Addition { public static void main (String args []) throwsI!"c#ption {

$u%%#r#dad#r stdin ' n#w $u%%#r#dad#r (n#w InputStr#amad#r(Sst#m.in));

doubl# # ' .+,; doubl# a- b; String string- string+; int num+- num;

Sst#m.out.println(#nt#r th# valu# o% /0); string ' stdin.r#ad1in#(); num ' Int#g#r.pars#Int (string);

%or(int c ' 2; c 3' num; c44){ Sst#m.out.println(#nt#r th# valu# o% t0); string+ ' stdin.r#ad1in#(); num+ ' Int#g#r.pars#Int (string+);

a '(doubl#)(+56ath.pow(#- num+)); b '(doubl#)(+56ath.pow(#- 7*num+)); Sst#m.out.println(8h# solution is0 4 (a 9 b)); :55%or loop:55m#thod main:55class Addition

-A.

%olution&

Network !tt!ins ! ste!d st!te

3herefore

i2(5-) # $5+1/ 2i2(5-) # +15 / = # +1= # 1+= :m.

v!(5/) # i2(5/)(2)

v!(5/) # (1+=)(=) # 1 $olt

for t 6

:ccording to kirchhoffs current l!w&


159/163



(v!6 $5)+1 / v!+2/ (1+9)v!dt # 5

utting 1# 15, 2# =, $5# 9 # J

(v!6 )+15/ v!+= / (1+(1+2))v!dt # 5

(v!6 )+15/ v!+= / 2v!dt # 5

v!+156 +15 / v!+= / 2v!dt # 5

v!+15/ 2

v!dt # +15

Differenti!ting with resect to Bt7

d+dtHv!+15/ 2v!dtI # d+dtH+15I

d+dtHv!+15I/ d+dtH2v!dtI # d+dtH+15I

(+15)d+dtHv!I/ 2v!# 5

(+15)d+dtHv!I# - 2v!

d+dtHv!I# - 2v!+(+15)

d+dtHv!I# - 25v!+

dv!+v! #- 25dt+

ntegr!ting both the sides

dv!+v! #

- 25dt+lnv! #- 25t+ / C

v! #e-25t+ / C

v! #e-25t+ eC

v! #ke-25t+


v!(5/) # ke-25(5/)+

v!(5/) # ke5

v!(5/) # k(1)

1 # k

3herefore

v! #ke-25t+

v! #(1)e-25t+

v! #e-25t+

efore switching


160/163

n c!se of D.C.


! short circuit



:fter switching

/

-

/

-

/

-


161/163


162/163

n c!se of D.C.c!!citor beh!ves

like !n oen circuit



:fter switching

-12.

%olution&

%witch closed !t t # 5

niti!l condition&-

i9(5-) # $+(1/ 2)

i9(5-) # i9(5/) # $+(1/ 2)

for t 5, G$9

1i / 9di+dt # $Dividing both the sides b B97

1i+9 / di+dt # $+9

%olving b integr!ting f!ctor method

* # 1+9 U # $+9

i(t) # e-*te*t.Udt / ke-*t

i(t) # e-(1+9)te(1+9)t($+9)dt / ke-(1+9)t


163/163



i(t) # e-(1+9)tHe(1+9)t+(1+9)I($+9) / ke-(1+9)t

i(t) # He5+(1+9)I($+9) / ke-(1+9)t

i(t) # H1+(1+9)I($+9) / ke-(1+9)t

i(t) # (9+1)($+9) / ke-(1+9)t

i(t) # $+1/ ke-(1+9)t


i(t) # $+1/ ke-(1+9)t

i(5/) # $+1/ ke-(1+9)(5/)

i(5/) # $+1/ ke5

i(5/) # $+1/ k(1)

$+(1/ 2) # $+1/ k

$+(1/ 2) - $+1# k

i(t) # $+1/ ke-(1+9)t

i(t) # $+1/ H$+(1/ 2) - $+1Ie-(1+9)t

Solutions Chapter 02, 03, 04 Network Analysis 3rd Edition - M. E. v. Valkenburg

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Transcript of Solutions Chapter 02, 03, 04 Network Analysis 3rd Edition - M. E. v. Valkenburg