Chapter 3, Solution 12
Therearetwo untnown nodes,as shownin the circuit below.
10c,
At node1,
vr:30
lo
-&
=o
*.v,:9 * vr
(1)
16Vr-10Vo = 30
At nodeo,
v'rut-rr"n50=o
(2)
-svl +6vo- 20rx= 0
But Ix = Vr/2. Substituting
this itr (2)leadsto
-l5vr + 6V.=0 or Vr =0.4Vo
(3)
Substitutilg(3) into 1,
16(0.4vo)-10V.=30
or vo =-8.333V.
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Chapter 3, Problem 13.
Calculate,1 and12itr the circuit of Fig. 3.62using nodal analysis.
2A
2V
Figure3.62
Chapter 3, Solution 13
At nodenumber2, [(v2+ 2) - O]l1O+ vzl4 = 3 or v2 = glp!ts
!u1, | = [(v, + 2) - 0]/10= (8 + 2)/10= 1 e.mpandvr = 8x1= 8volts
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Chapter3, Solution17
4cr
60v
8()
60v
v , v , - v ..
6 0 - v .' = r +
482
^. 60-Y" v,-v,
At node2, ' 1 0 2
12O='lvt - 4vz
At node1,
(l)
^
60-v,
Sutlo=-.
Hence
3(60-vt)*60-v, vr -vz =0 -------+ = 5\+ l2vz
lo2o
*
4102
e)
6n-v
Solving(llaDd(2)givesvr=53.08V. Hencei0 = ::----jl = 1.73A
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oit6s Manual nav & disola"ed reoroducedor distributedin anv form or b" anv means.without tbe orior
qrritten permissionof the publisher. or used b€yond the limil€d distribution to teachersand educarors
permittedbv Mccmv/-Hill for thel individual coursepreparation. If vou are a studentusinqthis Marual
\0u areusins it without temission.
-o
-v,
v,
v.
_i_o*_t'=0_-"
v^-0
v " - v ,,
+ 4+ _-___::
;
_0.t2sv^
=4
l.t25Vr
=0_r0.75V2_0.25V1- --4
v.-v^ v, -o
_T
*
r 2= 0 _+_0.25V:+0.75Vr= _2
i
v , - v , ___1____:
v, -o
_0.
_
_
___1_____jr
___r
2F
I r.tzs o
0
|
l0
=0
+
0;75
l 25vr
1.125v4 = 2
(1)
(2)
(3)
(4)
o -0.1251
f.l
-0.25 o
]"=|-,I
-o.2so;ls 0ll-tl
L-0.r25 0
0
r.12s
I
12)
NowwecanuseMATLAB to solvefor theunknownnodevoltages.
>> Y=[1.125,0,0,-0.125;0,0.75,-0.25,0;0,-0.25,0.75,0iO.125,0,0,1.1.251
v1.1250
0
0 -0.1250
0 0.7500-0.2500 0
0 -0.2500 0.7500 0
-0.12s0 0
0 1.1250
>> t=14,-4,-2,21'
I_
4
-4
-2
2
>> V=inv(Y)*I
3.8000
-7.0000
-5.0000
2.2000
vo = vl -v4 = 3.8-2.2 = 1.6V.
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lou areusins it without permission.
Chapter3, Problem31.
Findthenodevoltagesfor thecircuitin Fig. 3.80.
zrt
t,
.&.aa
\r
IA
I
{rl
.|7
i lr)
.t 2t
l4
''
.1r
{
l|)v
Figure3.80
Chapter3, Solution31
----_-_------_-rl.2,,i\2vo
lc)
4rr
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At the supemode,
r+z'"=f+f+!1la
(1)
But vo= vl- Yr. Hence(1)becomes,
4=-3yt+4vz+4vz
(2)
At node3,
^
v?
4-'2
- v 1 + -1 0 - v r
20 =4vr+Ovz-v:
A t f t e s u p e m o d ev.2 = r 1 + 4 i o . B u t i o = I l
(3)
Hence.
(4)
Solving(2) to (4) leadsto,
vt = !8!,
vz= 4.85V,vr = :O12Y.
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Chapter 3, Problem 67.
Obtainthenode-voltage
equations
for thecircuitin Fig.3.111by inspection.Thensolve
for Vo.
3V"
Fizure3.111For Prob.3.67.
Chapter 3, Solution 67
Considerthe circuit below.
3V.
| 0.35 -0.25
|
o I
0
L
6
0.e5-0.sv =l
l-0.25-0.5
L 0
0.5I
i/v oIl
|
l
1
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of this Manual mav be displaved.reproducedor distributedin anv form or bv anv means.without the prior
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p9!4I@! blMe&at LEilljbltlbei!.i!!!&Ldual coursepreparatiop. If you are a sludentusinq this MaNaL
vou areusins it without permissiotr.
\-
Sincewe actuallyhavefour unl<nownsandonly threeequations,we needa conshaint
equation.
V"=Vu-Vr
Substitutitrgthis backinto the matdx equation,the fust equationbecomes,
0.35Vr-3.25V:+3V:=-2
This now resultsin the following matrix equatioq
I o.3s -3.25 3 I
l-2)
L
L6l
oI
o.os-o.slv=l
l-0.:s
0 -0.s 0.s
I
Now we canuseMATLAB to solvefor V.
>> Y=[0.35,-3.25,3
;0,-0.5,0.5]
i0.25,0.95,-0.5
0.3500 -3.2500 3.0000
-0.2500 0.9500 -0.5000
0 -0.5000 0.5000
>> I=[-2,0,6]'
0
6
>> V=invCY)*I
-164.2105
-"t7.8947
-65.8947
Vo= V2-V3 =-77.89+ 65.89=_!2_Y.
Let us now do a quick checkat node 1.
-3C12) + 0.1( 164.21)+ 0.25(-164.21r'77.89)
+2=
+36- 16.421- 21.58+ 2 = -O.001;arswerchecks!
\.--
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witten pennission of the oublis
permittedbv McGraw-HiI for their individual coursep-!9paratj9!-Ifl&!-933-EEdgdlq!Ug
vou areu(ineir witiout permission.
Chapter3, Problem39.
Determine
themeshcurentsi1 andi2in thecircuitshownin Fig. 3.85.
21,
lt, ;-".
:oct
tr{l\r
':7
tlv
Figure3.85
Chapter3, Solution39
Formesh1,
- 1 0 - 2 1 ,+ 7 0 1-t 6 1 "= 0
But 1, = 1r -1r. Hence,
1O= lIt + 2Iz +lOIt - 6Iz
5= 4\-2Iz
For mesh2,
12+8Ix-611=O
6 =3It - 412
Solving(1) and(2) leadsto
1, =0.8A, 4 =-0.9A
o)
(2)
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written permissionof the publisher. or used bevond the li
permlttedbv Mcc{aw-Hin for their ii
co e
vou areusingit without permission.
Chapter3, Problem44.
Usemeshanalysisto obtaini, in thecircuitof Fig. 3.90.
6V
2A
4A
€
:r 1l
{
1:V
)rn
Figure3.90
Chapter 3, Solution 44
6V
i:
ir
Loop I and 2 folm a supemesh. For the supermesh,
6\+4iz-5iz+1'2=O
(1)
For loop3,
-ir-4iz+7i:+6 = 0
(2)
Also,
iz=3tir
(3)
Sotving(1)to (3), \ = -3.06'J,
\ = -7.3333;L = i1- i3- :1!:zEXA
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Chapter 3, Problem 54.
Find the meshcurents ir, ir, andi: in the circuit in Fig. 3.99.
|tat
: &tr!
Itn
I lif:
12v
10v
.
,.":\
ii. l
, Jn"l
tfv
Figure3.99
Chapter 3, Solution 54
Lei the meshcufientsbe in mA. For mesh1,
-12+10+2It- 12=0 -----) 2=2\-Iz
For mesh2,
. - 1 0 + 3 1 2I-t - 1 3 = O - - ' - r
For mesh3,
- 1 2 + 2 1 3 - 1 2= 0
(1)
10=-It+312-\
12=-Iz+2It
Q)
(3)
Putting (1) to (3) in matrix forin leadsto
(2
l o)f/,) f2)
-'
|( 0 -,1
tlttl
r /, r0 -------)4r=B
ll l=l |
2
)\r3)
\12)
Using MATI-AB,
f(rs'l
_tl
1=A-B=l
8.5 I
---)_1, =5.25mA,1z=8.5mA,1r=10.25rnA
Irous]
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of this Manual mav be displaved.reproduc€dor distributedin anv form or bv anv means.without theprior
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permittedby Mccraw-Hill for their individual corrse preparation.If vou are a studenrusing this Manual.
vop areus-|ngll w:ihour o!@j$i9!.
Chapter3, Problem60.
Calculate
thepowerdissipated
in eachresistorin thecircuitin Fig. 3.104.
0,sio
4{>
----------,4,l!,{
w
/\
EA
i i?!,{
,.I
kt
ito
{
i t0v
;tl
Figurc3.104
Chapter 3, Solution 60
lfi
10v
= (10-v1)/4,whichleadslovl=10/7
Atnode1, (vr/l) + (0.5v1/1)
At node2, (0.5vi/1)+ ((10- v)18) ='1212whichleadsto vz = 2217
P1o= (vr)2/1= 2_A!,1:watts,Prn = ('t)212 = 4.939 watts
Pao= (10 - vr)'?/4= 1&9!Lge!6, P8o= (10 - v2)2/8= 5.88watts
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Chapter 3, Problem 61.
Calculatethe cuirent gain i,[', in the circuit of Fig. 3.105.
294
10fi
Figure3.105
Chapter 3, Solution 61
10r,
io
/rco
At node1, L = (vr/30)+ ((vr - vz)/20)whichleadsto 60is= 5v1- 3v2
(l)
But v2= -5yo and vs=yl whichleadsto v2= -5vr
Hence,60i"=5vr+15vr=26utwhichleadsto vr = 3i" vz= -15i
i6 = vrl50 = -15i*r50 which leadsto rry'i,= -15i50 = -{.3
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of .his Manual rnav be displayed.reFoducedor distsibu
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vou areusing it wiihout permission.
Chapter3, Problem73.
Writethemesh-curentequations
for thecircuitin Fig.3.117.
5A
l tr J 3O
,ll
Q'o
Figure3.117
Chapter3, Solution73
R r 1= 2 + 3 + 4 = 9 , R 2 2 = 3
+ 5 = 8 , R 3 3= 1 + 1+ 4 = 6 , R 4 a = 1 + 1 = 2 ,
R12= -3, R13= -{, R1a= 0, R2 = Q, Rz=0, R34= -1
v1=6, v2= 4, v3 = 2, and va = -3
Hence,
le
l-3
4 -4
8
o
6
l-, o -1
Lo o
olf"l
o ll',I
-'lli l2 )Lio)
,l
_;l
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of this Manual mav be displayed.repmducedm dis
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pernitted bv Mccraw-Hill for their individual coursepreparation. If vou are a studentusinq this Manual.
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