- 1.Chapter 1, Problem 1 How many coulombs are represented by
these amounts of electrons: (a) (b)17 10482.6 18 1024.1 (c) (d)19
1046.2 20 10628.1 Chapter 1, Solution 1 (a) q = 6.482x1017 x
[-1.602x10-19 C] = -0.10384 C (b) q = 1. 24x1018 x [-1.602x10-19 C]
= -0.19865 C (c) q = 2.46x1019 x [-1.602x10-19 C] = -3.941 C (d) q
= 1.628x1020 x [-1.602x10-19 C] = -26.08 C Chapter 1, Problem 2.
Determine the current flowing through an element if the charge flow
is given by (a) ( ) ( ) mC83 += ttq (b) ( ) C2)48 2 t-t(tq += (c) (
) ( )nCe5e3tq t2-t = (d) ( ) pCtsin10 120tq = (e) ( ) Ct50cos20 4 t
etq = Chapter 1, Solution 2 (a) i = dq/dt = 3 mA (b) i = dq/dt =
(16t + 4) A (c) i = dq/dt = (-3e-t + 10e-2t ) nA (d) i=dq/dt = 1200
120 cos t pA (e) i =dq/dt = + e t tt4 80 50 1000 50( cos sin ) A
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2. Chapter 1, Problem 3. Find the charge q(t) flowing through a
device if the current is: (a) ( ) ( ) C10A,3 == qti (b)
0)0(,mA)52()( =+= qtti (c) C2(0)A,)6/10cos(20)( =+= qtti (d) ( )
0(0)A,40sin10 30 == qteti t Chapter 1, Solution 3 (a) C1)(3t +=+=
q(0)i(t)dtq(t) (b) mC5t)(t 2 +=++= q(v)dts)(2tq(t) (c) ( )q(t) 20
cos 10t / 6 q(0) (2sin(10 / 6) 1) Ct = + + = + + (d)
C40t)sin0.12t(0.16cos40e 30t- += + =+= t)cos40-t40sin30( 1600900
e10 q(0)t40sin10eq(t) -30t 30t- Chapter 1, Problem 4. A current of
3.2 A flows through a conductor. Calculate how much charge passes
through any cross-section of the conductor in 20 seconds. Chapter
1, Solution 4 q = it = 3.2 x 20 = 64 C Chapter 1, Problem 5.
Determine the total charge transferred over the time interval of 0
t 10s when 1 ( ) 2 i t t= A. Chapter 1, Solution 5 10 2 0 101 25 C
02 4 t q idt tdt= = = = PROPRIETARY MATERIAL. 2007 The McGraw-Hill
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3. Chapter 1, Problem 6. The charge entering a certain element is
shown in Fig. 1.23. Find the current at: (a) t = 1 ms (b) t = 6 ms
(c) t = 10 ms Figure 1.23 Chapter 1, Solution 6 (a) At t = 1ms, ===
2 80 dt dq i 40 A (b) At t = 6ms, == dt dq i 0 A (c) At t = 10ms,
=== 4 80 dt dq i 20 A PROPRIETARY MATERIAL. 2007 The McGraw-Hill
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4. Chapter 1, Problem 7. The charge flowing in a wire is plotted in
Fig. 1.24. Sketch the corresponding current. Figure 1.24 Chapter 1,
Solution 7 > I = inv(Z)*V I = 1.6196 mA 1.0202 mA 2.461 mA 3 mA
2.423 mA PROPRIETARY MATERIAL. 2007 The McGraw-Hill Companies, Inc.
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163. Chapter 3, Problem 54. Find the mesh currents i1, i2, and i3
in the circuit in Fig. 3.99. Figure 3.99 Chapter 3, Solution 54 Let
the mesh currents be in mA. For mesh 1, 2121 22021012 IIII ==++ (1)
For mesh 2, (2)321312 3100310 IIIIII +==+ For mesh 3, 3223 2120212
IIII +==+ (3) Putting (1) to (3) in matrix form leads to BAI I I I
= = 12 10 2 210 131 012 3 2 1 Using MATLAB, mA25.10,mA5.8,mA25.5
25.10 5.8 25.5 321 1 === == IIIBAI PROPRIETARY MATERIAL. 2007 The
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publisher, or used beyond the limited distribution to teachers and
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preparation. If you are a student using this Manual, you are using
it without permission. 164. Chapter 3, Problem 55. In the circuit
of Fig. 3.100, solve for i1, i2, and i3. Figure 3.100 PROPRIETARY
MATERIAL. 2007 The McGraw-Hill Companies, Inc. All rights reserved.
No part of this Manual may be displayed, reproduced or distributed
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Manual, you are using it without permission. 165. Chapter 3,
Solution 55 dI1 I2 I3 I4 4 A + + 10 V 6 2 4 12 8 V I3 I2 1A I4 i3
i1 4A 1A cb 0a i2 It is evident that I1 = 4 (1) For mesh 4, 12(I4
I1) + 4(I4 I3) 8 = 0 (2) For the supermesh 6(I2 I1) + 10 + 2I3 +
4(I3 I4) = 0 or -3I1 + 3I2 + 3I3 2I4 = -5 (3) At node c, I2 = I3 +
1 (4) Solving (1), (2), (3), and (4) yields, I1 = 4A, I2 = 3A, I3 =
2A, and I4 = 4A At node b, i1 = I2 I1 = -1A At node a, i2 = 4 I4 =
0A At node 0, i3 = I4 I3 = 2A PROPRIETARY MATERIAL. 2007 The
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preparation. If you are a student using this Manual, you are using
it without permission. 166. Chapter 3, Problem 56. Determine v1 and
v2 in the circuit of Fig. 3.101. Figure 3.101 PROPRIETARY MATERIAL.
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preparation. If you are a student using this Manual, you are using
it without permission. 167. Chapter 3, Solution 56 +12 V i1 i2 i3 +
v2 + v1 2 2 2 2 2 For loop 1, 12 = 4i1 2i2 2i3 which leads to 6 =
2i1 i2 i3 (1) For loop 2, 0 = 6i2 2i1 2 i3 which leads to 0 = -i1 +
3i2 i3 (2) For loop 3, 0 = 6i3 2i1 2i2 which leads to 0 = -i1 i2 +
3i3 (3) In matrix form (1), (2), and (3) become, = 0 0 6 i i i 311
131 112 3 2 1 = ,8 311 131 112 = 2 = 24 301 131 162 = 3 = 24 011
031 612 = , therefore i2 = i3 = 24/8 = 3A, v1 = 2i2 = 6 volts, v =
2i3 = 6 volts PROPRIETARY MATERIAL. 2007 The McGraw-Hill Companies,
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168. Chapter 3, Problem 57. In the circuit in Fig. 3.102, find the
values of R, V1, and V2 given that io = 18 mA. Figure 3.102 Chapter
3, Solution 57 Assume R is in kilo-ohms. VVVVmAxkV 2872100100,72184
212 ===== Current through R is R R RiVi R i RoR )18( 3 3 28 3 3 1,
+ == + = This leads to R = 84/26 = 3.23 k PROPRIETARY MATERIAL.
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it without permission. 169. Chapter 3, Problem 58. Find i1, i2, and
i3 the circuit in Fig. 3.103. Figure 3.103 Chapter 3, Solution 58 +
120 V 30 i1 i2 i3 10 30 10 30 For loop 1, 120 + 40i1 10i2 = 0,
which leads to -12 = 4i1 i2 (1) For loop 2, 50i2 10i1 10i3 = 0,
which leads to -i1 + 5i2 i3 = 0 (2) For loop 3, -120 10i2 + 40i3 =
0, which leads to 12 = -i2 + 4i3 (3) Solving (1), (2), and (3), we
get, i1 = -3A, i2 = 0, and i3 = 3A PROPRIETARY MATERIAL. 2007 The
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preparation. If you are a student using this Manual, you are using
it without permission. 170. Chapter 3, Problem 59. Rework Prob.
3.30 using mesh analysis. Chapter 3, Problem 30. Using nodal
analysis, find vo and io in the circuit of Fig. 3.79. Figure 3.79
PROPRIETARY MATERIAL. 2007 The McGraw-Hill Companies, Inc. All
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171. Chapter 3, Solution 59 i1 i2 i3 4v0 + + 120 V +100V + v0 80 40
20 10 i2 i3 I0 2I0 For loop 1, -100 + 30i1 20i2 + 4v0 = 0, where v0
= 80i3 or 5 = 1.5i1 i2 + 16i3 (1) For the supermesh, 60i2 20i1 120
+ 80i3 4 v0 = 0, where v0 = 80i3 or 6 = -i1 + 3i2 12i3 (2) Also,
2I0 = i3 i2 and I0 = i2, hence, 3i2 = i3 (3) From (1), (2), and
(3), 130 1231 3223 = 0 6 10 i i i 3 2 1 = ,5 130 1231 3223 = 2 =
,28 100 1261 32103 = 3 = 84 030 631 1023 = I0 = i2 = 2/ = -28/5 =
-5.6 A v0 = 8i3 = (-84/5)80 = -1.344 kvolts PROPRIETARY MATERIAL.
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of this Manual may be displayed, reproduced or distributed in any
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educators permitted by McGraw-Hill for their individual course
preparation. If you are a student using this Manual, you are using
it without permission. 172. Chapter 3, Problem 60. Calculate the
power dissipated in each resistor in the circuit in Fig. 3.104.
Figure 3.104 Chapter 3, Solution 60 0.5i0 v2 +10 V 10 V 2 8 4 1 v1
i0 At node 1, (v1/1) + (0.5v1/1) = (10 v1)/4, which leads to v1 =
10/7 At node 2, (0.5v1/1) + ((10 v2)/8) = v2/2 which leads to v2 =
22/7 P1 = (v1)2 /1 = 2.041 watts, P2 = (v2)2 /2 = 4.939 watts P4 =
(10 v1)2 /4 = 18.38 watts, P8 = (10 v2)2 /8 = 5.88 watts
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173. Chapter 3, Problem 61. Calculate the current gain io/is in the
circuit of Fig. 3.105. Figure 3.105 Chapter 3, Solution 61 + v0 20
v2 + is v1 30 40 10 5v0 i0 At node 1, is = (v1/30) + ((v1 v2)/20)
which leads to 60is = 5v1 3v2 (1) But v2 = -5v0 and v0 = v1 which
leads to v2 = -5v1 Hence, 60is = 5v1 + 15v1 = 20v1 which leads to
v1 = 3is, v2 = -15is i0 = v2/50 = -15is/50 which leads to i0/is =
-15/50 = 0.3 PROPRIETARY MATERIAL. 2007 The McGraw-Hill Companies,
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174. Chapter 3, Problem 62. Find the mesh currents i1, i2, and i3
in the network of Fig. 3.106. Figure 3.106 Chapter 3, Solution 62
i1 i2 i3 +100V B + 4 k 8 k 2 k 40 V A We have a supermesh. Let all
R be in k, i in mA, and v in volts. For the supermesh, -100 +4i1 +
8i2 + 2i3 + 40 = 0 or 30 = 2i1 + 4i2 + i3 (1) At node A, i1 + 4 =
i2 (2) At node B, i2 = 2i1 + i3 (3) Solving (1), (2), and (3), we
get i1 = 2 mA, i2 = 6 mA, and i3 = 2 mA. PROPRIETARY MATERIAL. 2007
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it without permission. 175. Chapter 3, Problem 63. Find vx, and ix
in the circuit shown in Fig. 3.107. Figure 3.107 Chapter 3,
Solution 63 +50 V 4ix + i1 i2 10 5 A For the supermesh, -50 + 10i1
+ 5i2 + 4ix = 0, but ix = i1. Hence, 50 = 14i1 + 5i2 (1) At node A,
i1 + 3 + (vx/4) = i2, but vx = 2(i1 i2), hence, i1 + 2 = i2 (2)
Solving (1) and (2) gives i1 = 2.105 A and i2 = 4.105 A vx = 2(i1
i2) = 4 volts and ix = i2 2 = 2.105 amp PROPRIETARY MATERIAL. 2007
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publisher, or used beyond the limited distribution to teachers and
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preparation. If you are a student using this Manual, you are using
it without permission. 176. Chapter 3, Problem 64. Find vo, and io
in the circuit of Fig. 3.108. Figure 3.108 PROPRIETARY MATERIAL.
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of this Manual may be displayed, reproduced or distributed in any
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preparation. If you are a student using this Manual, you are using
it without permission. 177. Chapter 3, Solution 64 40 i1 i2 i3
+100V + 4i0 0.2V0 50 10 10 2 A B Ai1 i0 i2 i3i1 + For mesh 2, 20i2
10i1 + 4i0 = 0 (1) But at node A, io = i1 i2 so that (1) becomes i1
= (16/6)i2 (2) For the supermesh, -100 + 50i1 + 10(i1 i2) 4i0 +
40i3 = 0 or 50 = 28i1 3i2 + 20i3 (3) At node B, i3 + 0.2v0 = 2 + i1
(4) But, v0 = 10i2 so that (4) becomes i3 = 2 + (2/3)i2 (5) Solving
(1) to (5), i2 = 0.11764, v0 = 10i2 = 1.1764 volts, i0 = i1 - i2 =
(5/3)i2 = 196.07 mA PROPRIETARY MATERIAL. 2007 The McGraw-Hill
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178. Chapter 3, Problem 65. Use MATLAB to solve for the mesh
currents in the circuit of Fig. 3.109. Figure 3.109 Chapter 3,
Solution 65 For mesh 1, 12 + 12I1 6I2 I4 = 0 or 421 61212 III = (1)
For mesh 2, 6I1 + 16I2 8I3 I4 I5 = 0 (2) For mesh 3, 8I2 + 15I3 I5
9 = 0 or 9 = 8I2 + 15I3 I5 (3) For mesh 4, I1 I2 + 7I4 2I5 6 = 0 or
6 = I1 I2 + 7I4 2I5 (4) For mesh 5, I2 I3 2I4 + 8I5 10 = 0 or 5432
8210 IIII += (5) PROPRIETARY MATERIAL. 2007 The McGraw-Hill
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179. Casting (1) to (5) in matrix form gives BAI 10 6 9 0 12 I I I
I I 82110 27011 101580 118166 010612 5 4 3 2 1 = = Using MATLAB we
input:
Z=[12,-6,0,-1,0;-6,16,-8,-1,-1;0,-8,15,0,-1;-1,-1,0,7,-2;0,-1,-1,-2,8]
and V=[12;0;9;6;10] This leads to >>
Z=[12,-6,0,-1,0;-6,16,-8,-1,-1;0,-8,15,0,-1;-1,-1,0,7,-2;0,-1,-1,-2,8]
Z = 12 -6 0 -1 0 -6 16 -8 -1 -1 0 -8 15 0 -1 -1 -1 0 7 -2 0 -1 -1
-2 8 >> V=[12;0;9;6;10] V = 12 0 9 6 10 >> I=inv(Z)*V I
= 2.1701 1.9912 1.8119 2.0942 2.2489 Thus, I = [2.17, 1.9912,
1.8119, 2.094, 2.249] A. PROPRIETARY MATERIAL. 2007 The McGraw-Hill
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displayed, reproduced or distributed in any form or by any means,
without the prior written permission of the publisher, or used
beyond the limited distribution to teachers and educators permitted
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180. Chapter 3, Problem 66. Write a set of mesh equations for the
circuit in Fig. 3.110. Use MATLAB to determine the mesh currents.
PROPRIETARY MATERIAL. 2007 The McGraw-Hill Companies, Inc. All
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limited distribution to teachers and educators permitted by
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student using this Manual, you are using it without permission. +
_12 V 10 I1 I2 6 Figure 3.110 For Prob. 3.66. Chapter 3, Solution
66 The mesh equations are obtained as follows. + + =1 2 3 412 24 30
4 6 2 0I I I I or 30I1 4I2 6I3 2I4 = 12 (1) + + =1 2 4 524 40 4 30
2 6 0I I I I or 4I1 + 30I2 2I4 6I5 = 16 (2) 6I1 + 18I3 4I4 = 30 (3)
2I1 2I2 4I3 + 12I4 4I5 = 0 (4) 6I2 4I4 + 18I5 = 32 (5) 4 24 V 4 I3
I4 40 V 8 + _ + _ 8 10 + _ 2 2 30 V 8 4 6 + _ I5 32 V 8 181.
Putting (1) to (5) in matrix form = 32 0 30 16 12 I 184060 412422
041806 620304 026430 ZI = V Using MATLAB, >> Z =
[30,-4,-6,-2,0; -4,30,0,-2,-6; -6,0,18,-4,0; -2,-2,-4,12,-4;
0,-6,0,-4,18] Z = 30 -4 -6 -2 0 -4 30 0 -2 -6 -6 0 18 -4 0 -2 -2 -4
12 -4 0 -6 0 -4 18 >> V = [-12,-16,30,0,-32]' V = -12 -16 30
0 -32 >> I = inv(Z)*V I = -0.2779 A -1.0488 A 1.4682 A
-0.4761 A -2.2332 A PROPRIETARY MATERIAL. 2007 The McGraw-Hill
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182. Chapter 3, Problem 67. Obtain the node-voltage equations for
the circuit in Fig. 3.111 by inspection. Then solve for Vo.
PROPRIETARY MATERIAL. 2007 The McGraw-Hill Companies, Inc. All
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Figure 3.111 For Prob. 3.67. 10 5 4 A + _ 3 Vo Vo 2 4 2 A Chapter
3, Solution 67 Consider the circuit below. V3 + Vo - + = 6 0 V32 V
5.05.00 5.095.025.0 025.035.0 o V2 10 5 4 A 3 Vo V1 2 4 2 A 183.
Since we actually have four unknowns and only three equations, we
need a constraint equation. Vo = V2 V3 Substituting this back into
the matrix equation, the first equation becomes, 0.35V1 3.25V2 +
3V3 = 2 This now results in the following matrix equation, = 6 0 2
V 5.05.00 5.095.025.0 325.335.0 Now we can use MATLAB to solve for
V. >> Y=[0.35,-3.25,3;-0.25,0.95,-0.5;0,-0.5,0.5] Y = 0.3500
-3.2500 3.0000 -0.2500 0.9500 -0.5000 0 -0.5000 0.5000 >>
I=[-2,0,6]' I = -2 0 6 >> V=inv(Y)*I V = -164.2105 -77.8947
-65.8947 Vo = V2 V3 = 77.89 + 65.89 = 12 V. Let us now do a quick
check at node 1. 3(12) + 0.1(164.21) + 0.25(164.21+77.89) + 2 = +36
16.421 21.58 + 2 = 0.001; answer checks! PROPRIETARY MATERIAL. 2007
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this Manual may be displayed, reproduced or distributed in any form
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publisher, or used beyond the limited distribution to teachers and
educators permitted by McGraw-Hill for their individual course
preparation. If you are a student using this Manual, you are using
it without permission. 184. Chapter 3, Problem 68. Find the voltage
Vo in the circuit of Fig. 3.112. PROPRIETARY MATERIAL. 2007 The
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Manual may be displayed, reproduced or distributed in any form or
by any means, without the prior written permission of the
publisher, or used beyond the limited distribution to teachers and
educators permitted by McGraw-Hill for their individual course
preparation. If you are a student using this Manual, you are using
it without permission. 40 20 24 V + _ 4A Vo 25 10 3 A + _ Figure
3.112 For Prob. 3.68. Chapter 3, Solution 68 Consider the circuit
below. There are two non-reference nodes. V1 Vo 40 20 24 V + _ 4 A
Vo 25 10 3 A + _ 185. = + ++ = 04.2 7 25/243 34 V 19.01.0 1.0125.0
Using MATLAB, we get, >> Y=[0.125,-0.1;-0.1,0.19] Y = 0.1250
-0.1000 -0.1000 0.1900 >> I=[7,-2.04]' I = 7.0000 -2.0400
>> V=inv(Y)*I V = 81.8909 32.3636 Thus, Vo = 32.36 V. We can
perform a simple check at node Vo, 3 + 0.1(32.3681.89) +
0.05(32.36) + 0.04(32.3624) = 3 4.953 + 1.618 + 0.3344 = 0.0004;
answer checks! PROPRIETARY MATERIAL. 2007 The McGraw-Hill
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186. Chapter 3, Problem 69. For the circuit in Fig. 3.113, write
the node voltage equations by inspection. Figure 3.113 Chapter 3,
Solution 69 Assume that all conductances are in mS, all currents
are in mA, and all voltages are in volts. G11 = (1/2) + (1/4) +
(1/1) = 1.75, G22 = (1/4) + (1/4) + (1/2) = 1, G33 = (1/1) + (1/4)
= 1.25, G12 = -1/4 = -0.25, G13 = -1/1 = -1, G21 = -0.25, G23 =
-1/4 = -0.25, G31 = -1, G32 = -0.25 i1 = 20, i2 = 5, and i3 = 10 5
= 5 The node-voltage equations are: = 5 5 20 v v v 25.125.01
25.0125.0 125.075.1 3 2 1 PROPRIETARY MATERIAL. 2007 The
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Manual may be displayed, reproduced or distributed in any form or
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publisher, or used beyond the limited distribution to teachers and
educators permitted by McGraw-Hill for their individual course
preparation. If you are a student using this Manual, you are using
it without permission. 187. Chapter 3, Problem 70. Write the
node-voltage equations by inspection and then determine values of
V1 and V2 in the circuit in Fig. 3.114. PROPRIETARY MATERIAL. 2007
The McGraw-Hill Companies, Inc. All rights reserved. No part of
this Manual may be displayed, reproduced or distributed in any form
or by any means, without the prior written permission of the
publisher, or used beyond the limited distribution to teachers and
educators permitted by McGraw-Hill for their individual course
preparation. If you are a student using this Manual, you are using
it without permission. Figure 3.114 For Prob. 3.70. 1 S 2 S 2 A4 A
V1 4ix ix 5 S V2 Chapter 3, Solution 70 + = 2I4 4I4 V 50 03 x x
With two equations and three unknowns, we need a constraint
equation, Ix = 2V1, thus the matrix equation becomes, = 2 4 V 58 05
This results in V1 = 4/(5) = 0.8V and V2 = [8(0.8) 2]/5 = [6.4 2]/5
= 0.88 V. 188. Chapter 3, Problem 71. Write the mesh-current
equations for the circuit in Fig. 3.115. Next, determine the values
of I1, I2, and I3. + _10 V + _ 5 V 1 3 4 I3 I1 I2 2 5 Figure 3.115
For Prob. 3.71. PROPRIETARY MATERIAL. 2007 The McGraw-Hill
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189. Chapter 3, Solution 71 = 0 5 10 I 915 174 549 We can now use
MATLAB solve for our currents. >> R=[9,-4,-5;-4,7,-1;-5,-1,9]
R = 9 -4 -5 -4 7 -1 -5 -1 9 >> V=[10,-5,0]' V = 10 -5 0
>> I=inv(R)*V I = 2.085 A 653.3 mA 1.2312 A PROPRIETARY
MATERIAL. 2007 The McGraw-Hill Companies, Inc. All rights reserved.
No part of this Manual may be displayed, reproduced or distributed
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of the publisher, or used beyond the limited distribution to
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individual course preparation. If you are a student using this
Manual, you are using it without permission. 190. Chapter 3,
Problem 72. By inspection, write the mesh-current equations for the
circuit in Fig. 3.116. Figure 3.116 Chapter 3, Solution 72 R11 = 5
+ 2 = 7, R22 = 2 + 4 = 6, R33 = 1 + 4 = 5, R44 = 1 + 4 = 5, R12 =
-2, R13 = 0 = R14, R21 = -2, R23 = -4, R24 = 0, R31 = 0, R32 = -4,
R34 = -1, R41 = 0 = R42, R43 = -1, we note that Rij = Rji for all i
not equal to j. v1 = 8, v2 = 4, v3 = -10, and v4 = -4 Hence the
mesh-current equations are: = 4 10 4 8 i i i i 5100 1540 0462 0027
4 3 2 1 PROPRIETARY MATERIAL. 2007 The McGraw-Hill Companies, Inc.
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191. Chapter 3, Problem 73. Write the mesh-current equations for
the circuit in Fig. 3.117. Figure 3.117 Chapter 3, Solution 73 R11
= 2 + 3 +4 = 9, R22 = 3 + 5 = 8, R33 = 1+1 + 4 = 6, R44 = 1 + 1 =
2, R12 = -3, R13 = -4, R14 = 0, R23 = 0, R24 = 0, R34 = -1 v1 = 6,
v2 = 4, v3 = 2, and v4 = -3 Hence, = 3 2 4 6 i i i i 2100 1604 0083
0439 4 3 2 1 PROPRIETARY MATERIAL. 2007 The McGraw-Hill Companies,
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192. Chapter 3, Problem 74. By inspection, obtain the mesh-current
equations for the circuit in Fig. 3.11. Figure 3.118 Chapter 3,
Solution 74 R11 = R1 + R4 + R6, R22 = R2 + R4 + R5, R33 = R6 + R7 +
R8, R44 = R3 + R5 + R8, R12 = -R4, R13 = -R6, R14 = 0, R23 = 0, R24
= -R5, R34 = -R8, again, we note that Rij = Rji for all i not equal
to j. The input voltage vector is = 4 3 2 1 V V V V = ++ ++ ++ ++ 4
3 2 1 4 3 2 1 85385 88766 55424 64641 V V V V i i i i RRRRR0 RRRR0R
R0RRRR 0RRRRR PROPRIETARY MATERIAL. 2007 The McGraw-Hill Companies,
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193. Chapter 3, Problem 75. Use PSpice to solve Prob. 3.58. Chapter
3, Problem 58 Find i1, i2, and i3 the circuit in Fig. 3.103. Figure
3.103 PROPRIETARY MATERIAL. 2007 The McGraw-Hill Companies, Inc.
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194. Chapter 3, Solution 75 * Schematics Netlist * R_R4 $N_0002
$N_0001 30 R_R2 $N_0001 $N_0003 10 R_R1 $N_0005 $N_0004 30 R_R3
$N_0003 $N_0004 10 R_R5 $N_0006 $N_0004 30 V_V4 $N_0003 0 120V v_V3
$N_0005 $N_0001 0 v_V2 0 $N_0006 0 v_V1 0 $N_0002 0 i1 i2 i3
Clearly, i1 = 3 amps, i2 = 0 amps, and i3 = 3 amps, which agrees
with the answers in Problem 3.44. Chapter 3, Problem 76.
PROPRIETARY MATERIAL. 2007 The McGraw-Hill Companies, Inc. All
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195. Use PSpice to solve Prob. 3.27. Chapter 3, Problem 27 Use
nodal analysis to determine voltages v1, v2, and v3 in the circuit
in Fig. 3.76. Figure 3.76 Chapter 3, Solution 76 PROPRIETARY
MATERIAL. 2007 The McGraw-Hill Companies, Inc. All rights reserved.
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individual course preparation. If you are a student using this
Manual, you are using it without permission. 196. * Schematics
Netlist * I_I2 0 $N_0001 DC 4A R_R1 $N_0002 $N_0001 0.25 R_R3
$N_0003 $N_0001 1 R_R2 $N_0002 $N_0003 1 F_F1 $N_0002 $N_0001 VF_F1
3 VF_F1 $N_0003 $N_0004 0V R_R4 0 $N_0002 0.5 R_R6 0 $N_0001 0.5
I_I1 0 $N_0002 DC 2A R_R5 0 $N_0004 0.25 Clearly, v1 = 625 mVolts,
v2 = 375 mVolts, and v3 = 1.625 volts, which agrees with the
solution obtained in Problem 3.27. Chapter 3, Problem 77. Solve for
V1 and V2 in the circuit of Fig. 3.119 using PSpice. PROPRIETARY
MATERIAL. 2007 The McGraw-Hill Companies, Inc. All rights reserved.
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Manual, you are using it without permission. 197. PROPRIETARY
MATERIAL. 2007 The McGraw-Hill Companies, Inc. All rights reserved.
No part of this Manual may be displayed, reproduced or distributed
in any form or by any means, without the prior written permission
of the publisher, or used beyond the limited distribution to
teachers and educators permitted by McGraw-Hill for their
individual course preparation. If you are a student using this
Manual, you are using it without permission. 2 A5 A ix V1 V2 2 ix 2
5 1 Figure 3.119 For Prob. 3.77. Chapter 3, Solution 77 198. As a
check we can write the nodal equations, = 2 5 V 2.12.1 2.07.1
Solving this leads to V1 = 3.111 V and V2 = 1.4444 V. The answer
checks! Chapter 3, Problem 78. PROPRIETARY MATERIAL. 2007 The
McGraw-Hill Companies, Inc. All rights reserved. No part of this
Manual may be displayed, reproduced or distributed in any form or
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publisher, or used beyond the limited distribution to teachers and
educators permitted by McGraw-Hill for their individual course
preparation. If you are a student using this Manual, you are using
it without permission. 199. Solve Prob. 3.20 using PSpice. Chapter
3, Problem 20 For the circuit in Fig. 3.69, find V1, V2, and V3
using nodal analysis. PROPRIETARY MATERIAL. 2007 The McGraw-Hill
Companies, Inc. All rights reserved. No part of this Manual may be
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Figure 3.69 Chapter 3, Solution 78 200. The schematic is shown
below. When the circuit is saved and simulated the node voltages
are displaced on the pseudocomponents as shown. Thus, ,V15V,5.4V,3
321 === VVV . Chapter 3, Problem 79. Rework Prob. 3.28 using
PSpice. PROPRIETARY MATERIAL. 2007 The McGraw-Hill Companies, Inc.
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201. Chapter 3, Problem 28 Use MATLAB to find the voltages at nodes
a, b, c, and d in the circuit of Fig. 3.77. Figure 3.77 Chapter 3,
Solution 79 The schematic is shown below. When the circuit is saved
and simulated, we obtain the node voltages as displaced. Thus,
V88.26VV,6944.0VV,28.10VV,278.5V dcba ==== Chapter 3, Problem 80.
PROPRIETARY MATERIAL. 2007 The McGraw-Hill Companies, Inc. All
rights reserved. No part of this Manual may be displayed,
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202. Find the nodal voltage v1 through v4 in the circuit in Fig.
3.120 using PSpice. Figure 3.120 Chapter 3, Solution 80 PROPRIETARY
MATERIAL. 2007 The McGraw-Hill Companies, Inc. All rights reserved.
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individual course preparation. If you are a student using this
Manual, you are using it without permission. 203. * Schematics
Netlist * H_H1 $N_0002 $N_0003 VH_H1 6 VH_H1 0 $N_0001 0V I_I1
$N_0004 $N_0005 DC 8A V_V1 $N_0002 0 20V R_R4 0 $N_0003 4 R_R1
$N_0005 $N_0003 10 R_R2 $N_0003 $N_0002 12 R_R5 0 $N_0004 1 R_R3
$N_0004 $N_0001 2 Clearly, v1 = 84 volts, v2 = 4 volts, v3 = 20
volts, and v4 = -5.333 volts Chapter 3, Problem 81. Use PSpice to
solve the problem in Example 3.4 PROPRIETARY MATERIAL. 2007 The
McGraw-Hill Companies, Inc. All rights reserved. No part of this
Manual may be displayed, reproduced or distributed in any form or
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preparation. If you are a student using this Manual, you are using
it without permission. 204. Example 3.4 Find the node voltages in
the circuit of Fig. 3.12. Figure 3.12 Chapter 3, Solution 81
PROPRIETARY MATERIAL. 2007 The McGraw-Hill Companies, Inc. All
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205. Clearly, v1 = 26.67 volts, v2 = 6.667 volts, v3 = 173.33
volts, and v4 = -46.67 volts which agrees with the results of
Example 3.4. This is the netlist for this circuit. * Schematics
Netlist * R_R1 0 $N_0001 2 R_R2 $N_0003 $N_0002 6 R_R3 0 $N_0002 4
R_R4 0 $N_0004 1 R_R5 $N_0001 $N_0004 3 I_I1 0 $N_0003 DC 10A V_V1
$N_0001 $N_0003 20V E_E1 $N_0002 $N_0004 $N_0001 $N_0004 3 Chapter
3, Problem 82. If the Schematics Netlist for a network is as
follows, draw the network. PROPRIETARY MATERIAL. 2007 The
McGraw-Hill Companies, Inc. All rights reserved. No part of this
Manual may be displayed, reproduced or distributed in any form or
by any means, without the prior written permission of the
publisher, or used beyond the limited distribution to teachers and
educators permitted by McGraw-Hill for their individual course
preparation. If you are a student using this Manual, you are using
it without permission. 206. R_R1 1 2 2K R_R2 2 0 4K R_R3 2 0 8K
R_R4 3 4 6K R_R5 1 3 3K V_VS 4 0 DC 100 I_IS 0 1 DC 4 F_F1 1 3
VF_F1 2 VF_F1 5 0 0V E_E1 3 2 1 3 3 Chapter 3, Solution 82 + v0 4 3
k 2 k 4 k 8 k 6 k + 4A +100V 2i0 3v0 32 1 0 This network
corresponds to the Netlist. Chapter 3, Problem 83. PROPRIETARY
MATERIAL. 2007 The McGraw-Hill Companies, Inc. All rights reserved.
No part of this Manual may be displayed, reproduced or distributed
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individual course preparation. If you are a student using this
Manual, you are using it without permission. 207. The following
program is the Schematics Netlist of a particular circuit. Draw the
circuit and determine the voltage at node 2. R_R1 1 2 20 R_R2 2 0
50 R_R3 2 3 70 R_R4 3 0 30 V_VS 1 0 20V I_IS 2 0 DC 2A Chapter 3,
Solution 83 The circuit is shown below. 0 2 A 30 +20 V 321 70 50 20
When the circuit is saved and simulated, we obtain v2 = 12.5 volts
Chapter 3, Problem 84. Calculate vo and io in the circuit of Fig.
3.121. Figure 3.121 Chapter 3, Solution 84 From the output loop, v0
= 50i0x20x103 = 106 i0 (1) From the input loop, 3x10-3 + 4000i0
v0/100 = 0 (2) From (1) and (2) we get, i0 = 0.5A and v0 = 0.5
volt. Chapter 3, Problem 85. PROPRIETARY MATERIAL. 2007 The
McGraw-Hill Companies, Inc. All rights reserved. No part of this
Manual may be displayed, reproduced or distributed in any form or
by any means, without the prior written permission of the
publisher, or used beyond the limited distribution to teachers and
educators permitted by McGraw-Hill for their individual course
preparation. If you are a student using this Manual, you are using
it without permission. 208. An audio amplifier with resistance 9
supplies power to a speaker. In order that maximum power is
delivered, what should be the resistance of the speaker? Chapter 3,
Solution 85 The amplifier acts as a source. Rs + Vs RL - For
maximum power transfer, == 9sL RR Chapter 3, Problem 86. For the
simplified transistor circuit of Fig. 3.122, calculate the voltage
vo. Figure 3.122 Chapter 3, Solution 86 Let v1 be the potential
across the 2 k-ohm resistor with plus being on top. Then, [(0.03
v1)/1k] + 400i = v1/2k (1) Assume that i is in mA. But, i = (0.03
v1)/1 (2) Combining (1) and (2) yields, v1 = 29.963 mVolts and i =
37.4 nA, therefore, v0 = -5000x400x37.4x10-9 = -74.8 mvolts Chapter
3, Problem 87. PROPRIETARY MATERIAL. 2007 The McGraw-Hill
Companies, Inc. All rights reserved. No part For the circuit in
Fig. 3.123, find the gain vo/vs. of this Manual may be displayed,
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209. Figure 3.123 Chapter 3, Solution 87 v1 = 500(vs)/(500 + 2000)
= vs/5 v0 = -400(60v1)/(400 + 2000) = -40v1 = -40(vs/5) = -8vs,
Therefore, v0/vs = 8 Chapter 3, Problem 88. Determine the gain
vo/vs of the transistor amplifier circuit in Fig. 3.124. Figure
3.124 Chapter 3, Solution 88 Let v1 be the potential at the top end
of the 100-ohm resistor. (vs v1)/200 = v1/100 + (v1 10-3 v0)/2000
(1) For the right loop, v0 = -40i0(10,000) = -40(v1 10-3
)10,000/2000, or, v0 = -200v1 + 0.2v0 = -4x10-3 v0 (2) Substituting
(2) into (1) gives, (vs + 0.004v1)/2 = -0.004v0 + (-0.004v1
0.001v0)/20 This leads to 0.125v0 = 10vs or (v0/vs) = 10/0.125 =
-80 Chapter 3, Problem 89. PROPRIETARY MATERIAL. 2007 The
McGraw-Hill Companies, Inc. All rights reserved. No part of this
Manual may be displayed, reproduced or distributed in any form or
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publisher, or used beyond the limited distribution to teachers and
educators permitted by McGraw-Hill for their individual course
preparation. If you are a student using this Manual, you are using
it without permission. 210. For the transistor circuit shown in
Fig. 3.125, find IB and VCE. Let = 100 and VBE = 0.7V. 3 V + _ 1 k
+_ 0.7 V 100 k | | 15 V Figure 3.125 For Prob. 3.89. Chapter 3,
Solution 89 Consider the circuit below. + _ 1 k +_ 3 V 0.7 V 100 k
| | 15 VC E + _ VCE IC For the left loop, applying KVL gives = + +
= =0.73 3 0.7 100 10 0 30 ABEV B BE Bx I V I For the right loop, 3
15 (1 10 ) 0CE cV I x + = But = =100 30 A= 3 mAC BI I x 3 3 15 3 10
10 12 VCEV x x = = Chapter 3, Problem 90. PROPRIETARY MATERIAL.
2007 The McGraw-Hill Companies, Inc. All rights reserved. No part
of this Manual may be displayed, reproduced or distributed in any
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publisher, or used beyond the limited distribution to teachers and
educators permitted by McGraw-Hill for their individual course
preparation. If you are a student using this Manual, you are using
it without permission. 211. Calculate vs for the transistor in Fig.
3.126, given that vo = 4 V, = 150, VBE = 0.7V. Figure 3.126 Chapter
3, Solution 90 + V0 500 - + 18V - + vs i1 i2+ VCE + VBE IB IE 10 k
1 k For loop 1, -vs + 10k(IB) + VBE + IE (500) = 0 = -vs + 0.7 +
10,000IB + 500(1 + )IB which leads to vs + 0.7 = 10,000IB +
500(151)IB = 85,500IB But, v0 = 500IE = 500x151IB = 4 which leads
to IB = 5.298x10-5 Therefore, vs = 0.7 + 85,500IB = 5.23 volts
Chapter 3, Problem 91. PROPRIETARY MATERIAL. 2007 The McGraw-Hill
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212. For the transistor circuit of Fig. 3.127, find IB, VCE, and
vo. Take = 200, VBE = 0.7V. Figure 3.127 Chapter 3, Solution 91 We
first determine the Thevenin equivalent for the input circuit. RTh
= 6||2 = 6x2/8 = 1.5 k and VTh = 2(3)/(2+6) = 0.75 volts + V0 400 -
+ 9 V - + i1 i2+ VCE + VBE IB IE 1.5 k 5 k 0.75 V IC For loop 1,
-0.75 + 1.5kIB + VB BE + 400IE = 0 = -0.75 + 0.7 + 1500IBB + 400(1
+ )IBB IB = 0.05/81,900 =B 0.61 A v0 = 400IE = 400(1 + )IB =B 49 mV
For loop 2, -400IE VCE 5kIC + 9 = 0, but, IC = IB and IB E = (1 +
)IBB PROPRIETARY MATERIAL. 2007 The McGraw-Hill Companies, Inc. All
rights reserved. No part of this Manual may be displayed,
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prior written permission of the publisher, or used beyond the
limited distribution to teachers and educators permitted by
McGraw-Hill for their individual course preparation. If you are a
student using this Manual, you are using it without permission. VCE
= 9 5kIB 400(1 + )IB BB = 9 0.659 = 8.641 volts Chapter 3, Problem
92. 213. Find IB and VC for the circuit in Fig. 3.128. Let = 100,
VBE = 0.7V. Figure 3.128 Chapter 3, Solution 92 + V0 4 k - + 12V +
VCE + VBE IB IE 10 k 5 k VC IC I1 I1 = IB + IC = (1 + )IB and IE =
IB + IC = I1 Applying KVL around the outer loop, 4kIE + VBE + 10kIB
+ 5kI1 = 12 12 0.7 = 5k(1 + )IB + 10kIB + 4k(1 + )IB = 919kIB IB =
11.3/919k = 12.296 A Also, 12 = 5kI1 + VC which leads to VC = 12
5k(101)IB = 5.791 volts Chapter 3, Problem 93 Rework Example 3.11
with hand calculation. PROPRIETARY MATERIAL. 2007 The McGraw-Hill
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214. In the circuit in Fig. 3.34, determine the currents i1, i2,
and i3. Figure 3.34 Chapter 3, Solution 93 + v0 i3v1 v2i (b)(a) 2
3v0 + +24V + + v1 + v2 3v0 8 1 2 2 4 4 i i2 i1 From (b), -v1 + 2i
3v0 + v2 = 0 which leads to i = (v1 + 3v0 v2)/2 At node 1 in (a),
((24 v1)/4) = (v1/2) + ((v1 +3v0 v2)/2) + ((v1 v2)/1), where v0 =
v2 or 24 = 9v1 which leads to v1 = 2.667 volts At node 2, ((v1
v2)/1) + ((v1 + 3v0 v2)/2) = (v2/8) + v2/4, v0 = v2 v2 = 4v1 =
10.66 volts Now we can solve for the currents, i1 = v1/2 = 1.333 A,
i2 = 1.333 A, and i3 = 2.6667 A. PROPRIETARY MATERIAL. 2007 The
McGraw-Hill Companies, Inc. All rights reserved. No part of this
Manual may be displayed, reproduced or distributed in any form or
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publisher, or used beyond the limited distribution to teachers and
educators permitted by McGraw-Hill for their individual course
preparation. If you are a student using this Manual, you are using
it without permission. 215. Chapter 4, Problem 1. Calculate the
current io in the circuit of Fig. 4.69. What does this current
become when the input voltage is raised to 10 V? Figure 4.69
Chapter 4, Solution 1. + 5 1 41 1 i = + ==+ 4)35(8 , === 10 1 i 2 1
io PROPRIETARY MATERIAL. 2007 The McGraw-Hill Companies, Inc. All
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0.1A Since the resistance remains the same we get i = 10/5 = 2A
which leads to io = (1/2)i = (1/2)2 = 1A. 216. Chapter 4, Problem
2. Find v PROPRIETARY MATERIAL. 2007 The McGraw-Hill Companies,
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prior written permission of the publisher, or used beyond the
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student using this Manual, you are using it without permission. o
in the circuit of Fig. 4.70. If the source current is reduced to 1
A, what is v ?o Figure 4.70 Chapter 4, Solution 2. A 2 1 ii 21
==,3)24(6 =+ , 4 1 i 2 1 i 1o == == oo i2v 0.5V 0.5VIf i = 1A, then
vs o = 217. Chapter 4, Problem 3. (a) In the circuit in Fig. 4.71,
calculate v and I PROPRIETARY MATERIAL. 2007 The McGraw-Hill
Companies, Inc. All rights reserved. No part of this Manual may be
displayed, reproduced or distributed in any form or by any means,
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beyond the limited distribution to teachers and educators permitted
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a student using this Manual, you are using it without permission. o
o when v = 1 V.s (b) Find vo and i when vo s = 10 V. (c) What are v
and Io o when each of the 1- resistors is replaced by a 10-
resistor and vs = 10 V? Figure 4.71 Chapter 4, Solution 3. + + + vo
(a) We transform the Y sub-circuit to the equivalent . ,R 4 3 R4 R3
R3R 2 == R 2 3 R 4 3 R 4 3 =+ 2 v v s o = independent of R io = v
/(R)o 0.5V 0.5AWhen vs = 1V, vo = , io = 5V 5A(b) When v = 10V, v
=s o , io = (c) When v = 10V and R = 10,s vo = 5V 500mA, i =
10/(10) =o 218. Chapter 4, Problem 4. Use linearity to determine i
in the circuit in Fig. 4.72.o Figure 4.72 Chapter 4, Solution 4. If
Io = 1, the voltage across the 6 resistor is 6V so that the current
through the 3 resistor is 2A. + v1 .A3 4 v i o 1 === 263 , v = 3(4)
= 12V,o Hence Is = 3 + 3 = 6A PROPRIETARY MATERIAL. 2007 The
McGraw-Hill Companies, Inc. All rights reserved. No part of this
Manual may be displayed, reproduced or distributed in any form or
by any means, without the prior written permission of the
publisher, or used beyond the limited distribution to teachers and
educators permitted by McGraw-Hill for their individual course
preparation. If you are a student using this Manual, you are using
it without permission. If Is = 6A Io = 1 Is = 9A Io = 9/6 = 1.5A
219. Chapter 4, Problem 5. For the circuit in Fig. 4.73, assume vo
= 1 V, and use linearity to find the actual value of v .o Figure
4.73 Chapter 4, Solution 5. + V21 3 1 V1 =+ =If v = 1V,o 3 10 v 3 2
2V 1s =+ = 3 10 If v = v = 1s o =15x 10 3 PROPRIETARY MATERIAL.
2007 The McGraw-Hill Companies, Inc. All rights reserved. No part
of this Manual may be displayed, reproduced or distributed in any
form or by any means, without the prior written permission of the
publisher, or used beyond the limited distribution to teachers and
educators permitted by McGraw-Hill for their individual course
preparation. If you are a student using this Manual, you are using
it without permission. Then vs = 15 vo = 4.5V 220. Chapter 4,
Problem 6. For the linear circuit shown in Fig. 4.74, use linearity
to complete the following table. Experiment V Vs o 1 12 V 4 V 2 --
16 V 3 1 V -- 4 -- -2V Linear Circuit + _Vs Vo + Figure 4.74 For
Prob. 4.6. Chapter 4, Solution 6. Due to linearity, from the first
experiment, 1 3 o sV V= Applying this to other experiments, we
obtain: VExperiment Vs o 2 48 16 V 0.333 V PROPRIETARY MATERIAL.
2007 The McGraw-Hill Companies, Inc. All rights reserved. No part
of this Manual may be displayed, reproduced or distributed in any
form or by any means, without the prior written permission of the
publisher, or used beyond the limited distribution to teachers and
educators permitted by McGraw-Hill for their individual course
preparation. If you are a student using this Manual, you are using
it without permission. 3 1 V 4 -6 V -2V 221. Chapter 4, Problem 7.
PROPRIETARY MATERIAL. 2007 The McGraw-Hill Companies, Inc. All
rights reserved. No part of this Manual may be displayed,
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prior written permission of the publisher, or used beyond the
limited distribution to teachers and educators permitted by
McGraw-Hill for their individual course preparation. If you are a
student using this Manual, you are using it without permission. Use
linearity and the assumption that Vo = 1V to find the actual value
of V in Fig. 4.75.o . + _ 4 3 2 4 V + _ Vo 1 Figure 4.75 For Prob.
4.7. Chapter 4, Solution 7. If Vo = 1V, then the current through
the 2- and 4- resistors is = 0.5. The voltage across the 3-
resistor is (4 + 2) = 3 V. The total current through the 1-
resistor is 0.5 +3/3 = 1.5 A. Hence the source voltage = + =1 1.5 3
4.5 Vsv x If 4.5 1sv V= 1 4 4 0.8889 V 4.5 sv x= = = 888.9 mVThen .
222. Chapter 4, Problem 8. in the circuit of Fig. 4.76.Using
superposition, find Vo + _ 1 3 9 V 3 V Vo 4 5 + _ Figure 4.76 For
Prob. 4.8. PROPRIETARY MATERIAL. 2007 The McGraw-Hill Companies,
Inc. All rights reserved. No part of this Manual may be displayed,
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223. Chapter 4, Solution 8. Let V = V PROPRIETARY MATERIAL. 2007
The McGraw-Hill Companies, Inc. All rights reserved. No part of
this Manual may be displayed, reproduced or distributed in any form
or by any means, without the prior written permission of the
publisher, or used beyond the limited distribution to teachers and
educators permitted by McGraw-Hill for their individual course
preparation. If you are a student using this Manual, you are using
it without permission. o 1 + V , where V and V2 1 2 are due to 9-V
and 3-V sources respectively. To find V , consider the circuit
below.1 + _ 3 9 V V1 9 1 1 1 1 1 9 27/13 2.0769 3 9 1 V V V V = + =
= To find V , consider the circuit below.2 3 3 V9 V1 + _ 2 2 2 2 3
27/13 2.0769 9 3 1 V V V V + = = = = 4.1538 VVo = V1 + V2 224.
Chapter 4, Problem 9. Use superposition to find v in the circuit of
Fig. 4.77.o 1 4 18 V + _ vo 2 + _ 6 A 2 Figure 4.77 For Prob. 4.9.
PROPRIETARY MATERIAL. 2007 The McGraw-Hill Companies, Inc. All
rights reserved. No part of this Manual may be displayed,
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prior written permission of the publisher, or used beyond the
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225. Chapter 4, Solution 9. Let v = v + v PROPRIETARY MATERIAL.
2007 The McGraw-Hill Companies, Inc. All rights reserved. No part
of this Manual may be displayed, reproduced or distributed in any
form or by any means, without the prior written permission of the
publisher, or used beyond the limited distribution to teachers and
educators permitted by McGraw-Hill for their individual course
preparation. If you are a student using this Manual, you are using
it without permission. o 1 2, where v and v1 2 are due to 6-A and
20-V sources respectively. We find v using the circuit below.1 1 4
1 (6 ) 4 4 2 v x A= = + V2//2 = 1 , We find v using the circuit
below.2 1 4 + _ v1 2 6 A 2 1 4 18 V + _ v2 2 + _ 2 2 1 (18) 3 V 1 1
4 v = = + + = 4 + 3 = 7 V= v + vvo 1 2 226. Chapter 4, Problem 10.
For the circuit in Fig. 4.78, find the terminal voltage Vab using
superposition. Figure 4.78 Chapter 4, Solution 10. Let v = v
PROPRIETARY MATERIAL. 2007 The McGraw-Hill Companies, Inc. All
rights reserved. No part of this Manual may be displayed,
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prior written permission of the publisher, or used beyond the
limited distribution to teachers and educators permitted by
McGraw-Hill for their individual course preparation. If you are a
student using this Manual, you are using it without permission. ab
ab1 + v where v and vab2 ab1 ab2 are due to the 4-V and the 2-A
sources respectively. + + vab1 + + + vab2 For vab1, consider Fig.
(a). Applying KVL gives, - vab1 3 v + 10x0 + 4 = 0, which leads to
vab1 ab1 = 1 V For vab2, consider Fig. (b). Applying KVL gives, - v
3v + 10x2 = 0, which leads to v = 5ab2 ab2 ab2 = 1 + 5 = 6 Vvab
227. Chapter 4, Problem 11. Use the superposition principle to find
i PROPRIETARY MATERIAL. 2007 The McGraw-Hill Companies, Inc. All
rights reserved. No part of this Manual may be displayed,
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prior written permission of the publisher, or used beyond the
limited distribution to teachers and educators permitted by
McGraw-Hill for their individual course preparation. If you are a
student using this Manual, you are using it without permission. o
and v in the circuit of Fig. 4.79.o 40 30 V6 A + 4 io vo + 10 20 io
Figure 4.79 For Prob. 4.11. Chapter 4, Solution 11. Let v = v + vo
1 2, where v and v1 2 are due to the 6-A and 80-V sources
respectively. To find v , consider the circuit below.1 vb 40 6 A +
_ 4 i1 V1 10 20 I1 va At node a, = + = 6 240 5 4 40 10 a a b a v v
v v vb (1) At node b, I1 4I + (v 0)/20 = 0 or v1 b b = 100I1 228. 1
10 av v i = b But which leads to 100(v PROPRIETARY MATERIAL. 2007
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this Manual may be displayed, reproduced or distributed in any form
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publisher, or used beyond the limited distribution to teachers and
educators permitted by McGraw-Hill for their individual course
preparation. If you are a student using this Manual, you are using
it without permission. avb)10 = vb or v = 0.9091vb a (2)
Substituting (2) into (1), 5va 3.636va = 240 or va = 175.95 and v =
159.96b However, v = v1 a v = 15.99 V.b To find v , consider the
circuit below.2 10 20 io vc 40 30 V + _ 4 io v2 + 0 ( 30 4 0 50 20
c c o v v i + + = ) (0 ) 50 c o v i =But 5 (30 ) 0 1 50 20 c c c v
v v + = = 0 V 2 0 0 10 1 50 50 5 cv i + = = = 2 210 2 Vv i= =
=15.99 + 2 = 17.99 V /10= 1.799 Av = v + vo 1 2 and io = vo . 229.
Chapter 4, Problem 12. Determine v in the circuit in Fig. 4.80
using the superposition principle.o Figure 4.80 PROPRIETARY
MATERIAL. 2007 The McGraw-Hill Companies, Inc. All rights reserved.
No part of this Manual may be displayed, reproduced or distributed
in any form or by any means, without the prior written permission
of the publisher, or used beyond the limited distribution to
teachers and educators permitted by McGraw-Hill for their
individual course preparation. If you are a student using this
Manual, you are using it without permission. 230. Chapter 4,
Solution 12. Let v = v PROPRIETARY MATERIAL. 2007 The McGraw-Hill
Companies, Inc. All rights reserved. No part of this Manual may be
displayed, reproduced or distributed in any form or by any means,
without the prior written permission of the publisher, or used
beyond the limited distribution to teachers and educators permitted
by McGraw-Hill for their individual course preparation. If you are
a student using this Manual, you are using it without permission. o
o1 + vo2 + vo3, where vo1, vo2, and vo3 are due to the 2-A, 12-V,
and 19-V sources respectively. For vo1, consider the circuit below.
+ v 1 + v 1 6||3 = 2 ohms, 4||12 = 3 ohms. Hence, io = 2/2 = 1, vo1
= 5io = 5 V For vo2, consider the circuit below. + + v 2 + + + v 2
3||8 = 24/11, v = [(24/11)/(6 + 24/11)]12 = 16/51 = (5/8)(16/5) = 2
Vvo2 = (5/8)v1 For vo3, consider the circuit shown below. + v 3 + +
++ v 3 7||12 = (84/19) ohms, v = [(84/19)/(4 + 84/19)]19 = 9.9752 v
= (-5/7)v2 = -7.125 = 5 + 2 7.125 = -125 mVvo 231. Chapter 4,
Problem 13. Use superposition to find v in the circuit of Fig.
4.81.o 10 5 + _ vo 12 V 8 4 A + 2 A Figure 4.81 For Prob. 4.13.
Chapter 4, Solution 13. PROPRIETARY MATERIAL. 2007 The McGraw-Hill
Companies, Inc. All rights reserved. No part of this Manual may be
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beyond the limited distribution to teachers and educators permitted
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a student using this Manual, you are using it without permission.
3v v= + +Let v v , where v1 2o 1, v , and v2 3 are due to the
independent sources. To find v , consider the circuit below.1 10 2
A 5 v1 8 + _ 1 10 5 2 4.3478 10 8 5 v x x= = + + 232. To find v ,
consider the circuit below.2 PROPRIETARY MATERIAL. 2007 The
McGraw-Hill Companies, Inc. All rights reserved. No part of this
Manual may be displayed, reproduced or distributed in any form or
by any means, without the prior written permission of the
publisher, or used beyond the limited distribution to teachers and
educators permitted by McGraw-Hill for their individual course
preparation. If you are a student using this Manual, you are using
it without permission. 2 8 5 4 6.9565 8 10 5 v x x= = + + To find v
, consider the circuit below.3 10 4 A 5 v2 + _ 8 10 5 v3 8 + _ + 12
V = = + + 3 5 12 2.6087 5 10 8 v 1 2 3 8.6956 Vov v v v= + + =
=8.696V. 233. Chapter 4, Problem 14. Apply the superposition
principle to find v in the circuit of Fig. 4.82.o Figure 4.82
PROPRIETARY MATERIAL. 2007 The McGraw-Hill Companies, Inc. All
rights reserved. No part of this Manual may be displayed,
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prior written permission of the publisher, or used beyond the
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234. Chapter 4, Solution 14. Let v = v PROPRIETARY MATERIAL. 2007
The McGraw-Hill Companies, Inc. All rights reserved. No part of
this Manual may be displayed, reproduced or distributed in any form
or by any means, without the prior written permission of the
publisher, or used beyond the limited distribution to teachers and
educators permitted by McGraw-Hill for their individual course
preparation. If you are a student using this Manual, you are using
it without permission. o o1 + vo2 + vo3, where vo1, vo2 , and vo3,
are due to the 20-V, 1-A, and 2-A sources respectively. For vo1,
consider the circuit below. 6||(4 + 2) = 3 ohms, vo1 = ()20 = 10 V
For vo2, consider the circuit below. + + + + + 3||6 = 2 ohms, vo2 =
[2/(4 + 2 + 2)]4 = 1 V For vo3, consider the circuit below. 6||(4 +
2) = 3, vo3 = (-1)3 = 3 + v 3 + = 10 + 1 3 = 8 Vvo 235. Chapter 4,
Problem 15. For the circuit in Fig. 4.83, use superposition to find
i. Calculate the power delivered to the 3- resistor. Figure 4.83
Chapter 4, Solution 15. Let i = i + i PROPRIETARY MATERIAL. 2007
The McGraw-Hill Companies, Inc. All rights reserved. No part of
this Manual may be displayed, reproduced or distributed in any form
or by any means, without the prior written permission of the
publisher, or used beyond the limited distribution to teachers and
educators permitted by McGraw-Hill for their individual course
preparation. If you are a student using this Manual, you are using
it without permission. 1 2 + i3, where i , i1 2 , and i3 are due to
the 20-V, 2-A, and 16-V sources. For i1, consider the circuit
below. + 4||(3 + 1) = 2 ohms, Then i = [20/(2 + 2)] = 5 A, i = i /2
= 2.5 Ao 1 o 236. For i , consider the circuit below.3 PROPRIETARY
MATERIAL. 2007 The McGraw-Hill Companies, Inc. All rights reserved.
No part of this Manual may be displayed, reproduced or distributed
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of the publisher, or used beyond the limited distribution to
teachers and educators permitted by McGraw-Hill for their
individual course preparation. If you are a student using this
Manual, you are using it without permission. 2||(1 + 3) = 4/3, vo =
[(4/3)/((4/3) + 4)](-16) = -4 i3 = v /4 = -1o For i , consider the
circuit below. + + vo 2 2||4 = 4/3, 3 + 4/3 = 13/3 Using the
current division principle. i2 = [1/(1 + 13/2)]2 = 3/8 = 0.375 i =
2.5 + 0.375 - 1 = 1.875 A p = i2 R = (1.875)2 3 = 10.55 watts 237.
Chapter 4, Problem 16. Given the circuit in Fig. 4.84, use
superposition to get i .o Figure 4.84 PROPRIETARY MATERIAL. 2007
The McGraw-Hill Companies, Inc. All rights reserved. No part of
this Manual may be displayed, reproduced or distributed in any form
or by any means, without the prior written permission of the
publisher, or used beyond the limited distribution to teachers and
educators permitted by McGraw-Hill for their individual course
preparation. If you are a student using this Manual, you are using
it without permission. 238. Chapter 4, Solution 16. Let i = i
PROPRIETARY MATERIAL. 2007 The McGraw-Hill Companies, Inc. All
rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the
prior written permission of the publisher, or used beyond the
limited distribution to teachers and educators permitted by
McGraw-Hill for their individual course preparation. If you are a
student using this Manual, you are using it without permission. o
o1 + io2 + io3, where io1, io2, and io3 are due to the 12-V, 4-A,
and 2-A sources. For io1, consider the circuit below. 10||(3 + 2 +
5) = 5 ohms, io1 = 12/(5 + 4) = (12/9) A For io2, consider the
circuit below. + 2 + 5 + 4||10 = 7 + 40/14 = 69/7 i1 = [3/(3 +
69/7)]4 = 84/90, io2 =[-10/(4 + 10)]i1 = -6/9 For io3, consider the
circuit below. 3 + 2 + 4||10 = 5 + 20/7 = 55/7 i2 = [5/(5 + 55/7)]2
= 7/9, io3 = [-10/(10 + 4)]i = -5/92 io = (12/9) (6/9) (5/9) = 1/9
= 111.11 mA 239. Chapter 4, Problem 17. Use superposition to obtain
v in the circuit of Fig. 4.85. Check your result using PSpice.x
Figure 4.85 PROPRIETARY MATERIAL. 2007 The McGraw-Hill Companies,
Inc. All rights reserved. No part of this Manual may be displayed,
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prior written permission of the publisher, or used beyond the
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240. Chapter 4, Solution 17. Let v = v PROPRIETARY MATERIAL. 2007
The McGraw-Hill Companies, Inc. All rights reserved. No part of
this Manual may be displayed, reproduced or distributed in any form
or by any means, without the prior written permission of the
publisher, or used beyond the limited distribution to teachers and
educators permitted by McGraw-Hill for their individual course
preparation. If you are a student using this Manual, you are using
it without permission. x x1 + vx2 + vx3, where vx1,vx2, and vx3 are
due to the 90-V, 6-A, and 40-V sources. For vx1, consider the
circuit below. + + + 20||30 = 12 ohms, 60||30 = 20 ohms By using
current division, io = [20/(22 + 20)]3 = 60/42, vx1 = 10io = 600/42
= 14.286 V For vx2, consider the circuit below. + v 2 + io =
[12/(12 + 30)]6 = 72/42, vx2 = 10i = 17.143 Vo For vx3, consider
the circuit below. + + + io = [12/(12 + 30)]2 = 24/42, vx3 = -10i =
-5.714= [12/(12 + 30)]2 = 24/42, vo x3 = -10i = -5.714o = [12/(12 +
30)]2 = 24/42, vx3 = -10io = -5.714 = 14.286 17.143 5.714 = -8.571
Vvx 241. Chapter 4, Problem 18. Use superposition to find V in the
circuit of Fig. 4.86.o 2 A 4 10 V + _ Vo 2 1 0.5 Vo + _ Figure 4.86
For Prob. 4.18. PROPRIETARY MATERIAL. 2007 The McGraw-Hill
Companies, Inc. All rights reserved. No part of this Manual may be
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242. Chapter 4, Solution 18. Let V = V + V PROPRIETARY MATERIAL.
2007 The McGraw-Hill Companies, Inc. All rights reserved. No part
of this Manual may be displayed, reproduced or distributed in any
form or by any means, without the prior written permission of the
publisher, or used beyond the limited distribution to teachers and
educators permitted by McGraw-Hill for their individual course
preparation. If you are a student using this Manual, you are using
it without permission. o 1 2, where V1 and V2 are due to 10-V and
2-A sources respectively. To find V , we use the circuit below.1 10
V + _ V1 2 1 0.5 V1 + _ 10 V _ V1 2 0.5 V1 + _ - + 1 + 4 i -10 + 7i
0.5V = 01 But V = 4i1 ` 110 7 2 5 2, 8 Vi i i i V= = = = 243. To
find V , we use the circuit below.2 2 A + _ V2 2 0.5 V2 4 V _ V2 2
0.5 V2 + _ - + 1 + 4 i 1 4 - 4 + 7i 0.5V =02 = 4iBut V2 24 7 2 5
0.8, 4 3.2i i i i V i= = = = = V PROPRIETARY MATERIAL. 2007 The
McGraw-Hill Companies, Inc. All rights reserved. No part of this
Manual may be displayed, reproduced or distributed in any form or
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publisher, or used beyond the limited distribution to teachers and
educators permitted by McGraw-Hill for their individual course
preparation. If you are a student using this Manual, you are using
it without permission. o = V1 + V2 = 8 +3.2 =11.2 V 244. Chapter 4,
Problem 19. Use superposition to solve for vx in the circuit of
Fig. 4.87. Figure 4.87 PROPRIETARY MATERIAL. 2007 The McGraw-Hill
Companies, Inc. All rights reserved. No part of this Manual may be
displayed, reproduced or distributed in any form or by any means,
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245. Chapter 4, Solution 19. Let v = v + v PROPRIETARY MATERIAL.
2007 The McGraw-Hill Companies, Inc. All rights reserved. No part
of this Manual may be displayed, reproduced or distributed in any
form or by any means, without the prior written permission of the
publisher, or used beyond the limited distribution to teachers and
educators permitted by McGraw-Hill for their individual course
preparation. If you are a student using this Manual, you are using
it without permission. x 1 2, where v and v are due to the 4-A and
6-A sources respectively.1 2 + + + v1 + v2 , consider the circuit
in Fig. (a).To find v1 v /8 4 + (v (4i1 1 x))/2 = 0 or (0.125+0.5)v
= 4 2i1 x or v1 = 6.4 3.2ix But, ix = (v1 (4ix))/2 or i = 0.5vx 1.
Thus, v = 6.4 + 3.2(0.5v1 1), which leads to v1 = 6.4/0.6 = 10.667
To find v , consider the circuit shown in Fig. (b).2 v /8 6 + (v
(4i2 2 x))/2 = 0 or v + 3.2i2 x = 9.6 But ix = 0.5v . Therefore,2 v
+ 3.2(0.5v ) = 9.6 which leads to v2 2 2 = 16 = 10.667 16 =
26.67VHence, vx . Checking, ix = 0.5v = 13.333Ax Now all we need to
do now is sum the currents flowing out of the top node. 13.333 6 4
+ (26.67)/8 = 3.333 3.333 = 0 246. Chapter 4, Problem 20. Use
source transformations to reduce the circuit in Fig. 4.88 to a
single voltage source in series with a single resistor. PROPRIETARY
MATERIAL. 2007 The McGraw-Hill Companies, Inc. All rights reserved.
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teachers and educators permitted by McGraw-Hill for their
individual course preparation. If you are a student using this
Manual, you are using it without permission. 3 A 10 20 40 + _ + _12
V 16 V Figure 4.88 For Prob. 4.20. 247. Chapter 4, Solution 20.
Convert the voltage sources to current sources and obtain the
circuit shown below. 10 0.6 3 A 20 0.4 40 1 1 1 1 0.1 0.05 0.025
0.175 5.7143 10 20 40 eq eq R R = + + = + + = = PROPRIETARY
MATERIAL. 2007 The McGraw-Hill Companies, Inc. All rights reserved.
No part of this Manual may be displayed, reproduced or distributed
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of the publisher, or used beyond the limited distribution to
teachers and educators permitted by McGraw-Hill for their
individual course preparation. If you are a student using this
Manual, you are using it without permission. R = 5.714 eq Ieq = 3 +
0.6 + 0.4 = 4 Thus, the circuit is reduced as shown below. Please
note, we that this is merely an exercise in combining sources and
resistors. The circuit we have is an equivalent circuit which has
no real purpose other than to demonstrate source transformation. In
a practical situation, this would need some kind of reference and a
use to an external circuit to be of real value. 5.714 4 A + _
18.285 V 5.714 248. Chapter 4, Problem 21. Apply source
transformation to determine v and i PROPRIETARY MATERIAL. 2007 The
McGraw-Hill Companies, Inc. All rights reserved. No part of this
Manual may be displayed, reproduced or distributed in any form or
by any means, without the prior written permission of the
publisher, or used beyond the limited distribution to teachers and
educators permitted by McGraw-Hill for their individual course
preparation. If you are a student using this Manual, you are using
it without permission. o o in the circuit in Fig. 4.89. Figure 4.89
Chapter 4, Solution 21. To get i , transform the current sources as
shown in Fig. (a).o + + + vo From Fig. (a), -12 + 9io + 6 = 0,
therefore io = 666.7 mA , transform the voltage sources as shown in
Fig. (b).To get vo i = [6/(3 + 6)](2 + 2) = 8/3 = 3i = 8 Vvo 249.
Chapter 4, Problem 22. Referring to Fig. 4.90, use source
transformation to determine the current and power in the 8-
resistor. PROPRIETARY MATERIAL. 2007 The McGraw-Hill Companies,
Inc. All rights reserved. No part of this Manual may be displayed,
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student using this Manual, you are using it without permission.
Figure 4.90 Chapter 4, Solution 22. We transform the two sources to
get the circuit shown in Fig. (a). + + We now transform only the
voltage source to obtain the circuit in Fig. (b). 10||10 = 5 ohms,
i = [5/(5 + 4)](2 1) = 5/9 = 555.5 mA 250. Chapter 4, Problem 23.
Referring to Fig. 4.91, use source transformation to determine the
current and power in the 8- resistor. Figure 4.91 Chapter 4,
Solution 23 If we transform the voltage source, we obtain the
circuit below. 8 PROPRIETARY MATERIAL. 2007 The McGraw-Hill
Companies, Inc. All rights reserved. No part of this Manual may be
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10 6 3 5A 3A 3//6 = 2-ohm. Convert the current sources to voltages
sources as shown below. 10 8 2 + + 10V 30V - - Applying KVL to the
loop gives A10)2810(1030 ==++++ II W82 === RIVIp 251. Chapter 4,
Problem 24. Use source transformation to find the voltage V in the
circuit of Fig. 4.92.x + _ 8 10 10 3 A 40 V + Vx 2 Vx Figure 4.92
For Prob. 4.24. PROPRIETARY MATERIAL. 2007 The McGraw-Hill
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252. Chapter 4, Solution 24. Transform the two current sources in
parallel with the resistors into their voltage source equivalents
yield, a 30-V source in series with a 10- resistor and a 20Vx-V
sources in series with a 10- resistor. We now have the following
circuit, I 8 10 10 + _40 V + Vx + 20Vx + 30 V We now write the
following mesh equation and constraint equation which will lead to
a solution for V ,x 28I 70 + 20V = 0 or 28I + 20V PROPRIETARY
MATERIAL. 2007 The McGraw-Hill Companies, Inc. All rights reserved.
No part of this Manual may be displayed, reproduced or distributed
in any form or by any means, without the prior written permission
of the publisher, or used beyond the limited distribution to
teachers and educators permitted by McGraw-Hill for their
individual course preparation. If you are a student using this
Manual, you are using it without permission. x x = 70, but V = 8I
which leads tox = 2.978 V28I + 160I = 70 or I = 0.3723 A or Vx .
253. Chapter 4, Problem 25. Obtain vo in the circuit of Fig. 4.93
using source transformation. Check your result using PSpice. Figure
4.93 Chapter 4, Solution 25. Transforming only the current source
gives the circuit below. + + + + + Applying KVL to the loop gives,
(4 + 9 + 5 + 2)i + 12 18 30 30 = 0 20i = 66 which leads to i = 3.3
= 2i = 6.6 V PROPRIETARY MATERIAL. 2007 The McGraw-Hill Companies,
Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the
prior written permission of the publisher, or used beyond the
limited distribution to teachers and educators permitted by
McGraw-Hill for their individual course preparation. If you are a
student using this Manual, you are using it without permission. vo
254. Chapter 4, Problem 26. Use source transformation to find i in
the circuit of Fig. 4.94.o + _ 2 5 4 3 A 6 A io 20 V Figure 4.94
For Prob. 4.26. Chapter 4, Solution 26. Transforming the current
sources gives the circuit below. + _ io 20 V + 2 15 V 5 4 + _12 V =
636.4 mA15 +20 = 0 or 11i = 7 or i PROPRIETARY MATERIAL. 2007 The
McGraw-Hill Companies, Inc. All rights reserved. No part of this
Manual may be displayed, reproduced or distributed in any form or
by any means, without the prior written permission of the
publisher, or used beyond the limited distribution to teachers and
educators permitted by McGraw-Hill for their individual course
preparation. If you are a student using this Manual, you are using
it without permission. 12 + 11io o o . 255. Chapter 4, Problem 27.
Apply source transformation to find vx in the circuit of Fig. 4.95.
Figure 4.95 Chapter 4, Solution 27. Transforming the voltage
sources to current sources gives the circuit in Fig. (a). 10||40 =
8 ohms Transforming the current sources to voltage sources yields
the circuit in Fig. (b). Applying KVL to the loop, -40 + (8 + 12 +
20)i + 200 = 0 leads to i = -4 12i = -48 V PROPRIETARY MATERIAL.
2007 The McGraw-Hill Companies, Inc. All rights reserved. No part
of this Manual may be displayed, reproduced or distributed in any
form or by any means, without the prior written permission of the
publisher, or used beyond the limited distribution to teachers and
educators permitted by McGraw-Hill for their individual course
preparation. If you are a student using this Manual, you are using
it without permission. vx + v + i + + v 256. Chapter 4, Problem 28.
Use source transformation to find I in Fig. 4.96.o 1 + _ 3 8 V +
_Vo 4 Vo Io Figure 4.96 For Prob. 4.28. Chapter 4, Solution 28.
Convert the dependent current source to a dependent voltage source
as shown below. 1 + _ 4 8 V Vo io 3 + _Vo+ PROPRIETARY MATERIAL.
2007 The McGraw-Hill Companies, Inc. All rights reserved. No part
of this Manual may be displayed, reproduced or distributed in any
form or by any means, without the prior written permission of the
publisher, or used beyond the limited distribution to teachers and
educators permitted by McGraw-Hill for their individual course
preparation. If you are a student using this Manual, you are using
it without permission. 0 = Applying KVL, 8 (1 4 3)o oi V + + + =
But V i4o o 8 8 4 0 2 Ao o oi i i + = = 257. Chapter 4, Problem 29.
Use source transformation to find v in the circuit of Fig. 4.93.o +
+ vo Figure 4.93 Chapter 4, Solution 29. Transform the dependent
voltage source to a current source as shown in Fig. (a). 2||4 =
(4/3) k ohms PROPRIETARY MATERIAL. 2007 The McGraw-Hill Companies,
Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the
prior written permission of the publisher, or used beyond the
limited distribution to teachers and educators permitted by
McGraw-Hill for their individual course preparation. If you are a
student using this Manual, you are using it without permission. It
is clear that i = 3 mA which leads to vo = 1000i = 3 V If the use
of source transformations was not required for this problem, the
actual answer could have been determined by inspection right away
since the only current that could have flowed through the 1 k ohm
resistor is 3 mA. + vo + + vo 258. Chapter 4, Problem 30. Use
source transformation on the circuit shown in Fig 4.98 to find ix.
Figure 4.98 PROPRIETARY MATERIAL. 2007 The McGraw-Hill Companies,
Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the
prior written permission of the publisher, or used beyond the
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student using this Manual, you are using it without permission.
259. Chapter 4, Solution 30 Transform the dependent current source
as shown below. i PROPRIETARY MATERIAL. 2007 The McGraw-Hill
Companies, Inc. All rights reserved. No part of this Manual may be
displayed, reproduced or distributed in any form or by any means,
without the prior written permission of the publisher, or used
beyond the limited distribution to teachers and educators permitted
by McGraw-Hill for their individual course preparation. If you are
a student using this Manual, you are using it without permission. x
24 60 10 + + 12V 30 7ix - - Combine the 60-ohm with the 10-ohm and
transform the dependent source as shown below. ix 24 + 12V 30 70
0.1ix - Combining 30-ohm and 70-ohm gives 30//70 = 70x30/100 =
21-ohm. Transform the dependent current source as shown below. ix
24 21 + + 12V 2.1ix - - Applying KVL to the loop gives mA8.254 1.47
12 01.21245 ===+ xxx iii 260. Chapter 4, Problem 31. Determine v in
the circuit of Fig. 4.99 using source transformation.x Figure 4.99
Chapter 4, Solution 31. Transform the dependent source so that we
have the circuit in Fig. (a). 6||8 = (24/7) ohms. Transform the
dependent source again to get the circuit in Fig. (b). + + + + +
From Fig. (b), = 3i, or i = v /3.vx x Applying KVL, -12 + (3 +
24/7)i + (24/21)vx = 0 = 84/23 = 3.652 V PROPRIETARY MATERIAL. 2007
The McGraw-Hill Companies, Inc. All rights reserved. No part of
this Manual may be displayed, reproduced or distributed in any form
or by any means, without the prior written permission of the
publisher, or used beyond the limited distribution to teachers and
educators permitted by McGraw-Hill for their individual course
preparation. If you are a student using this Manual, you are using
it without permission. 12 = [(21 + 24)/7]vx/3 + (8/7)vx, leads to
vx 261. Chapter 4, Problem 32. Use source transformation to find i
in the circuit of Fig. 4.100.x Figure 4.100 Chapter 4, Solution 32.
As shown in Fig. (a), we transform the dependent current source to
a voltage source, + + + + In Fig. (b), 50||50 = 25 ohms. Applying
KVL in Fig. (c), = 1.6 A PROPRIETARY MATERIAL. 2007 The McGraw-Hill
Companies, Inc. All rights reserved. No part of this Manual may be
displayed, reproduced or distributed in any form or by any means,
without the prior written permission of the publisher, or used
beyond the limited distribution to teachers and educators permitted
by McGraw-Hill for their individual course preparation. If you are
a student using this Manual, you are using it without permission.
-60 + 40ix 2.5ix = 0, or ix 262. Chapter 4, Problem 33. Determine R
and V at terminals 1-2 of each of the circuits of Fig. 4.101.Th Th
Figure 4.101 Chapter 4, Solution 33. = 10||40 = 400/50 = 8 ohms
PROPRIETARY MATERIAL. 2007 The McGraw-Hill Companies, Inc. All
rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the
prior written permission of the publisher, or used beyond the
limited distribution to teachers and educators permitted by
McGraw-Hill for their individual course preparation. If you are a
student using this Manual, you are using it without permission. (a)
RTh = (40/(40 + 10))20 = 16 VVTh = 30||60 = 1800/90 = 20 ohms(b)
RTh 2 + (30 v1)/60 = v /30, and v1 1 = VTh 120 + 30 v = 2v , or v =
50 V1 1 1 = 50 VVTh 263. Chapter 4, Problem 34. Find the Thevenin
equivalent at terminals a-b of the circuit in Fig. 4.102. Figure
4.102 Chapter 4, Solution 34. To find R , consider the circuit in
Fig. (a).Th + + = 20 + 10||40 = 20 + 400/50 = 28 ohms PROPRIETARY
MATERIAL. 2007 The McGraw-Hill Companies, Inc. All rights reserved.
No part of this Manual may be displayed, reproduced or distributed
in any form or by any means, without the prior written permission
of the publisher, or used beyond the limited distribution to
teachers and educators permitted by McGraw-Hill for their
individual course preparation. If you are a student using this
Manual, you are using it without permission. RTh To find V ,
consider the circuit in Fig. (b).Th At node 1, (40 v1)/10 = 3 +
[(v1 v )/20] + v /40, 40 = 7v 2v2 1 1 2 (1) At node 2, 3 + (v1-
v2)/20 = 0, or v1 = v2 60 (2) = 92 VSolving (1) and (2), v1 = 32 V,
v2 = 92 V, and VTh = v2 264. Chapter 4, Problem 35. Use Thevenins
theorem to find v in Prob. 4.12.o Chapter 4, Problem 12. Determine
v in the circuit in Fig. 4.80 using the superposition principle.o
Figure 4.80 PROPRIETARY MATERIAL. 2007 The McGraw-Hill Companies,
Inc. All rights reserved. No part of this Manual may be displayed,
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prior written permission of the publisher, or used beyond the
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student using this Manual, you are using it without permission.
265. Chapter 4, Solution 35. To find R , consider the circuit in
Fig. (a).Th R = R = 6||3 + 12||4 = 2 + 3 =5 ohmsTh ab To find V ,
consider the circuit shown in Fig. (b).Th PROPRIETARY MATERIAL.
2007 The McGraw-Hill Companies, Inc. All rights reserved. No part
of this Manual may be displayed, reproduced or distributed in any
form or by any means, without the prior written permission of the
publisher, or used beyond the limited distribution to teachers and
educators permitted by McGraw-Hill for their individual course
preparation. If you are a student using this Manual, you are using
it without permission. At node 1, 2 + (12 v )/6 = v /3, or v1 1 1 =
8 At node 2, (19 v2)/4 = 2 + v2/12, or v2 = 33/4 But, -v1 + V + v =
0, or V = v v = 8 33/4 = -0.25Th 2 Th 1 2 + + + v1 + v2 + + + /2 =
-0.25/2 = 125 mVvo = VTh 266. Chapter 4, Problem 36. Solve for the
current i in the circuit of Fig. 4.103 using Thevenins theorem.
(Hint: Find the Thevenin equivalent as seen by the 12- resistor.)
Figure 4.103 Chapter 4, Solution 36. Remove the 30-V voltage source
and the 20-ohm resistor. + + From Fig. (a), R = 10||40 = 8 ohmsTh
From Fig. (b), V = (40/(10 + 40))50 = 40VTh + + The equivalent
circuit of the original circuit is shown in Fig. (c). Applying KVL,
30 40 + (8 + 12)i = 0, which leads to i = 500mA PROPRIETARY
MATERIAL. 2007 The McGraw-Hill Companies, Inc. All rights reserved.
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of the publisher, or used beyond the limited distribution to
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individual course preparation. If you are a student using this
Manual, you are using it without permission. 267. Chapter 4,
Problem 37. Find the Norton equivalent with respect to terminals
a-b in the circuit shown in Fig. 4.100. Figure 4.100 PROPRIETARY
MATERIAL. 2007 The McGraw-Hill Companies, Inc. All rights reserved.
No part of this Manual may be displayed, reproduced or distributed
in any form or by any means, without the prior written permission
of the publisher, or used beyond the limited distribution to
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individual course preparation. If you are a student using this
Manual, you are using it without permission. 268. Chapter 4,
Solution 37 RN is found from the circuit below. 20 a PROPRIETARY
MATERIAL. 2007 The McGraw-Hill Companies, Inc. All rights reserved.
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Manual, you are using it without permission. 40 12 b =+=
10)4020//(12NR IN is found from the circuit below. 2A 20 a + 40
120V 12 - IN b Applying source transformation to the current source
yields the circuit below. 20 40 + 80 V - + 120V IN - Applying KVL
to the loop yields 666.7 mA===++ 60/40I0I608012