电路基础(第三版)Alexander and Sadiku著 课后题答案chapter 02

1、Chapter 2, Problem 1.The voltage across a 5-k resistor is 16 V. Find the current through the resistor.Chapter 2, Solution 1v = iRi = v/R = (16/5) mA = 3.2 mAChapter 2, Problem 2.Find the hot resistance of a lightbulb rated 60 W, 120 V.Chapter 2, Solution 2p = v2/R R = v2/p = 14400/60 = 240 ohmsChapt

2、er 2, Problem 3.A bar of silicon is 4 cm long with a circular cross section. If the resistance of the bar is 240 at room temperature, what is the cross-sectional radius of the bar?Chapter 2, Solution 3For silicon, = 6.4x102 -m.A = r 2 . Hence,R = L= L r2 = L=6.4x102 x 4x102= 0.033953 r2 R x240Ar = 0

3、.1843 mChapter 2, Problem 4.(a) Calculate current i in Fig. 2.68 when the switch is in position 1.(b) Find the current when the switch is in position 2.Chapter 2, Solution 4(a) i = 3/100 = 30 mA(b) i = 3/150 = 20 mAPROPRIETARY MATERIAL. 2007 The McGraw-Hill Companies, Inc. All rights reserved. No pa

4、rt 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 u

5、sing this Manual, you are using it without permission.Chapter 2, Problem 5.For the network graph in Fig. 2.69, find the number of nodes, branches, and loops.Chapter 2, Solution 5n = 9;l = 7; b = n + l 1 = 15Chapter 2, Problem 6.In the network graph shown in Fig. 2.70, determine the number of branche

6、s and nodes.Chapter 2, Solution 6n = 12;l = 8;b = n + l 1 = 19Chapter 2, Problem 7.Determine the number of branches and nodes in the circuit of Fig. 2.71.1 4 12 V+85 2 A_Figure 2.71 For Prob. 2.7.Chapter 2, Solution 76 branches and 4 nodes.PROPRIETARY MATERIAL. 2007 The McGraw-Hill Companies, Inc. A

7、ll 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 preparatio

8、n. If you are a student using this Manual, you are using it without permission.Chapter 2, Problem 8.Use KCL to obtain currents i1, i2, and i3 in the circuit shown in Fig. 2.72.Chapter 2, Solution 812 ACHAPTER 1 -AI18 AI2BI312 AC9 ADAt node a,8 = 12 + i1i1= - 4AAt node c,9 = 8 + i2i2= 1AAt node d,9 =

9、 12 + i3i3= -3AChapter 2, Problem 9.Findi1 , i 2 , and i3in Fig. 2.73.8 A2 Ai212 A10 ABAi3i114 A4 ACChapter 2, Solution 9Figure 2.73 For Prob. 2.9.At A,2 + 12 = i1 i1 =14 AAt B,12 = i2 + 14 i2 =2 AAt C,14 = 4 + i3 i3 =10 APROPRIETARY MATERIAL. 2007 The McGraw-Hill Companies, Inc. All rights reserved

10、. 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 st

11、udent using this Manual, you are using it without permission.Chapter 2, Problem 10.In the circuit in Fig. 2.67 decrease in R3 leads to a decrease of:(a) current through R3(b) voltage through R3(c) voltage across R1(d) power dissipated in R2(e) none of the aboveChapter 2, Solution 104A2-2A1I1I233AAt

12、node 1,4+ 3 = i1i1 =7AAt node 3,3+ i2 = -2i2 =-5AChapter 2, Problem 11.In the circuit of Fig. 2.75, calculate V1 and V2.+ 1 V + 2 V +V15 VV2_Chapter 2, Solution 11Figure 2.75 For Prob. 2.11.V1 + 1+ 5 = 0 V1 =6 V 5 + 2 + V2 = 0 V2 =3 VPROPRIETARY MATERIAL. 2007 The McGraw-Hill Companies, Inc. All rig

13、hts 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

14、you are a student using this Manual, you are using it without permission.Chapter 2, Problem 12.In the circuit in Fig. 2.76, obtain v1, v2, and v3.Chapter 2, Solution 12+ 15V - 25V+ 10V -LOOP+ V2 -+LOOP+20VV1LOOPV3-For loop 1,-20-25 +10 + v1 = 0v1= 35vFor loop 2,-10+15 -v2 = 0v2= 5vFor loop 3,-V1 + V

15、2 + V3 = 0v3= 30vPROPRIETARY MATERIAL. 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

16、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.Chapter 2, Problem 13.For the circuit in Fig. 2.77, use KCL to find the branch currents I1 to I4.2 AI27 AI4I13 AI34 AFigure 2.77Chapter 2, Solut

17、ion 132A1I227A3I444AI13AI3At node 2,3 + 7 + I 2 = 0 I2 = 10 AAt node 1,I 1 + I2 = 2 I 1 = 2 I 2 = 12 AAt node 4,2 = I 4 + 4 I4 = 2 4 = 2 AAt node 3,7 + I 4 = I 3 I3 = 7 2 = 5 AHence,I 1 = 12 A, I 2 = 10 A,I 3 = 5 A, I4 = 2 APROPRIETARY MATERIAL. 2007 The McGraw-Hill Companies, Inc. All rights reserv

18、ed. 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

19、student using this Manual, you are using it without permission.Chapter 2, Problem 14.Given the circuit in Fig. 2.78, use KVL to find the branch voltages V1 to V4.+3 VV1V2+ 2 V +4 VV3V45 V+Figure 2.78Chapter 2, Solution 14+-3VV1I4V2-I3-+2V -+-+V3 -4VI2-V4I15V+-For mesh 1,V4 + 2 + 5 = 0 V4 = 7VFor mes

20、h 2,+4 + V3 + V4 = 0V3 = 4 7 = 11VFor mesh 3,3 + V1 V3 = 0 V1 = V3 + 3 = 8VFor mesh 4,V1 V2 2 = 0 V2 = V1 2 = 6VThus,V1 = 8V , V2 = 6V , V3 = 11V ,V4 = 7VPROPRIETARY MATERIAL. 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed

21、 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.Chapter

22、 2, Problem 15.Calculate v and ix in the circuit of Fig. 2.79.12 + 8 V +v ix+3 ix12 V2 V_Figure 2.79 For Prob. 2.15.Chapter 2, Solution 15For loop 1, 12 + v +2 = 0,v = 10 V2 AFor loop 2, 2 + 8 + 3ix =0,ix =Chapter 2, Problem 16.DetermineVo in the circuit in Fig. 2.80.2 6 y+9 V+Vo_yFigure 2.80 For Pr

23、ob. 2.16.Chapter 2, Solution 16Apply KVL,-9 + (6+2)I + 3 = 0, 8I = 9-3=6 ,I = 6/8Also,-9 + 6I + Vo = 0Vo= 9- 6I =4.5 V+3 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, w

24、ithout 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.Chapter 2, Problem 17.Obtain v1 throug

25、h v3 in the circuit in Fig. 2.78.Chapter 2, Solution 17Applying KVL around the entire outside loop we get, 24 + v1 + 10 + 12 = 0 or v1 = 2VApplying KVL around the loop containing v2, the 10-volt source, and the 12-volt source we get,v2 + 10 + 12 = 0 or v2 = 22VApplying KVL around the loop containing

26、 v3 and the 10-volt source we get, v3 + 10 = 0 or v3 = 10VChapter 2, Problem 18.Find I and Vab in the circuit of Fig. 2.79.Chapter 2, Solution 18APPLYING KVL,-30 -10 +8 + I(3+5) = 08I = 32I = 4A-Vab + 5I+ 8 = 0Vab =28VPROPRIETARY MATERIAL. 2007 The McGraw-Hill Companies, Inc. All rights reserved. No

27、 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 studen

28、t using this Manual, you are using it without permission.Chapter 2, Problem 19.From the circuit in Fig. 2.80, find I, the power dissipated by the resistor, and the power supplied by each source.Chapter 2, Solution 19APPLYING KVL AROUND THE LOOP, WE OBTAIN-12 + 10 - (-8) + 3i = 0 i = 2APower dissipat

29、ed by the resistor:p 3 = i2R = 4(3) = 12WPower supplied by the sources:p12V = 12 (2) = 24Wp10V = 10 (2) = 20Wp8V = (8)(2) = 16WChapter 2, Problem 20.Determine io in the circuit of Fig. 2.81.Chapter 2, Solution 20APPLYING KVL AROUND THE LOOP,-36 + 4i0 + 5i0 = 0 i0 = 4APROPRIETARY MATERIAL. 2007 The M

30、cGraw-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

31、individual course preparation. If you are a student using this Manual, you are using it without permission.Chapter 2, Problem 21.Find Vx in the circuit of Fig. 2.85.1 2 Vx+15 V5 Vx_2 Figure 2.85 For Prob. 2.21.Chapter 2, Solution 21Applying KVL,-15+ (1+5+2)I + 2 Vx = 0But Vx = 5I,I = 5/6-15+8I + 10I

32、 =0,Vx= 5I = 25/6 =4.167 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 written permission of the publisher, or used beyond the limited distribution to

33、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.Chapter 2, Problem 22.Find Vo in the circuit in Fig. 2.85 and the power dissipated by the controlled source.Chapter 2, Solution 224 + V

34、0-6 10A2V0At the node, KCL requires thatv0+ 10 + 2v0 = 0v0 = 4.444V4The current through the controlled source isi = 2V0 = -8.888Aand the voltage across it isv = (6 + 4) i0 (where i0 = v0/4) = 10 v40 = 11.111Hence,p2 vi = (-8.888)(-11.111) = 98.75 WPROPRIETARY MATERIAL. 2007 The McGraw-Hill Companies

35、, 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 pr

36、eparation. If you are a student using this Manual, you are using it without permission.Chapter 2, Problem 23.In the circuit shown in Fig. 2.87, determine vx and the power absorbed by the 12- resistor.1 1.2 + vx 4 6 A2 8 12 3 6 Figure 2.87Chapter 2, Solution 238/12 = 4.8, 3/6 = 2, (4 + 2)/(1.2 + 4.8) = 6/6 = 3 The circuit is reduced to that shown below.ix1 + vx -6A 23Applying current division,i x =2(6 A) = 2 A, v x = 1i x =2V2+ 1+ 3The current through the 1.2- resistor is 0.5ix = 1A. The voltage across the 12- resistor is 1 x 4.8 = 4.8 V. Hence the power isp = v2 = 4.82