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电路基础(清华大学出版社)英文答案solution of fundamentals of electric circuits
Chapter 1, Problem 1 How many coulombs are represented by these amounts of electrons: (a) 6.482 × 1017 (b) 1.24 × 1018 (c) 2.46 × 1019 (d) 1.628 × 10 20Chapter 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 CChapter 1, Problem 2. Determine the current flowing through an element if the charge flow is given by (a) q(t ) = (3t + 8) mC (b) q(t ) = ( 8t 2 + 4t-2) C (c) q (t ) = 3e -t ? 5e ?2 t nC (d) q(t ) = 10 sin 120π t pC (e) q(t ) = 20e ?4 t cos 50t μC()Chapter 1, Solution 2 (a) (b) (c) (d) (e) i = dq/dt = 3 mA i = dq/dt = (16t + 4) A i = dq/dt = (-3e-t + 10e-2t) nA i=dq/dt = 1200π cos 120π t pA i =dq/dt = ? e ?4t (80 cos 50 t + 1000 sin 50 t ) μ 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 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 1, Problem 3. Find the charge q(t) flowing through a device if the current is: (a) i (t ) = 3A, q(0) = 1C (b) i ( t ) = ( 2t + 5) mA, q(0) = 0 (c) i ( t ) = 20 cos(10t + π / 6) μA, q(0) = 2 μ C (d) i (t ) = 10e ?30t sin 40tA, q(0) = 0Chapter 1, Solution 3 (a) q(t) = ∫ i(t)dt + q(0) = (3t + 1) C (b) q(t) = ∫ (2t + s) dt + q(v) = (t 2 + 5t) mC q(t) = ∫ 10e -30t sin 40t + q(0) = (c) q(t) = ∫ 20 cos (10t + π / 6 ) + q(0) = (2sin(10t + π / 6) + 1) μ C (d) 10e -30t ( ?30 sin 40 t - 40 cos t) 900 + 1600 = ? e - 30t (0.16cos40 t + 0.12 sin 40t) CChapter 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 CChapter 1, Problem 5. Determine the total charge transferred over the time interval of 0 ≤ t ≤ 10s when 1 i (t ) = t A. 2 Chapter 1, Solution 51 t 2 10 q = ∫ idt = ∫ tdt = = 25 C 2 4 0 0PROPRIETARY MATERIAL. ? 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.10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 msFigure 1.23Chapter 1, Solution 6 (a) At t = 1ms, i = (b) At t = 6ms, i =dq 80 = = 40 A dt 2dq = 0A dt dq 80 = = C20 A dt 4(c) At t = 10ms, i =PROPRIETARY MATERIAL. ? 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.Chapter 1, Problem 7. The charge flowing in a wire is plotted in Fig. 1.24. Sketch the corresponding current.Figure 1.24Chapter 1, Solution 7? 25A, dq ? i= = - 25A, dt ? ? 25A, ?0& t&2 2&t&6 6& t&8which is sketched below:PROPRIETARY MATERIAL. ? 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.Chapter 1, Problem 8. The current flowing past a point in a device is shown in Fig. 1.25. Calculate the total charge through the point.Figure 1.25Chapter 1, Solution 8q = ∫ idt = 10 × 1 + 10 × 1 = 15 μC 2PROPRIETARY MATERIAL. ? 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.Chapter 1, Problem 9. The current through an element is shown in Fig. 1.26. Determine the total charge that passed through the element at: (a) t = 1 s (b) t = 3 s (c) t = 5 sFigure 1.26Chapter 1, Solution 9 (a) q = ∫ idt = ∫ 10 dt = 10 C0 13 5 ×1? ? q = ∫ idt = 10 × 1 + ?10 ? ? + 5 ×1 0 (b) 2 ? ? = 15 + 7.5 + 5 = 22.5C(c) q = ∫ idt = 10 + 10 + 10 = 30 C05Chapter 1, Problem 10. A lightning bolt with 8 kA strikes an object for 15 μ s. How much charge is deposited on the object?Chapter 1, Solution 10 q = it = 8x103x15x10-6 = 120 mCPROPRIETARY MATERIAL. ? 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.Chapter 1, Problem 11. A rechargeable flashlight battery is capable of delivering 85 mA for about 12 h. How much charge can it release at that rate? If its terminals voltage is 1.2 V, how much energy can the battery deliver? Chapter 1, Solution 11 q= it = 85 x10-3 x 12 x 60 x 60 = 3,672 C E = pt = ivt = qv =
= 4406.4 JChapter 1, Problem 12. If the current flowing through an element is given by ? 3tA, 0 & t & 6s ? 18A, 6 & t & 10s ? i (t ) = ? ?- 12 A, 10 & t & 15s ? 0, t & 15s ? Plot the charge stored in the element over 0 & t & 20s. Chapter 1, Solution 12 For 0 & t & 6s, assuming q(0) = 0,q (t ) = idt + q (0 ) = 3tdt + 0 = 1.5t 2∫0t∫0tAt t=6, q(6) = 1.5(6)2 = 54 For 6 & t & 10s,q (t ) = idt + q (6 ) = 18 dt + 54 = 18 t ? 546 6∫t∫tAt t=10, q(10) = 180 C 54 = 126 For 10&t&15s,q (t ) =∫ idt + q(10) = ∫ (?12)dt + 126 = ?12t + 24610 10ttPROPRIETARY MATERIAL. ? 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.At t=15, q(15) = -12x15 + 246 = 66 For 15&t&20s,q (t ) = 0 dt + q (15) =6615∫tThus,? 1.5t 2 C, 0 & t & 6s ? ? 18 t ? 54 C, 6 & t & 10s q (t ) = ? ??12t + 246 C, 10 & t & 15s ? 66 C, 15 & t & 20s ?The plot of the charge is shown below.14012010080 q(t) 60 40 20 0 0 5 10 t 15 20PROPRIETARY MATERIAL. ? 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.Chapter 1, Problem 13. The charge entering the positive terminal of an element isq = 10 sin 4π t mCwhile the voltage across the element (plus to minus) isv = 2cos 4π t V(a) Find the power delivered to the element at t = 0.3 s (b) Calculate the energy delivered to the element between 0 and 0.6s.Chapter 1, Solution 13dq = 40π cos 4π t mA dt p = vi = 80π cos 2 4π t mW At t=0.3s, p = 80π cos 2 (4π x0.3) = 164.5 mW(a) i =(b) W = ∫ pdt = 80π ∫ cos 2 4π tdt = 40π ∫ [1 + cos8π t ]dt mJ0 00.60.6? 0.6 ? 1 W = 40π ?0.6 + sin 8π t ? = 78.34 mJ 0 ? 8π ?PROPRIETARY MATERIAL. ? 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.Chapter 1, Problem 14. The voltage v across a device and the current I through it are v (t ) = 5 cos 2t V, i (t ) = 10(1 ? e ?0.5t ) A Calculate: (a) the total charge in the device at t = 1 s (b) the power consumed by the device at t = 1 s. Chapter 1, Solution 14 (a)q = ∫ idt = ∫ 10(1 - e -0.5t )dt = 10(t + 2e -0.5t )1 1 0= 10(1 + 2e0-0.5? 2 ) = 2.131 C(b)p(t) = v(t)i(t) p(1) = 5cos2 ? 10(1- e-0.5) = (-2.081)(3.935) = -8.188 WChapter 1, Problem 15. The current entering the positive terminal of a device is i (t ) = 3e ?2 t A and the voltage across the device is v (t ) = 5 di / dt V . (a) Find the charge delivered to the device between t = 0 and t = 2 s. (b) Calculate the power absorbed. (c) Determine the energy absorbed in 3 s. Chapter 1, Solution 15 (a)2q = ∫ idt = ∫ 3e2-2t= ?1.5 e -4 ? 1 =1.4725 C(0)? 3 2t dt = e 2 0(b)5di = ?6e 2t ( 5) = ?30e -2t dt p = vi = ? 90 e ?4 t W v=3 -4t(c) w = ∫ pdt = -90 ∫ e0? 90 -4t dt = e = ? 22.5 J ?4 03PROPRIETARY MATERIAL. ? 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.Chapter 1, Problem 16.Figure 1.27 shows the current through and the voltage across a device. (a) Sketch the power delivered to the device for t &0. (b) Find the total energy absorbed by the device for the period of 0& t & 4s. i (mA) 600 v(V) 5 0 0 -524t(s)24t(s)Figure 1.27For Prob. 1.16.PROPRIETARY MATERIAL. ? 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.Chapter 1, Solution 16(a) ? 30t mA, 0 & t &2 i (t ) = ? ?120-30t mA, 2 & t&4 ?5 V, 0 & t &2 v(t ) = ? ? -5 V, 2 & t&4 ? 150t mW, 0 & t &2 p(t ) = ? ?-600+150t mW, 2 & t&4 which is sketched below. p(mW) 3001 -30024t (s)(b) From the graph of p,W = ∫ pdt = 0 J0 4PROPRIETARY MATERIAL. ? 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.Chapter 1, Problem 17.Figure 1.28 shows a circuit with five elements. Ifp1 = ?205 W, p2 = 60 W, p4 = 45 W, p5 = 30 W,calculate the power p3 received or delivered by element 3.Figure 1.28Chapter 1, Solution 17Σ p=0→ -205 + 60 + 45 + 30 + p3 = 0p3 = 205 C 135 = 70 W Thus element 3 receives 70 W.PROPRIETARY MATERIAL. ? 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.Chapter 1, Problem 18.Find the power absorbed by each of the elements in Fig. 1.29.Figure 1.29Chapter 1, Solution 18p1 = 30(-10) = -300 W p2 = 10(10) = 100 W p3 = 20(14) = 280 W p4 = 8(-4) = -32 W p5 = 12(-4) = -48 WPROPRIETARY MATERIAL. ? 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.Chapter 1, Problem 19.Find I in the network of Fig. 1.30. I 1A + + 3V 4A 9V C 9V C C+ C+6VFigure 1.30For Prob. 1.19.Chapter 1, Solution 19I = 4 C1 = 3 A Or using power conservation, 9x4 = 1x9 + 3I + 6I = 9 + 9I 4 = 1 + I or I = 3 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 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 1, Problem 20.Find V0 in the circuit of Fig. 1.31.Figure 1.31Chapter 1, Solution 20Since Σ p = 0 -30×6 + 6×12 + 3V0 + 28 + 28×2 - 3×10 = 0 72 + 84 + 3V0 = 210 or 3V0 = 54 V0 = 18 VChapter 1, Problem 21.A 60-W, incandescent bulb operates at 120 V. How many electrons and coulombs flow through the bulb in one day?Chapter 1, Solution 21 p 60 ?? → i= = = 0.5 A p = vi v 120 q = it = 0.5x24x60x60 = 43200 C N e = qx 6.24 x1018 = 2.696 x10 23 electronsPROPRIETARY MATERIAL. ? 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.Chapter 1, Problem 22.A lightning bolt strikes an airplane with 30 kA for 2 ms. How many coulombs of charge are deposited on the plane?Chapter 1, Solution 22 q = it = 30 x103 x 2 x10?3 = 60 C Chapter 1, Problem 23.A 1.8-kW electric heater takes 15 min to boil a quantity of water. If this is done once a day and power costs 10 cents per kWh, what is the cost of its operation for 30 days?Chapter 1, Solution 23W = pt = 1.8x(15/60) x30 kWh = 13.5kWh C = 10cents x13.5 = $1.35Chapter 1, Problem 24.A utility company charges 8.5 cents/kWh. If a consumer operates a 40-W light bulb continuously for one day, how much is the consumer charged?Chapter 1, Solution 24W = pt = 40 x24 Wh = 0.96 kWh C = 8.5 cents x0.96 = 8.16 centsChapter 1, Problem 25.A 1.2-kW toaster takes roughly 4 minutes to heat four slices of bread. Find the cost of operating the toaster once per day for 1 month (30 days). Assume energy costs 9 cents/kWh.Chapter 1, Solution 25 4 Cost = 1.2 kW × hr × 30 × 9 cents/kWh = 21.6 cents 60PROPRIETARY MATERIAL. ? 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.Chapter 1, Problem 26.A flashlight battery has a rating of 0.8 ampere-hours (Ah) and a lifetime of 10 hours. (a) How much current can it deliver? (b) How much power can it give if its terminal voltage is 6 V? (c) How much energy is stored in the battery in kWh?Chapter 1, Solution 260. 8A ? h = 80 mA 10h (b) p = vi = 6 × 0.08 = 0.48 W (c) w = pt = 0.48 × 10 Wh = 0.0048 kWh(a) i =Chapter 1, Problem 27.A constant current of 3 A for 4 hours is required to charge an automotive battery. If the terminal voltage is 10 + t/2 V, where t is in hours, (a) how much charge is transported as a result of the charging? (b) how much energy is expended? (c) how much does the charging cost? Assume electricity costs 9 cents/kWh.Chapter 1, Solution 27(a) Let T = 4h = 4 × 3600 q = ∫ idt = ∫ 3dt = 3T = 3 × 4 × 3600 = 43.2 kC0 TT T 0. 5t ? ? ( b) W = ∫ pdt = ∫ vidt = ∫ ( 3) ?10 + ?dt 0 0 3600 ? ?? 0. 25t 2 ? ? = 3?10t + ? 3600 ? 0 ? ? = 475.2 kJ ( c)4×3600= 3[40 × 3600 + 0. 25 × 16 × 3600]W = 475.2 kWs, (J = Ws) 475.2 Cost = kWh × 9 cent = 1.188 cents 3600PROPRIETARY MATERIAL. ? 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.Chapter 1, Problem 28.A 30-W incandescent lamp is connected to a 120-V source and is left burning continuously in an otherwise dark staircase. Determine: (a) the current through the lamp, (b) the cost of operating the light for one non-leap year if electricity costs 12 cents per kWh.Chapter 1, Solution 28(a) i =P 30 = = 0.25 A V 120( b) W = pt = 30 × 365 × 24 Wh = 262.8 kWh Cost = $0.12 × 262.8 = $31.54Chapter 1, Problem 29.An electric stove with four burners and an oven is used in preparing a meal as follows. Burner 1: 20 minutes Burner 3: 15 minutes Oven: 30 minutes Burner 2: 40 minutes Burner 4: 45 minutesIf each burner is rated at 1.2 kW and the oven at 1.8 kW, and electricity costs 12 cents per kWh, calculate the cost of electricity used in preparing the meal.Chapter 1, Solution 29(20 + 40 + 15 + 45) ? 30 ? hr + 1.8 kW? ? hr 60 ? 60 ? = 2.4 + 0.9 = 3.3 kWh Cost = 12 cents × 3.3 = 39.6 cents w = pt = 1. 2kWPROPRIETARY MATERIAL. ? 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.Chapter 1, Problem 30.Reliant Energy (the electric company in Houston, Texas) charges customers as follows: Monthly charge $6 First 250 kWh @ $0.02/kWh All additional kWh @ $0.07/kWh If a customer uses 1,218 kWh in one month, how much will Reliant Energy charge?Chapter 1, Solution 30Monthly charge = $6 First 250 kWh @ $0.02/kWh = $5 Remaining 968 kWh @ $0.07/kWh= $67.76 Total = $78.76Chapter 1, Problem 31.In a household, a 120-W PC is run for 4 hours/day, while a 60-W bulb runs for 8 hours/day. If the utility company charges $0.12/kWh, calculate how much the household pays per year on the PC and the bulb.Chapter 1, Solution 31Total energy consumed = 365(120x4 + 60x8) W Cost = $0.12x365x960/1000 = $42.05Chapter 1, Problem 32.A telephone wire has a current of 20 μ A flowing through it. How long does it take for a charge of 15 C to pass through the wire?Chapter 1, Solution 32i = 20 ?A q = 15 C t = q/i = 15/(20x10-6) = 750x103 hrsPROPRIETARY MATERIAL. ? 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.Chapter 1, Problem 33.A lightning bolt carried a current of 2 kA and lasted for 3 ms. How many coulombs of charge were contained in the lightning bolt?Chapter 1, Solution 33i= dq → q = ∫ idt = 2000 × 3 × 10 ? 3 = 6 C dtChapter 1, Problem 34.Figure 1.32 shows the power consumption of a certain household in one day. Calculate: (a) the total energy consumed in kWh, (b) the average power per hour.Figure 1.32Chapter 1, Solution 34(a)Energy = = 10 kWh∑ pt= 200 x 6 + 800 x 2 + 200 x 10 + 1200 x 4 + 200 x 2(b)Average power = 10,000/24 = 416.7 WPROPRIETARY MATERIAL. ? 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.Chapter 1, Problem 35.The graph in Fig. 1.33 represents the power drawn by an industrial plant between 8:00 and 8:30 A.M. Calculate the total energy in MWh consumed by the plant.Figure 1.33Chapter 1, Solution 35energy = (5x5 + 4x5 + 3x5 + 8x5 + 4x10)/60 = 2.333 MWhrChapter 1, Problem 36.A battery may be rated in ampere-hours (Ah). A lead-acid battery is rated at 160 Ah. (a) What is the maximum current it can supply for 40 h? (b) How many days will it last if it is discharged at 1 mA?Chapter 1, Solution 36(a)160A ? h =4A 40 160Ah 160, 000h ( b) t = = = 6,667 days 0.001A 24h / day i=PROPRIETARY MATERIAL. ? 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.Chapter 1, Problem 37.A 12-V battery requires a total charge of 40 ampere-hours during recharging. How many joules are supplied to the battery?Chapter 1, Solution 37W = pt = vit = 12x 40x 60x60 = 1.728 MJChapter 1, Problem 38.How much energy does a 10-hp motor deliver in 30 minutes? Assume that 1 horsepower = 746 W.Chapter 1, Solution 38P = 10 hp = 7460 W W = pt = 7460 × 30 × 60 J = 13.43 × 106 JChapter 1, Problem 39.A 600-W TV receiver is turned on for 4 hours with nobody watching it. If electricity costs 10 cents/kWh, how much money is wasted?Chapter 1, Solution 39W = pt = 600x4 = 2.4 kWh C = 10cents x2.4 = 24 centsPROPRIETARY MATERIAL. ? 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.Chapter 2, Problem 1. The voltage across a 5-kΩ resistor is 16 V. Find the current through the resistor. Chapter 2, Solution 1 v = iR i = 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 2 p = v2/R → R = v2/p = 14400/60 = 240 ohmsChapter 2, Problem 3. A bar of silicon is 4 cm long with a circular cross section. If the resistance of the bar is 240 Ω at room temperature, what is the cross-sectional radius of the bar? Chapter 2, Solution 3 For silicon, ρ = 6.4 x102 Ω-m. A = π r 2 . Hence, R=ρLA=ρL π r2?? →r2 =ρ L 6.4 x102 x 4 x10?2 = = 0.033953 πR π x 240r = 0.1843 m Chapter 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) (b) i = 3/100 = 30 mA i = 3/150 = 20 mAPROPRIETARY MATERIAL. ? 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.Chapter 2, Problem 5. For the network graph in Fig. 2.69, find the number of nodes, branches, and loops. Chapter 2, Solution 5 n = 9; l = 7; b = n + l C 1 = 15Chapter 2, Problem 6. In the network graph shown in Fig. 2.70, determine the number of branches and nodes. Chapter 2, Solution 6 n = 12; l = 8; b = n + l C1 = 19Chapter 2, Problem 7. Determine the number of branches and nodes in the circuit of Fig. 2.71. 1Ω 4Ω12 V+ _8Ω5Ω2AFigure 2.71 For Prob. 2.7. Chapter 2, Solution 7 6 branches and 4 nodes.PROPRIETARY MATERIAL. ? 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.Chapter 2, Problem 8. Use KCL to obtain currents i1, i2, and i3 in the circuit shown in Fig. 2.72. Chapter 2, Solution 8 12 A A 8A I2 C At node a, At node c, At node d, Chapter 2, Problem 9. Find i1 , i 2 , and i3 in Fig. 2.73. 2A 10 A A i1 C 8A i2 12 A B 14 A i3 8 = 12 + i1 9 = 8 + i2 9 = 12 + i3 9 AD i1 = - 4A i2 = 1A i3 = -3A B CHAPTER 1 -I1 I3 12 A4A Figure 2.73 For Prob. 2.9. Chapter 2, Solution 9 At A, At B, At C, 2 + 12 = i1 12 = i2 + 14 14 = 4 + i3 ?? → ?? → ?? → i1 = 14 A i2 = ?2 A i3 = 10 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 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 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 above Chapter 2, Solution 10 4A 1 I1 2 I2 -2A 33A At node 1, At node 3, Chapter 2, Problem 11. In the circuit of Fig. 2.75, calculate V1 and V2. + 1V C + 2V C 4 + 3 = i1 3 + i2 = -2 i1 = 7A i2 = -5A+ V1 _+ 5V _+ V2 _Figure 2.75 For Prob. 2.11. Chapter 2, Solution 11 ?V1 + 1 + 5 = 0 ?5 + 2 + V2 = 0 ?? → ?? → V1 = 6 V V2 = 3 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 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 12. In the circuit in Fig. 2.76, obtain v1, v2, and v3.Chapter 2, Solution 12 + 15V -LOOP C 25V + 20V For loop 1, For loop 2, For loop 3, + 10V + V1 + V2 + V3 -LOOPLOOP-20 -25 +10 + v1 = 0 -10 +15 -v2 = 0 -V1 + V2 + V3 = 0v1 = 35v v2 = 5v v3 = 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 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 I4.2AI27AI4I13AI34AFigure 2.77 Chapter 2, Solution 13 2A1 I1I27A 2 3A 3I44 4A I3At node 2,3 + 7 + I2 = 0 ?→ ? ?→ ? ?→ ? ?→ ? I 2 = ?10 A I 1 = 2 ? I 2 = 12 A I 4 = 2 ? 4 = ?2 A I3 = 7 ? 2 = 5 AAt node 1,I1 + I 2 = 2At node 4,2 = I4 + 4At node 3,7 + I4 = I3Hence,I 1 = 12 A, I 2 = ?10 A, I 3 = 5 A, I 4 = ?2 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 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 14. Given the circuit in Fig. 2.78, use KVL to find the branch voltages V1 to V4.+ 3V C C 4V + + V3 C + V1 C + V4 C C V2 + + 2V C + 5V CFigure 2.78 Chapter 2, Solution 14+ 3V 4V + For mesh 1,?V4 + 2 + 5 = 0 ?→ ? ?→ ? ?→ ? ?→ ?+ I3 + I2 V3 V1 + I4 2V -+ V2+ - V4 I1+ 5V -V4 = 7V V3 = ?4 ? 7 = ?11V V1 = V3 + 3 = ?8V V2 = ?V1 ? 2 = 6VFor mesh 2,+4 + V3 + V4 = 0For mesh 3,?3 + V1 ? V3 = 0For mesh 4,?V1 ? V2 ? 2 = 0Thus,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 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 2, Problem 15. Calculate v and ix in the circuit of Fig. 2.79. 12 Ω + v + _ C + 2V _ ix + _ 3 ix + 8V C12 VFigure 2.79 For Prob. 2.15. Chapter 2, Solution 15 For loop 1, C12 + v +2 = 0, v = 10 V For loop 2, C2 + 8 + 3ix =0, ix = Chapter 2, Problem 16. Determine Vo in the circuit in Fig. 2.80. 6Ω + 9V + _ Vo _ + _ 3V 2Ω C2 AFigure 2.80 For Prob. 2.16. Chapter 2, Solution 16 Apply KVL, -9 + (6+2)I + 3 = 0, 8I = 9-3=6 , Also, -9 + 6I + Vo = 0 Vo = 9- 6I = 4.5 V I = 6/8PROPRIETARY MATERIAL. ? 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.Chapter 2, Problem 17. Obtain v1 through v3 in the circuit in Fig. 2.78. Chapter 2, Solution 17 Applying KVL around the entire outside loop we get, C24 + v1 + 10 + 12 = 0 or v1 = 2V Applying KVL around the loop containing v2, the 10-volt source, and the 12-volt source we get, v2 + 10 + 12 = 0 or v2 = C22V Applying KVL around the loop containing v3 and the 10-volt source we get, Cv3 + 10 = 0 or v3 = 10VChapter 2, Problem 18. Find I and Vab in the circuit of Fig. 2.79.Chapter 2, Solution 18 APPLYING KVL, -30 -10 +8 + I(3+5) = 0 8I = 32 I = 4A Vab = 28V-Vab + 5I + 8 = 0PROPRIETARY MATERIAL. ? 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.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 19 APPLYING KVL AROUND THE LOOP, WE OBTAIN -12 + 10 - (-8) + 3i = 0 Power dissipated by the resistor: p 3Ω = i2R = 4(3) = 12W Power supplied by the sources: p12V = 12 ((C2)) = C24W p10V = 10 (C(C2)) = 20W p8V = (C8)(C2) = 16W i = C2AChapter 2, Problem 20. Determine io in the circuit of Fig. 2.81. Chapter 2, Solution 20 APPLYING KVL AROUND THE LOOP, -36 + 4i0 + 5i0 = 0 i0 = 4APROPRIETARY MATERIAL. ? 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.Chapter 2, Problem 21. Find Vx in the circuit of Fig. 2.85. 1Ω 2 Vx + + _ 2Ω C + 5Ω Vx _15 VFigure 2.85 For Prob. 2.21.Chapter 2, Solution 21 Applying KVL, -15 + (1+5+2)I + 2 Vx = 0 But Vx = 5I, -15 +8I + 10I =0, I = 5/6 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 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 22 4Ω + V0 6Ω 10A 2V0At the node, KCL requires thatv0 + 10 + 2v 0 = 0 4v0 = C4.444VThe current through the controlled source is i = 2V0 = -8.888A and the voltage across it is v = (6 + 4) i0 (where i0 = v0/4) = 10 Hence, p2 vi = (-8.888)(-11.111) = 98.75 Wv0 = ?11.111 4PROPRIETARY MATERIAL. ? 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.Chapter 2, Problem 23. In the circuit shown in Fig. 2.87, determine vx and the power absorbed by the 12Ω resistor.1Ω +v C x 6A 2Ω 3Ω 6Ω 4Ω 8Ω 12 Ω 1.2 ΩFigure 2.87 Chapter 2, Solution 23 8//12 = 4.8, 3//6 = 2, (4 + 2)//(1.2 + 4.8) = 6//6 = 3 The circuit is reduced to that shown below. ix + 6A 1Ω vx -2Ω3ΩApplying current division,ix = 2 (6 A) = 2 A, 2 +1+ 3 v x = 1i x = 2VThe current through the 1.2- Ω resistor is 0.5ix = 1A. The voltage across the 12- Ω resistor is 1 x 4.8 = 4.8 V. Hence the power isp= v 2 4.8 2 = = 1.92W R 12PROPRIETARY MATERIAL. ? 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.Chapter 2, Problem 24. For the circuit in Fig. 2.86, find Vo / Vs in terms of α, R1, R2, R3, and R4. If R1 = R2 = R3 = R4, what value of α will produce | Vo / Vs | = 10? Chapter 2, Solution 24 (a) I0 =Vs R1 + R2V0 = ?α I0 (R3 R4 ) = ?R 3R 4 αVs ? R1 + R 2 R 3 + R 4? αR3 R4 V0 = Vs (R1 + R2 )(R3 + R4 )(b)If R1 = R2 = R3 = R4 = R,V0 α R α = ? = = 10 VS 2R 2 4Chapter 2, Problem 25.α = 40For the network in Fig. 2.88, find the current, voltage, and power associated with the 20kΩ resistor. Chapter 2, Solution 25 V0 = 5 x 10-3 x 10 x 103 = 50V Using current division, I20 =5 (0.01x50) = 0.1 A 5 + 20V20 = 20 x 0.1 kV = 2 kV p20 = I20 V20 = 0.2 kWPROPRIETARY MATERIAL. ? 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.Chapter 2, Problem 26. For the circuit in Fig. 2.90, io =2 A. Calculate ix and the total power dissipated by the circuit. ix io 2Ω 4Ω 8Ω 16 ΩFigure 2.90 For Prob. 2.26. Chapter 2, Solution 26 If i16= io = 2A, then v = 16x2 = 32 Vi8 = v =4A, 8 i4 = v = 8 A, 4 i2 = v = 16 2ix = i2 + i4 + i8 + i16 = 16 + 8 + 4 + 2 = 30 AP = ∑ i 2 R = 162 x 2 + 82 x 4 + 42 x8 + 22 x16 = 960 W or P = ix v = 30 x32 = 960 WChapter 2, Problem 27. Calculate Vo in the circuit of Fig. 2.91.4Ω + Vo -16 V+ _6ΩFigure 2.91 For Prob. 2.27. Chapter 2, Solution 27 Using voltage division, 4 Vo = (16V) = 6.4 V 4 + 16PROPRIETARY MATERIAL. ? 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.Chapter 2, Problem 28. Find v1, v2, and v3 in the circuit in Fig. 2.91. Chapter 2, Solution 28 We first combine the two resistors in parallel15 10 = 6 ΩWe now apply voltage division, v1 =14 ( 40) = 28 V 14 + 6 6 ( 40) = 12 V 14 + 6v2 = v3 = Hence,v1 = 28 V, v2 = 12 V, vs = 12 VChapter 2, Problem 29. All resistors in Fig. 2.93 are 1 Ω each. Find Req.ReqFigure 2.93 For Prob. 2.29. Chapter 2, Solution 29 Req = 1 + 1//(1 + 1//2) = 1 + 1//(1+ 2/3) =1+ 1//5/3 = 1.625 ΩPROPRIETARY MATERIAL. ? 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.Chapter 2, Problem 30. Find Req for the circuit in Fig. 2.94.6Ω6ΩReq2Ω2ΩFigure 2.94 For Prob. 2.30.Chapter 2, Solution 30 We start by combining the 6-ohm resistor with the 2-ohm one. We then end up with an 8-ohm resistor in parallel with a 2-ohm resistor. (2x8)/(2+8) = 1.6 ? This is in series with the 6-ohm resistor which gives us, Req = 6+1.6 = 7.6 ?.PROPRIETARY MATERIAL. ? 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.Chapter 2, Problem 31. For the circuit in Fig. 2.95, determine i1 to i5. 3Ω i1 i3i2 40 V + _ 4Ω 1Ω i4 2Ωi5Figure 2.95 For Prob. 2.31.Chapter 2, Solution 31Req = 3 + 2 // 4 //1 = 3 + i1 = 40 = 11.2 A 3.5714 i2 = v1 = 1.6 A 4 i3 = i4 + i5 = 9.6 A 1 = 3. + 1/ 4 + 1v1 = 0.5714 xi1 = 6.4V, i4 =v1 v = 6.4 A, i5 = 1 = 3.2 A, uuuuuu 1 2PROPRIETARY MATERIAL. ? 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.Chapter 2, Problem 32. Find i1 through i4 in the circuit in Fig. 2.96.Chapter 2, Solution 32 We first combine resistors in parallel.20 30 = 10 40 =20 x30 = 12 Ω 50 10 x 40 = 8Ω 50Using current division principle, 8 12 i1 + i 2 = (20) = 8A, i 3 + i 4 = ( 20) = 12A 8 + 12 2020 (8) = 3.2 A 50i1 =i2 =30 (8) = 4.8 A 50 10 (12) = 2.4A 50 40 (12) = 9.6 A 50i3 =i4 =PROPRIETARY MATERIAL. ? 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.Chapter 2, Problem 33. Obtain v and i in the circuit in Fig. 2.97.Chapter 2, Solution 33 Combining the conductance leads to the equivalent circuit below i + v 4S + v i9A1S4S9A1S2S6S 3S =6x3 = 2S and 2S + 2S = 4S 9Using current division,i= 1 1 1+ 2 (9) = 6 A, v = 3(1) = 3 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 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 34. Using series/parallel resistance combination, find the equivalent resistance seen by the source in the circuit of Fig. 2.98. Find the overall dissipated power. 20 Ω 8Ω 10Ω12 V+ _40 Ω 12 Ω40 Ω20 Ω10 Ω Figure 2.98 For Prob. 2.34.Chapter 2, Solution 34 40//(10 + 20 + 10)= 20 Ω, 40//(8+12 + 20) = 20 ΩReq = 20 + 20 = 40 ΩV I= = 12 / 40, Req122 P = VI = = 3.6 W 40PROPRIETARY MATERIAL. ? 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.Chapter 2, Problem 35. Calculate Vo and Io in the circuit of Fig. 2.99. Chapter 2, Solution 35 i 70 Ω++ V1 i1-30 Ω I0 + i2 V0 5Ω b50V-a 20 ΩCombining the versions in parallel,70 30 =70 x 30 = 21Ω , 10020 5 =20x 5 =4 Ω 25i=50 =2 A 21 + 4vi = 21i = 42 V, v0 = 4i = 8 V v v i1 = 1 = 0.6 A, i2 = 2 = 0.4 A 70 20 At node a, KCL must be satisfied i1 = i2 + I0 0.6 = 0.4 + I0 I0 = 0.2 AHence v0 = 8 V and I0 = 0.2APROPRIETARY MATERIAL. ? 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.Chapter 2, Problem 36. Find i and Vo in the circuit of Fig. 2.100. I 10 Ω 24 Ω 50Ω25 Ω 15 V + _ 60 Ω 20 Ω 20 Ω 30 Ω + Vo _Figure 2.100 For Prob. 2.36.Chapter 2, Solution 36 20//(30+50) = 16, 24 + 16 = 40, 60//20 = 15 Req = 10 + (15 + 25) // 40 = 10 + 20 = 30i=vs 15 = = 0.5 A Req 30If i1 is the current through the 24-Ω resistor and io is the current through the 50-Ω resistor, using current division gives 40 20 i 1= i = 0.25 A, i o = i1 = 0.05 A 40 + 40 20 + 80 vo = 30io = 30 x0.05 = 1.5 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 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 37. Find R for the circuit in Fig. 2.101. R + 10 V C 20 V + _ C + 30 10 ΩFigure 2.101 For Prob. 2.37.Chapter 2, Solution 37 Applying KVL, -20 + 10 + 10I C 30 = 0, I = 410 = RI ?? R = → 10 = 2.5 Ω 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 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 38. Find Req and io in the circuit of Fig. 2.102. 60 Ω 12 Ωio5Ω6Ω 80 Ω40 V+ _15 Ω20 ΩReq Figure 2.102 For Prob. 2.38 Chapter 2, Solution 38 20//80 = 80x20/100 = 16, 6//12 = 6x12/18 = 4 The circuit is reduced to that shown below. 5Ω 4Ω 60 Ω 15 Ω 16 ΩReq (4 + 16)//60 = 20x60/80 = 15 Req = 15 //15 + 5 = 12.5 Ωio =40 = 3.2 A ReqPROPRIETARY MATERIAL. ? 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.Chapter 2, Problem 39. Evaluate Req for each of the circuits shown in Fig. 2.103. 6 kΩ 2 kΩ 1 kΩ 4 kΩ 12 kΩ2 kΩ1 kΩ12 kΩ(a) Figure 2.103 For Prob. 2.39. Chapter 2, Solution 39(b)(a) We note that the top 2k-ohm resistor is actually in parallel with the first 1k-ohm resistor. This can be replaced (2/3)k-ohm resistor. This is now in series with the second 2k-ohm resistor which produces a 2.667k-ohm resistor which is now in parallel with the second 1k-ohm resistor. This now leads to, Req = [(1x2.667)/3.667]k = 727.3 ?. (b) We note that the two 12k-ohm resistors are in parallel producing a 6k-ohm resistor. This is in series with the 6k-ohm resistor which results in a 12k-ohm resistor which is in parallel with the 4k-ohm resistor producing, Req = [(4x12)/16]k = 3 k?. Chapter 2, Problem 40. For the ladder network in Fig. 2.104, find I and Req. Chapter 2, Solution 40 REQ = 3 + 4 ( 2 + 6 3) = 3 + 2 = 5Ω I= 10 10 = = 2A Re q 5PROPRIETARY MATERIAL. ? 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.Chapter 2, Problem 41. If Req = 50 Ω in the circuit in Fig. 2.105, find R. Chapter 2, Solution 41 Let R0 = combination of three 12Ω resistors in parallel1 1 1 1 = + + R o 12 12 12Ro = 4R eq = 30 + 60 (10 + R 0 + R ) = 30 + 60 (14 + R )50 = 30 + 60(14 + R ) 74 + R74 + R = 42 + 3Ror R = 16 ΩChapter 2, Problem 42. Reduce each of the circuits in Fig. 2.106 to a single resistor at terminals a-b.Chapter 2, Solution 42 (a) Rab = 5 (8 + 20 30) = 5 (8 + 12) = 5x 20 = 4Ω 25(b)Rab = 2 + 4 (5 + 3) 8 + 5 10 4 = 2 + 4 4 + 5 2.857 = 2 + 2 + 1.8181 = 5.818 ΩPROPRIETARY MATERIAL. ? 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.Chapter 2, Problem 43 Calculate the equivalent resistance Rab at terminals a-b for each of the circuits in Fig.2.107.Chapter 2, Solution 43 (a) Rab = 5 20 + 10 40 = 5x 20 400 + = 4 + 8 = 12 Ω 25 50?1(b)1 1 ? ? 1 60 20 30 = ? + + ? ? 60 20 30 ?=60 = 10Ω 6Rab = 80 (10 + 10) =80 + 20 = 16 Ω 100PROPRIETARY MATERIAL. ? 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.Chapter 2, Problem 44. For each of the circuits in Fig. 2.108, obtain the equivalent resistance at terminals a-b.20 Ω 20 Ωa10 Ω b (a) 15 Ω5Ω20 Ω11 Ω a10 Ω 30 Ω20 Ω 40 Ω 21 Ω 50 Ω 30 Ωb (b)Figure 2.108PROPRIETARY MATERIAL. ? 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.Chapter 2, Solution 44 (a) Convert T to Y and obtainR1 = 20 x 20 + 20 x10 + 10 x 20 800 = = 80 Ω 10 10 800 R2 = = 40 Ω = R3 20The circuit becomes that shown below.a R2 b R1//0 = 0,Rab = R2 / /(0 + 4.444) = 40 / /4.444 = 4ΩR1 R35ΩR3//5 = 40//5 = 4.444 Ω(b) 30//(20+50) = 30//70 = 21 Ω Convert the T to Y and obtainR1 =20 x10 + 10 x 40 + 40 x20 1400 = = 35Ω 40 40
R2 = = 70 Ω , R3 = = 140 Ω 20 10 The circuit is reduced to that shown below. 15Ω11 Ω R2 30 ΩR1 R3 21 Ω 21 ΩPROPRIETARY MATERIAL. ? 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.Combining the resistors in parallel R1//15 =35//15=10.5, 30//R2=30//70 = 21 leads to the circuit below. 11 Ω 10.5 Ω 21 Ω 140 Ω 21 Ω 21 ΩCoverting the T to Y leads to the circuit below. 11 Ω 10.5 Ω R4 R5 R621 ΩR4 =21x140 + 140 x 21 + 21x 21 6321 = = 301Ω = R6 21 21 6321 = 45.15 140R5 =10.5//301 = 10.15, 301//21 = 19.63 R5//(10.15 +19.63) = 45.15//29.78 = 17.94Rab = 11 + 17 .94 = 28.94ΩPROPRIETARY MATERIAL. ? 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.Chapter 2, Problem 45. Find the equivalent resistance at terminals a-b of each circuit in Fig. 2.109.10 Ω 40 Ω 20 Ω a 30 Ω 50 Ω b (a) 5Ω30 Ω12 Ω 5Ω 20 Ω25 Ω 15 Ω (b) 10 Ω60 ΩFigure 2.109 Chapter 2, Solution 45 (a) 10//40 = 8, 20//30 = 12, 8//12 = 4.8 Rab = 5 + 50 + 4.8 = 59.8 Ω (b) 12 and 60 ohm resistors are in parallel. Hence, 12//60 = 10 ohm. This 10 ohm and 20 ohm are in series to give 30 ohm. This is in parallel with 30 ohm to give 30//30 = 15 ohm. And 25//(15+10) = 12.5. Thus Rab = 5 + 12.8 + 15 = 32.5ΩPROPRIETARY MATERIAL. ? 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.Chapter 2, Problem 46. Find I in the circuit of Fig. 2.110. 20 Ω I 4Ω 5Ω 48 V + _ 15 Ω 15 Ω 15 Ω 5Ω24 Ω8Ω Figure 2.110 For Prob. 2.46. Chapter 2, Solution 461 Req = 4 + 5 // 20 + x15 + 5 + 24 // 8 = 4 + 4 + 5 + 5 + 6 = 24 3 I = 48/24 = 2 AChapter 2, Problem 47. Find the equivalent resistance Rab in the circuit of Fig. 2.111. Chapter 2, Solution 47 5x 20 5 20 = = 4Ω 25 6 3=6x3 = 2Ω 910 Ω A 4Ω8Ω B 2ΩRab = 10 + 4 + 2 + 8 = 24 ΩPROPRIETARY MATERIAL. ? 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.Chapter 2, Problem 48. Convert the circuits in Fig. 2.112 from Y to Δ. Chapter 2, Solution 48 (A)R 1 R 2 + R 2 R 3 + R 3 R 1 100 + 100 + 100 = = 30 R3 10 Ra = Rb = Rc = 30 ΩRA =30 x 20 + 30 x 50 + 20 x 50 3100 = = 103.3Ω 30 30
Rb = = 155Ω, R c = = 62Ω 20 50 Ra = 103.3 Ω, Rb = 155 Ω, Rc = 62 Ω Ra =(b)Chapter 2, Problem 49. Transform the circuits in Fig. 2.113 from Δ to Y. Chapter 2, Solution 49 (A)RaRc 12 *12 = = 4Ω Ra + Rb + Rc 36 R1 = R2 = R3 = 4 ΩR1 =(b)60 x 30 = 18Ω 60 + 30 + 10 60 x10 R2 = = 6Ω 100 30 x10 R3 = = 3Ω 100 R1 = 18Ω, R2 = 6Ω, R3 = 3Ω R1 =PROPRIETARY MATERIAL. ? 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.Chapter 2, Problem 50. What value of R in the circuit of Fig. 2.114 would cause the current source to deliver 800 mW to the resistors.Chapter 2, Solution 50 Using R Δ = 3RY = 3R, we obtain the equivalent circuit shown below:30MA3R3R 3RR R 30MA 3R 3R/23RxR 3 = R 4R 4 3R (3RxR ) /(4R ) = 3 /(4R )3R R =3 ? 3 ?3 3R ? R + R ? = 3R R = 4 ? 2 ?4 P = I2R R = 889 Ω3 3Rx R 2 3 3R + R = R 2 800 x 10-3 = (30 x 10-3)2 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 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 51. Obtain the equivalent resistance at the terminals a-b for each of the circuits in Fig. 2.115. Chapter 2, Solution 51(a)30 30 = 15Ω and 30 20 = 30x 20 /(50) = 12ΩRab = 15 (12 + 12) = 15x 24 /(39) = 9.231 ΩA 30 Ω B 30 Ω 30 Ω 30 Ω 20 ΩA 12 Ω 15 Ω 20 Ω B (b) Converting the T-subnetwork into its equivalent Δ network gives Ra'b' = 10x20 + 20x5 + 5x10/(5) = 350/(5) = 70 Ω Rb'c' = 350/(10) = 35Ω, Ra'c' = 350/(20) = 17.5 Ω Also 12 Ω30 70 = 30x 70 /(100) = 21Ω and 35/(15) = 35x15/(50) = 10.5Rab = 25 + 17.5 (21 + 10.5) = 25 + 17.5 31.5 Rab = 36.25 Ω30 Ω 25 Ω 17.5 Ω30 Ω 25 Ω 10 Ω 5Ω 20 Ω 15 ΩAAA’ 70 Ω35 ΩB’15 ΩBBC’C’PROPRIETARY MATERIAL. ? 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.Chapter 2, Problem 52. For the circuit shown in Fig. 2.116, find the equivalent resistance. All resistors are 1Ω. .Req Figure 2.116 For Prob. 2.52.PROPRIETARY MATERIAL. ? 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.Chapter 2, Solution 52 Converting the wye-subnetwork to delta-subnetwork, we obtain the circuit below. 1Ω 3Ω 3Ω1Ω 1Ω 1Ω3Ω1Ω 1Ω 2Ω3//1 = 3x1/4 = 0.75, 2//1 =2x1/3 = 0.6667. Combining these resistances leads to the circuit below. 1Ω 0.75 Ω 3Ω0.75 Ω 1Ω1Ω0.6667 ΩWe now convert the wye-subnetwork to the delta-subnetwork. 0.75 x1 + 0.75 x1 + 0.752 Ra = = 2..0625 Rb = Rc = = 2.75 0.75 This leads to the circuit below. 1Ω 2. 1Ω 2.75 ?Ω 3Ω2 3 x 2.065 2.75 x 2 / 3 = + = 1..0625 2 / 3 + 2.75 2.75 x1.7607 Req = 1 + 1 + 2.75 //1.7607 = 2 + = 3.0734 Ω 2.75 + 1.7607R = 3 // 2.0625 + 2.75 //PROPRIETARY MATERIAL. ? 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.Chapter 2, Problem 53. Obtain the equivalent resistance Rab in each of the circuits of Fig. 2.117. In (b), all resistors have a value of 30 Ω. Chapter 2, Solution 53 (a) Converting one Δ to T yields the equivalent circuit below:30 Ω 4Ω 60 Ω 5ΩA B20 Ω20 Ω 80 Ω40 x10 10 x 50 40 x 50 = 4Ω, R b 'n = = 5Ω, R c 'n = = 20Ω 40 + 10 + 50 100 100 Rab = 20 + 80 + 20 + (30 + 4) (60 + 5) = 120 + 34 65Ra'n =Rab = 142.32 Ω (c) We combine the resistor in series and in parallel.30 (30 + 30) = 30 x 60 = 20Ω 90We convert the balanced Δ s to Ts as shown below: A30 Ω 30 Ω 30 Ω 20 Ω 30 Ω 30 Ω 10 Ω 10 Ω 10 Ω 20 Ω 10 Ω 30 ΩA10 Ω 10 ΩBB Rab = 10 + (10 + 10) (10 + 20 + 10) + 10 = 20 + 20 40 Rab = 33.33 ΩPROPRIETARY MATERIAL. ? 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.Chapter 2, Problem 54. Consider the circuit in Fig. 2.118. Find the equivalent resistance at terminals: (a) a-b, (b) c-d.a 50 Ω 15 Ω 100 Ω b 150 Ω 60 Ω 100 Ω d cFigure 2.118Chapter 2, Solution 54 (a) Rab = 50 + 100 / /(150 + 100 + 150 ) = 50 + 100 / /400 = 130 Ω (b) Rab = 60 + 100 / /(150 + 100 + 150 ) = 60 + 100 / /400 = 140 ΩPROPRIETARY MATERIAL. ? 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.Chapter 2, Problem 55. Calculate Io in the circuit of Fig. 2.119.Chapter 2, Solution 55 We convert the T to Δ .I0 24 V+A20 Ω 40 Ω 10 Ω 20 Ω 60 ΩI0 24 V50 ΩA140 Ω 60 Ω 70 Ω-+-35 Ω 70 ΩB RE B RE R R + R 2 R 3 + R 3 R 1 20x 40 + 40x10 + 10x 20 1400 = = = 35Ω Rab = 1 2 R3 40 40 Rac = 1400/(10) = 140Ω, Rbc = 1400/(20) = 70Ω 70 70 = 35 and 140 160 = 140x60/(200) = 42 Req = 35 (35 + 42) = 24.0625Ω I0 = 24/(Rab) = 997.4mAPROPRIETARY MATERIAL. ? 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.Chapter 2, Problem 56. Determine V in the circuit of Fig. 1.120. Chapter 2, Solution 56 We need to find Req and apply voltage division. We first tranform the Y network to Δ .30 Ω 16 Ω + 100 V 35 Ω 15 Ω 12 Ω 10 Ω 20 Ω + 100 V 16 Ω 35 Ω 30 ΩA37.5 Ω 30 Ω 45 ΩB20 ΩRE Rab =REC15x10 + 10 x12 + 12 x15 450 = = 37.5Ω 12 12 Rac = 450/(10) = 45Ω, Rbc = 450/(15) = 30ΩCombining the resistors in parallel, 30||20 = (600/50) = 12 Ω, 37.5||30 = (37.5x30/67.5) = 16.667 Ω 35||45 = (35x45/80) = 19.688 Ω Req = 19.688||(12 + 16.667) = 11.672Ω By voltage division, v =11.672 100 = 42.18 V 11.672 + 16PROPRIETARY MATERIAL. ? 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.Chapter 2, Problem 57. Find Req and I in the circuit of Fig. 2.121.Chapter 2, Solution 574 ΩA2Ω 27 Ω 1Ω18 Ω36 Ω 7Ω 28 ΩB D10 ΩC E14 Ω6 x12 + 12 x8 + 8x 6 216 Rab = = = 18 Ω 12 12 Rac = 216/(8) = 27Ω, Rbc = 36 Ω 4 x 2 + 2 x8 + 8x 4 56 Rde = = 7Ω 8 8 Ref = 56/(4) = 14Ω, Rdf = 56/(2) = 28 ΩFPROPRIETARY MATERIAL. ? 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.Combining resistors in parallel,10 28 = 280 36 x 7 = 7.368Ω, 36 7 = = 5.868Ω 38 43 27 x 3 27 3 = = 2 .7 Ω 304Ω4Ω18 Ω 5.868 Ω2.7 Ω 14 Ω1.829 Ω 3.977 Ω 0.5964 Ω14 Ω7.568 Ω7.568 ΩR an = R bn R cnR eq18x 2.7 18x 2.7 = = 1.829 Ω 18 + 2.7 + 5.867 26.567 18x 5.868 = = 3.977 Ω 26.567 5.868x 2.7 = = 0.5904 Ω 26.567 = 4 + 1.829 + (3.977 + 7.368) (0.5964 + 14)= 5.829 + 11.346 14.5964 = 12.21 Ω i = 20/(Req) = 1.64 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 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 58. The lightbulb in Fig. 2.122 is rated 120 V, 0.75 A. Calculate Vs to make the lightbulb operate at the rated conditions. Chapter 2, Solution 58 The resistor of the bulb is 120/(0.75) = 160Ω40 Ω 2.25 A 1.5 A+ 90 V - 0.75 A VS+-160 Ω+Proble 20 V80 ΩOnce the 160Ω and 80Ω resistors are in parallel, they have the same voltage 120V. Hence the current through the 40Ω resistor is 40(0.75 + 1.5) = 2.25 x 40 = 90 Thus vs = 90 + 120 = 210 V Chapter 2, Problem 59. Three lightbulbs are connected in series t