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 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. 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 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. 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 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. 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. 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. 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 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. 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 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. 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 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. 166. Chapter 3, Problem 56. Determine v1 and v2 in the circuit of Fig. 3.101. Figure 3.101 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. 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, 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. 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. 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. 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 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. 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 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. 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. 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. 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 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. 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, 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. 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 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. 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 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. 176. Chapter 3, Problem 64. Find vo, and io in the circuit of Fig. 3.108. Figure 3.108 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. 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 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. 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 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. 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 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. 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 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 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 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. 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 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 3.111 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 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. 184. Chapter 3, Problem 68. Find the voltage Vo in the circuit of Fig. 3.112. 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. 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 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. 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 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. 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 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. 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 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. 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. 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. 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, 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. 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, 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. 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. 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. 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 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. 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. 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. 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. 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. 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 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. 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 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 3.69 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. 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. 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, 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. 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. 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. 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 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. 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 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. 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 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. 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, 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. 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 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. 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 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. 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 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. 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, 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. 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 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. 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 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. 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 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. 0.1A 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, 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 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, 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 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, 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. 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, 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. 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, 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. 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, 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. 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, 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 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 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. 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 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. 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, 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. 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 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||(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, 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. 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 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. 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 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 = 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, 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. 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. 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 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 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. 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, 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 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 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. 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 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. 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 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. 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, 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. 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. 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. 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 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. 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. 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. 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