Chapter 1, Problem 1 How many coulombs are represented by these amounts of electrons: (a)(b) 1710 482 . 6 1810 24 . 1 (c) (d) 1910 46 . 2 2010 628 . 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)( ) ( ) mC 8 3 + = t t q(b)( ) C 2) 4 82t- t ( t q + =(c)( ) ( )nC e 5 e 3 t qt 2 -t =(d)( ) pC tsin 10 120 t q =(e)( ) C t50 cos 204te t q= 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 tpA (e)i =dq/dt = +e t tt 480 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators permitted by McGraw-Hill for their individual course preparation.If you are a student using this Manual, you are using it without permission. Chapter 1, Problem 3. Find the charge q(t) flowing through a device if the current is: (a)( ) ( ) C 1 0 A, 3 = = q t i(b)0 ) 0 ( , mA ) 5 2 ( ) ( = + = q t t i(c)C 2 (0) A, ) 6 / 10 cos( 20 ) ( = + = q t t i(d)( ) 0 (0) A, 40 sin 1030= =q t e t it Chapter 1, Solution 3 (a)C 1) (3t + = + =q(0) i(t)dt q(t)(b)mC 5t) (t2+ = + + = q(v) dt s) (2t q(t)(c)( ) q(t) 20 cos10t / 6 q(0) (2sin(10 / 6) 1) C t = + + = + + (d)C 40t) sin 0.12 t (0.16cos40 e30t -+ =+= + = t) cos 40 - t 40 sin 30 (1600 900e 10q(0) t 40 sin 10e q(t)-30t30t - 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 when1( )2i t t =A. Chapter 1, Solution 5 10 2010125 C0 2 4tq 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators permitted by McGraw-Hill for their individual course preparation.If you are a student using this Manual, you are using it without permission. 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,= = =280dtdqi40 A (b) At t = 6ms,= =dtdqi0 A (c) At t = 10ms,= = =480dtdqi20 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators permitted by McGraw-Hill for their individual course preparation.If you are a student using this Manual, you are using it without permission. 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 < 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators permitted by McGraw-Hill for their individual course preparation.If you are a student using this Manual, you are using it without permission. 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators 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 3.110For Prob. 3.66. Chapter 3, Solution 66 The mesh equations are obtained as follows. + + =1 2 3 412 24 30 4 6 2 0 I I I Ior 30I1 4I2 6I3 2I4 = 12(1) + + =1 2 4 524 40 4 30 2 6 0 I I I Ior 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 Putting (1) to (5) in matrix form = 320301612I18 4 0 6 04 12 4 2 20 4 18 0 66 2 0 30 40 2 6 4 30 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 -430 0-2-6 -6 018-4 0 -2-2-412-4 0-6 0-418 >> 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators permitted by McGraw-Hill for their individual course preparation.If you are a student using this Manual, you are using it without permission. 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators permitted by McGraw-Hill for their individual course preparation.If you are a student using this Manual, you are using it without permission. Figure 3.111For Prob. 3.67. 10 5 4 A + _ 3 Vo Vo 2 4 2 A Chapter 3, Solution 67 Consider the circuit below. V3 + Vo- + = 60V 3 2V5 . 0 5 . 0 05 . 0 95 . 0 25 . 00 25 . 0 35 . 0o V2 10 54 A 3 Vo V1 2 4 2 A 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, = 602V5 . 0 5 . 0 05 . 0 95 . 0 25 . 03 25 . 3 35 . 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 -3.25003.0000 -0.25000.9500 -0.5000 0 -0.50000.5000 >> 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators permitted by McGraw-Hill for their individual course preparation.If you are a student using this Manual, you are using it without permission. Chapter 3, Problem 68. Find the voltage Vo in the circuit ofFig. 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators 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 3.112For 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 + _ =+ + +=04 . 2725 / 24 33 4V19 . 0 1 . 01 . 0 125 . 0 Using MATLAB, we get, >> Y=[0.125,-0.1;-0.1,0.19] Y = 0.1250 -0.1000 -0.10000.1900 >> 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators permitted by McGraw-Hill for their individual course preparation.If you are a student using this Manual, you are using it without permission. 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: = 5520vvv25 . 1 25 . 0 125 . 0 1 25 . 01 25 . 0 75 . 1321 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators permitted by McGraw-Hill for their individual course preparation.If you are a student using this Manual, you are using it without permission. 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators permitted by McGraw-Hill for their individual course preparation.If you are a student using this Manual, you are using it without permission. Figure 3.114For Prob. 3.70. 1 S 2 S 2 A 4 A V1 4ix ix 5 S V2 Chapter 3, Solution 70 +=2 I 44 I 4V5 00 3xx With two equations and three unknowns, we need a constraint equation, Ix = 2V1, thus the matrix equation becomes, =24V5 80 5 This results in V1 = 4/(5) = 0.8VandV2 = [8(0.8) 2]/5 = [6.4 2]/5= 0.88 V. Chapter 3, Problem 71. Write the mesh-current equations for the circuit in Fig. 3.115.Next, determine the values of I1, I2, andI3. + _ 10 V + _ 5 V 1 3 4 I3 I1 I2 2 5 Figure 3.115For 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators permitted by McGraw-Hill for their individual course preparation.If you are a student using this Manual, you are using it without permission. Chapter 3, Solution 71

= 0510I9 1 51 7 45 4 9 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators permitted by McGraw-Hill for their individual course preparation.If you are a student using this Manual, you are using it without permission. 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 forall i not equal to j. v1 = 8,v2 = 4,v3 = -10,andv4 = -4 Hence the mesh-current equations are: = 41048iiii5 1 0 01 5 4 00 4 6 20 0 2 74321 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators permitted by McGraw-Hill for their individual course preparation.If you are a student using this Manual, you are using it without permission. 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,andv4 = -3 Hence, = 3246iiii2 1 0 01 6 0 40 0 8 30 4 3 94321 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators permitted by McGraw-Hill for their individual course preparation.If you are a student using this Manual, you are using it without permission. 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 = 4321VVVV =+ + + + + + + +432143218 5 3 8 58 8 7 6 65 5 4 2 46 4 6 4 1VVVV iiiiR R R R R 0R R R R 0 RR 0 R R R R0 R R R 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators permitted by McGraw-Hill for their individual course preparation.If you are a student using this Manual, you are using it without permission. 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators permitted by McGraw-Hill for their individual course preparation.If you are a student using this Manual, you are using it without permission. Chapter 3, Solution 75 * Schematics Netlist * R_R4 $N_0002 $N_000130 R_R2 $N_0001 $N_000310 R_R1 $N_0005 $N_000430 R_R3 $N_0003 $N_000410 R_R5 $N_0006 $N_000430 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 i1i2i3 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators permitted by McGraw-Hill for their individual course preparation.If you are a student using this Manual, you are using it without permission. 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators permitted by McGraw-Hill for their individual course preparation.If you are a student using this Manual, you are using it without permission. * Schematics Netlist * I_I2 0 $N_0001 DC 4A R_R1 $N_0002 $N_00010.25 R_R3 $N_0003 $N_00011 R_R2 $N_0002 $N_00031 F_F1 $N_0002 $N_0001 VF_F1 3 VF_F1$N_0003 $N_0004 0V R_R4 0 $N_00020.5 R_R6 0 $N_00010.5 I_I1 0 $N_0002 DC 2A R_R5 0 $N_00040.25 Clearly,v1 = 625 mVolts,v2 = 375 mVolts,andv3 = 1.625 volts,which agrees with the solution obtained in Problem 3.27. Chapter 3, Problem 77. Solve for V1 andV2 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators permitted by McGraw-Hill for their individual course preparation.If you are a student using this Manual, you are using it without permission. 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators permitted by McGraw-Hill for their individual course preparation.If you are a student using this Manual, you are using it without permission. 2 A 5 A ix V1 V2 2 ix 2 5 1 Figure 3.119For Prob. 3.77. Chapter 3, Solution 77 As a check we can write the nodal equations, =25V2 . 1 2 . 12 . 0 7 . 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators permitted by McGraw-Hill for their individual course preparation.If you are a student using this Manual, you are using it without permission. 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators 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 The schematic is shown below.When the circuit is saved and simulated the node voltages are displaced on the pseudocomponents as shown.Thus, , V 15 V, 5 . 4 V, 33 2 1 = = = V V V . 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators permitted by McGraw-Hill for their individual course preparation.If you are a student using this Manual, you are using it without permission. Chapter 3, Problem 28 Use MATLAB to find the voltages at nodes a, b, c, and d in the circuit ofFig. 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, V 88 . 26 V V, 6944 . 0 VV, 28 . 10 VV, 278 . 5 Vd c b a = = = = 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators permitted by McGraw-Hill for their individual course preparation.If you are a student using this Manual, you are using it without permission. 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators permitted by McGraw-Hill for their individual course preparation.If you are a student using this Manual, you are using it without permission. * Schematics Netlist * H_H1 $N_0002 $N_0003 VH_H1 6 VH_H10 $N_0001 0V I_I1 $N_0004 $N_0005 DC 8A V_V1 $N_0002 0 20V R_R4 0 $N_00034 R_R1 $N_0005 $N_000310 R_R2 $N_0003 $N_000212 R_R5 0 $N_00041 R_R3 $N_0004 $N_00012 Clearly,v1 = 84 volts,v2 = 4 volts,v3 = 20 volts,andv4 = -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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators permitted by McGraw-Hill for their individual course preparation.If you are a student using this Manual, you are using it without permission. 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators permitted by McGraw-Hill for their individual course preparation.If you are a student using this Manual, you are using it without permission. Clearly, v1 = 26.67 volts,v2 = 6.667 volts,v3 = 173.33 volts,andv4 = -46.67 voltswhich agrees with the results of Example 3.4. This is the netlist for this circuit. * Schematics Netlist * R_R1 0 $N_00012 R_R2 $N_0003 $N_00026 R_R3 0 $N_00024 R_R4 0 $N_00041 R_R5 $N_0001 $N_00043 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators permitted by McGraw-Hill for their individual course preparation.If you are a student using this Manual, you are using it without permission. R_R1122K R_R2204K R_R3208K R_R4346K R_R5133K V_VS40DC100I_IS01DC4 F_F113VF_F12 VF_F1500V E_E132133 Chapter 3, Solution 82 + v0 4 3 k 2 k 4 k8 k 6 k +4A + 100V 2i03v0 3 2 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators permitted by McGraw-Hill for their individual course preparation.If you are a student using this Manual, you are using it without permission. The following program is the Schematics Netlist of a particular circuit. Draw the circuit and determine the voltage at node 2. R_R11220 R_R22050 R_R32370 R_R43030 V_VS1020V I_IS20DC2A Chapter 3, Solution 83 The circuit is shown below. 02 A30 + 20 V 3 2 1 70 50 20 When the circuit is saved and simulated, we obtainv2 = 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 = 106i0(1) From the input loop,3x10-3 + 4000i0 v0/100 = 0(2) From (1) and (2) we get,i0 = 0.5Aandv0 = 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators permitted by McGraw-Hill for their individual course preparation.If you are a student using this Manual, you are using it without permission. 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 + VsRL - For maximum power transfer, = = 9s LR R 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 mVoltsandi = 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators permitted by McGraw-Hill for their individual course preparation.If you are a student using this Manual, you are using it without permission. Figure 3.123 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 10-3v0)/2000(1) For the right loop, v0 = -40i0(10,000) = -40(v1 10-3)10,000/2000, or,v0 = -200v1 + 0.2v0 = -4x10-3v0(2) Substituting (2) into (1) gives, (vs + 0.004v1)/2 = -0.004v0 + (-0.004v1 0.001v0)/20 This leads to 0.125v0 = 10vsor(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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators permitted by McGraw-Hill for their individual course preparation.If you are a student using this Manual, you are using it without permission. 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 3.125For Prob. 3.89. Chapter 3, Solution 89 Consider the circuit below. + _ 1 k + _ 3 V 0.7 V 100 k || 15 V C E + _ VCE IC For the left loop,applying KVL gives = + + = =0.7 33 0.7 100 10 0 30ABEVB BE Bx I V IFor the right loop, 315 (1 10 ) 0C E cV I x + =But = =10030A= 3 mAC BI I x 3 315 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators permitted by McGraw-Hill for their individual course preparation.If you are a student using this Manual, you are using it without permission. 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 +V0500 -+ 18V-+vsi1 i2+VCE+VBEIB 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 tovs + 0.7 = 10,000IB + 500(151)IB = 85,500IB But,v0 = 500IE = 500x151IB = 4which leads toIB = 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators permitted by McGraw-Hill for their individual course preparation.If you are a student using this Manual, you are using it without permission. 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 +V0400 -+ 9 V -+ i1 i2+VCE+VBEIBIE 1.5 k 5 k 0.75 V IC For loop 1,-0.75 + 1.5kIB + V BBE + 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 = IBandI BE = (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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators 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 + )I BBB = 9 0.659 = 8.641 voltsChapter 3, Problem 92. Find IB and VC for the circuit in Fig. 3.128. Let = 100, VBE = 0.7V. Figure 3.128 Chapter 3, Solution 92 +V04 k -+ 12V+VCE+VBEIB IE 10 k 5 k VCIC I1 I1 = IB + IC = (1 + )IBandIE = 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 + VCwhich leads toVC = 12 5k(101)IB = 5.791 voltsChapter 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators permitted by McGraw-Hill for their individual course preparation.If you are a student using this Manual, you are using it without permission. In the circuit in Fig. 3.34, determine the currents i1, i2, and i3. Figure 3.34 Chapter 3, Solution 93 +v0i3v1 v2i (b) (a)2 3v0 ++ 24V ++v1+v23v0 8 1 2 2 4 4 i i2i1 From (b),-v1 + 2i 3v0 + v2 = 0which leadstoi = (v1 + 3v0 v2)/2 At node 1 in (a),((24 v1)/4) = (v1/2) + ((v1 +3v0 v2)/2) + ((v1 v2)/1), where v0 = v2 or24 = 9v1which leads tov1 = 2.667 volts At node 2,((v1 v2)/1) + ((v1 + 3v0 v2)/2) = (v2/8) + v2/4,v0 = v2 v2 = 4v1 = 10.66 voltsNow 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators permitted by McGraw-Hill for their individual course preparation.If you are a student using this Manual, you are using it without permission. 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. + 514 11i =+= = + 4 ) 3 5 ( 8 ,= = =101i21ioPROPRIETARY 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators 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. Chapter 4, Problem 2. Find vPROPRIETARY 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators 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. A21i i2 1= = , 3 ) 2 4 ( 6 = + ,41i21i1 o= == =o oi 2 v0.5V 0.5V If i= 1A, then vs o = Chapter 4, Problem 3. (a)In the circuit in Fig. 4.71, calculate vand IPROPRIETARY 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators 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 iwhen vo s = 10 V. (c)What are vand 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 . , R43R 4R 3R 3 R2= = R23R43R43= +2vvso =independent of R io = v /(R) o 0.5V 0.5A When vs = 1V, vo =, io = 5V 5A (b)When v= 10V, v= s o, io = (c)When v= 10V and R = 10,svo =5V 500mA , i=10/(10) =oChapter 4, Problem 4. Use linearity to determine iin 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. A 34vio1= = = 2 6 3 , v= 3(4) = 12V, oHenceIs = 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators 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 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. + V 2 131V1= + = If v= 1V,o310v322 V1 s= += 310If v= v= 1 s o= 15 x103PROPRIETARY 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators 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 Chapter 4, Problem 6. For the linear circuit shown in Fig. 4.74, use linearity to complete the following table. ExperimentV Vso 112 V4 V 2--16 V 31 V-- 4---2V Linear Circuit + _ Vs Vo + Figure 4.74For Prob. 4.6. Chapter 4, Solution 6. Due to linearity, from the first experiment, 13o sV V = Applying this to other experiments, we obtain: VExperimentVso 24816 V 0.333 VPROPRIETARY 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators permitted by McGraw-Hill for their individual course preparation.If you are a student using this Manual, you are using it without permission. 31 V 4-6 V-2V 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators 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 Vin Fig. 4.75.o. + _ 4 3 2 4 V + _ Vo 1 Figure 4.75For 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 If4.5 1sv V = 14 4 0.8889V4.5sv x = == 888.9 mV Then. 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 4.76For 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators permitted by McGraw-Hill for their individual course preparation.If you are a student using this Manual, you are using it without permission. Chapter 4, Solution 8. LetV= VPROPRIETARY 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators 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 Vand 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 11927/ 13 2.07693 9 1V V VV= + = = To find V , consider the circuit below. 2 3 3 V9 V1 + _ 2 2 22327/ 13 2.07699 3 1V V VV+ = = = = 4.1538V Vo = V1 + V2 Chapter 4, Problem 9. Use superposition to find vin the circuit of Fig. 4.77. o 1 4 18 V + _ vo 2 + _ 6 A 2 Figure 4.77For 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators permitted by McGraw-Hill for their individual course preparation.If you are a student using this Manual, you are using it without permission. Chapter 4, Solution 9. Let v= v+ vPROPRIETARY 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators 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 vand v1 2 are due to 6-A and 20-V sources respectively.We find vusing the circuit below. 1 141 (6 ) 44 2v x A = =+ V 2//2= 1 , We find vusing the circuit below. 21 4 + _ v1 2 6 A 2 1 4 18 V + _ v2 2 + _ 2 21(18) 3 V1 1 4v = =+ + = 4 + 3= 7 V = v+ v vo 1 2 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 =vPROPRIETARY 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators 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 vand 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 =5 ab2 ab2 ab2 =1 + 5=6 V vab Chapter 4, Problem 11. Use the superposition principle to find iPROPRIETARY 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators 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 vin the circuit of Fig. 4.79. o 40 30 V 6 A + 4 io vo + 10 20 io Figure 4.79For Prob. 4.11. Chapter 4, Solution 11. Let v= v+ vo 1 2, where vand 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 5 440 10a a bav v vv vb(1) At node b, I1 4I+ (v 0)/20 = 0 or v1 b b = 100I1

110av vi=bButwhich leads to 100(vPROPRIETARY 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators 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= 159.96 b 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 + e f 0 ( 304 050 20c cov vi + + =) (0 )50covi= But 5 (30 )0 150 20c ccv vv+ = = 0 V20 0 10 150 50 5cvi += = = 2 210 2 V v i = = =15.99 + 2 = 17.99 V /10= 1.799 A v= v+ vo 1 2 and io = vo. Chapter 4, Problem 12. Determine vin 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators permitted by McGraw-Hill for their individual course preparation.If you are a student using this Manual, you are using it without permission. Chapter 4, Solution 12. Letv =vPROPRIETARY 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators 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. + v1 + v1 6||3=2 ohms,4||12=3 ohms.Hence, io=2/2=1, vo1=5io=5 V For vo2, consider the circuit below. + + v2 + + + v2 3||8=24/11,v =[(24/11)/(6 + 24/11)]12=16/5 1 =(5/8)(16/5)=2 V vo2=(5/8)v1 For vo3, consider the circuit shown below. + v3 + + + + v3 7||12=(84/19) ohms,v =[(84/19)/(4 + 84/19)]19=9.975 2 v=(-5/7)v2=-7.125 =5 + 2 7.125=-125 mV vo Chapter 4, Problem 13. Use superposition to find vin the circuit of Fig. 4.81. o10 5 + _ vo 12 V 8 4 A + 2 A Figure 4.81For 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators 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 = + + Letv v ,wherev1 2 o1,v , and v2 3 are due to the independent sources.To find v , consider the circuit below. 1 10 2 A 5 v1 8 + _ 1105 2 4.347810 8 5v x x = =+ + To findv , 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators permitted by McGraw-Hill for their individual course preparation.If you are a student using this Manual, you are using it without permission. 285 4 6.95658 10 5v x x = =+ + To find v , consider the circuit below. 310 4 A 5 v2 + _ 8 10 5 v3 8 + _ +12 V = = + + 3512 2.60875 10 8v 1 2 38.6956 Vov v v v = + + ==8.696V. Chapter 4, Problem 14. Apply the superposition principle to find vin 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators permitted by McGraw-Hill for their individual course preparation.If you are a student using this Manual, you are using it without permission. Chapter 4, Solution 14. Let v = vPROPRIETARY 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators 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 + v3+ =10 + 1 3=8 V vo 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+ iPROPRIETARY 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators 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,Theni =[20/(2 + 2)]=5 A,i =i /2=2.5 A o 1 o 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators 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=-1 o For i , consider the circuit below. + + vo2 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=i2R=(1.875)23=10.55 watts 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators permitted by McGraw-Hill for their individual course preparation.If you are a student using this Manual, you are using it without permission. Chapter 4, Solution 16. Let i =iPROPRIETARY 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators 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 =-5/9 2 io=(12/9) (6/9) (5/9)=1/9=111.11 mA Chapter 4, Problem 17. Use superposition to obtain vin 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators permitted by McGraw-Hill for their individual course preparation.If you are a student using this Manual, you are using it without permission. Chapter 4, Solution 17. Let v= vPROPRIETARY 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators 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. + v2 + io=[12/(12 + 30)]6=72/42,vx2=10i = 17.143 V o 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 =-5.714 o=[12/(12 + 30)]2=24/42,vx3=-10io=-5.714 =14.286 17.143 5.714=-8.571 V vx Chapter 4, Problem 18. Use superposition to find Vin the circuit of Fig. 4.86. o 2 A4 10 V + _ Vo 2 1 0.5 Vo + _ Figure 4.86For 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators permitted by McGraw-Hill for their individual course preparation.If you are a student using this Manual, you are using it without permission. Chapter 4, Solution 18. Let V= V+ VPROPRIETARY 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators 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= 0 1 But V= 4i 1`110 7 2 5 2, 8 V i i i i V = = = =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=0 2 = 4iBut V2 24 7 2 5 0.8, 4 3.2 i i i i V i = = = = = VPROPRIETARY 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators 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 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators permitted by McGraw-Hill for their individual course preparation.If you are a student using this Manual, you are using it without permission. Chapter 4, Solution 19. Let v= v+ vPROPRIETARY 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators 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 vand vare 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 /86 + (v (4i2 2 x))/2 = 0 or v+ 3.2i2 x = 9.6 Butix=0.5v .Therefore, 2 v+ 3.2(0.5v ) =9.6which leads tov2 2 2=16 =10.667 16 =26.67V Hence, vx. Checking, ix = 0.5v= 13.333A x 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 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators 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 V16 V Figure 4.88For Prob. 4.20. 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 10.1 0.05 0.025 0.175 5.714310 20 40eqeqRR= + + = + + = =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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators 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 Chapter 4, Problem 21. Apply source transformation to determine vand iPROPRIETARY 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators 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,thereforeio=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 V vo 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators 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 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators permitted by McGraw-Hill for their individual course preparation.If you are a student using this Manual, you are using it without permission.

1063 5A 3A 3//6 = 2-ohm.Convert the current sources to voltages sources as shown below. 1082 + + 10V 30V-- Applying KVL to the loop gives A 1 0 ) 2 8 10 ( 10 30 = = + + + + I I W 82= = = R I VI p Chapter 4, Problem 24. Use source transformation to find the voltage Vin the circuit of Fig. 4.92. x + _ 8 10 10 3 A 40 V +Vx 2 Vx Figure 4.92For 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators permitted by McGraw-Hill for their individual course preparation.If you are a student using this Manual, you are using it without permission. 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 + 20VPROPRIETARY 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators 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 to x = 2.978 V 28I + 160I = 70 or I = 0.3723 A or Vx. 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=66which leads toi=3.3 =2i=6.6 VPROPRIETARY 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators permitted by McGraw-Hill for their individual course preparation.If you are a student using this Manual, you are using it without permission. vo Chapter 4, Problem 26. Use source transformation to find iin the circuit of Fig. 4.94. o + _ 2 5 4 3 A 6 A io 20 V Figure 4.94For 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 mA 15 +20 = 0 or 11i= 7 or iPROPRIETARY 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators 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. 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=0leads toi=-4 12i=-48 VPROPRIETARY 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators 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 Chapter 4, Problem 28. Use source transformation to find Iin Fig. 4.96. o 1 + _ 3 8 V + _ Vo 4 Vo Io Figure 4.96For 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 io3 + _ 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators 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 i 4o o8 8 4 0 2 Ao o oi i i + = = Chapter 4, Problem 29. Use source transformation to find vin 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators 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 mAwhich 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 + + voChapter 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators permitted by McGraw-Hill for their individual course preparation.If you are a student using this Manual, you are using it without permission. Chapter 4, Solution 30 Transform the dependent current source as shown below. iPROPRIETARY 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators permitted by McGraw-Hill for their individual course preparation.If you are a student using this Manual, you are using it without permission. x24 60 10 ++ 12V30 7ix -- Combine the 60-ohm with the 10-ohm and transform the dependent source as shown below. ix24 + 12V30 70 0.1ix - Combining 30-ohm and 70-ohm gives 30//70 = 70x30/100 = 21-ohm.Transform the dependent current source as shown below. ix24 21 ++ 12V 2.1ix -- Applying KVL to the loop gives mA 8 . 2541 . 47120 1 . 2 12 45 = = = + x x xi i i Chapter 4, Problem 31. Determine vin 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 inFig. (a).6||8=(24/7) ohms.Transform the dependent source again to get the circuit in Fig. (b). + + + + + From Fig. (b), =3i,ori=v /3.vx x Applying KVL, -12 + (3 + 24/7)i + (24/21)vx=0 =84/23=3.652 VPROPRIETARY 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators 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 tovx Chapter 4, Problem 32. Use source transformation to find iin 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 APROPRIETARY 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators 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,orix Chapter 4, Problem 33. Determine Rand Vat 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 ohmsPROPRIETARY 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators 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 V VTh =30||60=1800/90=20 ohms (b)RTh 2 + (30 v1)/60=v /30,andv1 1=VTh 120 + 30 v =2v ,orv =50 V 1 1 1 =50 V VTh 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 ohmsPROPRIETARY 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators 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,orv1=v2 60 (2) =92 V Solving (1) and (2),v1=32 V,v2=92 V,and VTh=v2 Chapter 4, Problem 35. Use Thevenins theorem to find vin Prob. 4.12. o Chapter 4, Problem 12. Determine vin 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators permitted by McGraw-Hill for their individual course preparation.If you are a student using this Manual, you are using it without permission. Chapter 4, Solution 35. To find R , consider the circuit in Fig. (a). ThR =R =6||3+ 12||4=2 + 3=5 ohms Th 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators 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,orv1 1 1=8 At node 2,(19 v2)/4=2 + v2/12,orv2=33/4 But,-v1 + V+ v =0,orV =v v =8 33/4=-0.25 Th 2 Th 1 2 + ++ v1+ v2 + + + /2=-0.25/2=125 mV vo=VTh 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 ohms Th From Fig. (b),V =(40/(10 + 40))50=40V Th + + The equivalent circuit of the original circuit is shown in Fig. (c).Applying KVL, 30 40 + (8 + 12)i=0, which leads toi=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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators permitted by McGraw-Hill for their individual course preparation.If you are a student using this Manual, you are using it without permission. 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators permitted by McGraw-Hill for their individual course preparation.If you are a student using this Manual, you are using it without permission. 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators 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 ) 40 20 //( 12NRIN is found from the circuit below. 2A 20 a +40 120V12 -IN b

Applying source transformation to the current source yields the circuit below. 20 40+80 V- + 120VIN -

Applying KVL to the loop yields 666.7 mA = = = + + 60 / 40 I 0 I 60 80 120N N. Chapter 4, Problem 38. Apply Thvenin's theorem to find Vin the circuit of Fig. 4.105. o Figure 4.105 Chapter 4, Solution 38 We find Thevenin equivalent at the terminals of the 10-ohm resistor.For RTh, consider the circuit below.1 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators permitted by McGraw-Hill for their individual course preparation.If you are a student using this Manual, you are using it without permission. 5 RTh 16

= + = + + = 5 4 1 ) 16 4 //( 5 1ThR For V , consider the circuit below.Th1 V4 V 1 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators permitted by McGraw-Hill for their individual course preparation.If you are a student using this Manual, you are using it without permission. 5+ V3A 16Th +- 12 V - At node 1, 2 12 1 14 5 484 163 V VV V V = + = (1) At node 2, 2 12 2 19 5 48 05124V VV V V+ = =+(2) Solving (1) and (2) leads to 2 . 192 = = V VTh Thus, the given circuit can be replaced as shown below. 5 + + 10 19.2VVo-- Using voltage division, 8 . 12 ) 2 . 19 (5 1010=+=oVV Chapter 4, Problem 39. Obtain the Thevenin equivalent at terminals a-b of the circuit in Fig. 4.106. 5 24 V 16 10 3 A + _ 10 cca b Figure 4.106For Prob. 4.39. 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators permitted by McGraw-Hill for their individual course preparation.If you are a student using this Manual, you are using it without permission. Chapter 4, Solution 39. We obtain Rusing the circuit below. Th 10 16 5 10 RTh 20516 20/ / 5 16 2025ThxR = + = + = To find V , we use the circuit below. Th 5 24 16 10 3 A + _ 10 _ V2 + V2 V1 _ + VTh At node 1, 1 1 21 2243 5410 10V V VV V + = = 2 (1) At node 2, 1 2 213 60 210 5V V VV V= + = 26 (2) Substracting (1) from (2) gives 1 26 5 1.2 V V V = =But + + = = 2163 0 49.2 VTh ThV x V VPROPRIETARY 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators permitted by McGraw-Hill for their individual course preparation.If you are a student using this Manual, you are using it without permission. Chapter 4, Problem 40. Find the Thevenin equivalent at terminals a-b of the circuit in Fig. 4.107. + _ 70 V +Vo 4 Vo + 10 k 20 k ccb a Figure 4.107For Prob. 4.40. Chapter 4, Solution 40. To obtain V , we apply KVL to the loop. Th70 (10 20) 4 0okI 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators permitted by McGraw-Hill for their individual course preparation.If you are a student using this Manual, you are using it without permission. I 10oV k = But 70 70 1 kI I mA = =70 10 0 60 VTh ThkI V V + + = =To find RTh, we remove the 70-V source and apply a 1-V source at terminals a-b, as shown in the circuit below. 1 V b 20 4 Vo 10 k We notice that Vo = -1 V. +1 11 20 4 0 0.25 mAokI V I + = =2 110.35 mA10VI Ik= + = 21 12.857 k0.35ThVR kI= = = + _ + Vo I2 a I1 f c + Chapter 4, Problem 41. Find the Thvenin and Norton equivalents at terminals a-b of the circuit shown in Fig. 4.108. Figure 4.108 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators permitted by McGraw-Hill for their individual course preparation.If you are a student using this Manual, you are using it without permission. Chapter 4, Solution 41 , consider the circuit belowTo find RTh 14 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators permitted by McGraw-Hill for their individual course preparation.If you are a student using this Manual, you are using it without permission. 65 b N ThR R = = + = 4 ) 6 14 //( 5 Applying source transformation to the 1-A current source, we obtain the circuit below. 6 -14V +14VTh a + 6V 3A 5 - b At node a, V 85314 66 14 = + =+ +ThTh ThVV V A 2 4 / ) 8 ( = = =ThThNRVI Thus, A 2 V, 8 , 4 = = = =N Th N ThI V R R Chapter 4, Problem 42. For the circuit in Fig. 4.109, find Thevenin equivalent between terminals a and b. Figure 4.109 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators permitted by McGraw-Hill for their individual course preparation.If you are a student using this Manual, you are using it without permission. Chapter 4, Solution 42. , consider the circuit in Fig. (a).To find RTh 20||20=10 ohms.Transform the wye sub-network to a delta as shown in Fig. (b). =30||(7.5 + 7.5)=10 ohms 10||30=7.5 ohms.RPROPRIETARY 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators permitted by McGraw-Hill for their individual course preparation.If you are a student using this Manual, you are using it without permission. Th =Rab. To find VTh, we transform the 20-V and the 5-V sources.We obtain the circuit shown in Fig. (c). + + + + For loop 1,-30 + 50 + 30i1 10i2=0,or-2=3i1 i2(1) For loop 2, -50 10 + 30i2 10i1=0,or6=-i+ 3i (2) 1 2 Solving (1) and (2),i1=0,i2=2 A Applying KVL to the output loop,-v 10i+ 30 10iab 1 2=0,v =10 V ab =10 volts VTh=vab Chapter 4, Problem 43. Find the Thevenin equivalent looking into terminals a-b of the circuit in Fig. 4.110 and solve for i . x Figure 4.110 Chapter 4, Solution 43. To find R , consider the circuit in Fig. (a). Th + ++ va+ vb =10||10 + 5=10 ohmsPROPRIETARY 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators 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 =2x5=10 V,v vb a=20/2=10 V =0 volts But,-va + VTh + vb=0,orVTh=va vb Chapter 4, Problem 44. For the circuit in Fig. 4.111, obtain the Thevenin equivalent as seen from terminals(a)a-b (b) b-c Figure 4.111 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators permitted by McGraw-Hill for their individual course preparation.If you are a student using this Manual, you are using it without permission. Chapter 4, Solution 44. (a)For R , consider the circuit in Fig. (a). Th =1 + 4||(3 + 2 + 5)=3.857 ohmsPROPRIETARY 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators permitted by McGraw-Hill for their individual course preparation.If you are a student using this Manual, you are using it without permission. RTh For V ,consider the circuit in Fig. (b).Applying KVL gives, Th 10 24 + i(3 + 4 + 5 + 2),or i=1 =4i=4 V VTh + + + (b)For R , consider the circuit in Fig. (c). Th RTh=5||(2 + 3 + 4)=3.214 ohms To get VTh, consider the circuit in Fig. (d).At the node, KCL gives, [(24 vo)/9] + 2=vo/5,orvo=15 v+ + =vo=15 VPROPRIETARY 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators permitted by McGraw-Hill for their individual course preparation.If you are a student using this Manual, you are using it without permission. VTh Chapter 4, Problem 45. Find the Thevenin equivalent of the circuit in Fig. 4.112. Figure 4.112 Chapter 4, Solution 45. For RN, consider the circuit in Fig. (a). RN=(6 + 6)||4=3 ohms For IN, consider the circuit in Fig. (b).The 4-ohm resistor is shorted so that 4-A current is equally divided between the two 6-ohm resistors.Hence, IN=4/2=2 A Chapter 4, Problem 46. Find the Norton equivalent at terminals a-b of the circuit in Fig. 4.113. 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators permitted by McGraw-Hill for their individual course preparation.If you are a student using this Manual, you are using it without permission. Figure 4.113For Prob. 4.46. 10 4 A 10 20 a c c b Chapter 4, Solution 46. RN is found using the circuit below. 10 10 20 a ccb RN RN = 20//(10+10) = 10 To find IN, consider the circuit below. 10 4 A 20 10 b IN The 20- resistor is short-circuited and can be ignored. 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators permitted by McGraw-Hill for their individual course preparation.If you are a student using this Manual, you are using it without permission. IN = x 4 = 2 A Chapter 4, Problem 47. Obtain the Thvenin and Norton equivalent circuits of the circuit in Fig. 4.114 with respect to terminals a and b. Figure 4.114 Chapter 4, Solution 47 Since V= VTh ab = V , we apply KCL at the node a and obtain x V 19 . 1 126 / 150 260 1230= = + =Th ThTh ThV VV V To find R , consider the circuit below. Th 12 V a x

2Vx60 1A At node a, KCL gives 4762 . 0 126 / 6012 602 1 = = + + =xx xxVV VV5 . 2 4762 . 0 / 19 . 1 , 4762 . 01= = = = =ThThNxThRVIVRThus, A 5 . 2 , 4762 . 0 , 19 . 1 = = = =N N Th ThI R R V V Chapter 4, Problem 48. Determine the Norton equivalent at terminals a-b for the circuit in Fig. 4.115. Figure 4.115 Chapter 4, Solution 48. To get R , consider the circuit in Fig. (a). Th + + VTh+ + 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators permitted by McGraw-Hill for their individual course preparation.If you are a student using this Manual, you are using it without permission. From Fig. (a),Io=1,6 10 V=0,orV=-4 =V/1=-4 ohms RN=RTh To get V , consider the circuit in Fig. (b), Th Io=2,V =-10I+ 4ITh o o=-12 V =3A IN=VTh/RTh Chapter 4, Problem 49. Find the Norton equivalent looking into terminals a-b of the circuit in Fig. 4.102. Figure 4.102 Chapter 4, Solution 49. =28 ohmsPROPRIETARY 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators permitted by McGraw-Hill for their individual course preparation.If you are a student using this Manual, you are using it without permission. RN=RTh To find IN, consider the circuit below, + At the node,(40 vo)/10=3 + (v /40) + (v /20),orvo o o=40/7 + 3=3.286 A io=vo/20=2/7,butIN=Isc=io Chapter 4, Problem 50. Obtain the Norton equivalent of the circuit in Fig. 4.116 to the left of terminals a-b. Use the result to find current i Figure 4.116 Chapter 4, Solution 50. 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 writtenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducators permitted by McGraw-Hill for their individual course preparation.If you are a student using this Manual, you are using it without permission. From Fig. (a),RN=6 + 4=10 ohms + From Fig. (b), 2 + (12 v)/6=v/4,orv=9.6 V -IN=(12 v)/6=0.4,which leads toIN=-0.4 A Combining the Norton equivalent with the right-hand side of the original circuit produces the circuit in Fig. (c). i=[10/(10 + 5)] (4 0.4)=2.4 A Chapter 4, Problem 51. Given the circuit in Fig. 4.117, obtain the Norton equivalent as viewed from terminals(a)a-b(b) c-d Figure 4.117 Chapter 4, Solution 51. (a)From the ci