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,