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
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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&
'en our book see the figure (*+)
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&
'en our book see the figure (*+)
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&
'en our book see the figure (*+)
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&
'en our book see the figure (*+>)
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&
'en our book see the figure (*+>)
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&
'en our book see the figure (*+")
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&
'en our book see the figure (*+")
(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&
'en our book see the figure (*+")
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&
'en our book see the figure (*+A)
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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&
'en our book see the figure (*+A)
: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+
Node-v1:ccording to GC9
%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
Node-v1:ccording to GC9
%um of currents entering into the Eunction # %um of currents le!ving the
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
Node-v1:ccording to GC9
%um of currents entering into the Eunction # %um of currents le!ving the
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
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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".
(!)
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/
-
/
-
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v(t)+1
v2
v
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vv1(t)+2
:ccording to GC9
%um of currents entering into the Eunction # %um of currents le!ving the
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)
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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
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1+92v(t)dt # v1(1+92d t / 1+91d t / c1d+dt) 6 c1dv2+dt- 1+91vd t
:ccording to GC9
%um of currents entering into the Eunction # %um of currents le!ving the
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
%um of currents entering into the Eunction # %um of currents le!ving the
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)
/
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*1
/
-
/
-
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v2 v
C
1 C1
9C
Cdv(t)+dt
:ccording to GC9
%um of currents entering into the Eunction # %um of currents le!ving the
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
%um of currents entering into the Eunction # %um of currents le!ving the
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
%um of currents entering into the Eunction # %um of currents le!ving the
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
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/
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/
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291
/
-
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v1 v2
291
1+291v(t)dt
v v
:ccording to GC9
%um of currents entering into the Eunction # %um of currents le!ving the
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
%um of currents entering into the Eunction # %um of currents le!ving the
Eunction
Node-v2&
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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
%um of currents entering into the Eunction # %um of currents le!ving the
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
%um of currents entering into the Eunction # %um of currents le!ving the
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
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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.
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*1
1
*-
/
:ccording to GC9
-
/
+!
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%um of currents entering into the Eunction # %um of currents le!ving the
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
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(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&/
-
/
-
/
-
+!
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1 >
0
2 3
O?. $olt!ge i1 i2 i i i= i i> i" iA
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1
2 =
Coefficients of
O?. $olt!ge di1+dt di2+dt di+dt di+dt
/
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1 5 -1 -1 5
2 1 -1 5 -1
5 -1 5 -1
5 5 -1 -1
/
-
/
-
/
-
/
-
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2h 1h
/
-
/
-
/
-
/
-
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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 $
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1 Jdt 2dt -1+2dt -dt
2 5 -1+2
dt 2
dt -
dt
1+dt -dt -dt dt
$2
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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&
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O?. $olt!ge $1 $2 $ $
1 5 -1 -1 5
2 5 -1 5 -1
+
!
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- -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 $2 2$
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($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.
/
-
/
-
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1
:ccording to GC9
%um of currents entering into the Eunction # %um of currents le!ving the
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)
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# $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.
+!
+
!
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+
!
+
!
+
!
+
!
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i1
+
!
+
!
+!
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# M(=+2)(2) / 2 # >
-1 2
i2# >+>.>= # 5.A5 !meres. :ns.
-A. %olve for the four node-to-d!tum volt!ges.
$2$1
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i
i2
i1
1 17
e e?uiv!lent with resect to the !ir of termin!ls
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1 1
(!)
(b)
1 1
O?u!ting (!) (b)
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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 / $5sintHC5sintI
1-15.
t
w # vi dt
-
For !n inductor
v9# 9di+dt
utting the v!lue of volt!ge
t
w #
vi dt
-
t
w # (9di+dt)i dt
-
t
w # 9idi
-
t
w # 9i2+2
-
w # 9Mi2(t)+2 - i2(-)+2
w # 9Mi2(t)+2 6 (i(-))2+2
w # 9Mi2(t)+2 6 (5)2+2
w # 9Mi2(t)+2 Hec!use i(-) # 5 for !n inductorI
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:s we know
# 9i
2# 92i2
2+9 # 9i2
w # 9Mi2(t)+2
w # 9i2+2 utting the v!lue of 9i2
w # (2+9)+2
w # 2+29 Hwhere # flu0 link!geI
1-11.
t
w # vi dt
-
For ! c!!citor
i # Cdv+dt
utting the v!lue of current
t
w # vi dt
-
t
w # (Cdv+dt)v dt
-
tw # Cvdv
-
t
w # Cv2+2
-
w # CMv2(t)+2 - v2(-)+2
w # CMv2(t)+2 6 (v(-))2+2
w # CMv2(t)+2 6 (5)2+2
w # CMv2(t)+2 Hec!use v(-
) # 5 for !n inductorI
:s we know
U # C$
$ # U+C
w # CMv2(t)+2
w # CM(?+C)2+2
w # CM?2+2C2
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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
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for 1t
%tr!ight-line e?u!tion
-intercet # c # 5
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(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 $m
9et c!!cit!nce be C
%loe # m
%tr!ight-line e?u!tion
-intercet # c # 2$m
%loe # m
%tr!ight-line e?u!tion
-intercet # c # -$m
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i # C$m
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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$mt
for 1t
? # C$
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? # 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)
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5 1 2 time
for 5
t
1
(01, 1) # (1, m)
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(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
%tr!ight-line e?u!tion
-intercet # c # 5
%loe # m
%tr!ight-line e?u!tion
-intercet # c # 2m
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(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
%tr!ight-line e?u!tion
-intercet # c # -m
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v() # (1+C)M(2() 6 ()2+2) - (+2) / m+2C
Muhammad Irfan Yousuf
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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)
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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
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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)
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: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.
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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.> $olts
(g) dwc+dt #
wc# (1+2)e-2t
dwc+dt # d(1+2)e-2t+dt
dwc+dt # (1+2)e-2t(-2) # -e-2t
dwc(t5)+dt # (1+2)e-2t(-2) # -e-2t5
t5# 1 sec.dwc(1)+dt # (1+2)e-2t(-2) # -e-2(1)
dwc(1)+dt # (1+2)e-2t(-2) # -e-2
dwc(1)+dt # (1+2)e-2t(-2) # -e-2 # - 5.1= w!tts.
(h) * # i2 # (e-t)2(1) # e-2t
*(t5) # i2 # (e-t)2(1) # e-2t5
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5 + +2 =+
for vc-5.=$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
ic(t) # d(Cv)+dt # Cdv+dt / vdC+dt
ic(t) # Cdv+dt / vdC+dt
for + t =+
C # 5.=F
v # sintic(t) # (5.=)dsint+dt / sintd(5.=)+dt
ic(t) # (5.=)cost
ic(t) # d(Cv)+dt # Cdv+dt / vdC+dt
ic(t) # Cdv+dt / vdC+dt
for =+ t
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C # 1F
v # sint
ic(t) # (1)dsint+dt / sintd(1)+dt
ic(t) # cost
vc(t)
5 + +2 =+
time
ic(t)
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5 + +2 =+ time
1-5.
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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
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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
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n the ste!d st!te
inductor beh!ves like
! short circuit
Muhammad Irfan Yousuf
Dedicated to: Prof. Dr. Sohail Aftab Qureshi
: 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
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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 # $+1
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i # ke-(1 / 2)t+9
i # ($+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
:ling initi!l condition
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
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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
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C!!citor is ch!rging
$olt!ge !cross the
c!!citor # $5
C!!citor is disch!rging
$olt!ge !cross the c!!citor # - $5
Muhammad Irfan Yousuf
Dedicated to: Prof. Dr. Sohail Aftab Qureshi
- 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
Differenti!ting both sides with resect to Bt7
(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
:ling initi!l condition
i(5/) # ke(-1+Ce?)(5)
i(5/) # ke5
i(5/) # k(1)
i(5/) # k
i(5/) # ($16 $2)+
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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
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# (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
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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
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n c!se of D.C.
inductor beh!ves like
! short circuit
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Dedicated to: Prof. Dr. Sohail Aftab Qureshi
(1 / 1+2)v2dt / 2v2# 5
(+2)v2dt / 2v2# 5
Differenti!ting both sides with resect to Bt7
(+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
:ling initi!l condition
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
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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
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n c!se of D.C.
inductor beh!ves like! short circuit
Muhammad Irfan Yousuf
Dedicated to: Prof. Dr. Sohail Aftab Qureshi
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
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: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.
inductor beh!ves like
! short circuit
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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
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Dedicated to: Prof. Dr. Sohail Aftab Qureshi
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&
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(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+
:ling initi!l condition
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
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n c!se of D.C.
inductor beh!ves like
! short circuit
Muhammad Irfan Yousuf
Dedicated to: Prof. Dr. Sohail Aftab Qureshi
:fter switching
/
-
/
-
/
-
-
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n c!se of D.C.c!!citor beh!ves
like !n oen circuit
Muhammad Irfan Yousuf
Dedicated to: Prof. Dr. Sohail Aftab Qureshi
: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
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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
:ling initi!l condition
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