Resistive and Reactive Circuits

Citation preview

MALVINO DPeOrDDS!

*

)\k

a

alpha

t©si§©[pj

OTHER BOOKS BY ALBERT

P.

AAALVINO

Transistor Circuit Approximations

Fundamentals and Applications (with D. Leach)

Electronic Instrumentation Digital Principles

Electronic Principles

Experiments for Electronic Principles (with G. Johnson)

lDi]C°]

Albert Paul Malvino, Ph.D.

McGraw-Hill Book Company

Library of Congress Cataloging in Publication

Data

Malvino, Albert Paul. Resistive

1.

and reactive

Electric circuits.

I.

circuits.

Title.

621.3815'3 TK454.M24 ISBN 0-07-039856-9

RESISTIVE

Copyright

AND

©

REACTIVE CIRCUITS

1974 by McGraw-Hill, Inc. All rights

reserved. Printed in the United States of America. 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,

or otherwise, without the prior written permission of the publisher.

1234567890MURM7987654

The editors for this book were Alan W. Lowe and Cynthia Newby, the designer was Marsha Cohen, and its production was supervised by James E. Lee and Laurence Chamow. It was set in Palatino by Progressive Typographers.

was printed by The Murray Printing Company and bound by Rand McNally & Company. It

To my and to I

know

son or fourth daughter v\ ho is now convinced nothing about mathematics.

first

my

dear wife

With doubt and dismay you are smitten You think there's no chance for you son? Why the best books haven't been written The best race hasn't been run! Barton

©(o)D^^©[j^l§(

Preface

PART

1

.

1.

RESISTIVE CIRCUITS

Introduction

2-3.

Units and Prefixes

Come From Fluid Theory tric Field

Where Formulas 1-4.

Coulomb's Law

1-5. 1-7.

1-2.

Early Discoveries

1-3.

Electron Theory

1-6. 1-8.

The Elec-

Con-

ventional Versus Electron Flow

2.

25

Three Electrical Quantities

Voltage

2-1.

Current

2-4.

Ohm's Law

2-6.

American Wire Gage

2-2.

2-5.

Temperature Coefficient

2-3.

More About 2-7.

Resistance Resistance

Resistors

2-8.

3.

Series

and

3-3.

Series Circuit

Kirchhoff's Current

Law

Analysis

Resistances

4.

Parallel Circuit

3-2.

Law

3-4.

Paral-

3-6.

Shortcuts for Parallel

3-7.

3-8.

3-3.

Kirchhoff's Voltage

Series-circuit Analysis

3-5.

lel-circuit

61

Parallel Circuits

Combination Circuits

105

Three Theorems

4-1.

Voltage-divider Theorem

Quantities

Other Ways

4-8.

4-4.

Get Thevenin Resistance

to

Wheatstone Bridge position

Thevenin

4-2.

Thevenin's Theorem

4-3.

The

Basic Idea of Super-

4-6.

Theorem

4-5.

Superposition Theorem

4-7.

Review of the Three Theorems

157

More Theorems

5-1.

Conductance Theorem

5-3.

Including Current Sources in Earlier Theorems

5-4.

Norton's Theorem

Power Theorems

6.

Kirchhoff

6-1.

7-1.

5-5.

Power

5-6.

Matched-load Power

Methods

195

Simultaneous Equations

Method Method Method

Basic

5-7.

Ideal Sources

5-2.

6-3.

Loop Method

6-5.

The Chassis

6-7.

Summary

6-2.

6-4. 6-6.

Branch

Node Ladder

233

Measurements

The Moving-coil Meter

Ranges Sensitivity

7-3.

7-2.

Simple Voltmeters

7-5.

The Ohmmeter

Volt-ohm-milliammeter

Ammeter 7-4.

7-6.

Meter The

259 8-1.

Basic Ideas

The Time Theorem

8-2.

The Similarity Theorem Preview of Part

8-5.

PART

9.

8-4.

8-3.

The Sine Wave

2

REACTIVE CIRCUITS

2.

Capacitance

9-1.

The Basic Idea

9-3.

More About Capacitance

Dielectric

Linear Capacitance

9-2.

9-4.

The

Manufactured Capacitors

9-5.

Equivalent Capacitance

9-7.

9-6.

Unwanted

Capacitance

10.

321

Inductance

10-1.

Basic Ideas

tance

10-3.

Core Material

10-5.

Two

Induction

Definition of Induc10-4.

Nonlinearity and Reten-

Coefficient of Coupling

10-6.

tivity

ductance of

10-2.

More About Inductance

Inductors

10-8.

10-7.

The Ideal Transformer

10-9.

In-

Electromagnetic 10-10.

Lead Inductance

1 1

357

Transients

32-3.

The

KC

Switching Prototype

Capacitor Discharge

12.

11-3.

11-2.

Inductor Transients

379

Reactance

12-1. in

Values of a Sine

an Inductor

12-4.

12-3.

Wave

12-2.

Equivalent Inductive Reactance

Current

in a

Capacitor

AC

Current

Inductive Reactance

12-6.

12-5.

Capacitive Re-

AC

actance 12-8.

Resonance

13.

Equivalent Capacitive Reactance

12-7.

Coupling Capacitor and Choke 12-10.

Parallel

Series

Codes, Numbers, and Sinusoids

13-1.

Codes

413

Numbers

13-2.

and Vector Codes

gular, Polar,

Rectan-

13-3.

The

13-4.

Arithmetic of Complex Numbers

14.

12-9.

Resonance

Sinusoids

13-5.

437

Phasor Analysis

14-1.

Phasor Current

14-3.

Impedance

Kirchhoff's Phasor

14-2.

Phasor Voltage

Ohm's

14-4.

Laws

14-6.

ac

Laws

14-5.

Equivalent Im-

14-7. Lead Network pedance of Series Circuits 14-8. 14-9. Lag Network Choke Filter Circuit

14-10.

15.

Parallel

RC

Circuit

Advanced AC Topics

15-1.

471

Conductance, Susceptance, and Admittance

15-2.

Parallel Circuits

tion of

an Inductor

version 15-6.

15-5.

AC

15-3.

15-8.

Con-

Parallel-to-series Conversion

Wheatstone Bridges

a Resistor

Second Approxima-

Series-to-parallel

15-4.

RMS

15-7.

Values

Power

15-9.

in

AC

Load Power

16.

507

Resonance

26-1.

Series

Resonant Circuits

Resonant Circuits a Lossy Inductor Circuits

16-5.

Transformers

16-3. 16-4.

16-2.

Parallel

Q, of

Bandwidth

Parallel

Resonance with

Parallel

16-6.

Resonant Air-core

17.

18.

Instantaneous

AC

535

Analysis

Phasors and Sinusoids

17-1.

Phasors

17-3.

Instantaneous Voltage and Current

17-2.

545

Switching Circuits

R

18-1.

rent

Su'itching Circuits

and Voltage

18-4.

RC

18-3.

Switching Prototype

Switching Theorem 18-7.

18-2.

18-6.

Capacitor Cur-

The Capacitor Theorem 18-5.

The

RC

The Switching Formula

Inductor Current and Voltage

18-8.

The

RL Switching Theorem

Appendix

Answers

Index

to

581

Odd-Numbered Problems

583

587

^[p©{?@]@©

The

poor job of preparing the technician for time is wasted on magnetic circuits, intricate ac calculations, power-oriented topics, three-phase systems, etc. The topics that really count, such as Thevenin's theorem, superposition theorem, resistive ac circuits, etc., get a thin treatment, with almost no follow-up in succeeding chapters. To select the right topics for this dc-ac textbook, I used Electronic Principles as my guide, because it covers the whole field of analog electronics including bipolars, FETs, ICs, op amps, etc. My reasoning was this: if a dc-ac textbook prepares a technician to study a modern book like Electronic Principles, then the technician has learned all the vital dc-ac topics needed for everyday typical dc-ac textbook

modem

does

electronic courses; too

a

much

electronics.

Part

1

of this

book

is

about resistive

circuits

with dc or ac sources, the special

case so prominent in today's electronics because of direct-coupled circuits.

On

completion of this first half of the book, the technician will be ready to begin a basic electronics course. Part 2 of this book is about reactive circuits. Chaps. 9 through 12 cover transients and ac theory without trigonometry, complex numbers, and the classic phasor approach. Because of this, the reader can cover the first 12 chapters entirely with algebra. These first 12 chapters may be adequate for many programs that wish to cover practical dc-ac theory without complex

numbers and phasors. The final part of the book. Chaps.

knowledge of numbers and discuss complex numbers and

13 through 18, requires a

trigonometry. Here you will find the extensive use of complex

phasors that typifies in-depth ac analysis.

We

phasors in detail; however, trigonometry is assumed as a prerequisite for Chaps. 13 through 18. think you will enjoy the format of this book. It includes two or three key questions at the beginning of longer topics; these questions are designed to focus your attention on what's important, to point you in the right direction before turning you loose. To make sure you understand the key ideas, multiplechoice tests are included throughout each chapter; the answers to these tests I

are at the very

end

of the chapter.

Many thanks to Lloyd Temes of Staten Island Community College, New York. He reviewed the manuscript and offered many excellent suggestions. Albert Paul Malvino

^u DDi]SD^®(^Qra@l§D©DT] A

physical quantity

anything you can detect with your senses or with measuring

is

struments. Electricity

is

in-

easier to understand after learning basic ideas about physical

quantities.

1-1.

AND

UNITS

In this section,

PREFIXES

your job

is to

(^

learn

What

7 How

Units of

are units of measure? are prefixes used?

measure

Saying something has a length of 7 means nothing; but saying 7

ft

it

gives the exact size. To discuss any physical quantity, therefore,

specify a number and a unit of measure, an

amount

has a length of

we must always

of the physical quantity

used as

a

reference. British

measured

units of

measure are familiar

in inches, feet,

minutes, hours, days,

etc.

yards, miles,

As

in

English-speaking countries. Length

and so on. Time

a rule, British units are rarely

is

measured

used

is

in

seconds,

in electrical

work, es-

pecially not in electronics.

Metric units of measure are standard in electronics. Table 1-1 quantities, their metric units

lists

some

and abbreviations, and the equivalent amounts

physical

in British

4

RESISTIVE CIRCUITS

TABLE

1-1.

METRIC UNITS ABBREVIATION

Length

BRmSH EQUIVAIENT

INTRODUCTION

5

Here are more examples of millisecond conversion: 0.025

5

=

25(10-3) s

=

0.175

s=

175(10-^)

s=

0.0004

=

s

0.4(10

') s

25

=

ms

175

0.4

ms

ms

Microunits

Micro (abbreviated prefix,

ft)

stands for one-millionth, 0.000001, or 10

amount

converts a unit into a microunit, an

it

To convert any amount into microunits, move the decimal point and replace 10" by fi. As an example, given 0.000005

''.

When

used as

a

that is one-millionth of the unit. six places to the right

s

move

the decimal point six places to the right to get

Then

replace

5(10"'') s

10" by

/li

to get

5

This

final

answer

is

/us

read as five microseconds.

Here are more examples:

Incidentally,

used

/ix

(mu)

is

0.0000003

s

0.00006 s

=

0.000175

s

letters

used

=

0.3(10'') s

60(10-"*) s

=

= 175(10")

60

s

=

0.3 ;as /Lis

175

fjiS

Greek alphabet. Greek letters are For your convenience, the front of the book lists all the

the twelfth letter of the

a great deal in electronics.

Greek

=

in this book. Refer to these as

needed.

Prefixes

Table 1-2 shows

all

the prefixes used in electronics.

We

use these prefixes

dicate multiples or submultiples of a unit of measure. Learn each prefix, tion,

and

its

EXAMPLE An ohm

equivalent power of

its

to in-

abbrevia-

10.

1-1.

(abbreviated

chapter. Express 7000

SI,

fl

Greek

letter

in kilohms.

omega)

is

a unit of

measure discussed

in the next

6

RESISTIVE CIRCUITS

TABLE

PREFIXES

1-2.

POWER OF

ABBREVIATION

PREFIX

10

10"

mega

10»

kilo

lO*

milli

io-»

micro

io-«

nano

io-»

pico

10-'

SOLUTION. 7000

n=

7(io^)

n=

We moved

the decimal point three places to the

EXAMPLE

1-2.

Express 2,200,000

il

kn

7

left

and replaced

10''

by

10'*

by M.

k.

megohms.

in

SOLUTION. 2,200,000

Here

we move

EXAMPLE Convert

n=

=

2.2(10" )n

the decimal point six places to the

2.2

left

Mn

and replace

1-3.

6.8 kfi to

ohms. Also, 2 Mfl

to

ohms.

SOLUTION. 6.8

and

2

Test 1-1 1.

Which

2.

Which

3.

Which

4.

A

(a)

((7)

(a)

(fl)

(answers

at

end

kn = 6.8(10M n =

6800

n

Mn = 2(10") n = 2,000,000 n

of chapter)

of these is not a metric unit?

meter

(b)

foot

(c)

second

(d)

newton

(

)

(

)

(

)

(

)

of these does not belong?

inch

(h) foot

(c)

of these is closest in

meter force of 4 0.5 lb

(b)

N

is

(h)

second

yard

(d)

meaning (c)

meter

to unit of

reference

measure?

(d) prefix

closest to 1

lb

(c)

2 lb

(li)

4 lb

INTRODUCTION

5.

5 jus equals

6.

2

7.

47 kfl

((J)

(a)

(a)

1-2.

A

0.005

ms

s

0.00005

(f)

s

0.05 s

(-J

•.vj5*t3'«T

*//,','* *.'-ri i/'jr-.'i^

oi

jituM* are tpUt*(«My>

by cr-wmc^ vi-irywr. ;r

abfl2

|>/?3

w

l|«.

l|'?5

64

RESISTIVE CIRCUITS

short

and has

a resistance

much

resistances

much

greater than

less

than

1 fl.

Typical loads in electronic circuits have

For this reason,

1 il.

we

can approximate the resistance

of connecting wires as zero.

Equipotential points

Whenever points

is zero.

the resistance

This

is

between two points

is

zero, the voltage

between the

true because

V=RI = 0x1 = No

matter what the current, zero resistance results in zero voltage. In Fig. 3-3h

if

we treat A and

age between points

the resistance of connecting wires as zero, there

B and C, and so on. Because of

B, points

this,

is

zero volt-

charges have

the same potential energy anywhere along the upper path. Whenever charges have the same potential energy along a path, the entire path is called an equipotential point. In Fig. 3-3b, this means the tops of all resistors are connected to an equipotential point. Similarly, charges anywhere along the E-F-C-H path have the same potential

energy. Therefore, the bottoms of point. Because of this. Fig. 3-3b has

all

the resistors connect to another equipotential

two equipotential

points. In fact,

any

parallel circuit

has only two ecjulpotential points.

One

voltage

same voltage appears

In a parallel circuit the

across

all

loads, because these loads

between the same pair of equipotential points. If 5 V is across K, in Fig. 3-3b, 5 V must also be across R.>, R.,, and so on. The word "parallel" therefore is synonymous with "one voltage." A parallel circuit is a one-voltage circuit because the same voltage is across each load. Since each load in a parallel circuit has the same voltage across it, you can remove one of the loads without disturbing the current in other loads. Most Christmas tree are connected

lights are

wired in

parallel, so that

if

one bulb

bums

out, the others stay on.

Test 3-2 1.

A

always has

parallel circuit

(a)

one

(b)

2.

The voltage only two

3.

Series

(fl)

paths

is

two the

(())

(c)

same

one path

(rf)

none

across

(rf)

all

{b)

of these

as

one path

two paths

equipotential points?

more than

three

(

)

loads in a parallel circuit because the circuit has

equipotential points

is to parallel

(a)

how many

three

(r)

loads

(d)

currents

(

)

(

)

is to {c)

one voltage

SERIES

4.

The following words can be rearranged LEL VOLTAGE. Is the sentence true

{a) 5.

form

a sentence:

ONE

PARALLEL CIRCUITS

(

)

(

)

least?

one voltage

two equipotential points

(b)

(c)

parallel

(d) series

KIRCHHOFF'S CURRENT

3-3.

65

IMPLIES PARAL-

false

{h)

Which belongs (a)

to

AND

LAW

Before the twentieth century, electric fluid

was widely used because

the electron had

not yet been discovered. In this preelectron era, Kirchhoff found two important experi-

mental formulas; one for current, the other for voltage. This section

answers

is

about one of the laws Kirchhoff discovered. Your goal

What

Why

''

What

it

IS is

law always true?

says

sum

of currents into a point equals the

rents out of the point. For instance, in Fig. 3-4(3 charges flow into

shown. The current into law says

A

equals

/,,

and the current out

/,

Another example.

=

out of

A

+

/;).

still

A equals

In this case, Kirchhoff's current I,

As

A

is ij.

In Fig. 3-4b charges flow into point A,

along two different paths. The current into is I2

of

another example, in

of currents out of

A

is

+

Z^.

sum

of cur-

of point

A

as

Kirchhoff's current

/,,

where they

and the sum

split

and

of currents

law says

=h+h

Fig. 3-4f the /,,

and out

/.,

travel

sum

learn the

Kirchhoff's current law? the current

Kirchhoff's current law says the

the

is to

to:

sum

of currents into point

Kirchhoff's current law

tells

A

equals

/,

+

I^;

us

h+h= h + /j

h

N ^ /,

Figure 3-4. Kirchhoff's current taw.

66

RiSISTIVE CIRCUITS

We

can summarize Kirchhoff 's current law by the following formula:

1 currents

where the Greek

sum

S

letter

in

=

currents out

Hi

(sigma) is shorthand for "the

of currents in equals the

sum

(3-1)

says the

is

crucial in circuit analysis. For

one

allows us to write an equation for each point in an electric circuit; using these

it

we

equations,

Why

can derive formulas relating the currents and voltages

it,

in the circuit.

must be true

it

Kirchhoff's current law

proved

Equation

of."

of currents out.

Although simple, Kirchhoff's current law thing,

sum

(3-1)

and

electron,

we

point in

1

can see

s, 1

an experimental formula; countless experiments have

is

this is the reason

why

it is

called a law.

the current law

must be

must flow out

million electrons

would be created or destroyed

at

However, since the discovery true.

If 1

of the

million electrons flow into a

of this point in

1

s;

otherwise, electrons

the point. In an electric circuit, no electrons are

created or destroyed; therefore, the rate of flow into each point

must equal the

rate of

flow out of the point.

EXAMPLE What

is

3-2.

the value of

/

in Fig. 3-5a?

SOLUTION. Apply the current law

Now,

solve for

In Fig. 3-5a,

A and

/

point

A

to get

7

=

3

+

/

=

4

A

1

to get

when we

a total of 7

Figure 3-5. Examples of the current law.

to

visualize

A coming

/

equal to 4 A,

out of point A.

we

can see a

total of 7

A

going into point

SERIES

EXAMPLE Solve for

/

AND

67

PARALLEL CIRCUITS

3-3.

in Fig. 3-5b.

SOLUTION. With Kirchhoff's current law, 4 Solving for

What does

the negative sign

opposite that

is

shown

of conventional current

directions,

/

/,

/

rent

=9+

we

/

mean?

= -5 A means

It

the true direction of conventional cur-

words, in

in the circuit. In other

opposite that shown.

is

Here we see

get Fig. 3-5c.

If

a total of

Fig. 3-5b the true direction

we redraw the circuit showing 9 A flowing into the point and

true 9

A

flowing out of the point.

Whenever

possible,

we draw

the true direction of current in circuit diagrams,

using the rule that conventional current

beginning of

at the

For this reason, in

a

problem,

it's

is

from plus

impossible

to

to tell

some problems we may guess

minus

in

all

loads. Occasionally

which end of

wrong

the

a load is positive.

direction of current.

Kirchhoff's current law automatically catches errors like this because a negative sign

up when solving for the unknown current. Whenever you solve an equation and get a negative value

turns

to

do

of current,

all

you have

visualize or redraw the original circuit with this current in the opposite direc-

is

tion.

Test 3-3 1.

Kirchhoff's current law

defining

(b)

The current law

is

(a) 2.

3.

is

an example of which kind of formula?

experimental

(f)

derived

can disappear

(fl)

can be created

(f)

can flow into a point

{d)

can neither be created nor destroyed

at a

point

(b)

at a

at a

true

3-4.

(b)

point

COMES GOES

false

IN

(

)

OUT (

)

KIRCHHOFF'S VOLTAGE LAW

Kirchhoff's voltage law says the

words, point;

)

point

The following words can be rearranged to form a sentence: WHAT. With relation to the current law, the sentence is (fl)

(

true because electrons

start at

you

any point

will find

all

sum

in a circuit

of voltages

around

and go around

voltages add to zero.

a

closed path

is

zero. In other

a path that returns to the starting

68

RESISTIVE CIRCUITS

This section answers the following questions;

What '-'

How

to

Why

are the steps in applying the voltage law?

must the voltage law be true?

apply Kirchhoff's voltage law

Here are the key steps

1.

Select

any point

for

applying Kirchhoff's voltage law

to a circuit:

Imagine you are going

in the circuit as a starting point.

walk

to

clockwise around a path that returns to this starting point.

down

2.

As you

3.

magnitude of the voltage. When you arrive back at the starting point, equate

arrive at each source or load, write

the

first

sign you see (+ or

—

and

)

the

To understand each

One

source

In Fig. 3-6(3

we

of these steps,

will

voltages to zero.

all

go through

a

few examples.

and one load suppose we pick A

clockwise direction, the

first

sign

is

as a starting point.

Walking around the

plus and the magnitude of the voltage

circuit in a

is Vo.

So,

we

write

Continuing

in a clockwise direction,

minus and the magnitude

we

of voltage

next arrive

is

V,.

at the

source where the

Adding — V,

first

sign

is

to the other voltage gives

+ V.,-V, After walking through the source and arriving back

at

the starting point,

we

equate

all

voltages to zero to get

+v.-v, = This

is

the Kirchhoff voltage equation for the closed path of Fig.

ference of Vj and V, If

we

is

tells

3-6i!. It

says the dif-

zero.

transpose V, in the foregoing equation,

V,

This

o

=

we

get this relation

between

V,

and

V,

us the source voltage equals the load voltage in Fig. 3-6a. The result makes

and load are between the same pair of equipotential points. shows a simple way to visualize the equality of V, and V-,. The elevator lifts balls from the ground level to the upper level; this increases their potential energy. When the balls fall back to the ground, they lose their potential energy. The potential sense, because the source

Figure

3-6i'

SERIES

AND PARAUEL

CIRCUITS

69

Figure 3-6. (a) Kirchhoff's voltage low. (b) Mechatiical analog}/.

r-

•^

^—••

.

/7^^/ ////////////7/7777//////////77////////7/// Ground (a)

energy

lost

equals the potential energy gained

ground and back up

One

at

source

when

the balls travel from

to the

and two loads

As another example of the use of Kirchhoff's voltage law, look at point A and going around the circuit in a clockwise direction, we

then add +V:,

A

to A.

Fig. 3-7fl. Starting first

write

to get

+ V.,

V,,

-I-

then add —V, to get

We

then arrive back

at the starting

point and can equate

+v.,

Transposing

+

v.,

-

V,

all

voltages to zero to get

=

V, gives

V,

This says source voltage equals the

=

sum

v.,

+

V:,

of load voltages.

The

result

makes sense,

because the energy gained by charges passing through the source equals the energy lost

by these charges passing through the

Two

sources

In Fig. S-Zf),

loads.

and two loads

summing

voltages around the circuit gives V-i

+

v.,

-

V,

-

K,

=

70

RESISTIVE CIRCUITS

Figure 3-7. Examples of Kirchhoff's voltage law.

V,^

which rearranges

into V,

This says the

sum

of

all

+

V^

=

V3

+

^4

source voltages equals the

Again, the result makes sense. The

total

sum

of

all

load voltages.

energy gained by charges passing through

the sources should equal the total energy lost

by these charges passing through the

loads. (Charges passing through a source gain energy, because they

of

low potential energy

Why

to a

move from

a point

point of high potential energy.)

the voltage law must be true

In general, Kirchhoff's voltage law says

voltages around a closed path

2!

Because voltage ference

how

is

=

(3-2)

the difference in potential energy per unit charge,

complicated the circuit

is.

As long

as

we go around

it

makes no

dif-

a closed path, the volt-

we are returning to the same potential-energy point. combined with the current law allows us to write equa-

ages must add to zero because Kirchhoff's voltage law tions relating voltages

we

can derive

EXAMPLE What

is

all

and currents

in electric circuits. Starting

kinds of useful formulas.

3-4.

the value of

V

in Fig. 3-8fl?

SOLUTION. Summing voltages around

the circuit gives

3+4-U=0 Solving for V,

V=

7

V

with these equations,

SERIES

when we

In Fig. 3-8fl,

visualize

V equal

to 7 V,

we

AND

PARALLEL CIRCUITS

see that the source voltage

is

the

71

sum

of the load voltages.

This

is

a

simple example of using Kirchhoff's voltage law

to get

an equation which

is

any circuit, each loop (closed path) gives one voltage equation. Therefore, whenever you know all the voltages in a loop except one, then solved for the

you can solve

EXAMPLE

unknown

for the

voltage. In

unknown

voltage.

3-5.

Solve for V in Fig. 3-8b.

SOLUTION. With Kirchhoff's voltage law. 3

which rearranges

+

VV=

In Fig. 3-8b,

=

12

into

V equal

to 9

V means

9

V

the source voltage equals the

sum

of the load volt-

ages,

EXAMPLE

3-6.

Solve for V, and V2 in Fig. 3-8c.

SOLUTION. You can choose any path.

closed path, because the voltage law holds true for each closed

To minimize the work, always choose

a

path with the fewest unknowns. For in-

stance, to get an equation with V,, choose path

3

+

A-B-C-A which gives

V,~U =

Figure 3-8.

Summing

voltages

around a

12-=-

loop.

72

RESISTIVE CIRCUITS

Solving for V, gives V,

To

get V,, choose path

=

9

V

D-E-F-C-A-D which gives V,

+

5

-

=

12

Solving for Vj results in V2

We

will

=

apply Kirchhoff's voltage law

7

V

many

times in this book.

It is

important

to

closed path, usually the one with the fewest unknowns.

remember we can choose any

Test 3-4 1.

2.

Similar to the current law, the voltage law (a)

a

(c)

a

You

defined formula derived formula

an experimental formula

(rf)

none

sum

are also allowed to

of these

3.

4.

(c)

either sign, provided

first

Kirchhoff's voltage law

is

the second sign

and ending points correspond

the starting

(c)

voltages are neither created nor destroyed

load

(b)

to the

same

potential energy

(d)

none

of these

the voltage law as a point

circuit

(

)

(

)

(

)

true because

charges are neither created nor destroyed

is to

)

you wish.

you are consistent

(b)

The current law

if

down (b)

(«)

(a)

3-5.

sign of each voltage

the

(

voltages in a counterclockwise direction

This being the case, you must write (a)

is

(b)

(f)

is to a

equipotential point

{d)

loop

SERIES-CIRCUIT ANALYSIS

With Kirchhoff's current and voltage laws you can

learn

more about

series circuits. For

series circuits the key questions are

Why Why What

Same Here

current at is

all

is

current the same at

all

points?

do voltage ratios equal resistance ratios? is

equivalent resistance?

points

an important idea about a series

circuit: current

has the same value at any

point in the circuit. This follows logically from Kirchhoff's current law. In Fig. 3-9a, the

SERIES

AND

PARALLEL CIRCUITS

73

current law gives ;,

which says the current into current out of one point

is

same value

If

at all points.

every other point in

.4

=

/,

equals the current out of A. Furthermore, because the

the current into the next point, the current

is

5

A

at

point A,

it

we conclude must

also

current has the

be 5

A

at B, C,

and

Fig. 3-9(j.

Voltage ratios equal resistance ratios

Another important property of

a series circuit is this: the ratio of the load voltages

equals the ratio of the load resistances. tance,

it

will

If

much Or if a resistance is much voltage.

a resistance is twice as

have twice as much voltage across

as another resistance,

it

will

have 10 times as

it.

as another resis-

10 times as large

Here's the proof. In Fig. 3-9b, the current has the same value in R, and therefore, the voltages are

The

V,

=

V,

= RJ

ratio of these voltages is

V,

R,I

R-,;

74

RESISTIVE CIRCUITS

As an example, substituting

the values of Fig. 3-9r into Eq. (3-3) gives

_ ~

8 v.,

or

Vj

Visualizing Vj as 4

V

we

in Fig. 3-9c,

1000

500

=

see the l-kfl resistance has 8

and

resistance has 4 V, a 2-to-l ratio for voltages

As another example,

V

4

V and

the SOO-Q

resistances.

the SO-H resistance of Fig. 3-9d has

1

mV

across

it.

With

Eq. (3-3),

50

0.001

or

V,

In Fig. 3-9rf,

when

V,

equals 200

=

mV,

0.2

V=

200

mV

the voltage ratio

is

200, the

same

as the resistance

ratio.

Equivalent resistance

sum

of

the resistances between the points. In Fig. 3-lOrt the equivalent resistance between

A

The and B

equivalent resistance

between two points

in a series circuit equals the

is

R = 4000 + 2000 = 6000 n This equivalent resistance results in the same current as the two separate resistances. In other words, the current in Fig. 3-lOb has the same value as the current in Fig.

Here

is

the proof. In Fig. 3-lOf, V,

we

+

Vj

Furthermore, each load voltage in Fig. 3-lOf

and Substituting these expressions into Eq.

/

K,/.

V,

=

R,I

(3-4fl)

R,

This says the current in resistances.

Fig. 3-lOc

^

is

=

= --

+

(3-4fl)

V.,

V,

y=R,/ + RJ= or

3-lOfl.

voltages around the circuit to get

- V=

V=V, +

or

sum

can

gives (R,

+

R^}I

—

(3-4b)

R.

equals the source voltage divided by the

sum

of the

AND

SERIES

PARALLEL CIRCUITS

Fij^ure 3-10.

75

Example of equivalent resistance.

12V-=-

/

In Fig. 3-lOrf, the current is

'4 When

=

R the current in Fig. 3-lOrf is the as the current

If

a series circuit

what we have

all

just

gone through, we can show the following.

has n resistances, the equivalent resistance that results in the same

is

R= In

Ro

of Fig. 3-lOir,

a proof similar to

current

+

R,

as the current in Fig. 3-lOc. In other words, as far

concerned, the equivalent resistance of Fig. 3-lOd acts the same as the

is

two separate resistances By

same

+ R„

+ R.,+

R,

(3-5)

words, Eq. (3-5) says the equivalent resistance between two points equals the sum the resistances between the points. The symbol =^ is shorthand for "is equivalent to." Figure 3-lli! shows how

visualize Eq. (3-5) in circuit form.

the right

mean If

tance

when R

equals the

sum

The

circuit

a series circuit

80

ft. If

has K

,

=

left is

and

20

11,

R.,

=

(The dashes in

Fig. 3-11(7

R,,.)

50 Q, and R

the source voltage in this circuit

-f

to

equivalent to the circuit on

of the individual resistances.

there are resistances between R2

is

on the

of

0.1

A=

,,

=

10

ft,

the equivalent resis-

8 V, the current equals

is

100

mA

80

Generalization Figure 3-1 lu uses a battery for voltage sources.

The idea

its

voltage source. Later

of equivalent resistance

we

discuss other kinds of

does not depend on the kind of

volt-

76

RESISTIVE CIRCUITS

Figure 3-11. Equivalent resistance.

2n

ft2
)

•100

a >30on:

V

2kn,

86

RESISTIVE CIRCUITS

Figure 3-18. Three parallel resistances

combined

to tivo,

and

then one.

)

which rearranges

into 1

R= 1/R,

+

1/R.

+

+

(3-9)

1/R„

This derived formula says you add the reciprocals of the parallel resistances; then the reciprocal of the

sum

equals the equivalent resistance.

useful when you have an electronic slide rule or calculator to work with. You can quickly add the reciprocal of each resistance to get the denominator. The equivalent resistance equals the reciprocal of this sum. For instance, suppose a circuit has 2 Q., 3 il, and 6 n in parallel (Fig. 3-19fl). With Eq. (3-9),

Equation

(3-9)

is

1

R= 1/2 1

-I-

1/3

1

+

0.5

1/6

-H

0.333

-H

0.167

n

Therefore, the V/l ratio of Fig. 3-19b

is

equal

to the V/I ratio of Fig. 3-19fl.

Figure 3-19. Combining three parallel resistances by reciprocal rule.

'2n

1

3

kll

^ 9 kli

kH

'

1

kn

Woik

3-36-

Fift>-

^300kn

^ 2 kn ^ 3 kn ^ 4 kn

(e)

Ufl

3-35.

'lOOkn

out the equivalent resistamce between light bulbs, each

A and

with a resistance of 200

B in Fig. 3-32e.

ft,

are in parallel. VVTiat

is

the

equivalent resistance? 3-37.

Each load resistance in an ordinary house

is

in parallel across the

seven-room house has the following resistances for

15

its

rooms: 5

power ft,

10

kn

-AO-AA/V-*-

'

50

kn

^ 50 kn

'

10

kn

r—vW-t

4

^'i kn

kn

—VvA/ 12.5

kn

-O'

line.

ft,

20

A ft.

20 n. 30

30 n. and 30 it UTtat

ft,

is tt»e

equivaient resistance tor the entue

household? 3-38.

A 1-kft resistance is in paiaDd witti a l-Mfl resistance?

Does

ttie

lesistwice.

answer apfMoach the valoe of

tfte

What is

the equivalent

smafier or faoger resis-

tance?

and B in Fig. 3-33*? 6 in Fig. ^^S^. Work out the equivalent resistance between tt»e AB tenninals of Rg. What is the equivalent resistance fitom .4 to 6 in Fig. 3-3i3rf?

3-39.

What

3-40.

Calculate

5-41. 3-42.

is ttie

ANSWERS TO

TESTS

3-1.

I.

3-5.

b, c, a, d, c

3-6.

4. c, c. c.

3-7.

b,*.d.c

,>>.

equivalott resistance between

tfie

.''

b

equivalent resistance from

.

The

parallel

Therefore, the Thevenin voltage

V,„

^

fv^

1000 24

=

3000

connection of 6 kil and 3 kli

is

8

V

Figure 4-7. Applying Thevenin's

theorem.

1

kn


6kn

*

V,

:=:9v

J.

I

3kn

2kn

AAAv *6kn

OA

15V-=.

SOLUTION. This chapter has discussed three theorems: the voltage-divider theorem, Thevenin's

theorem, and the superposition theorem. In this example,

we

use

all

three theorems.

To get the Thevenin circuit for Fig. 4-36fl, we have to work out V/,/ and R;„. To find Vrii, add the voltages produced across the AB terminals when the sources are taken one at a time. In the first-source circuit of Fig. 4-36/),

6000 V,

In the

second-source circuit of

3000 + 6000

18

=

12

9

=

3

V

Fig. 4-36c,

v.,

=

3000

6000

-^

3000

V

RESISTIVE CIRCUITS

148

V,

and

V.J

have the same polarity as

therefore,

V/,,;

n+3

= v, + v., = = 15 V

Vri,

is the Thevenin voltage for the original circuit. To get Rrn, reduce all sources to zero as shown in tance between the AB terminals is

This

Rth

= 3000 = 2kn

II

Fig. 4-36if.

The equivalent

resis-

6000

Figure 4-36e shows the Thevenin circuit for Fig.

4-36rt.

Test 4-7 1.

2.

Which

same time

at the

all

(c)

divide-and-conquer strategy

Superposition

is

(a)

adding magnitudes

summing the load

(c)

any resistance

is

(fl)

the load

(c)

all

(d) all

{h)

(d) n

meaning

(/')

the load

linear

is

(d)

(

)

(

)

(

)

(

)

to

adding or subtracting

(b)

linear

superposition theorem

simpler circuits

the superposition theorem

(a)

You can use

4.

closest in

(d)

You cannot use

3.

5.

of these least belongs?

sources acting

(a)

is

(r)

algebraic

sum

if

independent

any resistance

is

nonlinear

the superposition theorem provided is

nonlinear

resistances are linear

(b)

sources are independent

all

and you reduce only independent sources

sources are independent and resistances are linear

The algebraic sum

of

two currents

greater than either current only

is

when

the cur-

rents are in the {a)

same

direction

(b)

opposite direction

4-8.

REVIEW OF THE THREE THEOREMS

The

three theorems of this chapter are

among

(c)

none

of these

(

)

the most practical circuit theorems

because they will help you solve an enormous range of problems.

The voltage-divider theorem is easy to understand and remember. It applies to any circuit, and says the output voltage across resistance R^. equals K..,/R

one-source series

times the input voltage.

Thevenin's theorem cuit.

Because of

on the

this,

it

is

outstanding.

eliminates

all

essential part of the problem.

It

reduces a complicated circuit

unnecessary information and

lets

to a series cir-

you concentrate

THREE THEOREMS

The superposition theorem

is

useful

when

the original circuit has

149

more than one

source. This theorem divides a many-source circuit into simpler one-source circuits.

SUMMARY OF FORMULAS DERIVED (4-1/))

(4-2) Rii,

R(

Rt

-^ = -^ /

=

;,

,

(balance)

-I-

/.,

+

+

(4-8)

(superposition)

/„

(4-10)

Problems 4-1

In Fig. 4-37(1,

the 4-2.

what

way down? At

is

the output voltage

when

the wiper

way up?

All

output voltage

for

is all

the

the middle position?

There are four switch positions

in Fig. 4-37)). Calculate the

each. 4-3.

Figure 4-37ir shows a step voltage divider used in

value of

V„u| in

A through

each switch position

4-4.

In the open-load circuit of Fig. 4-38n,

4-5.

Figure 4-38b shows an open-load

4-6.

What

4-7.

The wiper

is

the

Thevenin

what

voltmeters.

What

is

the

the value of V,,,?

K™?

What does Vrn equal? R,,,? AB terminals in Fig. 4-38f?

circuit.

circuit left of the

of Fig. 4-38 5 kH

)3rnA

5kn

EXAMPLE 5-10, A load resistance

of 5 kfi is

Calculate the load current

15

V^

connected between the AB terminals of

and voltage

Figs. S-lla

and

b.

in each circuit.

SOLUTION. Figure 5-120 shows the Norton circuit with a 5-kfl load resistance. The current out of the source

must

equally between the two resistances. So, the load current

split

/=1.5 and the load voltage

is

mA

is

y=

5000 X 0.0015

=

7.5

V

Figure 5-12rf shows the Thevenin circuit with a 5-kn load resistance. In this case, the load current equals

/

15

= 5000

and the load voltage

+

mA

is

V=5000 X The point

1.5

5000

is this.

Given an

0.0015

original circuit,

= we

7.5

V

this,

we have

the option of using whichever

is

it by a Thevenin circuit same answer. Because of

can replace

or a Norton circuit to find the load current; either gives the

easier in a particular problem.

MORE THEOREMS

173

Test 5-3 1.

2.

3.

4.

5.

5-5.

Norton

is to

Thevenin as

ideal current source is to

(a)

controlled source

(i/)

load-independent source

Which

(/>)

ideal voltage source

(c)

independent source (

)

(

)

of these belongs least?

((i)

load resistance

{ci)

Thevenin

black box

((')

(c)

Norton

circuit

circuit

The following words can be rearranged NORTON IS BETTER. The sentence is (fl)

true

An

ideal current source

(rt)

a short

(d)

infinite

(b)

to

form

a sentence:

CIRCUIT

ALWAYS

false

(h)

(

whose value

an open

(r)

is

zero

is

zero resistance

conductance

Zero voltage source

is to

(a)

zero conductance

(d)

all

zero current source as zero resistance (/))

)

identical to

infinite resistance

(c)

(

)

(

)

is to

open

the foregoing

POWER

Figure 5-13

shows atoms inside

a load resistance.

Chemical bonds

(ionic or covalent)

hold these atoms together. At absolute zero temperature (— 273°C) the atoms are motionless; but as the temperature rises, they spring back and forth about the positions

shown in Fig. 5-13. The higher the temperature, the faster the vibrations. As you read more about these vibrations, hunt for the answers to

Why How

docs current produce heat? is

pou'er defined?

Figure

.5-3,3.

Atoms and

their

bonds.

-tJUUU

174

RESISTIVE CIRCUITS

Free electrons produce heat in a resistance

When

free electrons

move through

collision

makes an atom vibrate

they collide with the atoms.

a load resistance,

Each collision deflects the electron and reduces a bit faster.

The

its

speed. At the

effect of

all

temperature of the load. In other words, heat

rise in the

is

same

time, each

atoms vibrating

faster is a

nothing more than faster

vibration of the atoms.

work and energy, free electrons lose energy as they move through a work done on the atoms. In turn, the work done on the atoms produces heat. So what it boils down to is this: the loss of energy by the elecIn terms of

load; the energy lost equals the

trons equals the heat generated in the resistance.

Definition of

Power

is

power

defined as the work done divided by the time in which

it's

done. As a

defining formula,

=

P

W -

(5-4)

= power W = work = time

where P t

If

free electrons passing

through

a resistor lose 5 5

2 s

Or

if

15

J

of heat are

produced

in 3

the

s,

of energy in 2

s,

the

power equals

I

= 7^ =

P

J

2.5J/s

power equals

the same as rate of work because its numerical value tells us the number work done in each second. In the first of the preceding examples, 2.5 J of work are done in each second. In the second example, 5 J of work are done during each second. The rate of work is greater in the second case than in the first.

Power

is

of joules of

The watt

The watt (abbreviated W) 1

is

the metric unit of measure for power;

watt

=

or

A

watt

is

the value of

power when

1

1

it's

defined as

joule per second

1

W=

J

of

1 J/s

work

is

done during each second. With

this

MORE THEOREMS

answers

definition, the

preceding examples become

in the

2.5 ]/s

5 J/s

175

= 2.5 X =5X 1

=

J/s

1

=

]/s

W = 2.5 W W=5W

X

2.5

5 X

1

1

The watt is standard in electrical and electronics work. For this reason, you always power in watts rather than joules per second. Since W is always in joules and

specify is

t

always in seconds, Eq.

automatically gives an answer in watts.

(5-4)

Incidentally, textbooks normally use italic letters

physical quantities (current, voltage, power, V,

W,

is

why

etc.).

are used as abbreviations for units of

etc.)

italic

V

represents voltage, but

resents work, but

Roman

W

On

(/,

V, P, etc.) as abbreviations for

measure (ampere,

Roman V

Roman

the other hand,

stands for

letters (A,

volt, watt, etc.).

volt; similarly, italic

This

W rep-

stands for watt. Watch for these differences in this and

other textbooks.

EXAMPLE

5-11.

Electrons passing through a resistor lose 30,000

]

in 15

What

s.

is

the power?

SOLUTION.

W = 30,000 = 2000 W = 2 kW — :, 15

P=

r

EXAMPLE 5-12. A house receives

electric

energy

at a rate of 2

kW.

How much

energy

is

received in

1

h?

SOLUTION. Multiply both sides of Eq. (5-4) by

(

to get

W=Pt This says the work done (equal to the energy received) equals power multiplied by

P

time. Since

=

2

kW

and

f

=

1

h

=

W=Pt = 2000 EXAMPLE

kilowatt-hour

is

sell

=

7,200,000

J

=

what

is

1

h

when

the

the electrical

power

MJ

of the electric energy

is

C=

1500 X 0.02

=

is 1

bill for a

during the month?

SOLUTION. The cost

7.2

energy by the kilowatt-hour (abbreviated kWh).

electric

the energy received in

costs 2 cents per kilowatt-hour,

kWh

s,

X 3600

5-13.

Power companies

1500

3600

30 dollars

kW.

house

If

electric

One

energy

that has received

176

RESISTIVE CIRCUITS

Test 5-4 1.

2.

3.

Current through a conductor produces heat because

atoms

atoms

(a)

electrons hit atoms

(d)

electrons gain energy with each collision

The

(b)

speed of the electrons

(c)

battery voltage

(b)

4.

Power most

5.

Watt

6.

P

=

(fl)

is to

(b)

power

ohm

(fl)

(d) rate

a resistor equals the

heat produced in the resistor

(b)

of charge flow

power means

(c)

watt

per second

(d) joule

closely

joules

(fl)

atoms give up energy

of these belongs least?

energy

(a)

(r)

by electrons passing through

potential energy lost

(a)

Which

hit

watts as

heat

(d)

is to

parallel

(b)

Wjt does not have

work

rate of

(c)

mho

conductance

(c)

defining formula

(b)

rate of

(rf)

be proved because

to

it is

work

a

experimental formula

derived formula

(c)

...

POWER THEOREMS

5-6.

As you may

suspect,

exist for calculating

power

relations

and

power power rules

is

related to voltage

in series

by

and

and

current. Furthermore, simple rules

parallel circuits.

few theorems.

stating a

We

can summarize these

Among

other things, these

theorems answer the following:

How Why How Relation to voltage

is

power

related to voltage

does each load IS

source

and

power add

power

and current?

to the total

related to load

power?

power?

current

Given a resistance with voltage V and current voltage and current. In symbols,

/,

the

power equals

the product of

P=V1

(5-5)

= power V = voltage = current

where P

/

So,

if

a resistance

has 10

V

across

it

P Here

is

how

to

and

=

10

derive Eq. (5-5) from

5

A

X 5

through

=

50

known

it,

the

power equals

W

formulas. First, recall that

MORE THEOREMS

)77

This rearranges into

W=VQ Substitute this expression into the defining formula for

t

because Q/t equals

/.

Therefore,

power

to get

t

we have proved

Eq. (5-5).

Alternative power formulas

With Eq.

(5-5)

we

can derive two alternative power formulas. Since

V=

Rl,

P=VI = RIX I=RP or as

it

usually

is

written,

P=I'R

(5-6)

The derived equation says power equals the square of the current times the resistance. This formula is useful when you have the values of the current and the resistance, but not the voltage.

Another alternative formula

Since

is this.

/

=

V/R,

V

P

or

=

^K

(5-7)

This says the power equals the square of the voltage divided by the resistance; use this

when you have Of the

the values of the voltage and the resistance, but not the current.

three formulas, the one to

remember

is

P

=

VI.

derive the other two. After a while, you automatically will

When

necessary, you can

remember

the alternative

formulas.

Power

in

a series

circuit

In the series circuit of Fig. 5-14(!, the

and These formulas

tell

power

P,

= VJ

P-2

= VJ

in

each resistance

is

us the rate of work done by electrons moving through each resis-

tance. Equivalently, they

tell

us the rate of heat production.

In the equivalent circuit of Fig. 5-14^, the

P

=

V1

power

is

(5-8fl)

178

Figure 5-14.

RESISTIVE CIRCUITS

Power

in series

circuit.

R,

S V, fl,

This formula total

tells

+flj

us the rate of work done on the equivalent resistance, or the rate of

heat production in the circuit. In Fig. 5-14rt,

V=V, + Substituting this into Eq.

(5-8rt)

gives

P= or

Equation

(5-8b)

is

makes

+ V,)/= V,/+ VJ

(V,

P

All this final result says

V2

the total

=

P,

+

(5-8b)

P.,

power equals

the

sum

of the individual powers.

sense. Free electrons lose potential energy passing through

load resistances. In a unit of time, the total energy lost should equal the energy lost to the

first

resistance, plus the energy lost to the

second resistance. Equivalently, the

heat produced in the circuit should equal the heat produced in the

first resistance,

total

plus

the heat produced in the second resistance. In general,

no matter how many resistances are P

To

get the total

Power

A

power

=

P,

+

in a series circuit,

p.,

+

in a series circuit.

+

P„ (5-9)

you add the powers

in the series resistances.

in parallel circuits

similar result occurs in parallel circuits: the total

individual powers. The proof for two resistances follows. In Fig. 5-15fl, the

power

to

each resistance

is

P=VI, P

and In Fig. 5-15b, the

power

to the

= Vh

equivalent resistance

P

=

V1

is

power equals the sum

of the

MORE THEOREMS

figure 5-15.

Power

in parallel

circuit.

Since

/

=

/,

+

/.,

=

P

V{I,

As we

see,

matter

how many

(5-10)

P.,

derived equation

this

identical

is

power equals the sum

Therefore, the total

+Vh

+ k) =VI,

P=P, +

to

parallel resistances there are, the total P,

+

the

one

for

a

series circuit.

of the individual powers. In general,

+

P-.+

power

no

is

P,,

Additive power theorem

Using proofs similar the total

power

tances. This

in

any

theorem

is

ones

to the

just given,

circuit equals the

sum

we

energy

lost

resistances. This

by moving charges has is

powers

easy to understand and believe. Power

time, or energy lost per unit of time. Since energy total

can prove the following theorem:

of the

equivalent

work done per

unit of

neither created nor destroyed, the

equal the energy lost to individual load

to

saying the

to

is

in the individual resisis

total

heat produced

must equal the sum of

the individual heats produced in the load resistances.

Total source

power equals

total load

power

The next theorem may be obvious, but should be supply energy

to the load resistances.

Since energy

As an example,

power equals

The voltage

across the

=1= R

upper resistance P,

=

V,;

=

The sources all

the sources. In theorem

the total load power.

the current in Fig. 5-16

1

stated just in case.

neither created nor destroyed,

come from

the energy received by the load resistances must

form: the total source

is

is

10

=

1

mA

10,000 is

6 V; therefore the

6 X 0.001

=

0.006

power

W = 6 mW

in this resistance is

RESISTIVE CIRCUITS

180

Figure 5-16. Total source

power

equals total load power.

>6kn

fl,

The lower

=

P.

The

total

V

resistance has 4

load

power

the total source

Vj/

=

4

P,

+

Pi

X 0.001

=

power

=

a

power

=

0.004

of

W = 4 mW

6

mW + 4 mW = 10 mW

is

Psourve

rate

and

it,

is

Pioad

And

across

=

P|,«d

=

10

HlW

The energy delivered to the load resistances comes from the source. Therefore, the of energy lost by the source equals the rate of energy absorbed by the loads. In this

particular example, the source loses energy at a rate of 10

absorbs

it

EXAMPLE

A

at the

same

mW,

and the equivalent load

rate.

5-14.

12-V battery produces

power? At what

rate

a current of 5

mA

in a series circuit.

What

is

the total load

does the battery lose energy?

SOLUTION. The

total

voltage

is

12 V,

P This

is

and the current

=

5

is

W= 12(0.005)

mA;

=0.06

the total load power. Equivalently,

it

is

so,

W = 60 mW the rate at

which the battery

loses

energy.

EXAMPLE

5-15.

Twenty Christmas tree lights are in parallel across a 120-V source. tance of 10 kJl, what is the power in each? The total power?

SOLUTION. Given voltage and

resistance,

we

can use

If

each has a

resis-

MORE THEOREMS

Alternatively,

we

can calculate the current in each bulb and then use P

we

Either way,

The power

get

=

VI

the same answers.

in the first

bulb

is

P,

=

-^-!-

19 bulbs

=

=

1.44

10,000

R,

'

The remaining

181

W

have the same value of power. The

total

power

is

the

sum

of

the individual powers:

P = 20x

EXAMPLE What

is

=

1.44

28.8

W

5-16.

power

the

in the 3-(l resistance of Fig. 5-17i!?

SOLUTION. Apply the voltage-divider theorem V=

The current through

to get

= ^V/ = |l8 = 6V

this resistor is

'

= I^ = Rj

And

the

power

3

is

Pj

EXAMPLE

^2A

=

V.J

=

6(2)

=

12

W

5-17.

Figure 5-l7b shows the schematic symbol of a fuse, a resistor that melts

when

the cur-

The 3-A fuse shown melts when the current exceeds 3 A; this opens the path for charge flow and protects the circuit from too much current. If the fuse has a resistance of 0.02 il, what is the fuse power at the melting point? rent

is

excessive.

What value

of R, causes the fuse to

open?

SOLUTION. The

fuse

power

at the

melting point

P = PR =

is

3^(0.02)

=

0.18

W=

180

mW

182

RESISTIVE CIRCUITS

Figure 5-17. Calculating power.

3A fuse

30.

V-=-

18

(6)

In Fig. 5-17b, the

only

the equivalent resistance

way

is

A

to get 3

6.02

fl,

is to

reduce R,

to zero.

When

this

happens,

and 18 3

A

6.02

So the only way

blow the fuse

to

is

by shorting the load terminals. is intact and offers only 0.02

When

R,, is

greater than zero, the fuse

Because of

this,

an intact fuse

approximated as

is

a short

and

a

ft

of resistance.

blown fuse

as an open.

Test 5-5 1.

When

2.

P

=

(a)

P

=

V/ was

resistive

(a)

VI

is

first stated,

{b)

4.

(n) never {b) sometimes The additive power theorem

6.

energy

(d)

experiment

5-7.

is

(c)

is

most

adds

of the time

derived formula to the total (if)

.

power:

always

based on

neither created nor destroyed

Which does not belong? (a) total source power

(b) total

load

power

(b)

heat

(r)

(c)

definition

sometimes equal

always equal

the current through a short

(fl)

(t)

in each resistor of a complicated circuit

(a)

If

specified as

capacitive

experimental formula

{b)

The power

(d)

(f)

an example of a

defining formula

3.

5.

was

the load

inductive

zero

(b)

3W

(c)

9

is 3

W

A, the power equals (if)

infinity

MATCHED-IOAD POWER

Figure

5-18fl

shows

occurs

when

R, equals the

a black

box driving an adjustable load resistance. A special case of the black box. The most important things to learn

Rm

MORE THEOREMS

183

Figure 5-18.

Circuit with

sources and linear resistances

Matched

load.

RESISTIVE CIRCUITS

184

And

the load current

is

12 I-

4000

The matched-load power P If

Ri

is

changed

to

+

1.5

4000

mA

is

=

VI

=

6(0.0015)

any value other than 4

=

0.009

W = 9 mW

kfl, the load

power

is

always

less

than 9 niW.

Misconception about the theorem

The maximum load-power theorem does not apply to the case of an adjustable Rthis, you don't adjust R™ to match the value of a fixed Kf If Rth is variable, as shown in Fig. 5-20a, the maximum load power occurs when Rrn is zero (Fig. 5-201'). An adjustable R™ like Fig. 5-20fl is rare. In practice, you normally get a black box or circuit whose Thevenin resistance is fixed and whose load resistance is adjustable. In this case, the maximum load power occurs when R, is adjusted to equal Rth-

That

Applications of the theorem

When of time,

it

a source is

may be

weak, that

is,

can deliver only a small amount of energy in a unit

necessary to match the load

to the source.

A TV

antenna, for in-

up an electrical signal from a transmitting station. The antenna acts like a voltage source and a series resistance, the Thevenin circuit of Fig. 5-21fl. The schematic symbol for the voltage source of Fig. 5-21a is different from a battery stance, picks

because the

TV

signal

is

not a fixed voltage. Instead, this signal varies in time. Chapter

8 discusses time-varying sources; the

main point

at the

moment

is

that a

TV

antenna

has a Thevenin voltage iVh and a Thevenin resistance Rth-

The quality

of a

TV

picture depends on the

the receiver has a resistance of R^, this

Figure 5-20. Variable Thevenin resistance.

is

power

to the

TV

receiver. In Fig. 5-21i',

the load resistance connected to the antenna.

MORE THEOREMS

Figure 5-21.

I8S

Maximum

load-power theorem.

^TH

186

RESISTIVE CIRCUITS

Figure 5-22. Thevenin circuit of signal generator.

ib)

should equal 50

17 (Fig. 5-22b).

For this matching resistance, the load voltage

V=

^=i=

and the load power

0.5V

2

2

is

V^ = -— = ^^^=- = 5^

P

EXAMPLE Prove the

0.005

50

R,.

W = 5 mW

5-19.

maximum load-power theorem

for Fig. 5-23fl.

SOLUTION. For the matched case, Rt equals

1 fl

(Fig. 5-23fc),

and the load voltage

v = i^ = i = iv 2

The load current

2

is

-i.-\-^^

is

is

MORE THEOREMS

1

187

n

Figure 5-23.

-AAA/2

V-=•

1

1

n

n

•I

The load power

is

W=

1

X

=

1

W

1

Suppose we move the wiper of Fig. 5-23a to get a case, the voltage-divider theorem gives a load voltage of

V = -2 =

1.33

R, of 2 fl (Fig. 5-23r). In this

V

The load current equals

V The load power

is

P which

is less

power always

On

1.33

than

1

=

W. For any

results. (Try

the other hand, R/

V7

=

R, greater than

other values of

may be

this case, the voltage-divider

less

than

— 0.5

2

1 fl

to

R/,

theorem gives

V=

=

X 0.667

1.33

0.887

W

in Fig. 5-23h, less

1 fl.

Figure 5-23d shows an

a load voltage of

=

0.667

V

The load current equals 0.667 0.5

than

1

W of load

convince yourself.)

=

1.33

A

K,,

of 0.5

fl.

In

RESISTIVE CIRCUITS

18S

The load power equals P

which

than

is less

always

5-23fl is

V/

=

0.667(1.33)

W. For any load 1 W.

1

less

=

=

0.887

W

resistance less than

1

(I,

power

the load

in Fig.

than

Test 5-6 In the following series,

1.

what

comes next?

logically

THEVENIN VOLTAGE, MATCHED LOAD (a)

current

When you

2.

((;)

voltage

use the

(c)

power

half of the

(rf)

maximum load-power

Thevenin voltage

(

)

theorem, the Thevenin resistance must

remain (a)

A

3.

in

ohms

(b)

fixed

(r)

varying

load resistance has a value of 10

(rf)

il.

To

none

of these

get the

maximum

(

)

load power, an

adjustable Thevenin resistance should equal

{d)20n (f)lOn (b)5il antenna has a Thevenin resistance

(rt)

A microwave

4.

{

of 50

fl.

To

get the

maximum

)

load

power, the load resistance should equal (b)

(fl)

An

5.

25

oscillator is

cillator

n

circuit

has a Thevenin resistance of

using a load resistance equal (fl)

(d)lOOn

(c)50n

an electronic

zero

(b)

half of the

you 1

(

will learn

kfl,

we

about

can get the

in a later course.

maximum

load

If

)

an os-

power by

to

Thevenin resistance

(i)

1000

S I

(d)

infinity

(

)

>

I

SUMAAARY OF FORMULAS

DEFINING G=

y

(5-lfl)

=^

(5-4)

G=\

(5-lb)

P

t

DERIVED

MORE THEOREMS

G=

189

RESISTIVE CIRCUITS

190

5-7.

Figure 5-24b shows a potentiometer with a total resistance of 10 k(l. load voltage

5-8.

5-9.

when

the wiper

at

is

What

is

the

the top? At the bottom? In the middle?

Suppose you have a stock of resistors with values from 10 il to 1 M(i. If you connect one of these between the AB terminals of Fig. 5-24c, what is the minimum possible current? The maximum possible current? (Three-digit accuracy is adequate for the answers.) To a close approximation, what value does the load current have for any load resistance between 10 fl and 1 MJl? What is the Thevenin voltage between the AB terminals of Fig. 5-25fl? The Thevenin resistance?

5-10.

Thevenize

5-11.

What

5-12.

Thevenize

5-13.

AB

Fig. 5-25b left of the

the V,,i Fig.

between the AB terminals

5-25d

left

of the

CD

of Fig. 5-25c?

terminals. Next,

5-17.

What is the Norton circuit left of the AB terminals in Fig. 5-26a? Work out the Norton circuit left of the AB terminals in Fig. 5-26b. In Fig. 5-26c, Thevenize the circuit left of the AB terminals. What circuit left of the AB terminals? Work out the Norton circuit to the left of the AB terminals in Fig.

5-18.

A

5-14.

5-15.

5-16.

car battery loses 24,000

many 5-19.

A

kilowatts

horsepower

British unit

is

is

J

is

the Norton

5-26d.

of energy during 10 s while starting an engine.

How

this?

a British unit of

and the metric unit 1

Figure 5-25.

terminals.

The Rjh'^ what is the Thevenin voltage between the AB terminals? The Thevenin resistance between AB? The current source of Fig. 5-25e has a value of Irun- If hao equals 1 /xA, what is the Thevenin voltage between the AB terminals? The Thevenin resistance? is

measure

for

is

hp

=

746

W

power. The relation between this

MORE THEOREMS

191

3kn

Figure 5-26.

6kn

-AAA >3kn

t)l2mA

If

5-20.

an

electric drill

The battery

What

is

does 1865

J

of

work

what

in 5 s

in a transistor radio loses 5.4

power from the battery? bulb consumes 180,000 J during an

J

is

when

the radio

is

on.

the

A

5-22.

The battery voltage in a transistor radio equals 9 V. tery is 10 mA, what is the power from the battery?

5-23.

A

light

the horsepower of the drill?

per minute

5-21.

light

oA

V

bulb has 120

across

it.

If its

hour.

resistance

What

is

If

is

the

power?

the current out of the bat-

144

(I,

what

the

is

power

in

the light bulb? 5-24.

An AWG-22 copper

5-25.

5-26.

wire 500

power in the wire? How much power is there in tance? The total load power? Calculate the power in each is

ft

in length

has a current of 0.5

A

through

it.

What

the

the S-kll resistance of Fig. 5-27«?

resistance of Fig. 5-27fc.

What

The

IS-kfl resis-

the total source

is

power? 5-27.

5-28.

5-29.

Work

out the

power

in

each resistance of

Fig.

5-27c.

What

the total load

is

power? The total source power? Ten circuits are in parallel across a 9-V source. Five of the circuits each have 1 mA of current. Three circuits have 2 mA each. And two circuits have 4 mA each. What is the total load power? Source power? A power supply is an instrument or circuit that produces a steady voltage similar to a battery.

A good power

supply produces an almost constant voltage and has

almost zero Thevenin resistance. In the

power supply

output voltage from 10

power with

to

a lOO-Jl load?

of Fig. 5-28fl there

is

an adjustment

30 V. This being the case, what

The maximum load power?

is

for

the

changing the

minimum

load

19i

RESISTIVE CIRCUITS

Figure 5-27.

'3kn

>3kn

5-30.

What a.

b. c.

5-31.

To is

>6kn

^4kn

^6kn

current rating should the fuse of Fig. 5-28b have

if

it

is to

blow when

A short is between the AB terminals. A 2-n load is between the AB terminals. A 75-fl load is between the AB terminals. get the

maximum

the resulting load

load power power?

in Fig. 5-28c,

Figure 5-28.

V

9kn

ath

of Fig. 6-16r.

Ground Ofter.. the

nected to

STi

power plug

of electronics

ac oudet. this third

at the dectiic

prong

power company. \Mien

equipment has

is

grounded (put

this

kind of plug

the ciiassis in contact with the earth; this

is

why

2 ir.

is

ruri rrcr-c

cr. it

\Vhen conback

ccr.rac: vr.ih the earth)

used, the third prong places

the chassis

is

often referred to as

ground.

Even if the jx>wer plug does not have a third prcmg, ifs stiD customary to refer to the chassis as ground. In describing the circuit of Fig. 6-16b. we'd say the negative ter-

UKHHOff MFTHOOS

minal of the 9-V battery

is

grounded, the bottom of the

the positive terminal of the 3-V battery

Because the chassis

is

ground

EXAMPLE Calculate

|7oints is

all

3-fl resistor is

grounded, and

grounded.

an excellent conductor,

equipotential point. In other words, in pair of

is

221

ground points represent the same

all

preliminary analysis the voltage between any

approximated as zero.

6-9.

node voltage V^ and current

/

in Fig. 6-17a.

SOLUTION. With the voltage-divider theorem.

The current equals ,

I

EXAMPLE

=

18 -^-
.

voltmeter has a meter accuracy of ±2.5 percent of

what

the 50-V range,

is

full-scale. If

it

reads 20

V

the lowest possible voltage across the voltmeter?

on The

highest? 7-15.

VVTiat is the voltage across the

30-kn

input resistance of 200 kfl

connected across this

is

resistor of Fig. 7-18j?

If

resistor,

a voltmeter

what

with an

viiU the

new

voltage be? 7-16.

A

voltmeter

loading error

is

to

is to

be connected across the 3-kn resistor of Fig. 7-18l>. If the be less than 3 percent, what input resistance should the volt-

meter have? 7-17.

If

the voltmeter of Fig. 7-\Sc were ideal,

-AB terminals?

If

what would be the voltage bet^veen the what is the

the voltmeter has an input resistance of 100 kfl,

voltage between these terminals?

An ohmmeter like

Fig. 7-l9a is called a series type

because the mo\-ing-coil meter

figure 7-17.

495 kn

—VSAr-O

4

20 mA

5kn

U»

BASIC MEASUREMINTS

1J7

ligure 7 18.

)

1mAU

(c)

is

in series

with the resistance being measured.

When

the

AH

terminals are

shorted, the zero-adjust can be varied to get full-scale deflection.

value of R. that zeroes the 7-19.

In Fig. 7-19(1,

What

is

the

ohmmeter?

what resistance does

half-scale deflection represent? Quarter-scale

deflection?

fiffure 7-19.

—

1.5 +

h|i|

— AAA,

i--^AA/

0-20kii

130 kSi

50 ",

24

kn

1.5 +

^_^

i5n

^^

-WA^ " ^ ^°. ^ .150

o

R X 100

a

.1.58kn

^AA/

(iA

2kU

2Sa

7-20.

RESISTIVE CIRCUITS

When measured

resistances are small, a shunt-ti/pe

ohmmeter

like Fig. 7-19b is

often used. a.

b.

What value does R. have when the ohmmeter is zeroed? With the AB terminals shorted, hov^r much current is there

in

the 15-fl

resistor? c.

If

the resistance being measured equals 15

Ji,

how much

current

is

there in

the moving-coil meter? 7-21.

AB terminals can be Thevenized to get a Vjh and an Rm that depends on the switch position. the R x 1 range, what does half-scale deflection represent? the R X 10 range, what does quarter-scale deflection represent? the R x 100 range, what resistance is being measured if the needle in-

In Fig. 7-\9b, the circuit right of the

of 1.5 V, a.

b. c.

On On On

dicates 5 /iA.

ANSWERS TO 7-1.

TESTS

TdDttQ©

^D In

many

chapter

circuits, time is just as

tells

how

to

extend

all

important as current, voltage, and resistance. This

you've learned

to time-varying signals,

currents or volt-

ages that change from one instant to the next.

BASIC IDEAS

8-1.

To understand how:

to

apply Ohm's law, Kirchhoff s laws, Thevenin's theorem,

time-varying signals, you have to

know more about

time. In

what

etc., to

follows, your job

is

to learn

What

How

is

time?

do you visualize

What

is

a

it?

waveform?

Nature of time

To begin with, time is a fundamental quantity, along with length, mass, temperaand charge. These five quantities are the foundations of the physical universe; all

ture,

other physical quantities can be defined in terms of the five fundamental quantities. it is a fundamental quantity, time cannot be defined mathematically. That no defining formula for in terms of other fundamental quantities. But we can measure time by defining a unit of measure. The metric unit of time is the second,

Because

is,

there

is

(

defined as

^^^^°"'*

=

Average solar day

SMOO

260

RESISTIVE CIRCUITS

An

average solar day

we

ing value,

is

the length of a day averaged over

many

years.

With the forego-

can build clocks and other instruments to measure time.'

Visualizing time

Time time

so elusive

is

with

is

we need

a visual aid to pin

a horizontal axis like Fig. 8-1

it

down. The usual way

Each point along

ij.

to visualize

this axis stands for a point

in time called an instant.

where

In Fig. 8-l«, the instant

because

all

earlier in time; those right of

When like

It's

f

=

it

up problems, we

setting

(

mark the

Usually,

we

instant of

let

f

=

Instants

left

of the origin occur

are free to locate the reference instant anywhere.

choosing a reference node in

=

as the origin or reference instant, it.

take place later.

a circuit;

node, and measure other voltages from let

known

is

other points in time are measured from

we

can pick any node as the reference

Likewise with the reference instant.

it.

any event, and measure other instants from

We

can

it.

coincide with something important, such as the blast-off of a

rocket, the start of a race, the closing of a switch, etc. For example, after the switch of Fig. 8-lb is closed, the current let

equals 2

mA.

the closing of the switch coincide with

circuit before the switch is closed,

switch

is

closed.

A

t

In discussing a circuit like this,

=

0.

Then, negative values of

and positive values represent the

point in time like

f

=

(

it

helps to

represent the

circuit after the

2 s represents the circuit 2 s after the switch is

closed.

At a particular photograph of

a

instant, time stops

moving

thing stops while

object.

we examine

and

When we

all

action ceases.

It's

analogous

to taking a

refer to a particular instant, therefore, every-

a circuit at the particular point in time.

Waveforms

TV signals are good examples; they have varisound and pictures being transmitted. Most currents and

Signals carry information. Radio and ations proportional to the

In 1967, the 13th General Conference of Weights and Measures met in Paris and redefined the second in terms of a cesium standard (an atomic clock). '

Figure 8-1. Visualizing time.

-•

•

•

-3-2-1

•

• 1

Seconds

•

•

2

3

ft

10V.=

Figure 8-2. Waveforms, (a)

Ramp.

(b) Triangular wave.

voltages in electronics are time-varying. For this reason,

we have

to

work with

instanta-

neous values, the values of current and voltage at each instant in time.

To help us

visualize signals in time,

we

use waveforms; these are graphs of instan-

taneous current or voltage versus time. For instance. voltage. Right of the origin,

we i;

i>

i;

Fig. 8-2a

shows

a

time-varying

read these instantaneous values:

= 5V = 10V = 15V

at

f

at

t

at

f

= = =

1

s

2 s 3 s

and so on. So the voltage increases 5 V during each second; this is why the graph of Fig. 8-2fl is linear. Any waveform that increases linearly with time is called a ramp. Figure 8-2b shows a triangular waveform, another example of a time-varying signal. and t = 1, a Notice it's made up of different ramps: an increasing ramp between t = decreasing ramp from = 1 to ( = 3, and an increasing ramp between t = 3 and t = 4. f

Think

stance, take Fig. 8-3c,

waveform as a sequence of values. When you look at Fig. 8-2a, for inthe waveform apart in your mind and see Fig. 8-3a, then Fig. 8-3b, then

of a

and

finally Fig. 8-3d.

Taking

a signal

apart like this

is

essential in circuit

analysis.

Figure 8-3. Succession of instantaneous values.

3s

id)

264

RESISTIVE CIRCUITS

Test 8-1 1.

Time is not measured

(fl)

(f) 2.

3.

4.

Which

origin

(d)

any point on

(a)

second

(d)

any instant

a fundamental quantity (d)

measurable

time

in

(c)

a

(

)

(

)

(

)

(

)

(

)

(

)

any instant

time axis

any node as origin is to waveform (c) reference instant

is to

closely

means

triangular signal

(c)

signal

graph of current or voltage versus time

A ramp

is

not

waveform

(b)

nonlinear

(c)

related to time

uniform increase or decrease

Which of these does not apply to an oscilloscope? (a) shows each successive voltage value (b)

displays voltage waveforms

(d) vertical axis

8-2.

a

(h)

Waveform most (b) («) ramp

{d) a 6.

any point

(b)

Reference node

(a)

(b)

of these belongs least?

(rt)

((».. Vth

The Thevenin

resistance

!'7,(

is

= —V

is

Rrn Notice

3000 v „„„„ 9000

=

7) w

one-third of

v.

=

R,

II

=

R..

Because of

6000 3000

=

II

2

kH

ramp

this, Vth is a

of voltage

whose values

are

one-third the values of the source ramp.

Figure S-Sb shows the Thevenized kil.

At the instant

At

=

f

=

1,

I'm

=

2

+

Rth f

2,

Vth

=

The equivalent

circuit.

=

=

4 t;:^:::'

=

=

3,

V;n

=

and

6

6

=

'

=1.5

4000 Figure

ramp

mA

niA

1

4000

f

0.5

4000

R,

4 and

'

At

series resistance equals 4

and

shows

8-8(-

mA

the current waveform. So a

ramp

of source voltage results in a

of load current.

EXAMPLE Draw

the

8-4.

waveform

through the 10-kfl resistor of

of current

Fig. 8-9a.

SOLUTION. With the lO-kO load disconnected, no charges flow through the 3-kfl resistor. Because of this, the voltage between the AB terminals is the same as the voltage between the CD terminals. With the voltage-divider theorem,

""'

The Thevenin

8-9(;

shows

f

=

1, Vtii

=

4000 8000

=

R:,

5

+

R,

II

=

Ro

3000

-I-

_v "'2

+

R,

=

5000

+

4000 4000

=

II

the Thevenized circuit. Rtii

At

_

"'

resistance equals

Rth Figure

__R^ ~ R

10,000

=

15 kJl

and

=

"^ Rth+Ri.

5 kfl

The equivalent

= _L^ = 0.333 mA 15,000

resistance equals

270

RESISTIVE CIRCUITS

4kn

Figure 8-9. Example of

3kn

c

A

instantaneous analysis.

(c)

Because of Ohm's law,

neous source voltage;

all

other instantaneous currents are proportional to the instanta-

this is

why

the current

waveform

is

the triangular

wave

of Fig.

8-9c.

Negative values of current represent a reversal of direction. Betw^een t

=

4,

the true direction of

EXAMPLE

i

is

opposite that

shown

(

=

2

and

in Fig. S-9a.

8-5.

The source waveform

of Fig. 8-lOfl

is

called a square wave.

What

is

the

waveform

of

voltage across the 20-kft resistor?

SOLUTION. Between t = and t = I, the current source produces conventional flow shovkTi by the arrow. The voltage at each instant therefore equals v

with the polarity

shovk^n.

= =

Ri

20

= 20,000(0.001) V

in the direction

272

4.

RESISTIVE CIRCUITS

Which (a)

5.

6.

7.

(b)

formulas and theorems apply

(f)

circuit

Which

must be

theorem?

resistive

each instant

at

works

(d)

for

any

((')

current

circuit

(

)

(

)

(

)

(

)

of these belongs least?

one direction but varies

(«)

flow

(i)

direct current

in

is

(rt)

similar to constant current

(f)

a sine

A ramp

is

wave

{d)

constant

is

current does not change in value

(d)

Alternating current has a waveform

(a)

8-3.

of these is not related to the time

each instant treated as separate problem

whose shape

(b)

is

non-time-varying

has only positive values

an example of which kind of current?

time-varying

direct

(b)

(c)

constant

alternating

(d)

THE SIMILARITY THEOREM

What In earlier examples, a

ramp

is

the simUarity theorem?

of source voltage

produced

a

ramp

of load current, a

tri-

angular wave of source voltage resulted in a triangular wave of load current, and so on.

We

can summarize these

voltage waveforms have the resistive, linear,

results'

with the simihmty theorem.

same shape

and has only

as the source

It

says

all

current and

waveform, provided the

circuit is

o»t' source.

Ohm's

law, current and voltage are proportional in a waveform and the voltage waveform have the same shape. If the current through a resistor is a ramp, the voltage must be a ramp; if the current is a sine wave, the voltage must be a sine wave, and so on. The only way to satisfy Kirchhoff's current and voltage laws everywhere in the circuit is for all currents and voltages to have the same shape as the source waveform. As an example. Fig. 8-11 n shows a triangular wave of source voltage. The similarity

Here's the proof. Because of

linear resistance; therefore, the current

Figure 8-11. Similarity f/ifurfm.

theorem says the load current

is

also a triangular

positive peak value (the highest point

R Figure

8-11/'

shows

this current.

negative peak of source voltage at

f

=

wave.

When

on the source waveform), 5

2000

Because

(maximum

all

the source reaches

t

=

1,

v

=

10,

its

and

mA

other current values are proportional, the

negative value) produces

—5 mA,

as

shown

3.

EXAMPLE

8-6.

Figure 8-12(z shows a source waveform called a sawtooth.

waveforms

Draw

all

current and voltage

in the circuit.

Figure 8-12. Example with

sawtooth.

9 200

100

300

Microseconds

100

200

Microseconds

300

274

RESISTIVE CIRCUITS

Figure 8-13. Sawtooth driving ladder.

^^^ Milliseconds

SOLUTION. Because of the similarity theorem, the sawtooth of source voltage produces a sawtooth of load current. At

=

t

100

'

/us,

"

u

=

V and

50

50 10 000

^^^^

'

P^^"^ current)

Figure 8-12h shows the current waveform.

The voltage waveforms

v,

and

u,

=

R,i

=

8000(0.005)

=40 V

v..

=

R,/

=

2000(0.005)

=

I'j

are sawtooths with peak values of

and

Figure 8-12f and d

A

sawtooth

show a

is

10

V

these voltage waveforms.

common waveform

TV

in

and other

receivers, oscilloscopes,

visual-display instruments.

EXAMPLE

8-7.

Describe the current and voltage waveforms for the ladder of Fig. 8-13.

SOLUTION. The

similarity

theorem

tells

us a sawtooth source produces a sawtooth current through

each resistance, and a sawtooth voltage tooths occur at the

ms,

t

=

3 ms,

same

at

each node. The peak values of these saw-

instants as the source peak values, that

is,

at

f

=

1

ms,

t

=

2

and so on.

The source has

a positive

stant the source equals 9 V,

we

peak of 9 V.

we

If

get I,

=

1.17

i2

=

0.5

I3

=

0.25

=

0.667

I.

A

A A A

apply the ladder method

at the in-

=

I.,

(These values were worked out in Sec.

A

0.25

6-6.)

Test 8-3

You cannot apply

1.

the similarity theorem

resistances are linear

if

used one resistance is nonlinear (d) a tc source is involved A sine-wave source drives a circuit with 100 linear resistances. The current resistance must have the shape of a (n)

(/')

a current source is

(f) 2.

(a)

dc signal

When

3.

8-4.

(b)

(ii)

one instant

(d)

many

wave

(r)

two instants

(b)

square wave

we have

to

at least three

(c)

ramp

(d)

analyze the circuit

THE SINE

many

)

(

)

(

)

at

or four instants

instants

The sine wave output of

sine

the similarity theorem applies,

(

in each

WAVE is

the most fundamental of

circuits.

all

waveforms, because

the natural

is

it

Furthermore, transduced physical quantities often result in

waves of current or voltage. As a result, the sine wave is the most important waveform studied in this book. The discussion that follows captures the basic ideas behind the sine wave. In parsine

ticular,

look for the answers to

What

How How

IS

the period of a sine

is

frequency defined?

is

frequency related

to

wave? period?

Period

Look

at

the sine

wave

of Fig. 8-14rt.

this is the starting point of the sine

peak point;

it

Here

wave. At

what you should

is /

=

1

ms, v

—

has the largest voltage in the positive direction. At

the crossover point

where the voltage reverses

its

polarity.

At

see.

10 V; this

t

=

t

=2

At is

t

=

0,

v

=

Q;

the positive

= 0; this is = —10 V; this

ms, v

3 ms, v

is maximum and of opposite polarity. At wave starts over and repeats all voltage values. A cycle is the smallest part of a waveform that establishes the size and shape of the waveform. In Fig. 8-14«, the waveform between t = and = 4 ms is a cycle, the part between = 4 ms and f = 8 ms is another cycle, the section from = 8 ms to f = 12 ms

is /

the negative peak point; the voltage here

=

4 ms, V

=

0; at this

point the sine

f

(

f

276

RESISTIVE CIRCUITS

Figure 8-14. Sine wave.

1

Figure 8-15. Peak value and period.

(6)

(0

Vf,

and T are the most important values of a sine wave; once you know what they

you have the particular sine wave pinned down. For instance. Fig. 8-15fl shows a sine wave with a V^ of 10 V and a T of 2 s; automatically, the negative peak voltage is — 10 V. Also, the positive peak occurs at t = 0.5 s, the crossover point at f = 1 s, the are,

negative peak

at

f

=

1.5 s,

As another example.

and so on. shows

Automatically, the negative peak voltage at

f

=

2

s,

a sine

Fig. 8-15b

the crossover point at

f

=

is

4

—5

s,

wave with

a

V;,

of 5

V and

V. Furthermore, the positive

the negative peak at

f

=

6

s,

a

T of 8

s.

peak occurs

and so on.

Frequency

The frequency

of a sine

wave

is

defined as the

number

of cycles divided

time in which they occur. The defining formula for frequency

(8-3)

f=l where

/

n (

If

= frequency = number of = time

by the

is

cycles

12 cycles (hereafter abbreviated c) occur in 4

s,

the frequency of the sine

wave

is

12 c

f=rAs another example, has a frequency of

Fig. 8-15f

shows

3 c/s

4

s

5 c occurring in 0.25

s;

this

means the sine wave

278

RESISTIVE CIRCUITS

These two examples indicate the nature tition,

because

it

tells

us

how many

of frequency;

cycles occur in

1

s.

it's

the

amples, 3 c occur during each second; in the second example, 20 second. Therefore, the second sine

wave has

same

as rate of repe-

In the first of the preceding ex-

each

c take place in

the higher repetition rate.

Hertz

The

hertz (abbreviated

Hz)

is

the metric unit of measure for frequency;

it's

de-

fined as 1

hertz

=

cycle per second

1

or 1

With

this definition, the

Hz. Since the hertz

is

answers

Hz =

in the

c/s

1

preceding examples are written as 3

Hz and

20

the standard unit for frequency, answers are specified in hertz

rather than cycles per second.

Frequency meters and electronic counters are instruments that measure frequency.

Put a sine wave into either, and

it

indicates the frequency in hertz.

Relation of frequency to period

Frequency

where

n is the

occurs in time

defined as

is

number T

of cycles in a time

(Fig. 8-16(j).

Since «

=

1

(.

From

and

/=i

Figure 8-16. Frequency and period.

t

=

the discussion of period,

we know

1

c

T,

(8-4)

Equation

(8-4) is

wave with an

EXAMPLE

widely used. For one thing, you can measure the period of

oscilloscope; then

you can

a sine

calculate the frequency using Eq. (8-4).

8-8.

Calculate the frequencies of the sine

waves shown

in Figs. 8-l6b

and

c.

SOLUTION. The sine wave

of Fig. 8-l6b

has a period of

/

The sine wave

f

An

therefore,

its

frequency

is

= f=o^ = 5"^

=-= T

—— =

ms, so

250

its

frequency

is

Hz

0.004

8-9.

wave of What does

oscilloscope displays the sine

resistor.

s;

of Fig. 8-16t" has a period of 4

'

EXAMPLE

0.2

What

is

the frequency?

Fig. 8-1 7rt

when connected

across a 1-kft

the current through this resistor look like?

SOLUTION. Figure 8-l7n shows a period of 10

/xs;

therefore, 1

/

The current through the The peak current equals

EXAMPLE

8-10.

the sine

wave

If

=

10(10-«)

resistor is a sine

of Fig. 8-17b

is

100

kHz

wave with

the

same frequency

across an 8-kfl resistance,

what

is

as the voltage.

the frequency and

peak value of the current through the resistor?

Figure 8-17. Calculating frequency and peak value.

280

RESISTIVE CIRCUITS

SOLUTION. The peak current

is

V„

and the frequency

0.02

,

is

T

EXAMPLE

^ ^

= ^=8000=2-5'^^

^"

0.005

8-11.

The voltage out of a typical ac outlet in a home is a sine wave with mately 165 V and a frequency of 60 Hz. Calculate the period.

a peaic of

approxi-

SOLUTION.

We

can rearrange Eq.

(8-4) to get

Substituting the given frequency,

T=

EXAMPLE What

^= 60

16.7

ms

8-12.

are the current

and voltage waveforms

of Fig. 8-18.

SOLUTION.

We first

analyzed this ladder for

a

dc source of 9

ladder for a sawtooth source with a peak of 9

analyze

it

for a

sine-wave source with

a

V (Sec. 6-6). Recently, we analyzed this V (Example 8-7). Now we are going to

peak of 9 V.

The first step is realizing the similarity theorem applies. Because of this, every current and voltage waveform throughout the ladder has the shape of a sine wave. The second step is realizing the peak values of the currents and voltages are identical to those worked out earlier. Therefore, the current and voltage sine waves in Fig. 8-18 have these peak values: ;,

=

h = /•,

=

74 =

1.17

0.25

A

0.667

=

0.25

V4

=

2

V

V«

=

l

V

/,

A

A

0.5

A

A

Figure 8-18. Sine

wave

driving ladder.

letters are used with time-varying currents and voltages, not with fixed valThe peak values of a sine wave do not change with time; this is why we use capital /'s and V's in the foregoing list.) The final step is to work out the frequency. Because of the similarity theorem, all sine waves have the same frequency as the source. Therefore, they all have a frequency of

(Lowercase ues.

1

/4T

=

5

kHz

wave,

it is

0.2(10-=)

Test 8-4

To

establish the size

(a)

one cycle

Which

((')

and shape a

of a sine

couple of cycles

many

(c)

sufficient to look at

cycles

(

of these belongs least?

duration of a cycle

{a)

Vio of a cycle

(rf)

time from beginning

(b)

end

to

(r)

period

of cycle

(

Frequency always equals

number number

(n) (d)

Charge

is to

time

(ij)

Which (a)

of cycles

current as

hertz

{b)

number

period

(c)

(c)

period

by corresponding time frequency

(d)

amperes

cycle

(i))

mho

frequency

is to

(b)

(r)

1/T

((/)

resistance as frequency

period

(c)

second

is

8-5.

The

(

repetition rate

true

((>)

(

to

{d) hertz

The following words can be rearranged to form FREQUENCY IS PERIOD OF THE. The sentence is (a)

(

of cycles is to

of these belongs least?

Conductance (a)

(/')

of cycles divided

(

a

sentence:

false

RECIPROCAL (

PREVIEW OF PART 2 first

half of this

electronics.

book has emphasized

resistive circuits,

by

far the

most important

in

But the two other properties of loads can be very important in some

282

RESISTIVE CIRCUITS

applications. Therefore, our theory will not

and inductance

in Part 2.

Here

is

be complete

a glimpse of

what

is

until

we

discuss capacitance

coming.

Capacitance

The second property (abbreviated

F).

When

a

of a load is its capnicitance,

measured

in a unit called the farad

a circuit

with capacitances, each

sine-wave source drives

capacitance opposes the flow of charge. The opposition to ac flow reactance.

The

larger this capacitive reactance, the smaller the sine

As derived

is

called capacitive

wave

of the current.

in Part 2, capacitive reactance equals 1

Xr

(8-5)

27r/C

= /=

where Xr

C=

capacitive reactance,

ohms

frequency of sine wave, hertz capacitance, farads

Therefore, given the frequency

f

and the capacitance

C,

we

can calculate the capacitive

The greater X, the smaller the current. Figure 8-19fl shows the schematic symbol for capacitance. On a schematic diagram, the value of capacitance is given in farads. Figure 8-1% shows a capacitance of reactance X,

,

.

C= which

is

= If

a sine

wave with

a

frequency of 60

Xr

=

Hz

Figure 8-19. Capacitance and

5(10-") F drives this capacitor, Eq. (8-5) gives

1

1

lirfC

277(60)5(10-")

= 530n

inductance.

5 fiF

equivalent to

Inductance

The

third property of

(abbreviated H).

When

a

all

loads

imiuctaiicc,

is

sine-wave source drives

measured

ductance opposes the flow of charge. This opposition reactance.

The

to

in a unit called the henry

with inductances, each

a circuit

flow

is

known

in-

as inductive

larger the inductive reactance, the smaller the current.

Part 2 derives this formula for inductive reactance:

X,

=

2nfL

(8-6)

where X, = inductive reactance, ohms / = frequency of sine wave, hertz L = inductance, henrys

Given the values

of

f

and

L,

we

can calculate X,; this inductive reactance

is

the opposi-

tion to ac flow.

Figure 8-19r shows the schematic symbol of inductance.

given in henrys. The inductance of Fig.

wave with

a

frequency of 60

Hz

8-19rf, for instance,

On

a diagram, the value

has a value of 20 H.

If

is

a sine

drives this inductance, Eq. (8-6) gives X,.

= =

27r/L

=

277(60)20

7.54 kfl

SUMMARY OF FORMULAS DEFINING

f=^

(8-3)

/=i

(8-4)

DERIVED

Problems 8-1.

For the ramp of Fig. t

=

2 s?

8-20rt,

what does the voltage equal when

f

=

1

s?

When

284

RESISTIVE CIRCUITS

Figure 8-20.

3-2.

In Fig. 8-20!i,

3-3.

The equation

what does equal when = 1 ms? When = 10 ms? for any straight line passing through the origin is t

i'

(

For a voltage ramp, this becomes i;

8-4.

a.

What

b.

For the ramp of

t

=

2 ms;

t

=

Switch

9.5

8-8.

8-9.

8-10.

of Fig. 8-20;!?

^

t

=

4 ms;

t

=

6 ms;

A

and

is

t'= 8

the voltage for each of these in-

ms?

each of these instants;

f

=

9

s; f

=

1 s;
. Also,

6kn

A

figure 8-21.

286

RESISnVE CIRCUITS

Figure S-22.

lo

-7V

_6 -1

mV

— 1

r

10V.=-

2

•-

-•

VV^v

-•

Vv'V^

:6

-A/W^

—-

-AA/V-

8-16.

A

dc source and a

position theorem,

tc

source are in series in Fig. 8-22c. By applying the super-

we

can reduce this two-source circuit to two separate one-

source circuits; then, the similarity theorem applies.

Knowing

this,

you can

answer these questions:

When

a.

the

source

tc

is

reduced

to zero,

what does the waveform

of current

what does the waveform

of current

through the lO-kH resistance look like?

When

b.

the dc source

is

reduced

to zero,

look like? 8-17.

In

Fig.

S-22d,

a

8-18.

suppose the

source

tc

is

a

and voltage in the sine wave, what does every waveform look

waveform

of every current

Because of the time theorem, p

=

vi

If

the

tc

the

source

is

like?

applies at each instant in a resistive circuit.

In Fig. 8-22a, calculate the total load

= 1 ms; and f = 3 ms. A sawtooth drives the series

What does

triangular wave. circuit look like?

power

at

each of these instants:

t

=

0;

(

8-19.

like a

pure resistance of 100

meter look

like?

circuit of Fig. 8-23fl.

If

the moving-coil meter looks

what does the waveform of current through the the inertia of moving parts, the needle of the

ft,

Because of

moving-coil meter cannot follow the current waveform through needle points to the average current through the meter.

What

it.

Instead the

value of current

does the needle indicate? 8-20.

In

Fig.

switch

8-23b,

A

automatically closes

and opens according

to

this

schedule: a.

It

closes at

f

b.

It

opens

at

f

c.

It

closes at

f

d.

It

opens

f

Draw

the

at

= 0. = 2 s. = 4 s. = 6 s.

waveform

of current

i

betw^een

f

=

and

t

=

8

s.

Figure 8-23.

»* KSBTWE

cmam

FtgMTt S-24.

f\^ e

2ldl

O' I

8-21.

Given the sine wave of Rg. S-24a, what is the voltage t = 0: i = 5 ms; f = 10 ms; f = 15 ms; and f = 20 ms?

8-22.

WTiat

8-23.

Seventy-five cydes of a sine

8-24.

8-25.

8-26.

at

each of these instants:

wave in Fig. 8-24j? What does T 2 equal? wave occur in 3 s. What is the frequency of the

the period of the sine

is

7/4? sine

the period of the sine t^ave?

Channel 4 of a TV receiva- operates this frequency,

8-28.

)30

wave? The period? If an oscilloscope shows 3 c of a sine wave occurring in 15 ixs, what is the frequency of the sine wave? The period? In Fig. 6-24b, what is the period of the sine vrave? The frequoicy? In Europe, the frequency of the sine-wave vohage out of an ac outlet is 50 Hz. Wliat

8-27.

is

•

what

In Rg. 8-24er and its meta] coating eliminate air gaps, improve up sheets of metallized pap>er, we get a paper capacitor with capacitors use metallized paper, miide

rFig- 9-lOc).

etc By rolling

and higher capacitance than the paper-foil type. a band or stripe on the end connected to the outside foil (Fig- 9-lOji). When a schematic diagram shows one side of a paper capacitor grounded, most people ground the banded end. This is not necessary, but it has this advantage: it grounds the outside foiL The effect is to shield the irmer foil from stray electric fields, unwanted signals, noise, etc As a guide, paper capacitors have values from less than 0.001 /i.F to more than 1 fif.

greater reliability

Paper capacitors have

Plostie-film copocitors Plastics are excellent dielectrics.

and

polystyrene.

The commonly used

These materials have high

very thin sheets like

plastic dielectrics are polyester

and can be produced

dielectric strength

Polyester-fihn cjipacitors use separate sheets of metal foil 9-llfl,

or they use metallized polyester film like Fig. 9-1

flexible film

in

film.'

lb. In

and polyester

like Fig.

either case, sheets of the

can be roUed into a cylindrical structure to get large capacitance v2Jues

(simUar to Fig. 9-lOb). Polyester film has a higher dielectric strength than paper. For the

same voltage

rating,

dunner sheets can be used

to get higher capacitance values than

with paper.

The main disadvantage

of pwlyester-film capacitors

large changes in capacitance occur less

when

temperature sensitive than polyester. This

preferred to polyester-film capacitors '

is their

temperature sensitivity;

the temperature changes. Polystyrene

when

is

why

is

much

polystyrene-film capacitors are

large temperature variations are expected.

Manufactmed under tradenames such as Mylar, Kodar, etc

CATACIIANCI

J07

Figurr 9-11. Pla$Ucfilm rapaeilor.

Polynlf r film

^' Pij|y«ntor film

M«Ul

foil

Polycstcr-fjim capacitors

from

h.ivi- v.ilui-s

polyslyriTif-dlm capacitors from less than

U-ss Ih.in

(H)I /il-

more than

pi- to

"i

to

more than

10 /iF;

0.?t fif-.

Ceramic capacitors Clay

IS

Iranslormfd by extreme heat. At temperatures near

bined water resulting

is

lOOO'^l-,

chemically com-

driven o(t and the clay undergoes an irreversible chemical change. The

product

is

called

a

i

Examples

crumic.

of

ceramics are porcelain, china,

brick, etc

The

mam

advantage

ceramic materials

of

is

their

tremendous range

in dielectric

properties. For instance, porcelain has a dielectric constant of 6 (approximate); barium-

strontium-titanate (another ceramic) has a dielectric constant of about 7500 By mixing

ceramic materials,

we

Using ceramic

can get a wide range in dielectric constant and other properties

dielectrics,

manufacturers can produce capacitors with positive,

negative, or zero lemfifriilurf loclfnifnls this

means

it

degree This

has is

a positive

you see

a

ceramic capacitor with I'lOO on ;'ijrjs

prr niillwn

if,

(ppm) per

equivalent to 100

.

1

A

per degree rise (Celsius). coefficient of 200

ppm

rise.

„^»^

=

10

*

=

,000,000

0.01 percent *

ceramic capacitor with N200 means a negative temperature

per degree, equivalent to

.

per degree

If

temperature coefficient of KM)

^°^ = -2(10

And with

the right

mix

«

)

= -0.02

percent

of materials, a ceramic capacitor can

have a

zero temperature coefficient, indicated by NPO; in this case, the capacitance doesn't

change with temperature. Ceramic capacitors have values from

less

than 10 pF to more than

1

ixV

Electrolytic capacitors Electrolytic capacitors

dielectric is

have more capacitance than any other type, because the of the plates. Mere's how an

an extremely thin layer of oxide grown on one

308

REACTIVE CIRCUITS

Figure 9-12. Electrolytic capacitor.

.Upper

^

'Aluminum

plate

oxide

l^^^^^^

-Electrolyte

-*

—Electrolyte

^

^. (a)

electrolytic capacitor is formed. Initially,

two aluminum plates and an

conducting liquid or paste) are in a container act like a capacitor

been applied

At

(Fig. 9-12n).

because the electrolyte shorts the two

for a while, electrochemical action

the positive plate (see Fig. 9-12b).

The process

produces

of

plates.

does not

But after voltage has

a layer of

growing

electrolyte (a

this device

first,

aluminum oxide on

this layer is a

slow one, and

referred to as forming the capacitor.

is

After the capacitor has been formed, the oxide layer acts like a dielectric between the upper

aluminum

electrolyte. The lower aluminum plate merely makes which functions as the negative plate. capacitor of Fig. 9-12b works fine as long as the upper plate is posi-

plate

and the

contact with the electrolyte,

The

electrolytic

and the lower one negative. But if you try to reverse the applied voltage, the oxide layer no longer acts like an insulator; instead you get a large current and a loss of normal capacitor action. For this reason, most electrolytic capacitors are polarized: they work properly only with one polarity of voltage. This is the reason a manufacturer puts tive

a plus sign (or other

When may

an

marking) on the positive end of an

deteriorate. In this case, the capacitor has to

reform the capacitor after long periods of storage

development

alumimum

minum The

it

layer

can be used in a

is

a nuisance,

and has led

to the

of tantalum electrolytic capacitors.

Tantalum capacitors can be stored than

aluminum oxide

be reformed before

This requires slow charging of the capacitor to regrow the oxide layer. Having

circuit.

to

electrolytic capacitor.

electrolytic capacitor is stored for long periods, the

indefinitely.

They

also

have more capacitance

ones. For this reason, tantalum electrolytic capacitors are replacing alu-

electrolytic capacitors in

many applications. why electrolytic

thin oxide layer is the key to

capacitors have

more capacitance

than other types. At the same time, however, the thin oxide layer has a disadvantage: its

resistance

like

an open,

is it

relatively

low compared with other

passes a small current (see Fig.

dielectrics. Instead of acting ideally

9-13fl).

We

call this

unwanted current

the leakage current.

To account capacitance, as

for leakage current, visualize a leakage resistance in parallel with the

shown

capacitor with 20

3-Mn

V

in Fig. 9-13b.

across

it;

The values given

are representative of a 250-/iF

in this case, the leakage current is equivalent to

having

a

resistance across the capacitance. In general, the larger the capacitance, the

CAPACITANCE

309

Figure 9-13. Leakage current.

1111 '

'

'

'

Leakage

250 mF

current

^3Mi2

smaller the leakage resistance; this leakage resistance can prevent

some

circuits

from

working properly. Electrolytic capacitors

have values from

less

than

1

fiF to

more than

10,000

fiF.

Summary

We have covered most of the types of modem capacitors. As a summary, Table 9-2 shows the approximate range, main advantage, and main disadvantage of each type. TABLE TYPE

9-2.

CAPACITOR TYPES

310

REACTIVE CIRCUITS

50

^° PP"" per degree

The

"

total

^°'^°

"

°-°°^

=

100°C

"

P^'""'

'

rise.

total rise in

temperature

is

T,-T.= The

"

1,000,000

percent change 50

125°C

0.005 percent x 100

Therefore, the total capacitance change

125°C

=

0.5 percent

is

0.5 percent at

25°C

is

ppm X {Ti-T.) =

and the capacitance

-

x 100 pF

=

0.5

pF

is

C=

100 pF

+

0.5

pF

=

100.5

pF

EQUIVALENT CAPACITANCE

9-6.

Capacitances in series or parallel act like a single equivalent capacitance. This section

answers Hoiv do parallel capacitances combine?

How

do series capacitances combine?

Parallel capacitances

Assume

the capacitors of Fig. 9-lia are initially uncharged. After the switch closes,

a total charge C-,.

Charges

Q

flows into the parallel connection of Fig. 9-14b. Q, goes to C,, and Qo to node B has less voltage than node A. For this reason,

will flow as long as

both capacitors charge until their voltages equal the source voltage.

Figure 9-14. Parallel capacitances.

A

1 {b)

CAPACITANCE

After the capacitors are fully charged, the capacitor law

and

The

charge equals the

total

With Eqs.

(9-7rt)

and

sum

us

Q,

=

C,V

(9-7fl)

Q-,

=

C,V

(9-7b)

of charges, so

becomes

(9-7/;), this

Q=

C,V + C.V

^ = C, + C

or

The

Q/V

ratio

is

the ratio of total charge to total voltage; therefore,

equivalent capacitance. Because of

this, the

C = C,+ This the

tells

311

sum

final result says the

of capacitances.

If

Q/V

is

the total or

preceding equation becomes

C,

(parallel)

(9-8)

equivalent capacitance of two parallel capacitances equals

pF and €,

C, is 500

is

700 pF, the equivalent capacitance

is

C = 500 pF + 700 pF = 1200 pF

Or

if

=

C,

10 /uF

and

Q = 30

/ixF,

C= Equation

10

mF +

=

30 /^F

40

mF

makes sense. When you connect two capacitors in parallel, you are two upper plates and the two lower plates. The effect is the same as

(9-8)

connecting the

increasing plate area. Since capacitance

is

directly proportional to plate area, the equiv-

alent capacitance increases.

By

a similar proof, a circuit

with u parallel capacitances has an equivalent capaci-

tance of

C=

C, -fC,

-I-

•

•

+

(9-9)

C„

This says you add parallel capacitances to get the equivalent capacitance.

Series capacitances

Figure 9-15 shows capacitors in series. Kirchhoff's current law says current has the

same value and equals

at all

points in a series circuit; therefore, each capacitor charge

is

the

same

the total input charge. In symbols,

Q=

Q,

=

Q,

(9-lOfl)

312

REACTIVE CIRCUITS

Q

Figure 9-25. Series capacitances.

Applying Kirchhoff

's

voltage law to Fig. 9-15 gives

V= Dividing

this

by

total

+

V,

Q~ Q^ Because of Eq.

(9-lOb)

V.,

charge Q,

Q

(9-lOfl)/

With the capacitor Jaw,

this

becomes

c

c

c,

Finally, algebra gives

^=c^ So,

we

we wind up

with the product-over-sum

*^-")

rule.

Given two capacitances

can lump them into a single equivalent capacitance equal

sum.

If

10

pF

is in parallel

to the

with 30 pF, the equivalent capacitance

pF X 30 pF 10 pF -h 30 pF

in series,

product over the

is

10

By rules:

a similar derivation,

^

more than two capacitances combine by

111

1

either of these

two

/« 1^

N

^

1/C,

+

(9-12b) I/C2

-^

•

•

•

-t-

1/C„

Either of these equations implies the equivalent capacitance of series capacitors

always smaller than the smallest capacitance.

is

i

EXAMPLE

313

the final charge on each capacitor?

The

9-9.

After the switch of Fig. 9-16u closes, total

CAPACITANCE

what

is

charge?

SOLUTION. The

capacitor law says the

first

Q,

capacitor has a charge of

= C,V=3(10

The

second capacitor has a charge of

The

total

Q.,

charge in

Q= EXAMPLE

=

C,V

a parallel circuit

Qi

=

'^)10

6(10-« )10

= 30mC

= 60

always equals the

+ Qa

=

30

mC +

60

fiC

sum

mC =

90

of the charges. So, fjiC

9-10.

After the switch closes in Fig. 9-16b, to the capacitors?

How much

what

voltage

is

is

the total charge transferred from the source

across each capacitor?

Figure 9-16. Examples of calculating equivalent capacitance.

o

—

-t(

\Q7

3mF=|=I/, 10

T

3mF

v-=-

6(iF 6

6MF=bV^2

(a)

1

+

4

V -

»

i

-1

o

C 4

1

314

REACTIVE CIRCUITS

SOLUTION. The equivalent capacitance

of the series capacitors is

C.Q _ + C,

_

C,

And

the total charge

As we know,

3

mF X mF +

6 6

mF mF

'^

is

Q = CV = to

3

Q=

Qi

=

mF X

2

10

V=

20

mC

in a series circuit. Therefore,

Q^

applying the capacitor law

each capacitor gives V,

Q, = 20(10-'^) = 6.67 V =-^ 3(10-

C,

and

02

20(10-")

C.

6(10-'*)

v..

These

up with more

3.33

V

When you

results bring out an important idea.

the smaller capacitance ends

=

charge two series capacitors,

voltage. In fact,

capacitive voltage divider of Fig. 9-16c has this formula for

c,

EXAMPLE

is

show

easy to

the

+

a

9-11.

Suppose the

3-fj.¥

capacitor of Fig. 9-17 has an initial voltage of 10 V.

open, this capacitor retains

its

charge indefinitely. But

capacitor will discharge into the other capacitor. parallel

it

output voltage:

C,

=

v.,

its

connection after the switch

is

What

when is

With the switch

the switch

is

closed, the

the final voltage across the

closed?

SOLUTION. First,

get the charge stored in the 3-/xF capacitor before the switch

is

closed.

With the

capacitor law,

Q = CV= This means -30 mC.

the upper plate has a charge of

After the switch

wise, the

Figure 9-17.

One

3(10-" )10

—30

is

closed, the -1-30

fxC spread over the

nC

+30

=

30 ixC

fi.C,

and the lower

are distributed onto both

two lower

plates.

Because of

y^

capacitor

discharging into another.

10V=^

3>iF

=±=6;iF

plate a charge of

upper

plates; like-

this, the total

charge on

CAPACITANCE

both capacitors

is still

30 fxC. This allows us to write

+ Q, =

Q,

Applying the capacitor law

V

is

the

same

mC

30

each capacitor, this becomes

to

C,V + (Voltage

31

QV = 30/[xC

across both capacitors, because they are

of equipotential points.) Solving the preceding equation for

„_30mC_

30

C,+C.

3

mC

mF +

6

between the same pair

V gives

_3_33^ mF

Test 9-5 1.

2.

3.

4.

5.

The

total

charge flowing into two parallel capacitors equals the

(ij)

sum

(r)

charge into either

of the charges into each capacitor (d) ratio of the first

(b)

difference of the charges

charge to the second charge

Equivalent capacitance of parallel capacitors has no relation (a)

kind of source used

(c)

ratio of total

Which

charge

sum

(b)

(

)

(

)

(

)

(

)

(

)

of capacitances

to parallel voltage

( = = -7^

1000

Wb

of Fig. 10-4^

is

different;

4 A. Therefore, the flux linkage

——^AWb — = „„„,.„Wb/A 1000

,,

200

5

/

it

has 250 turns,

a flux of 0.2

Wb, and

a current of

is

N

and the inductance

Wb =

100 X 10

is

L

The inductor

A

(see Fig. 10-4(7), the flux linkage equals

=

250 X 0.2

Wb =

50

Wb

is

I.

=

N-,

almost equal

to

4>t

INDUCTANCE OF TWO INDUCTORS

The inductance

of a single inductor is defined as N(j>/I, the ratio of its self-produced

flux linkage to its current.

When two

inductors are in series or in parallel, things get

complicated because the flux of one inductor calculate the equivalent inductance

answers

may

pass through the other inductor. To

under these conditions, you have

to

What

How How

is

mutual inductance?

do series inductances combine? do parallel inductances combine?

to

learn the

RCACTIVE CIRCUITS

338

Figure 20-11. Mutual inductance.

Mutual inductance Given two inductors age

in the

mutual inductance

like Fig. 10-11,

second inductor divided by the current in the

is

defined as the flux link-

first

inductor.

As

a defining

formula,

M=

^

(10-7)

'i

where

M = mutual Na 02 /,

If N.2

=

= = =

100,

flux

coupled into the second inductor

current in the (t>2

= 0.2 Wb,

In Eq. (10-7), of coupling

inductance

turns on the second inductor

k.

M

is

first

and

/,

inductor

=5

A, then by definition the mutual inductance equals

directly proportional to

M=k So

if

(j).,.

In turn,

({>,

is

related to the coefficient

By an advanced derivation,

two inductors have

a k of 0.95,

and

VLJ^

L, is

100

(10-8)

mH

and

L, is

400

mH,

the mutual in-

ductance equals

M = 0.95

VO.l X 0.4

=

190

mH

Series inductors

When two

inductors are in series

(Fig. 10-12rt),

inductor passes through the other. Because of

some

of the flux produced

this, the equivalent inductance

by each depends

on each separate inductance and on the mutual inductance. The derivation of equivalent inductance is too complicated to go into here, but it can be shown that equivalent inductance L

is

given by l

=

L,

+

U±2M

(10-9)

INDUCTANCE

339

Figure 10-12. Inductors in series.

U,

The

final

term

may be

+Z.2 ± 2/W

plus or minus, depending on whether the fluxes in the inductors

aid or oppose.

Figure 10-12b shows the case of aiding fluxes. Apply the right-hand rule the flux through each inductor

N

ductor

is

shown

in Fig.

a

pole.

On

upward.

In other

schematic diagram,

a

lO-Uc. So

is

if L,

=

100

mH,

L,

=

we 400

to see that

words, the upper end of each

indicate like poles

mH, and

M=

190

by using

mH

in-

dots, as

in Fig, 10-12c-,

Eq. (10-9) gives L

= L, + L, + 2M = 100 mH + 400 mH

-H

2(190

mH) = 880

mH

The right-hand rule shows point in opposite directions. Again, dots are used on a schematic diagram, as Fig. 10-12t', to indicate similar poles. With the same values given in the pre-

Figure 10-12d illustrates the other case, opposing fluxes. the fluxes

shown

in

ceding calculation, opposing inductors have an equivalent inductance of L

= =

L,

-)-

100

La

- 2M

mH

-I-

400

mH

-

2(190

mH) =

120

mH

340

REACTIVE CIRCUITS

Often, the series inductors are far apart and the coefficient of coupling

is

approxi-

mately zero. In this case, the equivalent inductance equals L

=

+

Lt

L.

(10-9(1)

when

This says you add the inductances of series inductors

between them.

If

there's negligible coupling

the two inductors of the previous calculations are

apart, k drops to zero

moved

far

enough

and the equivalent inductance becomes

= =

L

L,

L.

-I-

mH

100

+

mH = 500 mH

400

Parallel Inductors

Inductors in parallel have an equivalent inductance of L,L„ L,

The

2M

+

may be

term in the denominator

- M± 2M

(10-lOa)

L,

plus or minus, depending on whether the

fluxes aid or oppose.

When

the

two inductors

are far

enough

apart, k

drops

to zero

and the equivalent

inductance equals

(w-wb)

^"r+V If L,

=

100

mH and Lj = 400 mH, two parallel inductors with no mutual inductance have

an equivalent inductance

of

X

0.1

0.4

Test 10-7 1.

M=

a definition

(d)

a discovery

2.

Inductance

3.

When

(a)

(n) ((/)

4.

is

N.,

is

used

what

means. Figure

d4>/dt

shows two

10-14fl

the points are very close together, the special nota-

change between the points, and

to represent the flux

time change between the points. The

rate of flux

change

is

dt is

used

for the

defined as the flux change

divided by the time change. As a defining formula. Rate of flux change ^

=

^

(10-12)

dt

Here's an example. The flux change between the

first

pair of points in Fig. lO-lib

is

d4>

and the time change

=

dt

Therefore, the rate of flux change

0.02

As another example, the

flux

=

1

ms

Wb

0.001

dt

20

Wb/s

s

change between the second pair

of points in Fig.

is

d(t)

and the time change

=

rate of flux

0.001

Wb

is

dt

So the

Wb

is

d(t>

10-14b

0.02

is

change

=

I

ms

is

d\"itch of Fig. 11-14j has been in jjosition

A

discharging current through the inductor look Uke

for a long time, if

the switch

is

what does the

thrown

to posi-

tion B?

SOLUTION. Just before s^sitching, the current is 10

mA.

Just after sv^-itching the current is 10

because inductor current caivnot change instantaneously. This

waveform is at 10 mA, as sho\\Ti The discharging time constant equals

the current

-

7

So

after 0.4

EXAMPLE

first

mA,

point on

->

= 4- = ^^— = R

time constants

the

in Fig. ll-l-ic.

0.4

5000

ms. the current has decreased

Fig. ll-14i- After five

why

is

(2 ins),

to

ms

the 37 percent point as

the current

is

at the

1

shown

in

percent point

11-9.

Explain what happens

when

the switch of Fig. 11-15j

is

of>ened.

soLunos. Before the switch sho\%-n. If the

is

opened, the voltage across the inductor has the plus-minus polarity

magnetic

field

has reached

its final

value, the flux

no longer changes and

the induced voltage r is zero. Neglecting the resistance of the inductor, the current

equals 2

A

in Fig. ll-15fl.

\N"hen the s^s-itch is opened, the circuit looks like Fig. ll-15r. At the

the inductor current

still

equals 2

A because inductor current can

taneously. Because the magnetic field starts to collapse,

ad) dt is

first instant,

never change

iristan-

negative and the in-

duced voltage re\'erses {wlarit},-, as shown in Fig. ll-15tWith the switch open, how can there be current? Either of two things can happ>€n. First, the air between the switch contacts may break down if the induced voltage is enough to exceed the dielectric strength of air (Set 9-4). In this case, charges flow betvveen the op>en contacts, as

charge through

its

own

shown

in Fig. ll-15c. Second, the inductor

may

dis-

stray capacitance (see Fig. ll-15i). Either or both of the forego-

ing possibilities will occur

when you open

the switch in Fig. ll-15fl.

TIANSiBfrS

373

Figure II-J5. Interrupting inductor current, (a) Steady current, (b)

Sudden switch opening, (c) Switch breakdown (d) Discharge through stray capacitance.

5f2

—O2A

lOV^ -

1

T

^

2A

+

W

T

Ic)

EXAMPLE If

11-10.

5-H inductor

a

field

when

(rf)

is

used

the current

is

in Fig. ll-15fl,

2

how much

energy

is

stored in the magnetic

A?

SOLUTION. With Eq.

(11-5),

W = ^LP = ^ Test 1.

1 1

(5)2^

=

10

J

-3

Inductor current can't change instantaneously because (a)

voltage can't be infinite

(c)

the magnetic field can't change instantaneously

(b)

d)

(14-1)

instantaneous current

peak current phase angle

(Sometimes

/„,

is

used instead of

/,,.)

Figure

14-lrt

shows the graph

of this sinusoidal

current. is defined as the polar number whose magnitude equals the rms and whose angle equals the phase angle. As a defining formula,

Phaser current current,

1

where

I

/

4)

= phasor current = rms current

=

phase angle

=

1

l±

(14-2)

438

REACTIVE CIRCUITS

Figure 14-1. Phasor current. I

l±

(b)

Notice that phasor current

On

complex number.

is

printed as a boldface

the other hand, rms current

is

I;

these differences in later formulas. Since phasor current visualize

it

this is

done

printed as an is

a

to indicate it's a italic

/.

Watch

for

we

can

complex number,

as a vector (see Fig. 14-lb).

Here's an example. Suppose a sinusoidal current has an equation of 1

The

rms current

;

=

Therefore, the phasor current

0.707/p

this as 7.07

A

shows the phasor

Figure 14-2. Examples of plwfor current.

10 sin {0

+

30°)

at

an angle of

current.

=

0.707(10 A)

=

7.07

A

is

I

Read

=

is

=

/

30°.

/^ =

7.07

A

730°

Figure 14-2a shows the waveform, and Fig. 14-2b

PHASOR ANALYSIS

As another example, given

a sinusoidal current like Fig. 14-2c,

439

you can calculate an

rms current of

=

;

Then, the phasor current

0.707/,,

0.707(5

mA) =3.54 mA

is

1

how

Figure \4-2d shows

=

=

/

/^=3.54

to visualize this

mA

7-45°

phasor current as a vector.

PHASOR VOLTAGE

14-2.

The general formula y is

=

M

+

sin {0

4))

written as follows for a sinusoidal voltage: v

= V„ = (^ =

where v

V„ sin (0

+

(14-3)

(t>)

instantaneous voltage

peak voltage phase angle

(Sometimes

V,„

or

£„, is

Phasor voltage voltage,

=

used instead of

V,,.)

defined as the polar

is

and whose angle equals the phase

number whose magnitude equals

\ = V Z± V= V=

PHASOR ANALYSIS

4$7

4.

Which does not belong?

5.

At very low frequencies the output phase angle of a lead network approaches

6.

Which does not belong?

coupling capacitor

(a)

0"

{a)

(fl)

14-8.

A

lag

lag

large

(d)

impedance

U

=

4.

For a resistor, the power factor equals (fl)

60°, the

(b)

(fl)

0°

)

(

)

(

)

(

)

factor

3.

(f)

(

of the following belongs least?

(b)

is

mostly reactive

power 0.5

90°

factor (c)

(c)

is

0.707

1

id)

((i)

1

cos 90°

502

REACTrVE CIRCUITS

SUMMARY OF FORMULAS DEFINED

(15-2)

(15-3)

^=Z

(15-4)

Q=

(15-16)

^=Q

(15-17)

Q=

(series)

(15-18)

Q=4

(parallel)

(15-19)

DERIVED

Y«

=G

(15-5)

Y,.

= -;B,

(15-6)

Y,.

= /B,

z=

+Z2

Z,

Y=

(15-7)

Z.Z^

Y,

-I-

Y,

(15-8)

(15-10)

(15-20)

1 '

2nRC

P=VI

cos

(15-21)

(15-30)

ADVANCED AC

TOPICS

S03

Problems

15-1.

What

15-2.

Calculate the conductance of a 300-li resistor.

15-3.

Work

15-4.

the conductance of a S-kll resistor?

is

out the inductive susceptance for each of these:

= 8 kn = 100 mH and/= 1 MHz = 25 mH and f = 200 kHz

a.

X,

b.

L

c.

L

Calculate the capacitive susceptance for each of these: X,.

a.

=

2 kll

C = 10mF and/=120 Hz c. C = 500 fiF and f = 2 MHz What is the admittance for each a. R = 1 kn b.

15-5.

15-6.

=

20

b.

X,,

c.

C = 400 pFand/=5

d.

L

A

2-kn

=

250

of these:

11

mH

and

=

f

MHz

450

kHz

resistor is in parallel with a capacitive reactance of 3 kil.

What

the

is

equivalent admittance? 15-7.

A

1-kn resistor

admittance 15-8.

at

1

is

with a 200-pF capacitor. What

in parallel

The equivalent admittance

of a circuit is

Y= What 15-9.

A

is

What

An

is

An

+;0.25

has an admittance of 0.002

the parallel resistance?

inductor has an X, of 2500

The value 15-11.

0.1

the parallel circuit with this admittance.

parallel circuit

Y=

15-10.

the equivalent

is

MHz?

of

- yO.OOl

The parallel reactance? O and an R, of 125 fl. What

is

the value of

Q?

D?

inductor has an X, of 10

kn and

Q

a

of 125.

What

is

the value of R,?

The

value of D? 15-12.

What

15-13.

A A

15-14.

is

series

series

Q

the

RC RL

of a

circuit

parallel circuit at

15-15.

A

series

RL

200-mH inductor

circuit

1

has an R of 25

1

MHz,

given an R, of 10

fl

(2.

and an L of 10 mH. What

JI?

What

is

the

Q?

is

the equivalent

is

the equivalent

kHz?

circuit has

parallel circuit at 500

15-16.

at

has an X, of 2 kO. and an R of 100

an R of 100

U and

an

/,

of 25

mH. What

Hz?

R = 50 n and C = 500 pF RC circuit at 450 kHz?

in a series

RC

circuit.

What

is

the equivalent parallel

S04

REACTIVE CIRCUITS

15-17.

An

inductor has a

Q

of 100

15-18.

A series

RL

circuit

and an L of 200 nH. What

MHz? has R = 50 n and =

circuit for the inductor at

is

the equivalent parallel

1

Z.

1

mH.

If

Q equals

75,

what

is

the equiv-

alent parallel circuit? 15-19. 15-20.

15-21.

The Q of Fig. 15-21(i equals 100. What is the equivalent parallel circuit? Suppose the Q of Fig. 15-21^ equals 150. What is the equivalent parallel circuit? A parallel RL circuit has a resistance of 2 kf) and an inductance of 150 piH. Convert this circuit to

15-22.

A

parallel

RC

lent series circuit at

15-23.

A

equivalent series form at 250 kHz.

its

an R of 10 kil and a C of 1000 pF.

circuit has 1

measuring instrument indicates

resistance of 4 kfl.

If

What

is

the equiva-

MHz?

Q

is

that an inductor has an equivalent parallel

what

100,

is

the equivalent series resistance of the in-

ductor? 15-24.

At what frequency does the Wien bridge of

15-25.

A Wien

bridge has an R of 10 kfl and a

the bridge?

If

R

is

C

Fig. 15-22

of 100 pF.

increased by a factor of

10,

what

balance?

What frequency is

balances

the frequency that bal-

ances the bridge? 15-26.

A

sinusoidal voltage with a peak of 10

V

is

across a 500-n resistor.

What

is

the

average power in the resistor? 15-27.

The rms voltage across

15-28.

A

15-29.

The voltage

50-kn

rms. 15-30.

A

What

series

a

resistor has 20

the average

circuit

is

250

it.

mV rms,

and the current through

it is

0.5

A

power?

has an impedance of

Z= If

mV. What is the average power? What is the average power?

resistor is 100

rms through

across a resistor

is

RC

470-n

mA

the rms voltage across

it is

kn -

2

20

V

;3

kn

rms, what

is

the average

power

into the cir-

cuit.

15-31.

The voltage across

a series

What

is

the average

a.

=

30°

b. c.

can use the approximate equivalent circuit of

16-12b for everyday work. With this equivalent circuit, the resonant frequency

fr

=

'

and the rms voltage

at

resonance

Often you

will

know

K=

(16-11)

2nVLC

is

the value of

Q

and X, but not

for Q'-R,. Since

,

Q=

(16-12) R„

=

=

10

of

node voltage

V and

v,

resistor; therefore, v

waveform

=

of f is a negative step

0.

18-2.

the

waveform

v in Fig. 18-4fl?

SOLUTION. At

f

=

0" the current source has a value of zero, equivalent to an open. Therefore, the

circuit before the

At

/

=

switching instant

is

equivalent to Fig.

0* the current source has a value of 10

mA,

18-4f).

as

Clearly, v

shown

=

10 V.

in Fig. 18-4f. This 10

Figure 18-3.

(a)

Switching circuit. Negative step.

(b)

1

548

REACTIVE CIRCUITS

Figure 78-4. Example of generating a negative voltage step.

-Ov

i>-

I

) 10

mA

1 (rf)

Figure 18-5. Example of generating and negative steps.

positive

n n

12

3

4

r, 5

Seconds (a|

u w

SWITCHING CIRCUITS

mA

Node

flows through the l-kd resistor. t'

=

10

-

Ri

=

10

Figure 18-4i^ shows the waveform of

EXAMPLE

549

voltage v drops to zero, because

-

v,

1000(0.01

)

=

a negative step with d„

=

V and

10

y,

=

0.

18-3.

The switch

of Fig. 18-5rt closes at

What does

the

waveform

f

=

0,

opens

at

f

=

s,

1

closes at

f

=

2

s,

and so on.

of v look like?

SOLUTION. With the switch open, v

= 10 V. With the switch closed, v = 0. Therefore, A waveform like this is called a square wave.

the

waveform

of V looks like Fig. 18-5b.

EXAMPLE

18-4.

The switch

of Fig. 18-5f closes at

=

f

and opens

at

f

=

1

s.

What does

the

waveform

of

V look like?

SOLUTION.

When

the switch

produces at




is

This

is

why

a v of 10 V.

s (Fig. 18-5rf)-

When

zero.

A waveform

the switch

is

closed, voltage-divider action

waveform steps positively

the

at

f

=

and negatively

like Fig. \8-5d is called a rectangular pulse.

Test 18-1 1.

Which (n)

(b)

2.

Which

3.

Initial

(a)

(a) 4.

of the following belongs least?

V,

f

t

In an

=

f

=

0"

(f)

v^

(d) just

before switching instant

(

)

(

)

(

)

of these does not belong?

0*

(b)

voltage

=

(b)

(c)

v,,

is to

t

=

v„

R switching

lu

(d) just after

0* as preinitial voltage (c)

circuit

V,

(d)

f

=

switching instant is to

0"

with negligible stray capacitance and lead inductance, the

changing voltages and current are (a) 5.

sine

Which

waves

(h)

(fl)

step source

(d)

sinusoidal source

18-2.

sawtooth waves

of these has an infinite (b)

(c)

exponentials

(d) steps

...

(

)

(

)

Thevenin resistance?

step voltage source

(c)

step current source

CAPACITOR CURRENT AND VOLTAGE

Resistive-capacitive (RC) switching circuits use resistors

either step sources or dc sources with switches.

and capacitors; the sources are

Ohm's law

not to the capacitors. In what follows, find the answers to

applies to each resistor, but

REACTIVE CIRCUITS

SSO

VV/jflf IS

instantaneous current?

What

the capacitor formula?

is

Rate of voltage change Figure voltage

V,

shows conventional flow shown in Fig. 18-6/). Since

18-6fl

as

into a capacitor. This results in an increasing

= Cv

q

(18-1)

As discussed in the waveform of q looks the same as the waveform Sec. 10-8, dt is the special notation used for the time change between two nearby points in time. The corresponding voltage change is symbolized by dv, and the charge change of v (see Fig. 18-6c).

by

dq.

At t=

At

t,,

Eq. (18-1) gives

a slightly later instant

The

when

f

=

difference of these two equations (j2

-

(?i

i?i

=

Cv,

(fa

=

Cv-i

fj,

=

is

Cv.,

—

Cu,

= Civo-

I'l)

which may be written dq

Dividing by

dt

=C

dv

gives

(18-2) dt

Figure 18-6. Charging a capacitor. (a) Circuit, (b) Voltage, (c) Charge

waveform.

}

dt

SWITCHING CIRCUITS

The meaning way, dq/dt

The

similar to the

is

tells

it

how

charge change

steep the

tells

waveform

how

is.

how

dq/dt. Similarly, dv/dt indicates

meaning

fast the flux

the rate of charge change and dv/dt

is

rate of

how

lent to

and dv/dt

of dq/dt

the rate of flux change;

d)

Figure 18-26.

6

lOkiiX

5H