1、4Techniques of Circuit AnalysisAssessment ProblemsAP 4.1 a Redraw the circuit, labeling the reference node and the two node voltages:The two node voltage equations are15 + v160 + v115 + v1 v25 = 05+ v22 + v2 v15 = 0Place these equations in standard form:v1parenleftbigg 160 +115 +15parenrightbigg+ v2
2、parenleftbigg15parenrightbigg= 15v1parenleftbigg15parenrightbigg+ v2parenleftbigg12 +15parenrightbigg= 5Solving, v1 = 60 V and v2 = 10 V;Therefore, i1 = (v1 v2)/5 = 10 Ab p15A = (15 A)v1 = (15 A)(60 V) = 900 W = 900 W(delivered)c p5A = (5 A)v2 = (5 A)(10 V) = 50 W= 50 W(delivered)41 2010 Pearson Edu
3、cation, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical,
4、 photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.42 CHAPTER 4. Techniques of Circuit AnalysisAP 4.2 Redraw the circuit, choosing the node voltages and reference node as s
5、hown:The two node voltage equations are:4.5+ v11 + v1 v26+ 2 = 0v212 +v2 v16+ 2 +v2 304 = 0Place these equations in standard form:v1parenleftbigg1+ 18parenrightbigg+ v2parenleftbigg18parenrightbigg= 4.5v1parenleftbigg18parenrightbigg+ v2parenleftbigg 112 +18 +14parenrightbigg= 7.5Solving, v1 = 6 V v
6、2 = 18 VTo find the voltage v, first find the current i through the series-connected 6and 2 resistors:i = v1 v26 + 2 = 6188 = 1.5 AUsing a KVL equation, calculate v:v = 2i + v2 = 2(1.5) + 18 = 15 VAP 4.3 a Redraw the circuit, choosing the node voltages and reference node asshown:The node voltage equ
7、ations are:v1 506 +v18 +v1 v22 3i1 = 05+ v24 + v2 v12 + 3i1 = 0 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrie
8、val system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.Problems 43The dependent source requires
9、the following constraint equation:i1 = 50v16Place these equations in standard form:v1parenleftbigg16 +18 +12parenrightbigg+ v2parenleftbigg12parenrightbigg+ i1(3) = 506v1parenleftbigg12parenrightbigg+ v2parenleftbigg14 +12parenrightbigg+ i1(3) = 5v1parenleftbigg16parenrightbigg+ v2(0) + i1(1) = 506S
10、olving, v1 = 32 V; v2 = 16 V; i1 = 3 AUsing these values to calculate the power associated with each source:p50V = 50i1 = 150 Wp5A = 5(v2) = 80 Wp3i1 = 3i1(v2 v1) = 144 Wb All three sources are delivering power to the circuit because the powercomputed in (a) for each of the sources is negative.AP 4.
11、4 Redraw the circuit and label the reference node and the node at which thenode voltage equation will be written:The node voltage equation isvo40 +vo 1010 +vo + 20i20 = 0The constraint equation required by the dependent source isi = i10 + i30 = 10vo10 + 10 + 20i30Place these equations in standard fo
12、rm:voparenleftbigg 140 +110 +120parenrightbigg+ i(1) = 1voparenleftbigg 110parenrightbigg+ iparenleftbigg1 2030parenrightbigg= 1+ 1030Solving, i = 3.2 A and vo = 24 V 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written per
13、mission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Depa
14、rtment, Pearson Education, Inc., Upper Saddle River, NJ 07458.44 CHAPTER 4. Techniques of Circuit AnalysisAP 4.5 Redraw the circuit identifying the three node voltages and the reference node:Note that the dependent voltage source and the node voltages v and v2 form asupernode. The v1 node voltage eq
15、uation isv17.5 +v1 v2.5 4.8 = 0The supernode equation isv v12.5 +v10 +v22.5 +v2 121 = 0The constraint equation due to the dependent source isix = v17.5The constraint equation due to the supernode isv + ix = v2Place this set of equations in standard form:v1parenleftbigg 17.5 +12.5parenrightbigg+ vpar
16、enleftbigg 12.5parenrightbigg+ v2(0) + ix(0) = 4.8v1parenleftbigg 12.5parenrightbigg+ vparenleftbigg 12.5 +110parenrightbigg+ v2parenleftbigg 12.5 + 1parenrightbigg+ ix(0) = 12v1parenleftbigg 17.5parenrightbigg+ v(0) + v2(0) + ix(1) = 0v1(0) + v(1) + v2(1) + ix(1) = 0Solving this set of equations gi
17、ves v1 = 15 V, v2 = 10 V, ix = 2 A, andv = 8 V. 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or t
18、ransmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.Problems 45AP 4.6 Redraw the circuit identifying the refe
19、rence node and the two unknown nodevoltages. Note that the right-most node voltage is the sum of the 60 V sourceand the dependent source voltage.The node voltage equation at v1 isv1 602 +v124 +v1 (60 + 6i)3 = 0The constraint equation due to the dependent source isi = 60 + 6i v13Place these two equat
20、ions in standard form:v1parenleftbigg12 +124 +13parenrightbigg+ i(2) = 30+ 20v1parenleftbigg13parenrightbigg+ i(12) = 20Solving, i = 4 A and v1 = 48 VAP 4.7 a Redraw the circuit identifying the three mesh currents:The mesh current equations are:80 + 5(i1 i2) + 26(i1 i3) = 030i2 + 90(i2 i3) + 5(i2 i1
21、) = 08i3 + 26(i3 i1) + 90(i3 i2) = 0Place these equations in standard form: 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage
22、 in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.46 CHAPTER 4. Techniques of C
23、ircuit Analysis31i1 5i2 26i3 = 805i1 + 125i2 90i3 = 026i1 90i2 + 124i3 = 0Solving,i1 = 5 A; i2 = 2 A; i3 = 2.5 Ap80V = (80)i1 = (80)(5) = 400 WTherefore the 80 V source is delivering 400 W to the circuit.b p8 = (8)i23 = 8(2.5)2 = 50 W, so the 8 resistor dissipates 50 W.AP 4.8 a b = 8, n = 6, bn + 1
24、= 3b Redraw the circuit identifying the three mesh currents:The three mesh-current equations are25 + 2(i1 i2) + 5(i1 i3) + 10 = 0(3v) + 14i2 + 3(i2 i3) + 2(i2 i1) = 01i3 10 + 5(i3 i1) + 3(i3 i2) = 0The dependent source constraint equation isv = 3(i3 i2)Place these four equations in standard form:7i1
25、 2i2 5i3 + 0v = 152i1 + 19i2 3i3 + 3v = 05i1 3i2 + 9i3 + 0v = 100i1 + 3i2 3i3 + 1v = 0Solvingi1 = 4 A; i2 = 1 A; i3 = 3 A; v = 12 Vpds = (3v)i2 = 3(12)(1) = 36 WThus, the dependent source is delivering 36 W, or absorbing 36 W. 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved
26、. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information re
27、garding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.Problems 47AP 4.9 Redraw the circuit identifying the three mesh currents:The mesh current equations are:25+ 6(ia ib) + 8(ia ic) = 02ib + 8(ib ic) + 6(ib ia) = 05i + 8(ic ia) + 8(
28、ic ib) = 0The dependent source constraint equation is i = ia. We can substitute thissimple expression for i into the third mesh equation and place the equationsin standard form:14ia 6ib 8ic = 256ia + 16ib 8ic = 03ia 8ib + 16ic = 0Solving,ia = 4 A; ib = 2.5 A; ic = 2 AThus,vo = 8(ia ic) = 8(42) = 16
29、VAP 4.10 Redraw the circuit identifying the mesh currents:Since there is a current source on the perimeter of the i3 mesh, we know thati3 = 16 A. The remaining two mesh equations are30+ 3i1 + 2(i1 i2) + 6i1 = 08i2 + 5(i2 + 16) + 4i2 + 2(i2 i1) = 0 2010 Pearson Education, Inc., Upper Saddle River, NJ
30、. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewis
31、e. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.48 CHAPTER 4. Techniques of Circuit AnalysisPlace these equations in standard form:11i1 2i2 = 302i1 + 19i2 = 80Solving: i1 = 2 A, i2 = 4 A, i3 = 16 AThe curr
32、ent in the 2 resistor is i1 i2 = 6 A . . p2 = (6)2(2) = 72 WThus, the 2 resistors dissipates 72 W.AP 4.11 Redraw the circuit and identify the mesh currents:There are current sources on the perimeters of both the ib mesh and the icmesh, so we know thatib = 10 A; ic = 2v5The remaining mesh current equ
33、ation is75 + 2(ia + 10) + 5(ia 0.4v) = 0The dependent source requires the following constraint equation:v = 5(ia ic) = 5(ia 0.4v)Place the mesh current equation and the dependent source equation isstandard form:7ia 2v = 555ia 3v = 0Solving: ia = 15 A; ib = 10 A; ic = 10 A; v = 25 VThus, ia = 15 A. 2
34、010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electron
35、ic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.Problems 49AP 4.12 Redraw the circuit and identify the mesh currents:The 2 A current source is shared by t
36、he meshes ia and ib. Thus we combinethese meshes to form a supermesh and write the following equation:10 + 2ib + 2(ib ic) + 2(ia ic) = 0The other mesh current equation is6+ 1ic + 2(ic ia) + 2(ic ib) = 0The supermesh constraint equation isia ib = 2Place these three equations in standard form:2ia + 4i
37、b 4ic = 102ia 2ib + 5ic = 6ia ib + 0ic = 2Solving, ia = 7 A; ib = 5 A; ic = 6 AThus, p1 = i2c(1) = (6)2(1) = 36 WAP 4.13 Redraw the circuit and identify the reference node and the node voltage v1:The node voltage equation isv1 2015 2+v1 2510 = 0 2010 Pearson Education, Inc., Upper Saddle River, NJ.
38、All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise.
39、 For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.410 CHAPTER 4. Techniques of Circuit AnalysisRearranging and solving,v1parenleftbigg 115 +110parenrightbigg= 2+ 2015 + 2510 . . v1 = 35 Vp2A = 35(2) = 70 WThus
40、 the 2 A current source delivers 70 W.AP 4.14 Redraw the circuit and identify the mesh currents:There is a current source on the perimeter of the i3 mesh, so i3 = 4 A. Theother two mesh current equations are128 + 4(i1 4) + 6(i1 i2) + 2i1 = 030ix + 5i2 + 6(i2 i1) + 3(i2 4) = 0The constraint equation
41、due to the dependent source isix = i1 i3 = i1 4Substitute the constraint equation into the second mesh equation and placethe resulting two mesh equations in standard form:12i1 6i2 = 14424i1 + 14i2 = 132Solving,i1 = 9 A; i2 = 6 A; i3 = 4 A; ix = 94 = 5 A. . v4A = 3(i3 i2)4ix = 10 Vp4A = v4A(4) = (10)
42、(4) = 40 WThus, the 2 A current source delivers 40 W. 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system
43、, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.Problems 411AP 4.15 a Redraw the circuit with a hel
44、pful voltage and current labeled:Transform the 120 V source in series with the 20 resistor into a 6 Asource in parallel with the 20 resistor. Also transform the 60 V sourcein series with the 5 resistor into a 12 A source in parallel with the 5resistor. The result is the following circuit:Combine the
45、 three current sources into a single current source, usingKCL, and combine the 20, 5, and 6 resistors in parallel. Theresulting circuit is shown on the left. To simplify the circuit further,transform the resulting 30 A source in parallel with the 2.4 resistor intoa 72 V source in series with the 2.4
46、 resistor. Combine the 2.4 resistorin series with the 1.6 resisor to get a very simple circuit that stillmaintains the voltage v. The resulting circuit is on the right.Use voltage division in the circuit on the right to calculate v as follows:v = 812(72) = 48 Vb Calculate i in the circuit on the rig
47、ht using Ohms law:i = v8 = 488 = 6 ANow use i to calculate va in the circuit on the left:va = 6(1.6 + 8) = 57.6 VReturning back to the original circuit, note that the voltage va is also thevoltage drop across the series combination of the 120 V source and 20 2010 Pearson Education, Inc., Upper Saddl
48、e River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording,
49、 or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.412 CHAPTER 4. Techniques of Circuit Analysisresistor. Use this fact to calculate the current in the 120 V source, ia:ia = 120va20 = 12057.620 = 3.12 Ap120V = (120)ia = (120)(3.12)
