rinciples
`
`of Electric
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`of Electric
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`Library of Congress Cataloging-Publication Data
`
`Floyd, Thomas L.
`Principles of electric circuits/ Thomas L. Floyd.-5th ed.
`p. cm.
`Includes index.
`ISBN 0-13-232224-2
`1. Electric circuits.
`I. Title.
`TK454.F57 1997
`621.3815-dc20
`
`96-13443
`CIP
`
`Cover photo: © Superstock
`Editor: Linda Ludewig
`Developmental Editor: Carol Hink1in Robison
`Production Editor: Rex Davidson
`Text Designer: Anne Flanagan
`Cover Designer: Brian Deep
`Production Buyer: Patricia A. Tonneman
`Marketing Manager: Debbie Yarnell
`Illustrations: Jane Lopez and Steve Botts
`
`This book was set in Times Roman by The Clarinda Company and was printed and bound by Von
`Hoffmann Press, Inc. The cover was printed by Von Hoffmann Press, Inc.
`
`=-
`■ © 1997, 1993 by Prentice-Hall, Inc.
`Simon & Schuster/A Viacom Company
`=-
`Upper Saddle River, New Jersey 07458
`
`All rights reserved. No part of this book may be reproduced, in any form or by any means, with(cid:173)
`out permission in writing from the publisher.
`
`Earlier editions© 1989, 1985, 1981 by Merrill Publishing Company
`
`Printed in the United States of America
`
`10 9 8 7 6 5 4 3 2 1
`
`ISBN 0-13-232224-2
`
`Prentice-Hall International (UK) Limited, London
`Prentice-Hall of Australia Pty. Limited, Sydney
`Prentice-Hall Canada Inc., Toronto
`Prentice-Hall Hispanoamericana, S. A., Mexico
`Prentice-Hall of India Private Limited, New Delhi
`Prentice-Hall of Japan, Inc., Tokyo
`Simon & Schuster Asia Pte. Ltd., Singapore
`Editora Prentice-Hall do Brasil, Ltda., Rio de Janeiro
`
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`
`VOLTAGE DIVIDERS ®
`
`135
`
`Nowthat you know V,, use Ohm’s law to calculate R, as follows:
` 68.6V
`V,
`Ry=— = —— = 3430
`+="=300mA >?
`
`Related Exercise Determine R, in Figure 5—36 for Vs = 150 V and J = 200 mA.
`
`This is most likely a 330 Q resistor because 343 Q is within a standard tolerance
`range (+5%) of 330 Q.
`
`SECTION 5-6
`REVIEW
`
`1. State Kirchhoff’s voltage law in two ways.
`2. A50 V source is connected to a series resistive circuit. Whatis the total of the volt-
`age dropsin this circuit?
`3. Two equal-value resistors are connected in series across a 10 V battery. Whatis the
`voltage drop across each resistor?
`4. In a series circuit with a 25 V source, there are three resistors. One voltage dropis
`5 V, and the other is 10 V. What is the value of the third voltage drop?
`5. The individual voltage drops in a series string are as follows: 1 V, 3 V,5V, 8 V, and
`7 V. Whatis the total Voltage applied across the series string?
`
`5-7 @ VOLTAGE DIVIDERS
`
`
`A series circuit acts as a voltage divider. You will learn what this term means and why
`voltage dividers are an important application of series circuits.
`
`After completing this section, you should be able to
`
`@ Use a series circuit as a voltage divider
`[1 Apply the voltage-divider formula
`LI Use the potentiometer as an adjustable voltage divider
`C Describe some voltage-divider applications.
`
`
`
`Toillustrate howa series string of resistors acts as a voltage divider, we will examine Fig-
`ure 5-37 where there are two resistors in series. There are two voltage drops: one across
`R, and one across R2. These voltage drops are V, and V2, respectively, as indicated in the
`diagram.
`
`FIGURE 5-37
`
`Two-resistor voltage divider.
`
`Since each resistor has the same current, the voltage drops are proportional to the
`resistance values. For example, if the value of R, is twice that of R,, then the value of V,
`
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`136 »
`
`SERIES CIRCUITS
`
`is twice that of V,. In other words, the total voltage drop divides amongthe series resis-
`tors in amounts directly proportionalto the resistance values.
`For example, in Figure 5-37,if Vs is 10 V, R, is 50 Q, and R, is 100 Q, then V,is
`one-third the total voltage, or 3.33 V, because R, is one-third the total resistance (150 Q).
`Likewise, V> is two-thirds Vs, or 6.67 V.
`.
`
`Voltage-Divider Formula
`With a few steps, a formula for determining how the voltages divide among series
`resistors can be developed. Let’s assume that we have several resistors in series as
`shown in Figure 5-38. This figure showsfive resistors as an example, but there can be
`any number.
`
`FIGURE 5-38
`
`Five-resistor voltage divider.
`
`Let’s call the voltage drop across any one of the resistors V,, where x represents the
`numberof a particular resistor (1, 2, 3, and so on). By Ohm’slaw, the voltage drop across
`any of the resistors in Figure 5-38 can be written as follows:
`
`V, = IR,
`
`where x = 1, 2, 3, 4, or 5.
`The currentis equal to the source voltage divided by the total resistance (J = Vs/Rr).
`For the example circuit of Figure 5—38, the total resistance is R, + Rp + R3 + Ra + Rs. Sub- :
`stituting Vo/Ry for J in the expression for V, results in
`
`Rearranging the terms yields
`
`V.
`
`v.= (3)
`
`_ {Re
`
`v.=(ral
`
`_
`
`(5-5)
`
`Equation (5-5) is the general voltage-divider formula. It canbe stated as follows:
`
`“The voltage drop across any resistor or combination of resistors in a
`series circuit is equal to the ratio of that resistance value to the total resis-
`tance, multiplied by the source voltage.
`
`The following three examplesillustrate use of the voltage-divider formula.
`
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`306 ■ CIRCUIT THEOREMS AND CONVERSIONS
`
`SECTION 8-7
`REVIEW
`
`1. To what type of circuit does Millman's theorem apply?
`2. Write the Millman theorem formula for REQ·
`3. Write the Millman theorem formula for VEQ.
`4. Find the load current (h) and the load voltage (VL) in Figure 8-54.
`
`FIGURE 8-54
`
`8-8 ■ MAXIMUM POWER TRANSFER THEOREM
`
`The maximum power transfer theorem is important when you need to know the v4lue
`of the load at which the most power is delivered from the source.
`
`After completing this section, you should be able to
`· ■ Apply the maximum power transfer theorem
`□ State the theorem
`□ Determine the value of load resistance for which maximum power is transferred
`from a given circuit
`
`The maximum power transfer theorem states as follows:
`
`When a source is connected to a load, maximum power is delivered to the
`load when the load resistance is equal to the internal source resistance.
`
`The source resistance, Rs, of a circuit is the equivalent resistance as viewed from the out(cid:173)
`put terminals using Thevenin's theorem. An equivalent circuit with its output resistance
`and load is shown in Figure 8-55. When RL = Rs, the maximum power possible is trans(cid:173)
`ferred from the voltage source to RL.
`
`FIGURE 8-55
`Maximum power is transferred to the load
`when RL = Rs-
`
`Source
`Rs
`
`Practical applications of this theorem include audio systems such as stereo, radio,
`and public address. In these systems the resistance of the speaker is the load. The circuit
`th~ drives the speaker is a power amplifier. The systems are typically optimized for max(cid:173)
`imum power to the speakers. Thus, the resistance of the speaker must equal the internal
`source resistance of the amplifier.
`Example 8-16 shows that maximum power occurs when RL = Rs-
`
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`MAXIMUM POWER TRANSFER THEOREM ■ 307
`
`EXAMPLE 8-16
`
`The source in Figure 8-56 has an internal source resistance of 75--tl. Determine the
`load power for each of the following values of load resistance:
`(d) 100 Q
`(a) 25 Q
`(c) 75 Q
`(e) 125 Q
`(b) 50 Q
`Draw a graph showing the load power versus the load resistance.
`
`FIGURE 8-56
`
`75 n
`
`+_ Vs
`-iov
`
`Solution Use Ohm's law (/ = V/R) and the power formula (P = / 2R) to find the load
`power, PL> for each value of load resistance.
`(a) For RL = 25 n.
`
`I=
`
`=
`
`Vs
`lOV
`75 Q + 25 Q =
`Rs + RL
`PL= I2RL = (100 mA)2(25 Q) = 250 mW
`
`lOOmA
`
`(b) For RL = 50 n,
`
`(c) For RL = 75 Q,
`
`(d) For RL = 100 Q,
`
`I=
`
`Vs = 10 V = 80 mA
`Rs+RL 1250
`PL= I 2RL = (80 mA)2(50 Q) = 320 mW
`
`I=
`
`= 10 V = 66.7 mA
`Vs
`Rs +RL
`150 Q
`PL= I 2RL = (66.7 mA)2(75 0) = 334 mW
`
`=
`
`I=
`
`Vs
`lOV =57.1 mA
`175 Q
`Rs +RL
`PL= I2RL = (57.1 mA)2(100 0) = 326 mW
`
`(e) For RL = 125 0.
`
`I=
`
`Vs = 10 V = 50 mA
`200 n
`Rs +RL
`2
`2
`'
`PL= I RL = (50 mA) (125 Q) = 313 mW
`Notice that the load power is greatest when RL = 15 n, which is the same as the
`internal source resistance. When the load resistance is less than or greater than this
`value, the power drops off, as the curve in Figure 8-57 graphically illustrates.
`The calculator sequence for RL = 25 n is
`
`••••••••••••••••
`
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`308 ■ CIRCUIT THEOREMS AND CONVERSIONS
`
`400
`
`300
`
`200
`
`100
`
`25
`
`50
`
`75
`
`100
`
`125
`
`FIGURE 8-57
`Curve showing that the load power is maximum when RL = Rs.
`
`If the source resistance in Figure 8-56 is 600 n, what is the max(cid:173)
`Related Exercise
`imum power than can be delivered to a load?
`
`SECTION 8-8
`REVIEW
`
`1. State the maximum power transfer theorem.
`2. When is maximum power delivered from a source to a load?
`3. A given circuit has an internal source resistance of 50 n. What will be the value of
`the load to which the maximum power is delivered?
`
`8-9 ■ DELTA-TO-WYE (8-TO-Y) AND
`WYE-TO-DELTA (Y-TO-8) CONVERSIONS
`
`Conversions between delta-type and wye-type circuit arrangements are useful in certain
`specialized applications. One example is in the analysis of a loaded Wheatstone bridge
`circuit. In this section, the conversion formulas and rules for remembering them are
`given.
`
`After completing this section, you should be able to
`
`■ Perform ii-to-Y and Y-to-Ll conversions
`□ Apply ii-to-Y conversion to a bridge circuit
`
`A resistive delta(~) circuit has the form shown in Figure 8-58(a). A wye (Y) circuit is
`shown in Figure 8-58(b). Notice that letter subscripts are used to designate resistors in
`
`FIGURE 8-58
`Delta and wye circuits.
`
`(a)Delta
`
`(b)Wye
`
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