Problemes circuits Irwin, Ejercicios de Electrónica

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Problem-Solving Companion

To accompany

Basic Engineering Circuit Analysis

Ninth Edition

J. David Irwin

Auburn University

JOHN WILEY & SONS, INC.

Executive Editor Bill Zobrist Assistant Editor Kelly Boyle Marketing Manager Frank Lyman Senior Production Editor Jaime Perea

Copyright © 200 5 , 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, (508) 750-8400, fax (508) 750-

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ISBN 0-471- 74026 -

STUDENT PROBLEM COMPANION

To Accompany

BASIC ENGINEERING CIRCUIT ANALYSIS, NINTH EDITION By J. David Irwin and R. Mark Nelms

PREFACE

This Student Problem Companion is designed to be used in conjunction with Basic Engineering Circuit Analysis, 8e, authored by J. David Irwin and R. Mark Nelms and published by John Wiley & Sons, Inc.. The material tracts directly the chapters in the book and is organized in the following manner. For each chapter there is a set of problems that are representative of the end-of-chapter problems in the book. Each of the problem sets could be thought of as a mini-quiz on the particular chapter. The student is encouraged to try to work the problems first without any aid. If they are unable to work the problems for any reason, the solutions to each of the problem sets are also included. An analysis of the solution will hopefully clarify any issues that are not well understood. Thus this companion document is prepared as a helpful adjunct to the book.

CHAPTER 1 PROBLEMS

1.1 Determine whether the element in Fig. 1.1 is absorbing or supplying power and how much. -2A

12V

Fig. 1.

1.2 In Fig. 1.2, element 2 absorbs 24W of power. Is element 1 absorbing or supplying power and how much.

12V

6V

Fig. 1.

1.3. Given the network in Fig.1.3 find the value of the unknown voltage VX.

+ 4V^ - + 10V -

2A

2A

4A

12V 8V

6A

VX

Fig. 1.

i

v

Fig. S1.2(a)

If we now isolate the element 2 and examine it, since it is absorbing power, the current must enter the positive terminal of this element. Then P = VI 24 = 6(I) I = 4A

The current entering the positive terminal of element 2 is the same as that leaving the positive terminal of element 1. If we now isolate our discussion on element 1, we find that the voltage across the element is 6V and the current of 4A emanates from the positive terminal. If we reverse the current, and change its sign, so that the isolated element looks like the one in Fig. S1.2(a), then

P = (6) (-4) = -24W

And element 1 is supplying 24W of power.

1.3 By employing the sign convention for power, we can determine whether each element in the diagram is absorbing or supplying power. Then we can apply the principle of the conservation of energy which means that the power supplied must be equal to the power absorbed.

If we now isolate each element and compare it to that shown in Fig. S1.3(a) for the sign convention for power, we can determine if the elements are absorbing or supplying power. i

V

P = Vi

Fig. S1.3(a)

For the 12V source and the current through it to be arranged as shown in Fig. S1.3(a), the current must be reversed and its sign changed. Therefore

P12V = (12) (-6) = -72W

Treating the remaining elements in a similar manner yields

P 1 = (4) (6) = 24W P 2 = (2) (10) = 20W P 3 = (8) (4) = 32W PVX = (VX) (2) = 2VX

Applying the principle of the conservation of energy, we obtain

-72 + 24 + 20 + 32 + 2VX = 0 And VX = -2V

2.5 Find all the currents and voltages in the network in Fig. 2.5.

2kΩ 10kΩ 2kΩ

48V 4kΩ^ 3kΩ^ 4kΩ

6kΩ

I 1

V 1 V 2 V 3

I 2

I 4

A B

I 3 I 5

I 6

Fig. 2.

2.6 In the network in Fig. 2.6, the current in the 4kΩ resistor is 3mA. Find the input voltage VS.

2kΩ 1kΩ VS (^) 4kΩ 3mA

2kΩ 6kΩ^ 3kΩ

9kΩ +-

Fig. 2.

CHAPTER 2 SOLUTIONS

2.1 We recall that if the circuit is of the form

V 1

R 1

R 2 V 0

Fig. S2.1(a)

Then using voltage division

1 1 2

2 (^0) R R V

R

V ⎟⎟

That is the voltage V 1 divides between the two resistors in direct proportion to their resistances. With this in mind, we can draw the original network in the form

V 1

2kΩ

V 2

12V +-

3kΩ

4kΩ

2kΩ

Fig. S2.1(b)

The series combination of the 4kΩ and 2kΩ resistors and their parallel combination with the 3kΩ resistor yields the network in Fig. S2.1(c).

12V

2kΩ

V 1

2kΩ

Fig. S2.1(c)

Now voltage division can be sequentially applied. From Fig. S2.1(c).

6 V

2 k 2 k

2 k V 1

Then from the network in Fig. S2.1(b)

2.3 To provide some reference points, the circuit is labeled as shown in Fig. S2.3(a).

A

B

8k 10k 2k

4k 18k

6k 3k

3k 6k

12k

12k

A' A"

B' B"

Fig. S2.3(a)

Starting at the opposite end of the network from the terminals A-B, we begin looking for resistors that can be combined, e.g. resistors that are in series or parallel. Note that none of the resistors in the middle of the network can be combined in anyway. However, at the right-hand edge of the network, we see that the 6k and 12k ohm resistors are in parallel and their combination is in series with the 2kΩ resistor. This combination of 6k⎪⎢12k + 2k is in parallel with the 3kΩ resistor reducing the network to that shown in Fig. S2.3(b).

A

B

8k (^) 10k

2k = 3k (6k 12k + 2k)

4k 18k 6k 3k

12k

A' A"

B' B"

Fig. S2.3(b)

Repeating this process, we see that the 2kΩ resistor is in series with the 10kΩ resistor and that combination is in parallel with the12kΩ resistor. This equivalent 6kΩ resistor (2k + 10k)⎪⎢12k is in series with the 3kΩ resistor and that combination is in parallel with the 18kΩ resistor that (6k + 3k)⎪⎢18k = 6kΩ and thus the network is reduced to that shown in Fig. S2.3(c).

A

B

8k

4k 6k 6k

A'

B'

Fig. S2.3(c)

At this point we see that the two 6kΩ resistors are in series and their combination in parallel with the 4kΩ resistor. This combination (6k + 6k)⎪⎢4k = 3kΩ which is in series with 8kΩ resistors yielding A total resistance RAB = 3k + 8k = 11kΩ.

2.4 An examination of the network indicates that there are no series or parallel combinations of resistors in this network. However, if we redraw the network in the form shown in Fig. S2.4(a), we find that the networks have two deltas back to back. A

B

4k

2k (^) 12k 12k

12k

6k 18k

Fig. S2.4(a)

If we apply the ∆→Y transformation to either delta, the network can be reduced to a circuit in which the various resistors are either in series or parallel. Employing the ∆→Y transformation to the upper delta, we find the new elements using the following equations as illustrated in Fig. S2.4(b)

12k

R 1

6k 18k R (^2) R (^3)

Fig. S2.4(b)

= 3 k 6 k 12 k 18 k

6 k 18 k R (^1)

= 2 k 6 k 12 k 18 k

6 k 12 k R (^2)

= 6 k 6 k 12 k 18 k

12 k 18 k R (^3)

The network is now reduced to that shown in Fig. S2.4(c).

A B

2k

10k

V 1 4k

48V

6k

I 1 I 2

I 3

I 4

+- 2k

V 2

Fig. S2.5(a)

Proceeding, the 2k and 10k ohm resistors are in series and their combination is in parallel with both the 4k and 6k ohm resistors. The combination (10k + 2k)⎪⎢6k⎪⎢4k = 2kΩ. Therefore, this further reduction of the network is as shown in Fig. S2.5(b).

2k

(^48) 2k

V 1

I 1

Fig. S2.5(b)

Now I 1 and V 1 can be easily obtained.

12 mA 2 k 2 k

I 1 =

And by Ohm’s law

V 1 = 2kI (^1) = 24V or using voltage division

24 V

2 k 2 k

2 k V 1 48

once V 1 is known, I 2 and I 3 can be obtained using Ohm’s law

6 mA 4 k

4 k

V

I 2 =^1 = =

4 mA 6 k

6 k

V

I 3 =^1 = =

I 4 can be obtained using KCL at node A. As shown on the circuit diagram.

I 1 = I 2 + I 3 + I (^4)

k I^4

k

k

2 mA k

I 4 = =

The voltage V 2 is then

V 2 = V 1 - 10kI (^4)

k

24 10 k

= 4V

or using voltage division

4 V

10 k 2 k

2 k V 2 V 1

Knowing V 2 , I 5 can be derived using Ohm’s law

mA 3

3 k

V

I 5 2

and also

mA 3

2 k 4 k

V

I 6 2

current division can also be used to find I 5 and I 6.

mA 3

2 k 4 k 3 k

2 k 4 k I 5 I 4

and

6 mA

I I

k

I 312

Then using Ohm’s law

V 2 = I 3 (1k) = 6V

KVL can then be used to obtain V 3 i.e.

V 3 = V 2 + V 1 = 6 + 12 = 18V

Then

9 mA

2 k

V

I 4 3

And

15 mA

k

k

I 5 I 3 I 4

using Ohm’s law

V 4 = (2k) I (^5) = 30V

and finally

VS = V 4 + V 3 = 48V

CHAPTER 3 PROBLEMS

3.1 Use nodal analysis to find V 0 in the circuit in Fig. 3.1.

1kΩ 1kΩ 1kΩ 2kΩ

2mA

12V V 0

Fig. 3.

3.2 Use loop analysis to solve problem 3.

3.3 Find V 0 in the network in Fig. 3.3 using nodal analysis.

2kΩ

12V

2kΩ 1kIX 1kΩ V 0

IX

Fig. 3.

3.4 Use loop analysis to find V 0 in the network in Fig. 3.4.

4mA

1kΩ 1kΩ

1kΩ 2kΩ V 0

2IX

IX

Fig. 3.