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Full file at https://fratstock.euThe 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
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Problem 2-1
The current through a 33-kΩ 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
Full file at https://fratstock.euThe Analysis and Design of Linear Circuits, Sixth Edition Solutions Manual
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No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical,
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Problem 2-2
A 6.2-kΩ 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
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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
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Problem 2-4
In Figure P2-4 the resistor dissipates 25 mW. Find Rx.
RX15 V
P =25 mWX
Solution:
R
vp
2
clear all
p = 25e-3;
v = 15;
R = v^2/p R =
9.0000e+003
Answer:
R = 9 kΩ
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Problem 2-5
In Figure P2-5 find Rx and the power delivered to the resistor.
R X
10 mA
100 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 kΩ, p = 1 W
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No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical,
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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-Ω 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
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-40 -30 -20 -10 0 10 20 30 40-0.5
-0.4
-0.3
-0.2
-0.1
0
0.1
0.2
0.3
0.4
0.5
Voltage (V)
Curr
ent
(A)
Nonlinear
Linear
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
Full file at https://fratstock.euThe Analysis and Design of Linear Circuits, Sixth Edition Solutions Manual
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Problem 2-7
A 10-kΩ resistor has a power rating of ⅛W. 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
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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
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Problem 2-8
A certain type of film resistor is available with resistance values between 10 Ω and 100 MΩ.
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 MΩ, and that the power rating is the controlling limit
when R < 1 MΩ.
Solution: clear all
V = 500;
p = 1/4;
R = V^2/p R =
1.0000e+006
R
vp
2
At R = 1 MΩ, both p and v can take their maximum values and there are no issues. For R >
1 MΩ, 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 MΩ, 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.
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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
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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 Ae i v 1102 4016
(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
+
-
v
i
Full file at https://fratstock.euThe Analysis and Design of Linear Circuits, Sixth Edition Solutions Manual
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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
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-0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8-2
0
2
4
6
8
10
12
14
16x 10
-3
Voltage (V)
Curr
ent
(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 10−16 A and p = 1.00 10−15 W. The model is valid because the
current and power are both essentially zero.
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Problem 2-10
In Figure P2-10 i2 = –2 A and i3 = 5 A. Find i1 and i4.
BC
i3
i4
i2i1
A 1 2
3
4
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.
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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.
1
3i
4
31i 4i
BA
C
2
2i
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: 0321 iii ;
node B: 043 ii ;
node C: 0421 iii
KVL: loop 1-2: 021 vv ;
loop 2-3-4: 0432 vvv ;
loop 1-3-4: 0431 vvv
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Problem 2-12
In Figure P2-11, i2 = –10 mA and i4 = 20 mA. Find i1 and i3.
1
3i
4
31i 4i
BA
C
2
2i
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.
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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.
2
3i
6
32i 6i
BA
CD5
5i5
v
6v
3v
2v
1i1v
4v4
1
4i
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: 0432 iii ;
node B: 0631 iii ;
node C: 0521 iii ;
node D: 0654 iii
KVL: loop 1-3-2: 0231 vvv ;
loop 2-4-5: 0542 vvv ;
loop 3-6-4: 0463 vvv
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Problem 2-14
In Figure P2-13 v2 = 10 V, v3 = –10 V, and v4 = 3 V. Find v1, v5, and v6.
2
3i
6
32i 6i
BA
CD5
5i5
v
6v
3v
2v
1i1v
4v4
1
4i
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.
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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.
3i
6i
B
A
C
D
4i 5i
2i1i
3 61 2 4 5
Solution:
(a) There are four nodes and at least five loops.
(b) KCL: node A: 0432 iii ;
node B: 0631 iii ;
node C: 0521 iii ;
node D: 0654 iii
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
2
3i
6
32i 6i
BA
CD5
5i5
v
6v
3v
2v
1i1v
4v
4
1
4i
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No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical,
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(b) KCL: node A: 0432 iii ;
node B: 0631 iii ;
node C: 0521 iii ;
node D: 0654 iii
(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.
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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
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Problem 2-16
In Figure P2-16, v2 = 10 V, v3 = 10 V, and v4 = 10 V. Find v1 and v5.
v2 v4
v3 v5v1 1
2
3
4
5
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.
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No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical,
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Problem 2-17
In Figure P2-17 i2 = 10 mA, i3 = –15 mA, and i4 = 5 mA. Find i1 and i5.
A
B C
i
1
i3
i2
i4
i5
1
2
3
4
5
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
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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?
A
B C
i
1
i3
i2
i4
i5
1
2
3
4
5
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: 021 vv
loop 3-4: 043 vv
loop 2-4-5: 0542 vvv
(b) If v3 = 0 V, then v4 = 0 V. In addition, v2 = v5 and v1 = v5.
Answer:
Presented above.
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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 R2
R3
R4
R5
B
A
C
i1 i2 i4
i3
i5
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Problem 2-20
Find vx and ix in Figure P2-20.
vX2 mA
47 kΩ
iX
33 kΩ
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.
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Problem 2-21
Find vx and ix in Figure P2-21.
Rest of
the
circuit
v10 Ω
i
X
X
5 Ω
4 Ω
½ A
Solution:
Find the voltage across the 10-Ω resistor using Ohm's Law. The 10-Ω and 5-Ω resistors are in
parallel, so they have the same voltage. Find the current through the 5-Ω resistor. The current
through the 4-Ω resistor is the sum of the currents through the other two resistors. Find the
voltage across the 4-Ω resistor. Then vx is the sum of the voltages across the 4-Ω and 10-Ω
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
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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 Ω
5V 10V
50 Ω
5V
100 Ω
v4 v5
6v
A CB
6i
5i4i
2i1i 3i
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-Ω resistor.
Element 2: left 100-Ω resistor.
Element 3: right 100-Ω 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:
0S311S vvv
02S21S vvv
03S32S vvv
clear all
format short eng
vs1 = 5;
vs2 = 10;
vs3 = 5;
v1 = vs1-vs3
v2 = vs1-vs2
v3 = vs2-vs3
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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
01S21 iii
0S232 iii
0S331 iii
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
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Problem 2-23
Find the power dissipated in the 1.5 kΩ resistor in Figure P2-23.
5 mA
PL
500 Ω
1 kΩ 1.5 kΩ
Solution:
Label the elements.
Element 1: 1-kΩ resistor with current flowing down
Element 2: 500-Ω resistor with current flowing to the right
Element 3: 1.5-kΩ 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
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vx
5 k4 mA
2 k
6 mA
20 V
12 V
ix
Rest of the Circuit
8 k
vA
Problem 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-kΩ resistor with current flowing from left to right
Element 2: 2-kΩ 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.
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(b) iOUT = 0 mA.
(c) vA = 16 V.
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Problem 2-25
In Figure P2-25 ix = –0.5 mA. Find the value of R.
R
10 k
4 V 15 V
i x
Rest of the Circuit
10 k
Solution:
Label the circuit elements.
Element 1: Resistor R with current flowing down.
Element 2: 10-kΩ 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 kΩ
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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.
i100 kΩ
v
-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.
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Problem 2-27
Find the equivalent resistance REQ in Figure P2-27.
REQ
75
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 Ω
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Problem 2-28
Find the equivalent resistance REQ in Figure P2-28.
REQ3.3 k
1.5 k2.2 k4.7 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 kΩ
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Problem 2-29
Find REQ in Figure P2-29 when the switch is open. Repeat when the switch is closed.
REQ
200
100
50 50
Solution:
With the switch open, the 200-Ω and 50-Ω resistors are in parallel. With the switch closed, the
200-Ω and 50-Ω 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 Ω.
Switch closed: REQ = 150 Ω.
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Problem 2-30
Show how the circuit in Figure P2-30 could be connected to achieve a resistance of 100 Ω, 200
Ω, 150 Ω, 50 Ω, 25 Ω, 33.3 Ω, and 133.3 Ω.
A
D
100
B
C
50
100
Solution:
The required resistor combinations are described below.
100 Ω: A single 100-Ω resistor.
200 Ω: Two 100-Ω resistors in series.
150 Ω: A 100-Ω resistor in series with a 50-Ω resistor.
50 Ω: A single 50-Ω resistor.
25 Ω: A parallel combination of two 100-Ω resistors and a 50-Ω resistor
33.3 Ω: A parallel combination of a 100-Ω resistor and a 50-Ω resistor
133.3 Ω: A 100-Ω resistor in series with a parallel combination of a 100-Ω resistor and a 50-Ω
resistor.
Answer:
The following table summarizes the required connections. A plus sign indicates the nodes are
connected together at one of the terminals.
Resistance (Ω) 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
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133.3 A B+C
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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.
B C
D
40 Ω
40 Ω 30 Ω
10 Ω80 Ω60 Ω
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 Ω
RAC = 60 Ω
RAD = 40 Ω
RBC = 130 Ω
RBD = 110 Ω
RCD = 40 Ω
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Problem 2-32
Select a value of RL in Figure P2-32 so that REQ = 25 kΩ. Repeat for REQ = 20 kΩ.
REQ
10 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 kΩ, RL = 10 kΩ.
For REQ = 20 kΩ, RL = 0 kΩ.
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Problem 2-33
Using no more than four 1-kΩ resistors, show how the following equivalent resistors can be
constructed: 2 kΩ, 500 Ω, 1.5 kΩ, 333 Ω, 250 Ω, and 400 Ω.
Solution:
R = 2 kΩ: two resistors in series
R = 500 Ω: two resistors in parallel
R = 1.5 kΩ: one resistor in series with a parallel combination of two resistors
R = 333 Ω: three resistors in parallel;
R = 250 Ω: four resistors in parallel;
R = 400 Ω: two resistors in series in parallel with two resistors in parallel
clear all
R = 1000;
R2000 = R + R
R500 = 1/(1/R + 1/R)
R1500 = R + 1/(1/R + 1/R)
R333 = 1/(1/R + 1/R + 1/R)
R250 = 1/(1/R + 1/R + 1/R + 1/R)
R400 = 1/(1/(R+R) + 1/R + 1/R) R2000 =
2.0000e+003
R500 =
500.0000e+000
R1500 =
1.5000e+003
R333 =
333.3333e+000
R250 =
250.0000e+000
R400 =
400.0000e+000
Answer:
Presented above.
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Problem 2-34
Find the equivalent practical voltage source at terminals A and B in Figure P2-34
A
5 A
5 Ω
10 Ω
B
Solution:
A current source in series with a resistor is equivalent to just the current source, so we can
remove the 5-Ω resistor. That leaves a 5-A current source in parallel with a 10-Ω resistor. The
current source and parallel resistor can be converted into a voltage source in series with the same
resistor. The value for the voltage source follows Ohm's Law, so vS = (5 A)(10 Ω) = 50 V.
5
105Adc
105Adc
50Vdc
10
Answer:
vS = 50 V and RS = 10 Ω.
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Problem 2-35
In Figure P2-35 the i-v characteristic of network N is v + 50i = 5 V. Find the equivalent practical
current source for the network.
N
i
v
A
B
Solution:
When the circuit is open between nodes A and B, there is no current, i = 0 A, and the voltage
must be v = 5 V. When a short is placed between nodes A and B, the voltage is zero, v = 0 V,
and the current is i = 100 mA. The corresponding practical current source will have a current iS
= 100 mA and a parallel resistance R = (5 V)/(100 mA) = 50 Ω.
50100mA
Answer:
iS = 100 mA and RS = 50 Ω.
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Problem 2-36
Select the value of Rx in Figure P2-36 so that REQ = 60 kΩ.
REQ
47 k
22 k
RX
10 k
Solution:
Find an expression for REQ in terms of Rx and then solve for Rx in terms of REQ.
clear all
syms Rx Req positive
Eqn = 'Req - (47e3 + 1/(1/22e3 + 1/(Rx+10e3)))';
Soln = solve(Eqn,'Rx');
Rx_value = subs(Soln,Req,60e3)
Check_Req = 47e3 + 1/(1/22e3 + 1/(Rx_value+10e3)) Rx_value =
21.7778e+003
Check_Req =
60.0000e+003
Answer:
Rx = 21.78 kΩ.
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Problem 2-37
Two 10-kΩ potentiometers (a variable resistor whose value between the two ends is 10 kΩ and
between one end and the wiper – the third terminal – can range from 0 Ω to 10 kΩ) are
connected as shown in Figure P2-37. What is the range of REQ?
REQ
10 k
10 k
Solution:
At the limits of their settings, the two potentiometers are either in series or parallel. These
represent the maximum and minimum equivalent resistances that the combination can take.
clear all
R = 10e3;
Rmax = R + R
Rmin = 1/(1/R + 1/R) Rmax =
20.0000e+003
Rmin =
5.0000e+003
Answer:
5 kΩ REQ 20 kΩ
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Problem 2-38
Select the value of R in Figure P2-38 so that RA-B = RL.
A
B
R R
4R R L
RA-B
Solution:
Find an expression for RA-B in terms of R and RL. Set RA-B equal to RL. Solve for R in terms of
RL and choose the positive solution for the resistance.
clear all
syms R Rab RL positive
Eqn = 'RL - (R + 1/(1/4/R + 1/(R+RL)))';
Soln = solve(Eqn,'R')
R = RL/3;
Check_Rab = R + 1/(1/4/R + 1/(R+RL)) Soln =
-1/3*RL
1/3*RL
Check_Rab =
RL
Answer:
R = RL/3.
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Problem 2-39
What is the range of REQ in Figure P2-39?
4 kΩ
REQ
1 kΩ
4 kΩ
Solution:
The potentiometer varies between 0 and 4 kΩ. Find REQ with the potentiometer to its maximum
and minimum values.
clear all
Rp = [0 4e3];
Req = 1e3 + 1./(1./Rp + 1/4e3) Req =
1.0000e+003 3.0000e+003
Answer:
1 kΩ REQ 3 kΩ
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Problem 2-40
Find the equivalent resistance between terminals A and B in Figure P2-40
A BR R R
Solution:
Place a voltage source between Nodes A and B and redraw the circuit as the equivalent circuit
shown below
Vs
R1 R2
R3
0
Use KVL to show that each resistor experiences a voltage drop equal to the voltage at the source:
for R1, the drop is from left to right, for R2, the drop is from right to left and for R3, and the drop
is from top to bottom. Therefore, each resistor has current flowing in the direction of the voltage
drop equal to vS/R. Applying KCL at the node above the voltage source, the current flowing out
of the voltage source is iS = vS/R + vS/R + vS/R = 3 vS/R. The equivalent resistance is the ratio of
vS to iS, REQ = vS/iS = R/3.
Alternatively, rearrangement of the circuit shows that the three resistors are connected in
parallel, which yields the same final result.
Answer:
REQ = vS/iS = R/3.
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Problem 2-41
Use voltage division in Figure P2-41 to obtain an expression for vL in terms of R, RL, and vS.
vLR
R
RLv
S
Solution:
Find an equivalent resistance for the two resistors in parallel. The voltage vL appears across this
parallel combination, so use voltage division with the equivalent resistance and the resistor in
series with the source.
clear all
syms vs R RL Req vL
Req = 1/(1/R + 1/RL);
vL = simple(Req*vs/(Req+R)) vL =
vs*RL/(2*RL+R)
Answer:
L
SLL
2RR
vRv
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Problem 2-42
Use current division in Figure P2-42 to find vL if R =10 Ω, RL = 20 Ω, and iS = 2.5 mA.
vLR
R
RLiS
Solution:
Use current division to find the current through the series combination of R and RL and then use
Ohm's Law to compute the voltage vL.
clear all
R = 10;
RL = 20;
is = 2.5e-3;
Req = R + RL;
iL = (1/Req)*is/(1/Req + 1/R);
vL = iL*RL vL =
12.5000e-003
Answer:
vL = 12.5 mV.
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Problem 2-43
Find ix and iy in Figure P2-43.
20
5
6 30 i y2 A
ix
Solution:
Combine the 20-Ω and 5-Ω resistors in parallel and then combine that result in series with the 6-
Ω resistor. Use current division to find ix. Use KCL to find the current entering the parallel
combination of the 20-Ω and 5-Ω resistors and then use current division again to find the current
through the 5-Ω resistor, which is iy.
clear all
Req1 = 1/(1/20 + 1/5);
Req2 = Req1 + 6;
is = 2;
ix = (1/30)*is/(1/30 + 1/Req2)
iReq1 = is - ix;
iy = (1/5)*iReq1/(1/5 + 1/20) ix =
500.0000e-003
iy =
1.2000e+000
Answer:
ix = 500 mA and iy = 1.2 A.
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Problem 2-44
The 1-kΩ load in Figure P2-44 needs 5 V across it to operate correctly. Where should the wiper
on the potentiometer be set (RX) to obtain the desired output voltage?
5 V
5 kΩ24 V
1 kΩ}RX
Solution:
Redraw the circuit as shown below to clearly identify the potentiometer settings and the resistor
values on each side of the wiper.
5k - Rx
Rx
1k
24Vdc
+5 V
-
Find an equivalent resistance for the parallel combination. Use voltage division to find an
expression for the voltage across the parallel combination in terms of RX. Solve for RX when the
voltage across the parallel combination is 5 V. Choose the positive solution.
clear all
syms Rx RL Req vL
Eqn1 = 'Req - 1/(1/Rx + 1/1e3)';
Eqn2 = '5 - (Req*24/(Req + 5e3 - Rx))';
Soln = solve(Eqn1,Eqn2,'Req','Rx');
Rx = double(Soln.Rx)
Rx = max(Rx);
Req_Check = 1/(1/Rx + 1/1e3);
V_Check = Req_Check*24/(Req_Check + 5e3 - Rx) Rx =
2.3383e+003
-2.1383e+003
V_Check =
5.0000e+000
Answer:
RX = 2.3383 kΩ.
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Problem 2-45
Find the range of values of vO in Figure P2-45.
100 Ω
vO
100 Ω
100 Ω24 V
Solution:
Find an expression for vO in terms of the value of the potentiometer and then evaluate the
expression for the maximum and minimum values of the potentiometer.
clear all
Rp = [0,100];
Req = 1./(1./Rp + 1/100);
vs = 24;
vo = Req*vs./(Req + 100) vo =
0.0000e-003 8.0000e+000
Answer:
0 V vO 8 V
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Problem 2-46
Figure P2-46 shows a voltage bridge circuit, that is, two voltage dividers in parallel with a source
vS. One resistor RX is variable. The goal is often to “balance” the bridge by making vX = 0 V.
Derive an expression for RX in terms of the other resistors for when the bridge is balanced.
VS
RB
vX
RA RC
RX
Solution:
Let the node between resistors RA and RB have a voltage v1 and let the node between resistors RC
and RX have a voltage v2. The goal is to make v1 equal v2. Use voltage division to derive
expressions for the two voltages. Set the expressions equal and solve for RX.
BA
SB1
RR
vRv
and
XC
SX2
RR
vRv
clear all
syms vs Ra Rb Rc Rx
Eqn = 'Rb*vs/(Ra+Rb) - Rx*vs/(Rc+Rx)';
Soln = solve(Eqn,'Rx') Soln =
Rb*Rc/Ra
Answer:
A
CBX
R
RRR
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Problem 2-47
Ideally, a voltmeter has infinite internal resistance and can be placed across any device to read
the voltage without affecting the result. A particular digital multimeter (DMM), a common
laboratory tool, is connected across the circuit shown in Figure P2-47. The expected voltage was
5.73 V. However, the DMM reads 3.81 V. The large, but finite, internal resistance of the DMM
was “loading” the circuit and causing a wrong measurement to be made. Find the value of the
internal resistance of this DMM.
DMM4.7 MΩ
6.3 MΩ10 V RM
Solution:
The DMM is in parallel with the 6.3-MΩ resistor, so they share the same voltage. Calculate the
current through the 6.3-MΩ resistor. The remaining voltage must drop across the 4.7-MΩ
resistor, so use that voltage and Ohm's Law to calculate the current through the 4.7-MΩ resistor.
Use KCL to find the current through the DMM and then Ohm's Law to find the internal
resistance of the DMM.
clear all
R1 = 4.7e6;
R2 = 6.3e6;
vs = 10;
v2 = 3.81;
i2 = v2/R2;
v1 = vs - v2;
i1 = v1/R1;
iDMM = i1 - i2;
RDMM = v2/iDMM RDMM =
5.3492e+006
Answer:
RDMM = 5.3492 MΩ.
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Problem 2-48
Select values for R1, R2, and R3 in Figure P 2-48 so the voltage divider produces the two output
voltages shown.
10 V
15 V
R1
3 VR3
R2
Solution:
If we select the resistors such that the total equivalent resistance seen by the source is 15 kΩ,
then 1 mA will flow through each resistor. We can then see that R1 = 5kΩ, R2 = 7 kΩ, and R3 = 3
kΩ will solve the problem.
clear all
vs = 15;
R1 = 5e3;
R2 = 7e3;
R3 = 3e3;
v23 = (R2+R3)*vs/(R1+R2+R3)
v3 = R3*vs/(R1+R2+R3) v23 =
10.0000e+000
v3 =
3.0000e+000
Answer:
R1 = 5kΩ, R2 = 7 kΩ, and R3 = 3 kΩ. Other solutions are possible.
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Problem 2-49
Select a value of Rx in Figure P2-49 so that vL = 3 V.
12 V vL
1 k R X
1 k 1 k
Solution:
The voltage vL appears across a parallel combination of two 1-kΩ resistors, which have an
equivalent resistance of 500 Ω. The current through the equivalent resistor must be 6 mA. That
same current must flow through the series combination of the 1-kΩ resistor and Rx and the pair
must drop a total of 9 V. The 1-kΩ resistor with 6 mA of current drops 6 V, so Rx must drop 3
V. To carry 6 mA of current and drop 3 V, Rx = 500 Ω.
clear all
R = 1e3;
vs = 12;
Req = 1/(1/R+1/R);
vL = 3;
iReq = vL/Req;
is = iReq;
ix = is;
vx = vs-is*R-vL;
Rx = vx/ix Rx =
500.0000e+000
Answer:
Rx = 500 Ω.
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Problem 2-50
Select a value of Rx in Figure P2-50 so that vL = 2 V. Repeat for 4 V and 6 V. Caution: Rx must
be positive.
12 V vL50 Ω
100
R X
Solution:
clear all
vL = [2 4 6];
vs = 12;
R1 = 100;
RL = 50;
v1 = vs-vL;
i1 = v1/R1;
iL = vL/RL;
ix = i1-iL;
vx = vL;
Rx = vx./ix;
Rx = Rx' Rx =
33.3333e+000
Inf
-100.0000e+000
Answer:
For vL = 2 V, select Rx = 33.33 Ω.
For vL = 4 V, make Rx an open circuit.
For vL = 6 V, there is no positive value for Rx that will meet this specification.
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Problem 2-51
Use circuit reduction to find vx and ix in Figure P2-51.
vx3R
2R
RiS
ix
Solution:
Find the equivalent resistance of the entire circuit. Use Ohm's Law first to compute vx and then
again to compute ix.
clear all
syms R is ix vx Req1 Req2
Req1 = 2*R + R;
Req2 = 1/(1/3/R+1/Req1);
vx = is*Req2
ix = vx/3/R vx =
3/2*is*R
ix =
1/2*is
Answer:
vx = 1.5 R iS, ix = 0.5 iS
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Problem 2-52
Use circuit reduction to find vx and ix in Figure P2-52.
vx 2R R
ix
2R
2R
vS
R
Solution:
Find the equivalent resistance of the entire circuit and compute the current through the source.
Use the source current and the equivalent resistance seen by vx to compute vx. Use the source
current and current division to calculate ix.
clear all
syms R vs vx ix Req1 Req2
Req1 = 1/(1/R + 1/R/2);
Req2 = 2*R + 2*Req1;
is = vs/Req2;
vx = Req1*is
ix = (1/R/2)*(-is)/(1/R + 1/R/2) vx =
1/5*vs
ix =
-1/10*vs/R
Answer:
5
Sx
vv
R
vi
10
Sx
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Problem 2-53
Use circuit reduction to find vx and the power provided by the source in Figure P2-53.
10 Ωv
20 Ω 20 Ω
X
¾ A
20 Ω 10 Ω10 Ω
Solution:
First, find the power provided by the source by combining all of the resistors to determine an
overall equivalent resistance. We can then calculate p = i2R. To find vx, combine the resistors to
the right of vx to determine their equivalent resistance and then use a source transformation to
convert the problem into a voltage division problem. Note: You cannot find the power provide
by the original source by finding the power provided by an equivalent transformed source.
clear all
Req1 = 1/(1/10 + 1/10);
Req2 = 20+Req1;
Req3 = 1/(1/10 + 1/Req2);
Req4 = 20+Req3;
Req = 1/(1/20 + 1/Req4);
is = 3/4;
Rs = 20;
ps = is^2*Req
% Source transformation
vs = is*Rs;
vx = Req3*vs/(Rs + 20 + Req3) ps =
6.4773e+000
vx =
2.2727e+000
Answer:
vx = 2.2727 V and pS = 6.4773 W.
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Problem 2-54
Use circuit reduction to find vx and ix in Figure P2-54.
8 kΩ
iX
12 kΩ12 V
6 kΩ
4 kΩ vX
Solution:
Perform a source transformation and then use current division to find ix and the current
associated with vx. Use Ohm's Law to find vx.
clear all
vs = 12;
Rs = 6e3;
is = vs/Rs;
ix = (1/12e3)*is/(1/12e3 + 1/6e3 + 1/12e3)
ivx = (1/12e3)*is/(1/12e3 + 1/6e3 + 1/12e3);
vx = ivx*4e3 ix =
500.0000e-006
vx =
2.0000e+000
Answer:
vx = 2 V and ix = 500 A.
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Problem 2-55
Use source transformation to find ix in Figure P2-55.
50 Ω
30 Ω 200 mA10 V
ix
Solution:
Perform a source transformation on the current source and the 30-Ω resistor and then use Ohm's
Law to find ix.
clear all
vs1 = 10;
Rs1 = 50;
is2 = 200e-3;
Rs2 = 30;
vs2 = is2*Rs2;
ix = (vs1-vs2)/(Rs1+Rs2) ix =
50.0000e-003
Answer:
ix = 50 mA.
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Problem 2-56
Select a value for RX so that ix = 0 A in Figure P2-56.
30 ΩRX
20 Ω-24 ViX=0
12 V
Solution:
Perform source transformations on both voltage sources to get the following equivalent circuit:
24/Rx400mAdc
Rx 20 30
All three resistors share the same voltage because they are in parallel. If ix = 0 A, then there is
no voltage drop across any of the resistors and no current flowing through them. All of the
current from the right source must flow through the left source, so we can compute RX.
clear all
vs1 = 24;
vs2 = 12;
Rs2 = 30;
is2 = vs2/Rs2
Rx = vs1/is2 is2 =
400.0000e-003
Rx =
60.0000e+000
Answer:
RX = 60 Ω.
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Problem 2-57
Use source transformations in Figure P2-57 to relate vO to v1, v2, and v3.
R
v
v
1
R
v2
R
v3
O
Solution:
Convert each voltage source and resistor pair into a current source in parallel with a resistor to
get the following circuit:
We can then add the three current
sources, since they are in parallel, and
combine the three parallel resistors to get
the following circuit:
The output voltage is then the product of the equivalent current source
and the equivalent resistance
33
321321O
vvvR
R
vvvv
clear all
syms v1 v2 v3 R i1 i2 i3 Req ieq vo
i1 = v1/R;
i2 = v2/R;
i3 = v3/R;
Req = 1/(1/R + 1/R + 1/R);
ieq = i1 + i2 + i3;
vo = simple(ieq*Req) vo =
1/3*v1+1/3*v2+1/3*v3
Answer:
3
321O
vvvv
v1/RR R R
v2/R v3/R
(v1+v2+v3) /RR/3
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Problem 2-58
The current through RL in Figure P2-58 is 40 mA. Use source transformations to find RL.
RL
50 Ω100 Ω
100 Ω10 V RL
50 Ω
100 Ω100 m A 100 Ω
50 Ω
RL
50 Ω50 Ω
5 V
Solution:
Perform a source transformation and then combine the two 100-Ω resistors in parallel to get a
50-Ω equivalent resistor. The resulting current source has a value of 100 mA. Since 40 mA
flows through RL, the remaining 60 mA must flow through the equivalent 50-Ω resistor and it
causes a voltage drop of 3 V. The series path with the original 50-Ω resistor and RL must also
drop a total of 3 V, so the total equivalent resistance of the path must be (3 V)/(40 mA) = 75 Ω.
Therefore RL = 25 Ω.
clear all
iL = 40e-3;
vs = 10;
Rs = 100;
is = vs/Rs;
Req = 1/(1/100 + 1/100);
iReq = is - iL;
vReq = iReq*Req;
v50L = vReq;
R50L = v50L/iL;
RL = R50L - 50 RL =
25.0000e+000
Answer:
RL = 25 Ω.
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Problem 2-59
Select RX so that 25 V are across it in Figure P2-59.
500 Ω RX
200 Ω100 V
800 Ω
1 kΩ 1 kΩ
25 V
Solution:
Reduce the circuit on both sides of RX. To the right of RX, the equivalent resistance is 500 Ω. To
the left of RX, after a source transformations, a parallel resistor combination, and another source
transformation, we have a voltage source vS =
66.7 V in series with a 333-Ω resistor. The
resulting equivalent circuit is shown below:
The two known resistors can be combined in series to reduce the circuit further:
If RX drops 25 V, then the 833-Ω resistor must drop (66.7 25) V
= 41.7 V. The current through the circuit must be 41.7/833 = 50
mA. We can then calculate RX = (25 V)/(50 mA) = 500 Ω.
clear all
vx = 25;
Req_right = 1/(1/1000 + 1/(800+200));
vs = 100;
Rs = 500;
is = vs/Rs;
Req_left = 1/(1/500 + 1/1000);
vs2 = is*Req_left;
Req = Req_left + Req_right;
vReq = vs2 - vx;
iReq = vReq/Req;
ix = iReq;
Rx = vx/ix Rx =
500.0000e+000
Answer:
RX = 500 Ω.
333 Rx
50066.7Vdc
Rx
83366.7Vdc
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Problem 2-60
The box in the circuit in Figure P2-60 is a resistor whose value can be anywhere between 8 kΩ
and 80 kΩ. Use circuit reduction to find the range of values of vx.
10 kΩ
50 V 10 kΩ 10 kΩ vx
Solution:
The solution is presented in the commented MATLAB script below.
clear all
R = 10e3;
% Source transformation to a current source
vs = 50;
Rs = 10e3;
is = vs/Rs
% Combine parallel resistors
Req = 1/(1/Rs + 1/R)
% Source transformation back to a voltage source
vs2 = is*Req
% Unknown resistor range
Rp = [8e3 80e3];
% Voltage division to calculate vx
vx = R*vs2./(R+Rp+Req) is =
5.0000e-003
Req =
5.0000e+003
vs2 =
25.0000e+000
vx =
10.8696e+000 2.6316e+000
Answer:
2.6316 V vx 10.8696 V
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Problem 2-61
A circuit is found to have the following element and connection equations:
v1 = 24 V
v2 = 8k i2
v3 = 5k i3
v4 = 4k i4
v5 = 16k i5
–v1 + v2 + v3 = 0
–v3 + v4 + v5 = 0
i1 + i2 = 0
–i2 + i3 +i4 = 0
–i4 + i5 = 0
Use MATLAB to solve for all of the unknown voltages and currents associated with this circuit.
Sketch one possible schematic that matches the given equations.
Solution:
Write the equations in matrix form using a vector of unknowns as:
x = [v1, v2, v3, v4, v5, i1, i2, i3, i4, i5].
clear all
A = [1 0 0 0 0 0 0 0 0 0;
0 1 0 0 0 0 -8000 0 0 0;
0 0 1 0 0 0 0 -5000 0 0;
0 0 0 1 0 0 0 0 -4000 0;
0 0 0 0 1 0 0 0 0 -16000;
-1 1 1 0 0 0 0 0 0 0;
0 0 -1 1 1 0 0 0 0 0;
0 0 0 0 0 1 1 0 0 0;
0 0 0 0 0 0 -1 1 1 0;
0 0 0 0 0 0 0 0 -1 1];
B = [24 0 0 0 0 0 0 0 0 0]';
x = A\B x =
24.0000
16.0000
8.0000
1.6000
6.4000
-0.0020
0.0020
0.0016
0.0004
0.0004
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One possible circuit design that matches the given equations is shown below.
24Vdc
8k
5k
4k
16k
Answer:
[v1, v2, v3, v4, v5, i1, i2, i3, i4, i5]
= [24 V, 16 V, 8 V, 1.6 V, 6.4 V, 2 mA, 2 mA, 1.6 mA, 0.4 mA, 0.4 mA]
The schematic is shown above.
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Problem 2-62
Consider the circuit of Figure P2-62. Use MATLAB to find all of the voltages and currents in
the circuit.
33 kΩ
22 kΩ120 V
68 kΩ
10 kΩ 15 kΩ
15 kΩ
Solution:
Use KVL, KCL, and Ohm's Law to create a set of 14 equations to solve for all of the voltages
and currents. Also compute the powers for each element. The unknown vector is:
x = [vS, v1, v2, v3, v4, v5, v6, iS, i1, i2, i3, i4, i5, i6]
The equations and solution are presented in the following MATLAB code.
clear all
format short eng
% Set the matrix to be all zeros and then assign individual values directly
A = zeros(14,14);
A(1,1) = 1;
A(2,2) = 1; A(2,9) = -15e3;
A(3,3) = 1; A(3,10) = -10e3;
A(4,4) = 1; A(4,11) = -33e3;
A(5,5) = 1; A(5,12) = -15e3;
A(6,6) = 1; A(6,13) = -68e3;
A(7,7) = 1; A(7,14) = -22e3;
A(8,8) = 1; A(8,9) = 1;
A(9,9) = 1; A(9,10) = -1; A(9,11) = -1;
A(10,11) = 1; A(10,12) = -1; A(10,13) = -1;
A(11,13) = 1; A(11,14) = -1;
A(12,1) = -1; A(12,2) = 1; A(12,3) = 1;
A(13,3) = -1; A(13,4) = 1; A(13,5) = 1;
A(14,5) = -1; A(14,6) = 1; A(14,7) = 1;
% Display the A matrix
%A
B = zeros(14,1);
B(1) = 120;
% Display the B matrix
%B
% Solve for the output vector
x = A\B;
Results = x';
% Create vectors of the voltages and currents
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Voltages = x(1:7)
Currents = x(8:14)
% Compute the associated powers
Powers = Voltages.*Currents
PTotal = sum(Powers) Voltages =
120.0000e+000
77.5537e+000
42.4463e+000
30.5455e+000
11.9008e+000
8.9917e+000
2.9091e+000
Currents =
-5.1702e-003
5.1702e-003
4.2446e-003
925.6198e-006
793.3884e-006
132.2314e-006
132.2314e-006
Powers =
-620.4298e-003
400.9720e-003
180.1687e-003
28.2735e-003
9.4420e-003
1.1890e-003
384.6732e-006
PTotal =
-101.2103e-018
Answer:
The results are summarized in the table below:
Element Voltage (V) Current (mA) Power (mW)
Source 120.0000 -5.1702 -620.4298
1 77.5537 5.1702 400.9720
2 42.4463 4.2446 180.1687
3 30.5455 0.9256 28.2735
4 11.9008 0.7934 9.4420
5 8.9917 0.1322 1.1890
6 2.9091 0.1322 0.3847
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Problem 2-63
Consider the circuit of Figure P2-62 again. Use OrCAD to find all of the voltages and currents
in the circuit.
Solution:
The results of the OrCAD simulation are shown below.
Answer:
Presented above.
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Problem 2-64
Use OrCAD to find all of the power dissipated or provided in the circuit of Figure P2-62. Verify
that the sum of all power in the circuit is zero.
Solution:
The results of the OrCAD simulation are shown below.
pTOTAL = 620.4 + 401.0 + 180.2 + 28.27 + 9.442 + 1.189 + 0.3847 = 0 mW
Answer:
The schematic is shown above. The sum of the powers is p = 0 mW.
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Problem 2-65
Nonlinear Device Characteristics (A)
The circuit in Figure P2-65 is a parallel combination of a 50-Ω linear resistor and a varistor
whose i-v characteristic is iV = 2.6 10-5v3. For a small voltage, the
varistor current is quite small compared to the resistor current. For
large voltages, the varistor dominates because its current increases
more rapidly with voltage.
(a) Plot the i-v characteristic of the parallel combination.
(b) State whether the parallel combination is linear or nonlinear,
active or passive, and bilateral or nonbilateral.
(c) Find the range of voltages over which the resistor current is at
least 10 times as large as the varistor current.
(d) Find the range of voltages over which the varistor current is at least 10 times as large as the
resistor current.
Solution:
The solution is presented in the following commented MATLAB code.
(a) clear all
disp('Part (a)')
% Set the range of voltages to plot
v = -200:1:200;
% Compute the current through the resistor
iR = v/50;
% Compute the current through the varistor
iV = 2.6e-5*v.^3;
% Sum the two path currents to get the total current
iTotal = iR + iV;
% Plot the i-v characteristic
plot(v,iTotal,'b','LineWidth',3)
xlabel('Voltage (V)')
ylabel('Current (A)')
grid on Part (a)
50 Ωv
i i v
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-200 -150 -100 -50 0 50 100 150 200-250
-200
-150
-100
-50
0
50
100
150
200
250
Voltage (V)
Curr
ent
(A)
(b) The plot created in Part (a) shows that the parallel combination is nonlinear, passive, and
bilateral.
(c) Solve the equations for positive voltages and realize that the full range will include negative
voltages.
clear all
syms v
Eqn = 'v/50 - 10*(2.6e-5*v^3)';
Soln = double(solve(Eqn,'v')) Soln =
0
8.7706
-8.7706
8.7706 V < v < 8.7706 V
(d) Solve the equations for positive voltages and realize that the full range will include negative
voltages.
clear all
syms v
Eqn = '2.6e-5*v^3 - v/5';
Soln = double(solve(Eqn,'v')) Soln =
0
87.7058
-87.7058
|v| > 87.7058 V
Answer:
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(a) The plot is shown above.
(b) The parallel combination is nonlinear, passive, and bilateral.
(c) 8.7706 V < v < 8.7706 V
(d) |v| > 87.7058 V
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Problem 2-66
Center Tapped Voltage Divider (A)
Figure P2-66 shows a voltage divider with the center tap connected to ground. Derive equations
relating vA and vB to vS, R1, and R2.
v
R
R
1
2
S
vA
vB
A
B
i
Solution:
Using the passive sign convention, iS = iA = iB. We can calculate the magnitude of the current
by combining the resistors in series and using Ohm's Law.
21
SA
RR
vi
Now apply Ohm's Law to each resistor to find vA and vB.
21
S11AA
RR
vRRiv
21
S22BB
RR
vRRiv
Answer:
21
S1A
RR
vRv
21
S2B
RR
vRv
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Problem 2-67
Active Transducer (A)
Figure P2-67 shows an active transducer whose resistance R(VT) varies with the transducer
voltage VT as R(VT) = 0.5 VT2 + 1. The transducer supplies a current to a 10-Ω load. At what
voltage will the load current equal 100 mA?
iLR(VT)
10 ΩVT
Transducer
Solution:
Develop an expression for the current in terms of the transducer voltage and solve for the voltage
that will make the current equal 100 mA.
clear all
format short eng
syms iL Vtr
Eqn = 'iL - Vtr/(0.5*Vtr^2 + 1 + 10)';
Soln = solve(Eqn,'Vtr');
Vtr_Soln = double(subs(Soln,iL,100e-3))
iL_Check = Vtr_Soln./(0.5*Vtr_Soln.^2+1+10) Vtr_Soln =
18.8318e+000
1.1682e+000
iL_Check =
100.0000e-003
100.0000e-003
Answer:
There are two answers: VT = 1.1682 V or VT = 18.8318 V.
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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 2-68
Programmable Voltage Divider (A)
Figure P2-68 shows a programmable voltage divider in which digital inputs b0 and b1 control
complementary analog switches connecting a multitap voltage divider to the analog output vO.
The switch positions in the figure apply when digital inputs are low. When inputs go high the
switch positions reverse. Find the analog output voltage for (b1,b0) = (0,0), (0,1), (1,0), and (1,1)
when VREF = 12 V.
O
v
b
O
vREF
1b
R
R
R
R
Solution:
There are four equal resistors in series with a voltage source, so each drops one quarter of the
total voltage, or 3 V in this case. As we cycle through the four combinations of the digital
inputs, the switches connect the output voltage to be across zero, one, two, or three resistors, in
that order. The output voltages are therefore 0 V, 3 V, 6 V, and 9 V.
Answer:
For (b1,b0) = (0,0), vO = 0 V.
For (b1,b0) = (0,1), vO = 3 V.
For (b1,b0) = (1,0), vO = 6 V.
For (b1,b0) = (1,1), vO = 9 V.
Full file at https://fratstock.euThe 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 2-69
Analog Voltmeter Design (A, D, E)
Figure P2-69(a) shows a voltmeter circuit consisting of a D'Arsonval meter, two series resistors,
and a two-position selector switch. A current of IFS = 400 μA produces full-scale deflection of
the D'Arsonval meter, whose internal resistance is RM = 25 Ω.
(a) (D) Select the series resistance R1 and R2 so a voltage vx = 100 V produces full-scale
deflection when the switch is in position A, and voltage vx = 10 V produces full-scale deflection
when the switch is in position B.
(b) (A) What is the voltage across the 20-kΩ resistor in Figure P2-69(b)? What is the voltage
when the voltmeter in part (a) is set to position A and connected across the 20-kΩ resistor?
What is the percentage error introduced connecting the voltmeter?
(c) (E) A different D'Arsonval meter is available with an internal resistance of 100 Ω and a full-
scale deflection current of 100 μA. If the voltmeter in part (a) is redesigned using this
D'Arsonval meter, would the error found in part (b) be smaller or larger? Explain.
1
B
R
A
R
R2
M
vx
30 k
50 V
20 k VM
(b)(a)
Solution:
(a) Solve for R2 first, such that a 10-V input causes 400 A to flow through the two resistors.
Then solve for R1, such that a 100-V input causes 400 A to flow through all three resistors.
clear all
IFS = 400e-6;
RM = 25;
Req10 = 10/IFS;
R2 = Req10 - RM
Req100 = 100/IFS;
R1 = Req100-Req10 R2 =
24.9750e+003
R1 =
225.0000e+003
Full file at https://fratstock.euThe 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) Using voltage division, the voltage across the 20-kΩ resistor is 20 V when the voltmeter is
not connected. When the voltmeter is set in position A and connected in parallel to the 20-kΩ
resistor, it is equivalent to placing a 250-kΩ resistor in parallel with the 20-kΩ resistor. We can
then find the voltage using voltage division and compute the error.
clear all
Req = 1/(1/20e3 + 1/250e3);
vM = Req*50/(Req + 30e3)
ErrorPercent = 100*(20-vM)/20 vM =
19.0840e+000
ErrorPercent =
4.5802e+000
(c) With a full-scale deflection current of 100 A for an applied voltage of 100 V, (switch in
position A,) the total resistance of the meter must be 1 MΩ. The increased meter resistance will
draw less current when it is connected to the 20-kΩ resistor and have a smaller impact on the
voltage. The error will decrease. The following calculations verify the results with numerical
values.
clear all
Req = 1/(1/20e3 + 1/1e6);
vM = Req*50/(Req + 30e3)
ErrorPercent = 100*(20-vM)/20 vM =
19.7628e+000
ErrorPercent =
1.1858e+000
Answer:
(a) R1 = 225 kΩ and R2 = 24.975 kΩ.
(b) vACTUAL = 20 V, vMEAS = 19.084 V, Error = 4.58%
(c) The error will be smaller as explained and verified above.
Full file at https://fratstock.euThe 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 2-70
MATLAB Function for Parallel Equivalent Resistors (A)
Create a MATLAB function to compute the equivalent resistance of a set of resistors connected
in parallel. The function has a single input, which is a vector containing the values of all of the
resistors in parallel, and it has a single output, which is the equivalent resistance. Name the
function “EQparallel” and test it with at least three different resistor combinations. At least one
test should have three or more resistor values.
Solution:
Create a MATLAB script (m-file) named EQparallel.m that contains the following code:
function Zp = EQparallel(Z)
% Compute the equivalent parallel impedance of a list of impedances
Zinv = 1./Z; Zp = 1/sum(Zinv);
Save the function file and change the MATLAB path to include the location of the function file.
Run the following commands to check the function.
R1 = EQparallel([1000 1000]) R2 = EQparallel([5e3 20e3]) R3 = EQparallel([4e3 5e3 20e3])
The results are:
R1 = 500
R2 = 4000
R3 = 2000
Answer:
The MATLAB function and examples are presented above. Run the code outside of this Word
document.