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CHAPTER Exercises E2.1 (a) R2, R3, and R4 are in parallel Furthermore R1 is in series with the combination of the other resistors Thus we have: Req R1 3 / R2 / R3 / R4 (b) R3 and R4 are in parallel Furthermore, R2 is in series with the combination of R3, and R4 Finally R1 is in parallel with the combination of the other resistors Thus we have: Req 5 / R1 /[R2 /(1 / R3 / R4 )] (c) R1 and R2 are in parallel Furthermore, R3, and R4 are in parallel Finally, the two parallel combinations are in series 1 Req 52.1 / R1 / R2 / R3 / R4 (d) R1 and R2 are in series Furthermore, R3 is in parallel with the series combination of R1 and R2 Req 1.5 k / R3 /(R1 R2 ) E2.2 (a) First we combine R2, R3, and R4 in parallel Then R1 is in series with the parallel combination Req 9.231 / R2 / R3 / R4 i1 20 V 20 1.04 A R1 Req 10 9.231 i2 v eq / R2 0.480 A v eq Req i1 9.600 V i4 v eq / R4 0.240 A i3 v eq / R3 0.320 A © 2014 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, 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 (b) R1 and R2 are in series Furthermore, R3, and R4 are in series Finally, the two series combinations are in parallel Req R1 R2 20 v eq Req 20 V Req R3 R4 20 Req i1 v eq / Req A / Req 1 10 / Req i2 v eq / Req A (c) R3, and R4 are in series The combination of R3 and R4 is in parallel with R2 Finally the combination of R2, R3, and R4 is in series with R1 Req R3 R4 40 Req v i1Req 20 V E2.3 (a) v v s 20 / Req / R2 i1 vs 1A R1 Req i2 v / R2 0.5 A i3 v / Req 0.5 A R1 10 V v v s R2 20 V R1 R2 R3 R4 R1 R2 R3 R4 Similarly, we find v 30 V and v 60 V © 2014 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, 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 (b) First combine R2 and R3 in parallel: Req (1 / R2 R3 ) 2.917 Then we have v v s v2 vs E2.4 R1 6.05 V Similarly, we find R1 Req R4 Req 5.88 V and v 8.07 V R1 Req R4 (a) First combine R1 and R2 in series: Req = R1 + R2 = 30 Then we have Req R3 15 30 i1 is A and i3 is A R3 Req 15 30 R3 Req 15 30 (b) The current division principle applies to two resistances in parallel Therefore, to determine i1, first combine R2 and R3 in parallel: Req = Req 1/(1/R2 + 1/R3) = Then we have i1 is 1A R1 Req 10 Similarly, i2 = A and i3 = A E2.5 E2.6 E2.7 Write KVL for the loop consisting of v1, vy , and v2 The result is -v1 - vy + v2 = from which we obtain vy = v2 - v1 Similarly we obtain vz = v3 - v1 v1 v3 v1 v2 v v1 v2 v2 v3 Node 2: ia 0 R1 R2 R2 R3 R4 v v v2 v3 v1 Node 3: ib R5 R4 R1 Node 1: Following the step-by-step method in the book, we obtain 1 R1 R2 R2 E2.8 R2 R2 R3 R4 v is v R4 1 v is R4 R5 R4 Instructions for various calculators vary The MATLAB solution is given in the book following this exercise © 2014 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, 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 E2.9 (a) Writing the node equations we obtain: v v3 v1 v1 v2 Node 1: 0 20 10 v v3 v v1 Node 2: 10 0 10 v v1 v3 v3 v2 Node 3: 0 20 10 (b) Simplifying the equations we obtain: 0.35v 0.10v 0.05v 0.10v 0.30v 0.20v 10 0.05v1 0.20v2 0.35v3 (c) and (d) Solving using Matlab: >>clear >>G = [0.35 -0.1 -0.05; -0.10 0.30 -0.20; -0.05 -0.20 0.35]; >>I = [0; -10; 0]; >>V = G\I V= -27.2727 -72.7273 -45.4545 >>Ix = (V(1) - V(3))/20 Ix = 0.9091 E2.10 Using determinants we can solve for the unknown voltages as follows: 0.2 v1 0.5 0.7 0.2 0.2 0.5 0.7 v2 0.2 10.32 V 0.35 0.04 0.2 0.7 1.2 6.129 V 0.7 0.2 0.35 0.04 0.2 0.5 Many other methods exist for solving linear equations © 2014 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, 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 E2.11 First write KCL equations at nodes and 2: Node 1: v 10 v1 v1 v2 0 10 v 10 v v v Node 2: 0 10 10 Then, simplify the equations to obtain: 8v v 50 and v 4v 10 Solving manually or with a calculator, we find v1 = 6.77 V and v2 = 4.19 V The MATLAB session using the symbolic approach is: >> clear [V1,V2] = solve('(V1-10)/2+(V1)/5 +(V1 - V2)/10 = 0' , '(V2-10)/10 +V2/5 +(V2-V1)/10 = 0') V1 = 210/31 V2 = 130/31 Next, we solve using the numerical approach >> clear G = [8 -1; -1 4]; I = [50; 10]; V = G\I V= 6.7742 4.1935 E2.12 The equation for the supernode enclosing the 15-V source is: v3 v2 v3 v1 v1 v2 R3 R1 R2 R4 This equation can be readily shown to be equivalent to Equation 2.37 in the book (Keep in mind that v3 = -15 V.) © 2014 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, 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 E2.13 Write KVL from the reference to node then through the 10-V source to node then back to the reference node: v 10 v Then write KCL equations First for a supernode enclosing the 10-V source, we have: v1 v1 v3 v2 v3 1 R1 R2 R3 Node 3: v3 v3 v1 v3 v2 0 R4 R2 R3 Reference node: v1 v3 1 R1 R4 An independent set consists of the KVL equation and any two of the KCL equations E2.14 (a) Select the reference node at the left-hand end of the voltage source as shown at right Then write a KCL equation at node v v 10 1 R1 R2 Substituting values for the resistances and solving, we find v1 = 3.33 V 10 v Then we have ia 1.333 A R2 (b) Select the reference node and assign node voltages as shown Then write KCL equations at nodes and © 2014 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, 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 v 25 v v v 0 R2 R4 R3 v 25 v v v 0 R1 R3 R5 Substituting values for the resistances and solving, we find v1 = 13.79 V v v2 and v = 18.97 V Then we have ib -0.259 A R3 E2.15 (a) Select the reference node and node voltage as shown Then write a KCL equation at node 1, resulting in v v 10 2ix 5 Then use ix (10 v ) / to substitute and solve We find v1 = 7.5 V 10 v Then we have ix 0.5 A (b) Choose the reference node and node voltages shown: Then write KCL equations at nodes and 2: v1 v 2iy 3 v2 v 2iy 10 3 © 2014 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, 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 Finally use iy v / to substitute and solve This yields v 11.54 V and iy 2.31 A E2.16 >> clear >> [V1 V2 V3] = solve('V3/R4 + (V3 - V2)/R3 + (V3 - V1)/R1 = 0', 'V1/R2 + V3/R4 = Is', 'V1 = (1/2)*(V3 - V1) + V2' ,'V1','V2','V3'); >> pretty(V1), pretty(V2), pretty(V3) R2 Is (2 R3 R1 + R4 R1 + R4 R3) R3 R1 + R4 R1 + R1 R2 + R4 R3 + R3 R2 R2 Is (3 R3 R1 + R4 R1 + R4 R3) R3 R1 + R4 R1 + R1 R2 + R4 R3 + R3 R2 Is R2 R4 (3 R1 + R3) R3 R1 + R4 R1 + R1 R2 + R4 R3 + R3 R2 E2.17 Refer to Figure 2.33b in the book (a) Two mesh currents flow through R2: i1 flows downward and i4 flows upward Thus the current flowing in R2 referenced upward is i4 - i1 (b) Similarly, mesh current i1 flows to the left through R4 and mesh current i2 flows to the right, so the total current referenced to the right is i2 - i1 (c) Mesh current i3 flows downward through R8 and mesh current i4 flows upward, so the total current referenced downward is i3 - i4 (d) Finally, the total current referenced upward through R8 is i4 - i3 E2.18 Refer to Figure 2.33b in the book Following each mesh current in turn, we have R1i1 R2 (i1 i4 ) R4 (i1 i2 ) v A R5i2 R4 (i2 i1 ) R6 (i2 i3 ) R7i3 R6 (i3 i2 ) R8 (i3 i4 ) R3i4 R2 (i4 i1 ) R8 (i4 i3 ) © 2014 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, 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 In matrix form, these equations become R4 R2 (R1 R2 R4 ) i1 v A i R4 (R4 R5 R6 ) R6 i3 R6 (R6 R7 R8 ) R8 R2 R8 (R2 R3 R8 ) i4 E2.19 We choose the mesh currents as shown: Then, the mesh equations are: 5i 10(i1 i2 ) 100 and 10(i2 i1 ) 7i2 3i2 Simplifying and solving these equations, we find that i1 10 A and i2 A The net current flowing downward through the 10-Ω resistance is i1 i2 A To solve by node voltages, we select the reference node and node voltage shown (We not need to assign a node voltage to the connection between the 7-Ω resistance and the 3-Ω resistance because we can treat the series combination as a single 10-Ω resistance.) © 2014 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, 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 The node equation is (v 10) / v / 10 v / 10 Solving we find that v1 = 50 V Thus we again find that the current through the 10-Ω resistance is i v / 10 A Combining resistances in series and parallel, we find that the resistance “seen” by the voltage source is 10 Ω Thus the current through the source and 5-Ω resistance is (100 V)/(10 Ω) = 10 A This current splits equally between the 10-Ω resistance and the series combination of Ω and Ω E2.20 First, we assign the mesh currents as shown Then we write KVL equations following each mesh current: 2(i1 i3 ) 5(i1 i2 ) 10 5i2 5(i2 i1 ) 10(i2 i3 ) 10i3 10(i3 i2 ) 2(i3 i1 ) Simplifying and solving, we find that i1 = 2.194 A, i2 = 0.839 A, and i3 = 0.581 A Thus the current in the 2-Ω resistance referenced to the right is i1 - i3 = 2.194 - 0.581 = 1.613 A E2.21 Following the step-by-step process, we obtain R3 R2 i1 v A (R2 R3 ) R i v (R3 R4 ) B R2 (R1 R2 ) i3 v B © 2014 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, electronic, mechanical, photocopying, 10 recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 Zeroing the source, we have: Combining resistances in series and parallel we find Rt 12 Then the Thévenin voltage is vt isc Rt 19.2 V The Thévenin and Norton equivalents are: P2.88 The Thévenin voltage is equal to the open-circuit voltage which is 12.5 V The equivalent circuit with the 0.1- load connected is: We have 12.5 /(Rt 0.1) 500 from which we find Rt 0.15 The Thévenin and Norton equivalent circuits are: © 2014 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, electronic, mechanical, photocopying,49 recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 The short-circuit current is 83.33 A Because no energy is converted from chemical form to heat in a battery under open-circuit conditions, the Thévenin equivalent seems more realistic from an energy conversion standpoint P2.89 The Thévenin voltage is equal to the open-circuit voltage, which is V The circuit with the load attached is: 35 mA and v x Vt V Thus, the Thévenin 200 2V resistance is Rt 57.14 35 mA We have iL P2.90 The equivalent circuit with a load attached is: For a load of kΩ, we have iL / 1000 mA , and we can write v L Vt Rt iL Substituting values this becomes Vt 0.008Rt (1) © 2014 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, electronic, mechanical, photocopying,50 recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 Similarly, for the 2-kΩ load we obtain 10 Vt 0.005Rt (2) Solving Equations (1) and (2), we find Vt 13.33 V and Rt 666 P2.91 Open-circuit conditions: Using KVL, we have 25 5(1.5ix ) 10ix 10ix Solving, we find ix 0.90909 A and then we have Vt v oc 10ix 9.0909 V Under short-circuit conditions, we have ix and the controlled source becomes an open circuit: 25 1.667 A Then, we have Rt v oc isc 5.45 Thus, the 15 equivalents are: isc © 2014 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, electronic, mechanical, photocopying, 51 recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 P2.92 As is Problem P2.83, we find the Thévenin equivalent: Then maximum power is obtained for a load resistance equal to the Thévenin resistance vt 22 Pmax 3.333 W Rt P2.93 As in Problem P2.86, we find the Thévenin equivalent: Then, maximum power is obtained for a load resistance equal to the Thévenin resistance Vt 22 Pmax 10.8 W Rt P2.94 For maximum power conditions, we have RL Rt The power taken from the voltage source is Ps Vt 2 Rt RL Vt 2 2Rt Then, half of Vt appears across the load and the power delivered to the load is PL 0.5Vt 2 Rt Thus, the percentage of the power taken from the source that is delivered to the load is © 2014 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, electronic, mechanical, photocopying,52 recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 PL 100 % 50% Ps On the other hand, for RL 9Rt , we have η Ps PL η Vt 2 Rt RL Vt 2 10Rt 0.9Vt 2 9Rt PL 100 % 90% Ps Design for maximum power transfer is relatively inefficient Thus, systems in which power efficiency is important are almost never designed for maximum power transfer P2.95* To maximize the power to RL , we must maximize the voltage across it Thus, we need to have Rt The maximum power is Pmax P2.96 20 80 W The circuit is By the current division principle: iL I n Rt RL Rt The power delivered to the load is PL iL RL I n 2 Rt 2 RL RL Rt 2 Taking the derivative and setting it equal to zero, we have 2 dPL Rt Rt RL 2Rt RL Rt RL I n dR L Rt RL 4 © 2014 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, electronic, mechanical, photocopying,53 recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 which yields RL Rt The maximum power is PL max I n 2 Rt P2.97* First, we zero the current source and find the current due to the voltage source iv 30 15 A Then, we zero the voltage source and use the current-division principle to find the current due to the current source 10 2A 10 Finally, the total current is the sum of the contributions from each source i iv ic A ic P2.98* The circuits with only one source active at a time are: Req 3.75 Ω 15 is ,c 1 10 0.667 A 10 © 2014 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, electronic, mechanical, photocopying,54 recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 is ,v 10 V Req 2.667 A Then the total current due to both sources is is is ,v is ,c 3.333 A P2.99 Zero the 2-A source and use the current-division principle: i1,1A 20 0.5714 A 15 20 Then zero the A source and use the current-division principle: i1,2A 2 Finally, 0.2857 A 30 i1 i1,1A i1,2A 0.2857 A P2.100 The circuits with only one source active at a time are: © 2014 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, electronic, mechanical, photocopying,55 recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 i1.4A 3A 15 15 i1,2A 2 1.5 15 15 Finally, we add the components to find the current with both sources active i1 i1, 4A i1,2A 1.5 A P2.101 The circuit, assuming that v V is: i2 v 5 0.2 A v1 30i2 V i10 v 10 0.6 A i30 v 30 0.2 A is i2 i10 i30 A v s 12is v 6is 24 V We have established that for v s 24 V , we have v V Thus, for v s 12 V , we have: v2 12 0.5 V 24 P2.102 We start by assuming i2 A and work back through the circuit to determine the value of vs The results are shown on the circuit diagram © 2014 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, electronic, mechanical, photocopying,56 recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 However, the circuit actually has vs = 10 V, so the actual value ofi2 is 10 (1 A) 0.5 A 20 P2.103 We start by assuming i6 A and work back through the circuit to determine the value of Vs This results in Vs 30 V However, the circuit actually has Vs = 10 V, so the actual value ofi6 is 10 (1 A) 0.3333 A 30 P2.104 (a) With only the 2-A source activated, we have i2 and v 2i2 3 16 V (b) With only the 1-A source activated, we have i1 1 A and v 2i1 3 2 V (c) With both sources activated, we have © 2014 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, electronic, mechanical, photocopying,57 recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 i A and v 2i 3 V Notice that i i1 i2 Superposition does not apply because device A has a nonlinear relationship between v and i P2.105 From Equation 2.91, we have R2 kΩ R3 3419 3419 R1 kΩ R 100 kΩ (b) Rx R3 3419 34.19 kΩ R1 10 kΩ (a) Rx P2.106* (a) Rearranging Equation 2.91, we have R3 R1 10 Rx 5932 5932 R2 10 (b) The circuit is: The Thévenin resistance is Rt 1 7447 R3 R1 R2 Rx The Thévenin voltage is vt v s R3 R1 R3 vs Rx Rx R2 0.3939 mV Thus, the equivalent circuit is: © 2014 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, electronic, mechanical, photocopying,58 recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 idetector Vt 31.65 10 9 A Rt Rdetector Thus, the detector must be sensitive to very small currents if the bridge is to be accurately balanced P2.107 If R1 and R3 are too small, large currents are drawn from the source If the source were a battery, it would need to be replaced frequently Large power dissipation could occur, leading to heating of the components and inaccuracy due to changes in resistance values with temperature If R1 and R3 are too large, we would have very small detector current when the bridge is not balanced, and it would be difficult to balance the bridge accurately P2.108 With the source replaced by a short circuit and the detector removed, the Wheatstone bridge circuit becomes The Thévenin resistance seen looking back into the detector terminals is 1 Rt R3 R1 R2 Rx The Thévenin voltage is zero when the bridge is balanced P2.109 Using the voltage-division principle, the voltage at node a is © 2014 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, electronic, mechanical, photocopying,59 recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 R0 R R R Vs R0 R R0 R 2R0 Similarly at node b, we have R R v b Vs 2R0 v a Vs Then, the output voltage is v o v ab v a v b Vs R R0 Finally using Equation 2.92 to substitute for R , we have L v o Vs G L P2.110 Before strain is applied, the resistance is R0 L A After strain is applied, the length becomes L L L(1 L / L), and the cross sectional area becomes A /( (1 L / L) so the volume is constant Thus, the resistance becomes R L(1 L / L) R0 R R0 (1 ) R0 (1 L / L) R0 A /(1 L / L) R R0 (1 ) R0 (1 2L / L (L / L) ) R0 However, we have L / L so we can neglect the (L / L) term to a good approximation This results in R R0 (1 ) R0 (1 2L / L) R0 R / R0 G 2 L / L P2.111 In this case, the bridge would be balanced for any value of R and the output voltage vo would be zero regardless of the strain Practice Test T2.1 (a) 6, (b) 10, (c) 2, (d) 7, (e) 10 or 13 (perhaps 13 is the better answer), (f) or (perhaps is the better answer), (g) 11, (h) 3, (i) 8, (j) 15, (k) 17, (l) 14 © 2014 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, electronic, mechanical, photocopying,60 recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 T2.2 The equivalent resistance seen by the voltage source is: Req R1 16 / R2 / R3 / R4 is vs 6A Req Then, using the current division principle, we have G4 / 60 i4 is 61 A G2 G3 G4 / 48 / 16 / 60 T2.3 Writing KCL equations at each node gives v1 v1 v2 v1 v3 0 v2 v1 v2 2 10 v3 v3 v1 2 In standard form, we have: 0.95v 0.20v 0.50v 0.20v 0.30v 0.50v 1.50v 2 In matrix form, we have GV I 0.95 0.20 0.50 v 0.20 0.30 v 0.50 1.50 v 2 The MATLAB commands needed to obtain the column vector of the node voltages are G = [0.95 -0.20 -0.50; -0.20 0.30 0; -0.50 1.50] I = [0; 2; -2] V = G\I % As an alternative we could use V = inv(G)*I Actually, because the circuit contains only resistances and independent current sources, we could have used the short-cut method to obtain the G and I matrices T2.4 We can write the following equations: KVL mesh 1: R1i1 Vs R3 (i1 i3 ) R2 (i1 i2 ) KVL for the supermesh obtained by combining meshes and 3: © 2014 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, electronic, mechanical, photocopying, 61 recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 R4i2 R2 i2 i1 R3 (i3 i1 ) R5i3 KVL around the periphery of the circuit: R1i1 Vs R4i2 R5i3 Current source: i2 i3 I s A set of equations for solving the network must include the current source equation plus two of the mesh equations The three mesh equations are dependent and will not provide a solution by themselves T2.5 Under short-circuit conditions, the circuit becomes Thus, the short-circuit current is A flowing out of b and into a Zeroing the sources, we have Thus, the Thévenin resistance is Rt 24 / 40 /(30 30) and the Thévenin voltage is Vt I sc Rt 24 V The equivalent circuits are: Because the short-circuit current flows out of terminal b, we have oriented the voltage polarity positive toward b and pointed the current source reference toward b © 2014 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, electronic, mechanical, photocopying,62 recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 T2.6 With one source active at a time, we have Then, with both sources active, we have We see that the 5-V source produces 25% of the total current through the 5- resistance However, the power produced by the 5-V source with both sources active is zero Thus, the 5-V source produces 0% of the power delivered to the 5- resistance Strange, but true! Because power is a nonlinear function of current (i.e., P Ri ), the superposition principle does not apply to power © 2014 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, electronic, mechanical, photocopying,63 recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 ... In a similar fashion to the solution for Problem P2.13, we can write the following expression for the resistance seen by the 16-V source k Req / Req / The solutions to this equation are... 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... storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying,22 recording, or likewise For information regarding permission(s), write to: Rights
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