The Analysis and Design of Linear Circuits, Sixth Edition Solutions Manual Copyright 2008 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 646-8600. Problem 1-1 Express the following quantities to the nearest standard prefix using no more than three digits. (a) 1,000,000 W (b) 92.5 109 Hz (c) 256.7 107 s (d) 25 109 F Solution: Use the standard decimal prefixes in Table 1-2 to find the appropriate prefix. MATLAB converts numbers into standard engineering notation. format short eng a = 1000000 b = 92.5e9 c = 256.7e-7 d = 25e-9 a = 1.0000e+006 b = 92.5000e+009 c = 25.6700e-006 d = 25.0000e-009 Answer: (a) 1 MW (b) 92.5 GHz (c) 25.7 s (d) 25 nF The Analysis and Design of Linear Circuits, Sixth Edition Solutions Manual Copyright 2008 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 646-8600. Problem 1-2 Express the following quantities to the nearest standard prefix using no more than three digits. (a) 0.000111 A (b) 200 105 C (c) 5.02 103 J (d) 3,264,000 Solution: Use the standard decimal prefixes in Table 1-2 to find the appropriate prefix. MATLAB converts numbers into standard engineering notation. format short eng a = 0.000111 b = 200e5 c = 5.02e3 d = 3264000 a = 111.0000e-006 b = 20.0000e+006 c = 5.0200e+003 d = 3.2640e+006 Answer: (a) 111 A (b) 20 MC (c) 5.02 kJ (d) 3.26 M The Analysis and Design of Linear Circuits, Sixth Edition Solutions Manual Copyright 2008 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 646-8600. Problem 1-3 An ampere-hour (Ah) meter measures the time-integral of the current in a conductor. During an 8-hour period, a certain meter records 3300 Ah. Find the number of coulombs that flowed through the meter during the recording period. Solution: 1 ampere = 1 coulomb/second 1 hour = 3600 seconds 3300 Ah = 3300 ampere-hour = 3300 (coulomb/second)(hour)(3600 second/hour) = 11.88 MC format short eng Ah = 3300; C = 3300*3600 C = 11.8800e+006 Answer: 11.88 MC The Analysis and Design of Linear Circuits, Sixth Edition Solutions Manual Copyright 2008 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 646-8600. Problem 1-4 Electric power companies measure energy consumption in kilowatt-hours, denoted kWh. One kilowatt-hour is the amount of energy transferred by 1 kW of power in a period of 1 hour. A power company billing statement reports a user's total energy usage to be 2150 kWh. Find the number of joules used during the billing period. Solution: 1 kWh = 1000 watt-hours 1 watt = 1 joule/second 1 hour = 3600 seconds 2150 kWh = 2150 kilowatt-hours = 2150000 watt-hours = 2150000 (joules/second)(hours)(3600 second/hour) = 7.74 GJ format short eng kWh = 2150; J = kWh*1000*3600 J = 7.7400e+009 Answer: 7.74 GJ The Analysis and Design of Linear Circuits, Sixth Edition Solutions Manual Copyright 2008 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 646-8600. Problem 1-5 (a) To convert capacitance from picofarads to microfarads, multiply by ______. (b) To convert resistance from megohms to kilohms, multiply by _____. (c) To convert voltage from millivolts to volts, multiply by_____. (d) To convert energy from megajoules to joules, multiply by _____. Solution: Assume we have X of the original units. Use identities to determine the multiplier for X in the new units. (a) X pF = X 1012 F = (X 106) 106 F = X 106 F (b) X M = X 106 = (X 103) 103 = X 103 k (c) X mV = X 103 V (d) X MJ = X 106 J Answer: (a) 106 (b) 103 (c) 103 (d) 106 The Analysis and Design of Linear Circuits, Sixth Edition Solutions Manual Copyright 2008 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 646-8600. Problem 1-6 A wire carries a constant current of 300 A. How many coulombs flow past a given point in the wire in 2 s? Solution: format short eng coulomb_per_second = 300e-6; sec = 2; coulomb = coulomb_per_second*sec coulomb = 600.0000e-006 Answer: 600 C The Analysis and Design of Linear Circuits, Sixth Edition Solutions Manual Copyright 2008 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 646-8600. Problem 1-7 The net positive charge flowing through a device is q(t) = 10 + 3t mC. Find the current through the device. Solution: ( ) ( )3mAdq ti tdt= = syms t real qt = 10 + 3*t; it = diff(qt,t) it = 3 Answer: i(t) = 3 mA The Analysis and Design of Linear Circuits, Sixth Edition Solutions Manual Copyright 2008 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 646-8600. Problem 1-8 Figure P1-8 shows a plot of the net positive charge flowing in a wire versus time. Sketch the corresponding current during the same period of time. Solution: Create the original waveform for the charge. heaviside(t) = u(t) = unit step function. syms t qt = (10+10*t)*(heaviside(t)-heaviside(t-2))... + (110-40*t)*(heaviside(t-2)-heaviside(t-3))... + (5-5*t)*(heaviside(t-3)-heaviside(t-5))... + (-270+50*t)*(heaviside(t-5)-heaviside(t-6)); tt = 0:0.01:6; qtt = subs(qt,t,tt); figure plot(tt,qtt,'b','LineWidth',3) xlabel('Time (s)') ylabel('Charge (C)') grid on Take the derivative of the charge waveform to find the current. it = diff(qt,t); itt = subs(it,t,tt); figure plot(tt,itt,'g','LineWidth',3) xlabel('Time (s)') ylabel('Current (A)') grid on axis([0 6 -60 60]) 0 1 2 3 4 5 6-60-40-200204060Time (s)Current (A) Answer: Shown above 0 1 2 3 4 5 6-20-15-10-5051015202530Time (s)Charge (C)The Analysis and Design of Linear Circuits, Sixth Edition Solutions Manual Copyright 2008 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 646-8600. Problem 1-9 The net positive charge flowing through a device varies as q(t) = 3t2 C. Find the current through the device at t = 0 s, t = 1 s, and t = 3 s. Solution: ( ) ( )( ) ( ) ( ) 6 A, 0 0A, 1 6A, 3 18Adq ti t t i i idt= = = = = syms t qt = 3*t^2; it = diff(qt,t) tt = [0, 1, 3]; itt = subs(it,t,tt) it = 6*t itt = 0 6 18 Answer: i(0) = 0 A; i(1) = 6 A; i(3) = 18 A The Analysis and Design of Linear Circuits, Sixth Edition Solutions Manual Copyright 2008 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 646-8600. Problem 1-10 An appliance is protected by a 15 A fuse. What is the maximum number of coulombs that can flow through the device before the breaker trips? Solution: The maximum current is 15 A, which is 15 coulombs per second. As long as the rate of coulombs flowing through the device remains below 15 coulombs per second, the total number of coulombs that can flow through it is theoretically infinite. Answer: Infinite. The Analysis and Design of Linear Circuits, Sixth Edition Solutions Manual Copyright 2008 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 646-8600. Problem 1-11 For 0 t 5 s, the current through a device is i(t) = 4t A. For 5 < t 10 s, the current is i(t) = 40 4t A, and i(t) = 0 A for t > 10 s. Sketch i(t) versus time and find the total charge flowing through the device between t = 0 s and t = 10 s. Solution: syms t it = 4*t*(heaviside(t)-heaviside(t-5))... + (40-4*t)*(heaviside(t-5)-heaviside(t-10)); qTotal = int(it,t,0,10) tt = 0:0.01:10; itt = subs(it,t,tt); plot(tt,itt,'b','LineWidth',3) grid on xlabel('Time (s)') ylabel('Current (A)') qTotal = 100 0 1 2 3 4 5 6 7 8 9 1002468101214161820Time (s)Current (A) Answer: i(t) is sketched above. ( )5 100 514 40 4 10 20 100C2TOTALq t dt t dt = + = = qTOTAL = 100 C The Analysis and Design of Linear Circuits, Sixth Edition Solutions Manual Copyright 2008 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 646-8600. Problem 1-12 The charge flowing through a device is q(t) = 1 e500t C. How long will it take the current to reach 200 A? Solution: ( ) ( )6 500 6 500 65ln2500 10 A, 500 10 200 10 , 1.8326ms500t tdq ti t e e tdt | | |\ = = = = = syms t qt = 1e-6*(1-exp(-500*t)); it = diff(qt,t) Time200 = solve(it-200e-6,t); Time200 = vpa(Time200,5) it = 1/2000*exp(-500*t) Time200 = .18326e-2 Answer: t = 1.8326 ms The Analysis and Design of Linear Circuits, Sixth Edition Solutions Manual Copyright 2008 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 646-8600. Problem 1-13 The 12-V automobile battery in Figure P1-13 has an output capacity of 100 ampere-hours (Ah) when connected to a head lamp that absorbs 200 watts of power. Assume the battery voltage is constant. (a) Find the current supplied by the battery. (b) How long can the battery power the headlight? i Solution: % Part (a) v = 12; p = 200; i = p/v % Part (b) w = 100; t = w/i i = 16.6667e+000 t = 6.0000e+000 Answer: (a) i = 16.67 A (b) t = 6 hours The Analysis and Design of Linear Circuits, Sixth Edition Solutions Manual Copyright 2008 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 646-8600. Problem 1-14 An incandescent lamp absorbs 100 W when connected to a 120-V source. A fluorescent lamp producing the same amount of light absorbs 12 W when connected to the same source. How much cheaper is it to operate the fluorescent bulb over the incandescent bulb over 1000 hours when electricity costs 7.2 cents/kW-h? Solution: V = 120; T = 1000; P_incand = 100; kWh_incand = P_incand*T/1000; cost_incand = 0.072*kWh_incand P_fluor = 12; kWh_fluor = P_fluor*T/1000; cost_fluor = 0.072*kWh_fluor cost_difference = cost_incand - cost_fluor cost_incand = 7.2000e+000 cost_fluor = 864.0000e-003 cost_difference = 6.3360e+000 Answer: $6.34 The Analysis and Design of Linear Circuits, Sixth Edition Solutions Manual Copyright 2008 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 646-8600. Problem 1-15 The current through a device is zero for t < 0 and is i(t) = 3e2t A for t 0 . Find the charge q(t) flowing through the device for t 0. Solution: ( ) ( ) ( )22 2 2 0033 33 1 1 C, 02 2 2ttt teq t e d e e t = = = = syms t tau it = 3*exp(-2*t)*heaviside(t); itau = subs(it,t,tau) qt = simple(int(itau,tau,-inf,t)) itau = 3*exp(-2*tau)*heaviside(tau) qt = -3/2*heaviside(t)*(exp(-2*t)-1) Answer: [ ] 0 C 123) (2 = t e t q t The Analysis and Design of Linear Circuits, Sixth Edition Solutions Manual Copyright 2008 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 646-8600. Problem 1-16 The current through and voltage across a two-terminal device are i(t) = 20 sin (1000t) mA and v = 100 sin (1000t) V. Find the maximum and minimum power delivered to the device. Solution: ( ) ( ) ( ) ( ) ( )21( ) 2sin 1000 2 1 cos 2000 1 cos 2000 W2p t i t v t t t t = = = = ( Since 1 cos(2000t) 1, pMAX = 2 W, pMIN = 0 W syms t it = 20e-3*sin(1000*t); vt = 100*sin(1000*t); pt = simplify(it*vt) tt = 0:0.0001:0.005; ptt = subs(pt,t,tt); ptMax = max(ptt) ptMin = min(ptt) plot(tt,ptt,'b','LineWidth',3) xlabel('Time (s)') ylabel('Power (W)') grid on pt = 2*sin(1000*t)^2 ptMax = 1.9997e+000 ptMin = 0.0000e-003 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5x 10-300.20.40.60.811.21.41.61.82Time (s)Power (W) Answer: The Analysis and Design of Linear Circuits, Sixth Edition Solutions Manual Copyright 2008 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 646-8600. pMAX = 2 W pMIN = 0 W The Analysis and Design of Linear Circuits, Sixth Edition Solutions Manual Copyright 2008 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 646-8600. Problem 1-17 When illuminated the i-v relationship for a photocell is i = ev 10 A. For v = 2, 2 and 3 V find the device power and state whether it is absorbing or delivering power. Solution: v = [-2 2 3]; iv = exp(v)-10; p = v.*iv pAbsorbs = p>0 p = 19.7293e+000 -5.2219e+000 30.2566e+000 pAbsorbs = 1 0 1 Answer: For v = 2 V, p = 19.73 W, absorbing For v = 2 V, p = 5.22 W, delivering For v = 3 V, p = 30.26 W, absorbing The Analysis and Design of Linear Circuits, Sixth Edition Solutions Manual Copyright 2008 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 646-8600. Problem 1-18 A new 1.5 V AA battery delivers 40 kJ of energy during its lifetime. How long will the battery last in an application that draws 10 mA continuously. Assume the battery voltage is constant. Solution: V = 1.5; w = 40e3; Ia = 10e-3; p = V*Ia; Tsec = w/p Tmin = Tsec/60; Thours = Tsec/3600; Tdays = Thours/24 Days = floor(Tdays) Hours = floor((Tsec - Days*3600*24)/3600) Minutes = floor((Tsec - Days*3600*24 - Hours*3600)/60) Seconds = Tsec - Days*3600*24 - Hours*3600 - Minutes*60 Tsec = 2.6667e+006 Tdays = 30.8642e+000 Days = 30.0000e+000 Hours = 20.0000e+000 Minutes = 44.0000e+000 Seconds = 26.6667e+000 Answer: 30.8642 days or 30 days, 20 hours, 44 minutes, and 26.7 seconds The Analysis and Design of Linear Circuits, Sixth Edition Solutions Manual Copyright 2008 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 646-8600. Problem 1-19 The maximum power the device can dissipate is 0.5 W. Determine the maximum current allowed by the device power rating when the voltage is 9 V. Solution: Pmax = 0.5; V = 9; Imax = Pmax/V Imax = 55.5556e-003 Answer: iMAX = 55.56 mA The Analysis and Design of Linear Circuits, Sixth Edition Solutions Manual Copyright 2008 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 646-8600. Problem 1-20 A constant current of 2 A charges a battery for 4 hours. During the charging period the battery voltage is v(t) = 12 2et, where t is in hours. Determine the energy stored in the battery. Solution: ( ) ( ) ( ) ( ) ( ) ( )4 440 024 4 W, 24 4 96 4 1 92.07Jt tp t i t v t e w p t dt e dt e = = = = = + = syms t it = 2; vt = 12-2*exp(-t); pt = vt*it; w = double(int(pt,t,0,4)) w = 92.0733e+000 Answer: w = 92.07 J The Analysis and Design of Linear Circuits, Sixth Edition Solutions Manual Copyright 2008 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 646-8600. Problem 1-21 Two electrical devices are connected as shown in Figure P1-21. Using the reference marks shown in the figure, find the power transferred and state whether the power is transferred from A to B or B to A when (a) v = + 11 V and i = -1.1 A (b) v = +80 V and i = +20 mA (c) v = -120 V and i = -12 mA (d) v = -1.5 V and i = -600 A Solution: The passive sign convention applies to Element B, so if the power is positive the transfer is from A to B, and if the power is negative, the transfer is from B to A. v = [11 80 -120 -1.5]; i = [-1.1 20e-3 -12e-3 -600e-6]; p = v.*i AtoB = p>0 p = -12.1000e+000 1.6000e+000 1.4400e+000 900.0000e-006 AtoB = 0 1 1 1 Answer: (a) p = 12.1 W, from B to A (b) p = 1.6 W, from A to B (c) p = 1.44 W, from A to B (d) p = 900 W, from A to B A B viThe Analysis and Design of Linear Circuits, Sixth Edition Solutions Manual Copyright 2008 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 646-8600. Problem 1-22 Figure P1-22 shows an electric circuit with a voltage and a current variable assigned to each of the six devices. The device voltages and currents are observed to be v(V) i(A) v(V) i(A) Device 1 15 1 Device 4 10 1 Device 2 5 1 Device 5 20 3 Device 3 10 2 Device 6 20 2 Find the power associated with each device and state whether the device is absorbing or delivering power. Use the power balance to check your work. vv v125ii i 125v4v3i3i4v6i621 3 5 64 Solution: v = [15 5 10 -10 20 20]; ia = [-1 1 2 -1 -3 2]; p = v.*ia Absorbing = p>0 Balance = sum(p) p = -15.0000e+000 5.0000e+000 20.0000e+000 10.0000e+000 -60.0000e+000 40.0000e+000 Absorbing = 0 1 1 1 0 1 Balance = 0.0000e-003 Answer: Device p (W) Absorbing or Delivering 1 15 Delivering 2 5 Absorbing 3 20 Absorbing 4 10 Absorbing 5 60 Delivering 6 40 Absorbing The Analysis and Design of Linear Circuits, Sixth Edition Solutions Manual Copyright 2008 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 646-8600. Power balance is zero, so the values check. The Analysis and Design of Linear Circuits, Sixth Edition Solutions Manual Copyright 2008 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 646-8600. Problem 1-23 Figure P1-22 shows an electric circuit with a voltage and a current variable assigned to each of the six devices. Use power balance to find v4 when v1 = 20 V, i1 = 2 A, p2 = 20 W, p3 = 10 W, i4 = 1 A, and p5 = p6 = 2.5 W. Is device 4 absorbing or delivering power? vv v125ii i 125v4v3i3i4v6i621 3 5 64 Solution: v1 = 20; i1 = -2; i4 = 1; p1 = v1*i1; p2 = 20; p3 = 10; p5 = 2.5; p6 = 2.5; p4 = -p1-p2-p3-p5-p6 v4 = p4/i4 Absorbing = p4>0 p4 = 5.0000e+000 v4 = 5.0000e+000 Absorbing = 1 Answer: v4 = 5 V, absorbing power The Analysis and Design of Linear Circuits, Sixth Edition Solutions Manual Copyright 2008 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 646-8600. Problem 1-24 Suppose in Figure P1-22 a ground is connected to the minus () side of element 6 and another to the junction of elements 2, 3 and 4. Further, assume that the voltage v4 is 5 V and v1 is 10 V. What are the voltages v2, v3, v5 and v6? vv v125ii i 125v4v3i3i4v6i621 3 5 64 Solution: v4 = 5; v1 = 10; v5 = -v4 v6 = v5 v3 = 0 v2 = v1 v5 = -5.0000e+000 v6 = -5.0000e+000 v3 = 0.0000e-003 v2 = 10.0000e+000 Answer: v2 = 10 V v3 = 0 V v5 = 5 V v6 = 5 V The Analysis and Design of Linear Circuits, Sixth Edition Solutions Manual Copyright 2008 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 646-8600. Problem 1-25 For t 0 the voltage across and power absorbed by a two-terminal device are v(t) = 2et V and p(t) = 40e2t mW. Find the total charge delivered to the device for t 0. Solution: ( ) ( )( )3 23040 1020 mA, 20 10 20mC2tt tTOTAL tp t ei t e q e dtv t e = = = = = syms t vt = 2*exp(-t)*heaviside(t); pt = 40e-3*exp(-2*t)*heaviside(t); it = simple(pt/vt) q = simple(int(it,t,0,inf)) it = 1/50/exp(t) q = 1/50 Answer: qTOTAL = 20 mC The Analysis and Design of Linear Circuits, Sixth Edition Solutions Manual Copyright 2008 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 646-8600. Problem 1-26 Repeat Problem 1-22 using MATLAB to perform the calculations. Create a vector for the voltage values, v = [15 5 10 -10 20 20], and a vector for the current values, i = [-1 1 2 -1 -3 2]. Compute the corresponding vector for the power values, p, using element-by-element multiplication (.*) and then use the sum command to verify the power balance. Solution: v = [15 5 10 -10 20 20]; ia = [-1 1 2 -1 -3 2]; p = v.*ia Absorbing = p>0 Balance = sum(p) p = -15 5 20 10 -60 40 Absorbing = 0 1 1 1 0 1 Balance = 0 Answer: Device p (W) Absorbing or Delivering 1 15 Delivering 2 5 Absorbing 3 20 Absorbing 4 10 Absorbing 5 60 Delivering 6 40 Absorbing Power balance is zero, so the values check. The Analysis and Design of Linear Circuits, Sixth Edition Solutions Manual Copyright 2008 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 646-8600. Problem 1-27 Using the passive sign convention, the voltage across a device is v(t) = 5 cos(10t) V and the current through the device is i(t) = 0.5 sin(10t) A. Using MATLAB, create a short script (m-file) to assign a value to the time variable, t, and then calculate the voltage, current, and power at that time. Run the script for t = 0.2 s and t = 0.4 s and for each result state whether the device is absorbing or delivering power. Solution: t = 0.2 vt = 5*cos(10*t) it = 0.5*sin(10*t) pt = vt*it Absorbing = pt>0 t = 0.4 vt = 5*cos(10*t) it = 0.5*sin(10*t) pt = vt*it Absorbing = pt>0 t = 0.2000 vt = -2.0807 it = 0.4546 pt = -0.9460 Absorbing = 0 t = 0.4000 vt = -3.2682 it = -0.3784 pt = 1.2367 Absorbing = 1 Answer: t v(t) i(t) p(t) Absorbing or Delivering 0.2 s 2.08 V 454.6 mA 946 mW Delivering 0.4 s 3.27 V 378.4 mA 1.237 W Absorbing The Analysis and Design of Linear Circuits, Sixth Edition Solutions Manual Copyright 2008 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 646-8600. Problem 1-28 In complete darkness the voltage across and current through a two-terminal light detector are +6.5 V and +7 nA. In full sunlight the voltage and current are +1.1 V and +4.5 mA. Express the light/dark power ratio of the device in decibels (dB), where the power ratio in dB is PRdB = 10 Log10 (p2/p1). Solution: v_dark = 6.5; i_dark = 7e-9; p_dark = v_dark*i_dark; v_sun = 1.1; i_sun = 4.5e-3; p_sun = v_sun*i_sun; PR_dB = 10*log10(p_sun/p_dark) PR_dB = 50.3659 Answer: PRdB = 50.3659 dB The Analysis and Design of Linear Circuits, Sixth Edition Solutions Manual Copyright 2008 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 646-8600. Problem 1-29 A manufacturer's data sheet for the converter in Figure P1-29 states that the output voltage is vdc = 5 V when the input voltage vac = 120 V. When the load draws a current idc = 40 A the input power is pac = 300 W. Find the efficiency of the converter. AC to DCLoadviConverterdcdc vac.aci Solution: p_ac = 300; v_dc = 5; i_dc = 40; p_dc = v_dc*i_dc; Efficiency = p_dc/p_ac Efficiency = 0.6667 Answer: Efficiency = 66.67% The Analysis and Design of Linear Circuits, Sixth Edition Solutions Manual Copyright 2008 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 646-8600. Problem 1-30 A capacitor is a two-terminal device that can store electric charge. In a linear capacitor the amount of charge stored is proportional to the voltage across the device. For a particular device the proportionality is q(t) = 107v(t). If v(t) = 0 for t < 0 and v(t) = 10(1 e5000t) for t 0, find the energy stored in the device at t = 200 s. Solution: ( ) ( ) ( ) ( )( ) ( ) ( ) ( )6 5000 5000 5000 1000010 1 C, 5 mA, 0.05 Wt t t tdq tq t e i t e p t i t v t e edt = = = = = ( ) ( ) ( )6 6200 10 200 10 1 25000 100000 01 1200 0.05 0.055000 100001.9979Jt t e ew s p t dt e e dt ( = = = ( = syms t tt = 200e-6; C = 1e-7; vt = 10*(1-exp(-5000*t)); qt = C*vt; it = diff(qt,t); pt = it*vt; wt = double(int(pt,t,0,tt)) wt = 1.9979e-006 Answer: w(200 s) = 1.9979 J The Analysis and Design of Linear Circuits, Sixth Edition Solutions Manual Copyright 2008 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 646-8600. Problem 1-31 A manufacturers data sheet for a desktop computer lists the power supply requirements as 30 A @ 5 V, 5 A @ 15 V, 5 A @ 15 V, 3.25 A @ 5 V and 0.5 A @ 12 V. The data sheet also states that the overall power consumption is 325 W. Are these data consistent? Explain. Solution: it = [30 5 5 3.25 0.5]; vt = [5 15 -15 -5 12]; pt = vt.*it Balance1 = sum(pt) Balance2 = sum(abs(pt)) pt = 150.0000 75.0000 -75.0000 -16.2500 6.0000 Balance1 = 139.7500 Balance2 = 322.2500 Answer: These data are consistent if the power supply is always supplying power at the stated currents and voltages and the overall power consumption is rounded up to the nearest 5 W. If we calculate the individual powers by multiplying current times voltage for each supply value, we will have positive and negative powers. It is not reasonable that the computer would supply power to a power supply, so the negative power values are not a correct interpretation of the passive sign convention. If we sum the magnitudes of the products of currents and voltages, the total power consumption is 322.25 W, which is very close to the stated value of 325 W. The Analysis and Design of Linear Circuits, Sixth Edition Solutions Manual Copyright 2008 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 646-8600. Problem 1-32 MATLAB is extremely powerful when linear operations are formatted using vector or matrix notation. The colon operator (:) is a special MATLAB operator that allows many calculations to be simplified or expressed in a compact form. For example, we can quickly generate a vector of time values ranging from 0 s to 1 s with increments of 0.1 s using the following command: t = 0:0.1:1. We can then substitute this vector into standard MATLAB functions, such as sin() or exp(), to generate the results of evaluating those functions for every value in the time vector: v = 5*sin(2*pi*t). or i = 3*exp(-4*t). Create the time vector described above and evaluate v = 5*sin(2*pi*t) using that time vector. Find the maximum and minimum values for the vector v using the commands max(v) and min(v), respectively. Repeat the problem with a time increment that is smaller by a factor of 10 and comment on the results. Solution: t = 0:0.1:1; v = 5*sin(2*pi*t); vMax = max(v) vMin = min(v) t2 = 0:0.01:1; v2 = 5*sin(2*pi*t2); v2Max = max(v2) v2Min = min(v2) vMax = 4.7553 vMin = -4.7553 v2Max = 5 v2Min = -5 Answer: For t = 0.1, vMAX = 4.7553 V and vMIN = 4.7553 V For t = 0.01, vMAX = 5 V and vMIN = 5 V The smaller time step provides better resolution and allows us to sample the voltage signal at its true maximum and minimum. The Analysis and Design of Linear Circuits, Sixth Edition Solutions Manual Copyright 2008 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 646-8600. Problem 2-1 The current through a 33-k resistor is 1.2 mA. Find the voltage across the resistor. Solution: v = iR clear all R = 33e3; ii = 1.2e-3; v = ii*R v = 39.6000e+000 Answer: v = 39.6 V The Analysis and Design of Linear Circuits, Sixth Edition Solutions Manual Copyright 2008 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 646-8600. Problem 2-2 A 6.2-k resistor dissipates 12 mW. Find the current through the resistor. Solution: p = i2R clear all format short eng R = 6.2e3; p = 12e-3; ii = sqrt(p/R) ii = 1.3912e-003 Answer: i = 1.3912 mA The Analysis and Design of Linear Circuits, Sixth Edition Solutions Manual Copyright 2008 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 646-8600. Problem 2-3 The conductance of a particular resistor is 1 mS. Find the current through the resistor when connected across a 9 V source. Solution: v = iR clear all G = 1e-3; R = 1/G; v = 9; ii = v/R ii = 9.0000e-003 Answer: i = 9 mA The Analysis and Design of Linear Circuits, Sixth Edition Solutions Manual Copyright 2008 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 646-8600. Problem 2-4 In Figure P2-4 the resistor dissipates 25 mW. Find Rx. RX15 VP =25 mWX Solution: Rvp2= clear all p = 25e-3; v = 15; R = v^2/p R = 9.0000e+003 Answer: R = 9 k The Analysis and Design of Linear Circuits, Sixth Edition Solutions Manual Copyright 2008 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 646-8600. Problem 2-5 In Figure P2-5 find Rx and the power delivered to the resistor. RX10 mA100 V Solution: clear all v = 100; ii = 10e-3; R = v/ii p = v*ii R = 10.0000e+003 p = 1.0000e+000 Answer: Rx = 10 k, p = 1 W The Analysis and Design of Linear Circuits, Sixth Edition Solutions Manual Copyright 2008 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 646-8600. Problem 2-6 The i-v characteristic of a nonlinear resistor are v = 75i + 0.2i3. (a) Calculate v and p for i = 0.5, 1, 2, 5, and 10 A. (b) Find the maximum error in v when the device is treated as a 75- linear resistance on the range |i| < 0.5 A. Solution: clear all format short eng ii = [-10, -5, -2, -1, -0.5, 0.5, 1, 2, 5, 10]; v = 75*ii + 0.2*ii.^3; p = v.*ii; Results = [ii' v' p'] syms i1 v1 = 75*i1+0.2*i1^3; v2 = 75*i1; ii1 = -0.5:0.01:0.5; vv1 = subs(v1,i1,ii1); vv2 = subs(v2,i1,ii1); plot(vv1,ii1,'b','LineWidth',3) hold on plot(vv2,ii1,'g','LineWidth',1) grid on xlabel('Voltage (V)') ylabel('Current (A)') legend('Nonlinear','Linear') MaxError = max(vv1)-max(vv2) MaxError2 = subs(v1-v2,i1,0.5) Results = -10.0000e+000 -950.0000e+000 9.5000e+003 -5.0000e+000 -400.0000e+000 2.0000e+003 -2.0000e+000 -151.6000e+000 303.2000e+000 -1.0000e+000 -75.2000e+000 75.2000e+000 -500.0000e-003 -37.5250e+000 18.7625e+000 500.0000e-003 37.5250e+000 18.7625e+000 1.0000e+000 75.2000e+000 75.2000e+000 2.0000e+000 151.6000e+000 303.2000e+000 5.0000e+000 400.0000e+000 2.0000e+003 10.0000e+000 950.0000e+000 9.5000e+003 MaxError = 25.0000e-003 MaxError2 = 25.0000e-003 The Analysis and Design of Linear Circuits, Sixth Edition Solutions Manual Copyright 2008 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 646-8600. -40 -30 -20 -10 0 10 20 30 40-0.5-0.4-0.3-0.2-0.100.10.20.30.40.5Voltage (V)Current (A) NonlinearLinear Answer: (a) i (A) v (V) p (W) -10 -950 9500 -5 -400 2000 -2 -151.6 303.2 -1 -75.2 75.2 -0.5 -37.525 18.7625 0.5 37.525 18.7625 1 75.2 75.2 2 151.6 303.2 5 400 2000 10 950 9500 (b) ERRORMAX = 25 mV The Analysis and Design of Linear Circuits, Sixth Edition Solutions Manual Copyright 2008 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 646-8600. Problem 2-7 A 10-k resistor has a power rating of W. Find the maximum voltage that can be applied to the resistor. Solution: clear all R = 10e3; p = 1/8; v_max = sqrt(p*R) v_max = 35.3553e+000 Answer: vmax = 35.36 V The Analysis and Design of Linear Circuits, Sixth Edition Solutions Manual Copyright 2008 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 646-8600. Problem 2-8 A certain type of film resistor is available with resistance values between 10 and 100 M. The maximum ratings for all resistors of this type are 500 V and 1/4 W. Show that the voltage rating is the controlling limit for R > 1 M, and that the power rating is the controlling limit when R < 1 M. Solution: clear all V = 500; p = 1/4; R = V^2/p R = 1.0000e+006 Rvp2= At R = 1 M, both p and v can take their maximum values and there are no issues. For R > 1 M, with a maximum voltage, the power must be less than 0.25 W, so the voltage rating on a particular resistor will control the maximum allowable value for the power. For R < 1 M, with a maximum voltage, the power will be greater than 0.25 W, so the power rating on a particular resistor will control the maximum allowable value for the voltage. Answer: Presented above. The Analysis and Design of Linear Circuits, Sixth Edition Solutions Manual Copyright 2008 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 646-8600. Problem 2-9 Figure P2-9 shows the circuit symbol for a class of two-terminal devices called diodes. The i-v relationship for a specific pn junction diode is ( ) A e i v1 10 240 16 = (a) Use this equation to find i and p for v = 0, 0.1, 0.2, 0.4, and 0.8 V. Use these data to plot the i-v characteristic of the element. (b) Is the diode linear or nonlinear, bilateral or nonbilateral, and active or passive? (c) Use the diode model to predict i and p for v = 5 V. Do you think the model applies to voltages in this range? Explain. (d) Repeat (c) for v = 5 V. Solution: clear all v = [-0.8, -0.4, -0.2, -0.1, 0 0.1, 0.2, 0.4, 0.8]; ii = 2e-16*(exp(40*v)-1); p = v.*ii; Results = [v' ii' p'] plot(v,ii,'b','LineWidth',3) xlabel('Voltage (V)') ylabel('Current (A)') grid on v = 5 i5 = 2e-16*(exp(40*v)-1) v = -5 iNeg5 = 2e-16*(exp(40*v)-1) Results = -800.0000e-003 -200.0000e-018 160.0000e-018 -400.0000e-003 -200.0000e-018 80.0000e-018 -200.0000e-003 -199.9329e-018 39.9866e-018 -100.0000e-003 -196.3369e-018 19.6337e-018 0.0000e-003 0.0000e-003 0.0000e-003 100.0000e-003 10.7196e-015 1.0720e-015 200.0000e-003 595.9916e-015 119.1983e-015 400.0000e-003 1.7772e-009 710.8888e-012 800.0000e-003 15.7926e-003 12.6341e-003 v = 5.0000e+000 i5 = 144.5195e+069 v = -5.0000e+000 iNeg5 = -200.0000e-018 +-viThe Analysis and Design of Linear Circuits, Sixth Edition Solutions Manual Copyright 2008 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 646-8600. -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8-20246810121416x 10-3Voltage (V)Current (A) Answer: (a) v (V) i (A) p (W) -0.8 -2.00E-16 1.60E-16 -0.4 -2.00E-16 8.00E-17 -0.2 -2.00E-16 4.00E-17 -0.1 -1.96E-16 1.96E-17 0 0 0 0.1 1.07E-14 1.07E-15 0.2 5.96E-13 1.19E-13 0.4 1.78E-09 7.11E-10 0.8 1.58E-02 1.26E-02 (b) The plot in Part (a) shows that the device is nonlinear and nonbilateral. The power for the device is always positive, so it is passive. (c) For v = 5 V, i = 1.45 1071 A and p = 7.23 1071 W. The model is not valid because the current and power are too large. (d) For v = 5 V, i = 2.00 1016 A and p = 1.00 1015 W. The model is valid because the current and power are both essentially zero. The Analysis and Design of Linear Circuits, Sixth Edition Solutions Manual Copyright 2008 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 646-8600. Problem 2-10 In Figure P2-10 i2 = 2 A and i3 = 5 A. Find i1 and i4. BCi3i4i2i1A1 234 Solution: Apply KCL at Nodes B and C. clear all i2 = -2; i3 = 5; i1 = -i2 i4 = i2+i3 i1 = 2.0000e+000 i4 = 3.0000e+000 Answer: i1 = 2 A and i4 = 3 A. The Analysis and Design of Linear Circuits, Sixth Edition Solutions Manual Copyright 2008 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 646-8600. Problem 2-11 For the circuit in Figure P2-11: (a) Identify the nodes and at least two loops. (b) Identify any elements connected in series or in parallel. (c) Write KCL and KVL connection equations for the circuit. 13i431i4iBAC22i Solution: There are three nodes and three loops. Answer: (a) nodes: A, B, C; loops: 1-2; 2-3-4; 1-3-4 (b) series: 3 and 4; parallel: 1 and 2 (c) KCL: node A: 03 2 1 = + + i i i ; node B: 04 3 = + i i ; node C: 04 2 1 = i i i KVL: loop 1-2: 02 1 = + v v ; loop 2-3-4: 04 3 2 = + + v v v ; loop 1-3-4: 04 3 1 = + + v v v The Analysis and Design of Linear Circuits, Sixth Edition Solutions Manual Copyright 2008 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 646-8600. Problem 2-12 In Figure P2-11, i2 = 10 mA and i4 = 20 mA. Find i1 and i3. 13i431i4iBAC22i Solution: Use the KCL equations developed in the solution to Problem 2-11. clear all i2 = -10e-3; i4 = 20e-3; i3 = i4 i1 = -i2-i3 i3 = 20.0000e-003 i1 = -10.0000e-003 Answer: i1 = 10 mA and i3 = 20 mA. The Analysis and Design of Linear Circuits, Sixth Edition Solutions Manual Copyright 2008 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 646-8600. Problem 2-13 For the circuit in Figure P2-13: (a) Identify the nodes and at least three loops in the circuit. (b) Identify any elements connected in series or in parallel. (c) Write KCL and KVL connection equations for the circuit. 23i632i6iBACD 55i5v6v3v2v1i1v4v414i Solution: There are four nodes and at least five loops. There are only three independent KVL equations. Answer: (a) nodes: A, B, C, D; loops: 1-3-2; 2-4-5; 3-6-4; 1-6-5; 2-3-6-5; 1-6-4-2; 1-3-4-5 (b) series: none; parallel: none (c) KCL: node A: 04 3 2 = + + i i i ; node B: 06 3 1 = + i i i ; node C: 05 2 1 = i i i ; node D: 06 5 4 = + i i i KVL: loop 1-3-2: 02 3 1 = + v v v ; loop 2-4-5: 05 4 2 = + + v v v ; loop 3-6-4: 04 6 3 = + v v v The Analysis and Design of Linear Circuits, Sixth Edition Solutions Manual Copyright 2008 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 646-8600. Problem 2-14 In Figure P2-13 v2 = 10 V, v3 = 10 V, and v4 = 3 V. Find v1, v5, and v6. 23i632i6iBACD 55i5v6v3v2v1i1v4v414i Solution: Use the KVL equations developed in the solution to Problem 2-13. clear all v2 = 10; v3 = -10; v4 = 3; v1 = v2 - v3 v5 = v2 - v4 v6 = v4 - v3 v1 = 20.0000e+000 v5 = 7.0000e+000 v6 = 13.0000e+000 Answer: v1 = 20 V, v5 = 7 V, and v6 = 13 V. The Analysis and Design of Linear Circuits, Sixth Edition Solutions Manual Copyright 2008 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 646-8600. Problem 2-15 The circuit in Figure P2-15 is organized around the three signal lines A, B, and C. (a) Identify the nodes and at least three loops in the circuit. (b) Write KCL connection equations for the circuit. (c) If i1 = 20 mA, i2 = 12 mA, and i3 = 50 mA, find i4, i5, and i6 (d) Show that the circuit in Figure P2-15 is identical to that in Figure P2-13. 3i6iBACD4i5i2i 1 i3 61 2 4 5 Solution: (a) There are four nodes and at least five loops. (b) KCL: node A: 04 3 2 = + + i i i ; node B: 06 3 1 = + i i i ; node C: 05 2 1 = i i i ; node D: 06 5 4 = + i i i clear all i1 = -20e-3; i2 = -12e-3; i3 = 50e-3; i4 = -i2-i3 i5 = -i1-i2 i6 = i3-i1 i4 = -38.0000e-003 i5 = 32.0000e-003 i6 = 70.0000e-003 Answer: (a) nodes: A, B, C, D; loops: 1-3-2; 2-4-5; 3-6-4; 1-6-5; 2-3-6-5; 1-6-4-2; 1-3-4-5 23i632i6iBACD 55i5v6v3v2v1i1v4v414iThe Analysis and Design of Linear Circuits, Sixth Edition Solutions Manual Copyright 2008 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 646-8600. (b) KCL: node A: 04 3 2 = + + i i i ; node B: 06 3 1 = + i i i ; node C: 05 2 1 = i i i ; node D: 06 5 4 = + i i i (c) i4 = 38 mA; i5 = 32 mA; i6 = 70 mA (d) The circuits have the same nodes, connections, and current directions, so they must be equivalent. The Analysis and Design of Linear Circuits, Sixth Edition Solutions Manual Copyright 2008 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 646-8600. Problem 2-16 In Figure P2-16, v2 = 10 V, v3 = 10 V, and v4 = 10 V. Find v1 and v5. v2v4v3v5v1 12345 Solution: Apply KVL to the circuit. clear all v2 = 10; v3 = 10; v4 = 10; v1 = v2+v3 v5 = v3-v4 v1 = 20.0000e+000 v5 = 0.0000e-003 Answer: v1 = 20 V and v5 = 0 V. The Analysis and Design of Linear Circuits, Sixth Edition Solutions Manual Copyright 2008 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 646-8600. Problem 2-17 In Figure P2-17 i2 = 10 mA, i3 = 15 mA, and i4 = 5 mA. Find i1 and i5. ABCi1i3i2i4i512345 Solution: Apply KCL to the circuit. clear all i2 = 10e-3; i3 = -15e-3; i4 = 5e-3; i1 = i2-i3+i4 i5 = i2-i1 i1 = 30.0000e-003 i5 = -20.0000e-003 Answer: i1 = 30 mA and i5 = 20 mA The Analysis and Design of Linear Circuits, Sixth Edition Solutions Manual Copyright 2008 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 646-8600. Problem 2-18 (a) Use the passive sign convention to assign voltage variables consistent with the currents in Figure P2-17. Write three KVL connection equations using these voltage variables. (b) If v3 = 0 V, what can be said about the voltages across all the other elements? ABCi1i3i2i4i512345 Solution: (a) Voltage signs: Elements 1 and 3: plus on bottom and minus on top Elements 2 and 4: plus on top and minus on bottom Element 5: plus on left and minus on right Write the KVL equations for the loops formed by 1-2, 3-4, and 2-4-5 loop 1-2: 02 1 = + v v loop 3-4: 04 3 = + v v loop 2-4-5: 05 4 2 = + v v v (b) If v3 = 0 V, then v4 = 0 V. In addition, v2 = v5 and v1 = v5. Answer: Presented above. The Analysis and Design of Linear Circuits, Sixth Edition Solutions Manual Copyright 2008 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 646-8600. Problem 2-19 The KCL equations for a three-node circuit are: Node A i1 + i2 i4 = 0 Node B i2 i3 + i5 = 0 Node C i1 + i3 + i4 i5 = 0 Draw the circuit diagram and indicate the reference directions for the element currents. Answer: R1 R2R3R4R5BACi1 i2 i4 i3 i5 The Analysis and Design of Linear Circuits, Sixth Edition Solutions Manual Copyright 2008 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 646-8600. Problem 2-20 Find vx and ix in Figure P2-20. vX2 mA47 kiX33 k Solution: Use KCL to find the current and Ohm's Law to find the voltage. clear all format short eng is = 2e-3; ix = -is vx = 47e3*ix ix = -2.0000e-003 vx = -94.0000e+000 Answer: vx = 94 V and ix = 2 mA. The Analysis and Design of Linear Circuits, Sixth Edition Solutions Manual Copyright 2008 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 646-8600. Problem 2-21 Find vx and ix in Figure P2-21. Rest ofthecircuitv10 iXX5 4 A Solution: Find the voltage across the 10- resistor using Ohm's Law. The 10- and 5- resistors are in parallel, so they have the same voltage. Find the current through the 5- resistor. The current through the 4- resistor is the sum of the currents through the other two resistors. Find the voltage across the 4- resistor. Then vx is the sum of the voltages across the 4- and 10- resistors. clear all i10 = 1/2; v10 = 10*i10; v5 = v10; i5 = v5/5; ix = i5 i4 = i10+i5; v4 = 4*i4; vx = v4+v10 ix = 1.0000e+000 vx = 11.0000e+000 Answer: vx = 11 V and ix = 1 A The Analysis and Design of Linear Circuits, Sixth Edition Solutions Manual Copyright 2008 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 646-8600. Problem 2-22 In Figure P 2-22: (a) Assign a voltage and current variable to every element. (b) Use KVL to find the voltage across each resistor. (c) Use Ohm's law to find the current through each resistor. (d) Use KCL to find the current through each voltage source. 100 5V 10V50 5V100 v4v56vACB6i5i4i2i1i3i Solution: (a) For each of the three resistors, the voltage positive sign is on the left and the negative sign is on the right. The current flows from left to right through each element. Element 1: 50- resistor. Element 2: left 100- resistor. Element 3: right 100- resistor. The left voltage source is vS1, with iS1 flowing down. The center voltage source is vS2, with iS2 flowing down. The right voltage source is vS3, with iS3 flowing down. (b) KVL equations: 0S3 1 1 S = + + v v v 02 S 2 1 S = + + v v v 03 S 3 2 S = + + v v v clear all format short eng vs1 = 5; vs2 = 10; vs3 = 5; v1 = vs1-vs3 v2 = vs1-vs2 v3 = vs2-vs3 The Analysis and Design of Linear Circuits, Sixth Edition Solutions Manual Copyright 2008 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 646-8600. v1 = 0.0000e-003 v2 = -5.0000e+000 v3 = 5.0000e+000 (c) v = iR i1 = v1/50 i2 = v2/100 i3 = v3/100 i1 = 0.0000e-003 i2 = -50.0000e-003 i3 = 50.0000e-003 (d) KCL equations 01 S 2 1 = + + i i i 0S2 3 2 = + + i i i 0S3 3 1 = + i i i is1 = -i1-i2 is2 = i2-i3 is3 = i1+i3 is1 = 50.0000e-003 is2 = -100.0000e-003 is3 = 50.0000e-003 Answer: (a) Presented above. (b) v1 = 0 V, v2 = 5 V, and v3 = 5 V (c) il = 0 mA, i2 = 50 mA, and i3 = 50 mA (d) iS1 = 50 mA, iS2 = 100 mA, and iS3 = 50 mA The Analysis and Design of Linear Circuits, Sixth Edition Solutions Manual Copyright 2008 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 646-8600. Problem 2-23 Find the power dissipated in the 1.5 k resistor in Figure P2-23. 5 mAPL500 1 k 1.5 k Solution: Label the elements. Element 1: 1-k resistor with current flowing down Element 2: 500- resistor with current flowing to the right Element 3: 1.5-k resistor with current flowing down Write KCL, KVL, and Ohm's Law equations: i1 + i2 5 mA = 0 i2 + i3 = 0 v1 + v2 + v3 = 0 v1 = 1000i1 v2 = 500i2 v3 = 1500i3 Solve the equations for v3 and i3 and then compute the power p3 = v3 i3. clear all format short eng Eqn1 = 'i1+i2-5e-3'; Eqn2 = '-i2+i3'; Eqn3 = '-v1+v2+v3'; Eqn4 = 'v1-1000*i1'; Eqn5 = 'v2-500*i2'; Eqn6 = 'v3-1500*i3'; Soln = solve(Eqn1,Eqn2,Eqn3,Eqn4,Eqn5,Eqn6,'v1','v2','v3','i1','i2','i3'); v3 = Soln.v3; i3 = Soln.i3; p3 = double(v3*i3) p3 = 4.1667e-003 Answer: PL = 4.167 mW The Analysis and Design of Linear Circuits, Sixth Edition Solutions Manual Copyright 2008 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 646-8600. vx5 k 4 mA2 k 6 mA20 V12 VixRest of the Circuit8 k vAProblem 2-24 Figure P2-24 shows a subcircuit connected to the rest of the circuit at four points. (a) Use element and connection constraints to find vx and ix. (b) Show that the sum of the currents into the rest of the circuit is zero. (c) Find the voltage vA with respect to the ground in the circuit. Solution: (a) Label the elements. Element 1: 5-k resistor with current flowing from left to right Element 2: 2-k resistor with positive voltage sign on the bottom Use Ohm's Law to compute i1. Use KCL at the center node to find ix. Use Ohm's Law to find vx. clear all v1 = 20; R1 = 5e3; i1 = v1/R1; ix = i1+4e-3 - 6e-3 Rx = 8e3; vx = ix*Rx ix = 2.0000e-003 vx = 16.0000e+000 (b) The sum of the currents into the rest of the circuit is i1 + i2 4 mA + ix. i2 = 6e-3; Current_Out = -i1+i2-4e-3+ix Current_Out = 0.0000e-003 (c) vA = 12 + vx v2 R2 = 2e3; v2 = i2*R2; vA = 12+vx-v2 vA = 16.0000e+000 Answer: (a) vx = 16 V and ix = 2 mA. The Analysis and Design of Linear Circuits, Sixth Edition Solutions Manual Copyright 2008 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 646-8600. (b) iOUT = 0 mA. (c) vA = 16 V. The Analysis and Design of Linear Circuits, Sixth Edition Solutions Manual Copyright 2008 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 646-8600. Problem 2-25 In Figure P2-25 ix = 0.5 mA. Find the value of R. R10 k 4 V15 VixRest of the Circuit10 k Solution: Label the circuit elements. Element 1: Resistor R with current flowing down. Element 2: 10-k resistor with current flowing from right to left Compute the voltage vx using Ohm's Law. Compute the voltage v1 using KVL. Compute the voltage v2 using KVL. Compute the current i2 using Ohm's Law. Compute the current i1 using KCL. Compute the resistance R1 = R using Ohm's Law. clear all Rx = 10e3; R2 = 10e3; ix = -0.5e-3; vx = ix*Rx; v1 = 4-vx; v2 = 15-v1; i2 = v2/R2; i1 = ix+i2; R1 = v1/i1; R = R1 R = 90.0000e+003 Answer: R = 90 k The Analysis and Design of Linear Circuits, Sixth Edition Solutions Manual Copyright 2008 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 646-8600. Problem 2-26 Figure P2-26 shows a resistor with one terminal connected to ground and the other connected to an arrow. The arrow symbol is used to indicate a connection to one terminal of a voltage source whose other terminal is connected to ground. The label next to the arrow indicates the source voltage at the ungrounded terminal. Find the voltage across, current through, and power dissipated in the resistor. i 100 kv-15 V Solution: The voltage across the resistor is the voltage on the right side (15 V) minus the voltage on the left side (0 V), so vx = 15 0 = 15 V. Using Ohm's Law, the current is ix = vx / Rx = 150 A. The power px = vx ix = 2.25 mW. clear all vx = -15-0 ix = vx/100e3 px = vx*ix vx = -15.0000e+000 ix = -150.0000e-006 px = 2.2500e-003 Answer: vx = 15 V, ix = 150 A, and px = 2.25 mW. The Analysis and Design of Linear Circuits, Sixth Edition Solutions Manual Copyright 2008 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 646-8600. The Analysis and Design of Linear Circuits, Sixth Edition Solutions Manual Copyright 2008 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 646-8600. Problem 2-27 Find the equivalent resistance REQ in Figure P2-27. REQ75 300 100 200 Solution: Combine the resistors working from right to left in the circuit. clear all R1 = 75; R2 = 300; R3 = 100; R4 = 200; % Combine R3 and R4 in series R34 = R3+R4; % Combine R2 in parallel with the series combination of R3 and R4 R234 = 1/(1/R2 + 1/R34); % Combine R1 in series with the other combination R1234 = R1 + R234; Req = R1234 Req = 225.0000e+000 Answer: REQ = 225 The Analysis and Design of Linear Circuits, Sixth Edition Solutions Manual Copyright 2008 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 646-8600. Problem 2-28 Find the equivalent resistance REQ in Figure P2-28. REQ3.3 k 1.5 k 2.2 k 4.7 k Solution: Combine the resistors working from right to left in the circuit. clear all R1 = 4.7e3; R2 = 3.3e3; R3 = 1.5e3; R4 = 2.2e3; R34 = 1/(1/R3 + 1/R4); R234 = R2 + R34; R1234 = 1/(1/R1 + 1/R234); Req = R1234 Req = 2.2157e+003 Answer: REQ = 2.2157 k The Analysis and Design of Linear Circuits, Sixth Edition Solutions Manual Copyright 2008 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 646-8600. Problem 2-29 Find REQ in Figure P2-29 when the switch is open. Repeat when the switch is closed. REQ200 100 50 50 Solution: With the switch open, the 200- and 50- resistors are in parallel. With the switch closed, the 200- and 50- resistors are in parallel with a short circuit, so their equivalent resistance is zero. clear all R1 = 100; R2 = 200; R3 = 50; R4 = 50; disp('Switch Open') Req_open = R1 + 1/(1/R2 + 1/R3) + R4 disp('Switch Closed') Req_closed = R1 + 0 + R4 Switch Open Req_open = 190.0000e+000 Switch Closed Req_closed = 150.0000e+000 Answer: Switch open: REQ = 190 . Switch closed: REQ = 150 . The Analysis and Design of Linear Circuits, Sixth Edition Solutions Manual Copyright 2008 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 646-8600. Problem 2-30 Show how the circuit in Figure P2-30 could be connected to achieve a resistance of 100 , 200 , 150 , 50 , 25 , 33.3 , and 133.3 . AD100 BC50 100 Solution: The required resistor combinations are described below. 100 : A single 100- resistor. 200 : Two 100- resistors in series. 150 : A 100- resistor in series with a 50- resistor. 50 : A single 50- resistor. 25 : A parallel combination of two 100- resistors and a 50- resistor 33.3 : A parallel combination of a 100- resistor and a 50- resistor 133.3 : A 100- resistor in series with a parallel combination of a 100- resistor and a 50- resistor. Answer: The following table summarizes the required connections. A plus sign indicates the nodes are connected together at one of the terminals. Resistance () Terminal 1 Terminal 2 100 A D 200 A B 150 A C 50 C D 25 A+B+C D 33.3 B+C D The Analysis and Design of Linear Circuits, Sixth Edition Solutions Manual Copyright 2008 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 646-8600. 133.3 A B+C The Analysis and Design of Linear Circuits, Sixth Edition Solutions Manual Copyright 2008 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 646-8600. Problem 2-31 In Figure P2-31 find the equivalent resistance between terminals A-B, A-C, A-D, B-C, B-D, and C-D. BCD40 40 30 10 80 60 RC-D is shown. Solution: For each pair of end terminals, combine the appropriate resistors in series and parallel to get the equivalent resistance. clear all Rab = 1/(1/40 + 1/(40+80)) + 60 Rac = 1/(1/40 + 1/(40+80)) + 30 Rad = 1/(1/40 + 1/(40+80)) + 10 Rbc = 60 + 1/(1/(40+40) + 1/80) + 30 Rbd = 60 + 1/(1/(40+40) + 1/80) + 10 Rcd = 30 + 0 + 10 Rab = 90.0000e+000 Rac = 60.0000e+000 Rad = 40.0000e+000 Rbc = 130.0000e+000 Rbd = 110.0000e+000 Rcd = 40.0000e+000 Answer: RAB = 90 RAC = 60 RAD = 40 RBC = 130 RBD = 110 RCD = 40 The Analysis and Design of Linear Circuits, Sixth Edition Solutions Manual Copyright 2008 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 646-8600. The Analysis and Design of Linear Circuits, Sixth Edition Solutions Manual Copyright 2008 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 646-8600. Problem 2-32 Select a value of RL in Figure P2-32 so that REQ = 25 k. Repeat for REQ = 20 k. REQ10 k 10 k 10 k RL Solution: Find an expression for RL in terms of REQ and then solve for RL for both values of REQ. clear all syms RL Req positive Eqn = 'Req - (10e3 + 1/(1/10e3 + 1/RL) + 10e3)'; Soln = solve(Eqn,'RL') Req_values = [25e3 20e3]; RL_values = subs(Soln,Req,Req_values) Soln = -10000.*(Req-20000.)/(Req-30000.) RL_values = 10.0000e+003 0.0000e-003 Answer: For REQ = 25 k, RL = 10 k. For REQ = 20 k, RL = 0 k. The Analysis and Design of Linear Circuits, Sixth Edition Solutions Manual Copyright 2008 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act,