PoEC 16 RC Circuits

8/21/2019 PoEC 16 RC Circuits

1/64

16

RC

CIRCUITS

PART :

16-1

16-2

16-3

PART :

16-4

16-5

PART :

16-6

PART :

16-7

16-8

16-9

16-10

SERIES

EACTIVEIRCUITS

Sinusoidal

esponse

f

RC

Circuits

lmpedance nd

Phase ngle

of

Series

C

Circuits

Analysis f Series CCircuits

PARALLELEACTIVEIRCUITS

lmpedance nd

Phase ngle

of

Parallel C

Circuits

Analysis f Parallel CCircuits

SERI S- ARALLEL

EACTIVE

CIRCUITS

Analysis

f Series-Parallel

RCCircuits

SPECIAL

OPICS

Power n RCCircuits

Basic ppl icat ions

Troubleshooting

Technology

heory nto Practice

ElectronicsWorkbench

EWB)

and

PSpiceTutorials t

http //www.

enh

al

.com/f oyd

r

INTRODUCTION

An RC circuit containsboth resistance nd capaci-

tance. t is one of the basic ypes of reactive

circuits that

you

will study. n this chapter,basic senes

and

parallel

RC circuits and their responses o sinusoi -

dal ac voltages are presented.Series-parallelcombina-

tions are also analyzed.True, reactive, and apparent

power

n RC circuits are discussed nd somebasicRC

applications are introduced. Applications of RC

circuits include filters, amplifler coupling, oscillators,

and wave-shapingcircuits.

Troubleshooting s

also

covered n this chapter.

The frequency response of the RC input net-

work in an amplifier circuit is similar to the one

you

worked

with in

Chapter

13 and is the subjectof

this chapter's TECH TIP.

I COVERAGE PTIONS

This chapter s divided into four

parls:

Series Reactive

Circuits,ParallelReactiveCircuits, Series-Parallel

ReactiveCircuits, and SpecialTopics.

The

purpose

f

this organization s to facilitate either of two

approaches o the coverageof reactive circuits in

Chapters

6-17.and 18.

In the first approach,all RC circuits

(Chapter

l6)

are

covered irst, followed by all R,L circuits

(Chapter

17), and hen all RIC circuits

(Chapter

18).

Using this approach,

ou

simply cover Chapters 6,

ll, and 18 n sequence.

In the secondapproach, a7lseries reactive

circuits are covered

irst. Then

all

parallel

reactive

circuits are covered

next, followedby

series-parallel

reactive circuits and

finally

special roplcs. Using

this

approach,

you

cover

Part l:

Series

Reactive

Circuits

in

Chapters

16, l'7

and

18; then Part 2: ParallelReac-

tive Circuits

n

Chapters

16, 17,

and

18; then Part3:

Series-Parallel

eactive

Circuits

n

Chapters

16, 17,

and 18. Finally,

Pafi 4:

SpecialTopics can be covered

in each of the chapters.


2/64

TECHnology

Theory

Into

Practice

CHAPTER

BfECTTVES

1:

SERIESREACTIVE

CIRCUITS

Describehe relationship

betweencurrent

and

voltase n

an RC circuit.

Determinempedance

and phase

angle

n a series

RCcircuit

Analyze series

RC circuit

2: PARALLEL

REACTIVE

CIRCUITS

Determine

mpedance

and

phase

angle n a

parallel

RC

circuit

Analyze parallel

RC

circuit

PART

3: SERIES-PARALLEL

EACTIVE

CIRCUITS

tr

Analyze

series-parallel

C circuits

PART4:

SPECIALTOPICS

O

Deterrnine

ower

n RC

circuits

tr

Discuss

ome asicRC

applications

tr

Troubleshoot

C circuits


3/64

16-1

r

SINUSOIDAL

ESPONSE

F

RCCIRCUITS

When

a sinusoidul

voltage

is applied

to any

type of

RC circuit,

each resulting

voltage

drop and

the current

in

the circuit are

also sinusoi.dal

snd

have

the same

requenc

as the applied

voltage.

The

capacitance

causes

a

phase

shift

between

the voltage

and

current

that depends

on

the relative values

of the

resistance

and the

capacitive

reac'

tsnce.

ffier

completing

this section,

you

should

be uble

to

I

Describe the

relationship

between

current

and voltage

in an

RC circuit

. Discuss voltage and current waveforms

.

Discuss

hase hilt

.

Describe

types

of signal

generators

As shown

in Figure

16-1, the

resistor voltage

(Vn),

the capacitor

voltage

(Vc),

and

he

current

(/)

are all sine

waves with

the frequency

ofthe

source.

Phaseshifts

are ntroduced

because

of the

capacitance.

As

you

will learn,

the

resistor voltage

and current

lead

the

source

voltage,

and the

capacitor

voltage

lags the source

voltage.

The

phase

angle

between

he current

and

the capacitor

voltage

s always 90'.

These

generalized

phase

ela-

tionships

are ndicated

n

Figure 16-1.

FIGURE

16-1

Illustration

of sinusoidal

response with

general phase

relationships

of

Vp, Vs, and

I relative

to the source

voltage.

Vp

leads V,, Vs

lags V", and

I lead's V,. Vp

and

I

are

in

phase

while Vp and.

Vg are

90" out of

phase.

".,ffi/

600


4/64

IMPEDANCE

ND PHASE NCLE

OF SERIESC

CIRCUITS 601

The

amplitudes and the

phase

relationships

of the voltages

and current depend

on

the values

of the resistanceand the

capacitive reactance.

When a circuit is

purely

resis-

tive, the

phase

angle between the

applied

(source)

voltage

and the total current is zero.

When a circuit is

purely

capacitive,

he

phase

angle between

the applied voltage

and the

total current is

90",

with the

current leading the voltage.

When there is a combination

of

both resistance

and capacitive reactance

n a circuit, the

phase

angle between he

applied

voltage and the total current is somewherebetween 0o and 90', dependingon the relative

values

of the resistance nd he reactance.

SignalGenerators

When a circuit is hooked

up for a laboratory experiment

or for troubleshooting,

a signal

generator

similar to those

shown n Figure 16-2 is

used o

provide

the source

voltage.

These nstruments, epending

on their capability,are

classifiedas sine wave

generators,

which

produce

only sine waves;

sine/square

generators,

which

produce

either

sine waves

or squarewaves;or function generators,

hich

produce

sine waves,

pulse

waveforms,or

triangular

(ramp)

waveforms.

t _ _ "

FIGURE 6-2

Typical

signal

(function) generators

sed n

circuit testingand troubleshooting,

Photography

courtesyof B&K Precision

Corp.).

A

60 Hz sinusoidal voltage s applied

to an RC circuit.

What

is

the

frequency

of the

capacitor voltage? What is the frequency

of the current?

What causes he

phase

shift between

V,

and in a

seriesRC circuit?

When the

resistance

n an RC

circuit

is

greater

han the

capacitive

reactance,

s the

phase

angle between he applied

voltage and the total current

closer to 0' or to

90o?

16-2 IMPEDANCEND

PHASE NCLE

OF SERIESC

CIRCUITS

ii

il

. ..

..,.-.....

:

(b )

sEcTtoN

6-1

1.

REVIEW

)

3.

The impedance

of any type of RC

circuit is the total opposition to

sinusoidal current

und its unit is the ohm. The

phase

angle is the

phase

dffirence

between he total cur-

rent

und the source voltage.

After completing this

section,

you

should be able to

I

Determine impedance

and

phase

angle in a

series

RC

circuit

.

Define impedance

.

Express capacitive reactance

n complex form

.

Express otal impedance n

complex form

.

Draw an impedance

riangle

.

Calculate mpedance

magnitude and the

phase

angle


5/64

602

I

RC

CIRCUITS

In a

purely

capacitive circuit, the

impedance is

equal to the total

capacitive reactance.

The impedanceof a seriesRC circuit

is

determined by both the

resistance

and

the capac-

itive reactance.These casesare illustrated in Figure 16-3.

The magnitude

of the

imped-

ance

s

symbolizedby

Z.

G"

51.

I t * ;

( a ) z = R

FIGURE 6-3

Threecases f impedance,

(c)

Z includes

both

R

and

X6

Recall from

Chapter

13 that capacitive eactance s expressedas a complex number

in rectansular

form

as

X6

=

-jXa

(16_l)

where boldface X6'designatesa

phasor quantity

(representing

both

magnitude and angle)

urd

X6 is

just

the magnitude.

In

the series

RC circuit of Figure 164, the total impedance s the

phasor

sum of R

and

-jX6

and

is

expressedas

Z = R - j X c

(16-2)

FICURE 6_4

SeriesRC circuit.

The mpedance riangle

In ac analysis,both

R

and

X,

are treated as

phasor quantities,

as shown n the

phasor

dia-

gram

of Figure 16-5(a), with X6' appearingat a

-90o

angle

with respect o R. This relation-

ship

comes rom the fact that the capacitor voltage

in

a series

RC circuit lags the current,

z

=

"/P

*xZ

\

_L

(a)

FICURE 6-5

Develbpmentof the impedance riangle

for

a

seriesRC circuit.

(b)

( b ) z = x c

R

(9

(c )


6/64

IMPEDANCEND PHASE NCLE

OF SERIESC CIRCUITS

603

and hus he

resistor

voltage,by

90'. Since

Z is the

phasor

sum ofR and

jX6,

its

phasor

representation

s shown n Figure 16-5(b).A repositioning

fthe

phasors,

sshown n

part

(c),

forms

a right triangle. This is called the impedance

riangle. The length of each

phasor

represents he magnitude n ohms,

and

the

angle0

is

the

phase

angle of the RC circuit

and

represents he

phase

difference between

he applied

voltage

and the current.

From

right-angle trigonometry

(Pythagorean

heorem), the magnitude

(length)

of

the impedancecan be expressed n terms of the resistanceand reactanceas

z=lFi4

(16-3)

The italic letter Z represents

he magnitude of the

phasor quantity

Z

and

s

expressed n

ohms.

The

phase

angle,0, s

expressed s

0

=

*tan

(16-4)

The

symbol an

I

stands or inverse angentand canbe found by

pressing

@, then [6I].

Combining the magnitude and angle, he

phasor

expression or impedance

n

polar

form is

z=f * +*x|z- tun

(1

6-5)

EXAMPLE

6-1

[:ffi1#TJ,'*#t#iJ?*write

the

phasor

xpressionor

he

mpedancen both

- t lXc\

\ ^ /

- , /X . \

\ R /

(a )

FIGURE 6-6

Solution

For

the circuit

in Figure

16-6(a), the

Z=

R

-

j0

=

R

=

56 O in rectangular orm

(X.=

Q)

Z = RZjo = 5610' O in polar form

The impedances simply the resistance,

nd he

phase

angle s zerobecause

ure

resis-

tance does not causea

phase

shift

between he voltage and current.

For the circuit in Figure 16-6(b),

the

impedance

s

Z

=

0

-

jXc

=

-i100

O

in rectangular orm

(R

=

0)

Z

=

Xcl-90"

=

l00Z-90" O in polar

form

The impedances simply the capacitive eactance,

nd he

phase

angle s

-90'

because

the capacitance auses he current to lead the voltage by

90'.

For

the circuit in Figure l6-6(c), the impedance n rectangular orm is

Z = R -

j X c = 5 6 O

j 1 0 0 O


7/64

604

I

RC CIRCUITS

sEcTroN

6-2

REVIEW

The

impedance n

polar

form is

z

=

Vrp'* r'..-tan

t(4c\

'

\ R /

=

\,/G6Of +

(t00el

z_tan-,(J445 )

l15z_60.8.

\ ) o r 2 /

In

this case, the impedance is the phasor

sum of the resistance

and the capacitive

reactance.

The

phase

angle is flxed

by the relative values

of

X6

and R. Rectangu-

lar to

polar

conversion

using a calculator was illustrated in

Chapter 12 and can be

used in

problems

like

this one.

Related Problem

Use

your

calculator to convefi

the

impedance

n Figure 16-6(c)

from rectangular

o

polar

form. Draw the impedance

phasor

diagram.

l. The impedance

of a certain RC circuit is

150 Q

-

j220

f2. What is

the value of the

resistance?he capacitive eactance?

A

series

RC

circuit has

a

total resistance

of 33

kO

and a capacitive reactance

of

50 kQ. Write the

phasor

expressionor the impedance n

rectangular orm.

For

the circuit in

Question

2, whatis

the magnitude of the impedance?

What is the

phase

ngle?

16_3

r

ANATYSIS

F SERIES

C

CIRCUITS

)

3.

In the

previous

section,

you

learned how to

express he impedance of a series RC

circuit. Now,

Ohm's law und Kirchhoff\ voltage law

are used n the analysis of RC

circuits.

After

completing this section,

you

should be uble to

I

Analyze a series RC circuit

.

Apply Ohm's law and Kirchhoff's

voltage law to seriesRC

circuits

.

Express

he voltages and current

as

phasor

quantities

.

Show how impedance and

phase

angle vary with frequency

Ohm's Law

The

application of Ohm's law to

series

RC

circuits involves the use

of the

phasor quanti-

ties of Z,Y, and I. Keep in mind that the use of boldface nonitalic letters indicatespha-

sor

quantities

where both magnitude

and angle are included. The three equivalent orms

of

Ohm's

law

are as follows:

Y

= l Z

(1

6-6)

(16-7)

(1

6-8)

r = I

Z

z =

I


8/64

ANALYSIS

F SERIESC

CIRCUITS 605

From

your

study of

phasor

algebra

n

Chapter 72,

you

should recall

that multipli-

cation and division are most easily

accomplishedwith the

polar

forms.

SinceOhm's law

calculations involve multiplications

and divisions, the voltage,

current, and impedance

should be expressed n

polar

form.

The following two

examples show the relationship

between he

source

voltage

and source current. In Example

16-2, the cuffent is

the

refer-

ence and n Example 16-3,

the voltage s the reference.Notice

that the reference s drawn

alons the X-axis n both cases.

EXAMPLE6-2

If

the cur:rent

n

Figure 16-7 is expressed n

polar

form as |

=

0.220" mA,

determine

the

source voltage expressed n

polar

form,

and draw a

phasor

diagram

showing the

relation

betweensourcevoltase and cunent.

FIGURE 6-7

C

0.01

pF

Solution The magnitude

of the capacitive reactance s

r - =

|

= r s q k ( )

" (

2nfC 2n(1000

Hztt0.Ol

Fl

The total impedance n rectangular orm

is

Z = R -

j X c

=

1 0 k A

-

j 1 5 . 9 k A

Converling to

polar

form

yields

z

=

\/F\

xrrz-run

(\\

-

\ R /

=

fiz-ra,'(ffi)

=

ra.ar-sl.S"rl

\

- - - -

,

Use

Ohm's

aw

to determinehe

source oltage.

Y,

=

IZ

=

(0.220"

mAXl

8.82-57.8' O)

=

3.7 Z- 57

8"Y

Themagnitude

f thesource oltages

3.76

V

at anangle f

-57.8"

with respect

o the

current,hat s,

the

voltage

ags hecur:rent

y 57.8', asshownn the

phasor

iagram

of Figure 16-8.

FIGURE 6-B

1 = 0 . 2 m A

=

3 7 6 V

Related

Problem Determine

. in Fieure 16-1 f

f

=

2 kHz and

=

0.210"

A.


9/64

606

r

RC CIRCUITS

EXAMPLE 6-3

Determine

the current

in

the circuit of Figure 16-9,

and draw a

phasor

diagram show-

ing the relation

between source voltage and current.

FIGURE 6-9

Figure 6.10 .

FtcuRE 6-10

% = 1 0 v

Related

Problem Determine I in Figure 16-9 if

the

frequency

s increased o 5 kHz.

Solution

The magnitude

of the capacitive reactance s

Y - =

I

=

I

- s ? r - ( )

--(

2nfC

2n(1.5kHzt(0

02

p.Fl

The total impedance n rectangular orm is

Z = R -

j X c = 2 . 2 k { l

- j 5 . 3

k O

Convefiing to

polar

form

yields

z= f R ' z x ) z -nn - t ( k \

\ R i

=

@

.-r^-'

(

E\

=

5.7 -67.s. kO

\ 2 . 2 k a l

'

Use Ohm's law to determine he cur:rent.

r= I=

loz9 'Y

=1 ,74167 .5 "mA

z 5.742.-61.5"k{L

The

magnitude of the current is | .74 mA. The positive phase

angle of 6'l .5" indicates

that the current leads the voltage

by that amount, as shown in the

phasor

diagram of

v

1020"

Relationships

f the Current

and

Voltagesn

a Series C

Circuit

In

a series circuit, the current is the same though

both the resistor and the capacitor

Thus, the resistor

voltage is in

phase

with the current,

and the capacitor voltage lags he

current

by 90'.

Therefore,

there is a

phase

difference

of 90' between the resistor volt-

age, Vp, and the capacitor voltage,

V6,

as shown in

the waveform diagram of Figure

16-1 .


10/64

flGURE 6-11

Phase

elation of voltages

nd

current in

a

series

C

circuit.

vR

v,=f vA*vlz-tun

where the magnitude of the source voltage is

v,=f

v 'o* l

and the

phase

angle between he resistor voltage and the source voltage is

. / v _ \

0 - - t a n ' l

' '

l

\

Yo l

ANALYSIS F SERIESC CIRCUITS 607

(16-10)

(16 -11 )

v(

FIGURE6-13

Volnge and current

phasor

diagram

or

the waveforms

n

Figure 16-11.

You know from Kirchhoff's voltage law that the sum of the voltage drops must

equal the applied voltage. However, since Vp andV6 are not in

phase

with each other, they

must be added as

phasor quantities,

with V6' lagging

Vp

by

90",

as shown in Figure

16-12(.a). s shown n Figure l6-12(b),

V,

is

the

phasor

sum of Va and V6',as expressed

in rectangular orm in the following

equation:

Y , = V p * j V s

This

equation can be expressed n

polar

form as

(1

6-9)

'(#)

(16-12)

Since the

resistor

voltage and the current are

in

phase,

d also represents he

phase

angle between he sourcevoltage and the current.

Figure 16-13

shows a complete voltage

and current

phasor

diagram that represents he waveform diagram of Figure 16-l l.

VR

,,7

-N,

a) (b)

f lGURE

6-12

Voltage

hasor

diagram

or

a seriesRC circuit.


11/64

608

T

RCCIRCUITS

Variation

f lmpedance

ith Frequency

As

you

know,

capacitive eactancevaries nversely

with frequency.

Since Z

=

f R' **r,

you

can

see

hat

when X6 increases,

he entire term under the

square oot sign increases

and thus the magnitude

of the total impedance

also ncreases;

and when X. decreases,

he

magnitude of the

total impedance also

decreases.Therefore,

in an RC circuit, Z

is

inversely dependent

on

frequency.

Figure 16-14 illustrates

how the voltages

and current in

a seriesRC circuit vary

as

the frequency ncreases

or decreases,

with the source voltage held

at a constant value. n

part

(a),

as the

frequency

s increased,X.

decreases; o less

voltage is dropped

across he

capacitor. Also, Z

decreases s X.

decreases, ausing he

cuffent to increase.An increase

in

the current causesmore

voltage acrossR.

In

Figure l6-14(b),

as he frequency s decreased,

6increases;

so

more

voltages

dropped

across he capac itor. Also, Z increases

as X6 increases,

causing the

current o

decrease.

decreasen the current

causesessvoltaee

across

R.

(a)

As frequency s increased,

andVp increase

and Yc decreases.

(b)

As frequency s

decreased, and Vlq ecrease

and Va ncreases.

FIGURE

6-14

An illustration of

how the variation of impedance

affects the voltages

and current as the

source

requency

s varied.The

sourcevoltage s held

at a constontamplitude.

The

effect of changes n Z

and X6 can be

observed as shown in Figure 16-15.

As

the frequency increases,

he voltage acrossZ remains

constant because

V"

is

constant.

Also, the voltage

across C decreases. he increasing

cunent indicates

that

Z

is decreas-

ing. It does

so because f the inverse elationship

stated n

Ohm's

law

(Z=VzlD.The

increasing

culrent also indicates that Xs is

decreasing

Xc

=

VclD.

The

decrease n

Vc

conesponds o the decrease n X6,.

Variation

f the

Phase

nglewith Frequency

Since

X6' s

the factor that introduces

the

phase

angle in

a seriesRC circuit, a changen

Xg

produces

a change in the

phase

angle. As the frequency is

increased,X. becomes

smalleq

and thus the

phase

angle decreases. s the frequency

is decreased,X. becomes

larger,

and thus the

phase

angle

increases.

This variation is illustrated

in Figure 16-16

where a

"phase

meter"

is connected

across he capacitor to illustrate

the change n angle

between

V" and Vp. This angle s

the

phase

angle of the circuit

because1 is in

phase

with

Vn.

By measuring

he

phase

of

Va,

you

are effectively measuring

he

phase

of L An

oscil-

loscope is normally

used to

observe he

phase

angle. The meter

is used for convenient

illustration of the concept.


12/64

\

%i t

constant.

r-t

ffir

\qY,

G

A o

E C

Frequency is

increasing.

ANALYSIS

F

SERIESC CIRCUITS

T

609

By watching these

wo

meters,

you

can tell

whatZis doing:1is

increasing and V

2is

constant.

Thus, Z is decreasing.

By watching these wo

meters,

you

can tell

what X6, s doing:

I is increasing and Vc'

s

decreasing.

Thus, X6'

is decreasing.

l t ,

=

FIGURE

6-15

An illustration of how

Z

anil

X6

change

with

frequency.

FIGURE

6-16

The

"phase

meter" shows he effectof

frequenc

on the

phase

angle of a circuit,

The

"phase

meter" indicates he

phase

angle

changebetween wo voltages, ,

and Vp. Since Vp

and

I are in

phase,

his angle

s the sameas the angle betweenV,

and L

Figure

16-17 uses he impedance riangle

to illustrate the variations

inX6, Z, and 0

as he frequency changes.

Of course,

R

remains constant.

The key

point is

that

becauseX6'

varies nversely with

the frequency, so also do the

magnitude of the total

impedance and

the

phase

angle. Example

16-4 illustrates this.

vr l

7 f

f lGURE16-17

As he

requency

increases, Xg decreases,

Z decreases,and

0 lecreases.

ach value of

frequency

can

be visualized as

forming

a dffirent

impedance triangle.

Increasing

/

ft

c3

^c2

J 2

J 1

Phasemeter

Phasemeter

^c l


13/64

610

r

RC

CIRCUITS

SECTION 1

6-3

1.

In a

certain series

RC circuit,

Vn

=

4

V, and

Vc

=

6 V. What is

the magnitude

of

the

REVIEW

source oltage?

2. In

Question

1,

what is the phase

angle

between

the source voltage

and the

current?

3.

What

is

the

phase

difference

between

he

capacitor

voltage

aad the resistor

voltage

in

a seriesRC

circuit?

4.

When

the frequency

of

the applied

voltage in

a series

RC circuit

is increased,

what

happens

o the

capacitive reactance?

what happens

o the magnitude

of the total

impedance?

What

happens

o the

phase

angle?

EXAMPLE

6-4

For

the seriesRC

circuit

in Figure

16-18,

determine

he magnitude

of the total

imped-

ance

and the

phase

angle for

each of the

following

values

of input

frequency:

(a)

10 kHz

(b)

20

kHz

(c)

30 kHz

F IGURE

6-18

Solution

(a)

For/=

10kHz,

_ _ |

I

=

1.59

Q

2nfC

2n(r0kHz)(Q.01 F)

0.01

F

z=f R\

x/z- tan- ' l I t \

\ R /

=

@z-t"''(ffi)

=

.ssr-sz.e.

o

Thus, =

1.88 O

and

0

=

-57.9'.

(b )

For f=20kHz,

"'=;rzo

**0o'

uo,

=

7e62

z

=

@

Lan-,

J9SP)

r.zz.-zs.s.

o

Thus,

Z

=

1.28

kO

and0

=

-38.5'.

(c)

For/=

30kHz,

x'=

t(30 kHhJl

,F)

=

531

z

=

@.2-tan-r( 11

I

\

=

t.tzt-zs.o"

ko

Thus,

Z= 1.13

kO and

0

=

-28.0".

Notice

that

as the frequency

ncreases,

X6,

Z, and

d decrease.

Related

Prohlem

Find the

magnitude

of the

total impedance

and the

phase

angle

n

Figure

16-18

for

=

| kJlz.

I

coverage of

serics reuctive

circuits

continues

in chapter 17, Part r, on page 664.


14/64

r

IMPEDANCE

ND

PHASE NGTEOF

PARATTEL

C CIRCUITS

In this section,

you

will

learn how

to determine the

impedance and

phase

angle of a

parallel

RC circait.

Also, capacitive susceptance

and admittance

of a

parallel

RC cir-

cuit are

introduced,

After compkting

this section,

you

should

be able to

I

Determine impedance

and

phase

angle

in a

parallel

RC circuit

.

Express otal

impedance n complex

form

.

Define and calculate

conductance,

capacitive susceptance,

and admittance

Figure l6-19 shows

a basic

parallel RC circuit connected

o an ac voltage

source.

FIGURE

6-19

Basic

parallel RC circuit.

The expression

or the total impedance

s developed

as follows, using

the rules of

phasor

algebra.

Since there

are only two components,

he

total impedance

can be found

from the

product-over-sum ule.

z

(R/0")dcz-90")

t =

R a x ,

By multiplying the

magnitudes, adding

the angles

n the numerator,

and converting

the

denominator o

polar

form, we

get

Rxcl(o"

-

90')

Z -

\/n\x'rz.--

(+)

611


15/64

612

r

RC CIRCUITS

Now,

dividing

the magnitude

expression

n the numerator

by

that in the

denominatoq

and

by subtracting

he angle n

the denominator

rom

that in the

numerator,

we

get

z 1

RX6

\ / . /Y - \ \

L

=

t_ - __ t . / t

_90o

+

tan_ ' l * l l

( 16_13)

\VR'

+

x(

l - \

' "

"* '

\

R

/

Equation (16-13)

is

the expression

or the

total impedance

or

a basic

parallel

RC cu-

cuit. where

the masnitude

is

Z _

and the phase

angle

between

he applied

voltage and

the total current

is

0=-900+r--, l ,+)

\ R /

Equivalently,

this expressioncan be written as

^

- ' l R \

\x'l

EXAMPTE

16-5

For

each circuit

in Figure 16-20,

determine

he magnitude

of

the total impedance

and

the

phase

angle.

R

1.0 o

Solution

For

the circuit in Figure

l6-20(a),

the

total impedance

s

z = ( - : X . \ z - t u n - , / 4 \

\V

n')

xl-

\x./

t ( l 0 0 Q x 5 0 Q ) - t , r t o o o r

= l

_ l l - t a n '

l = J 4 . 1 1 _ 6 3 . 1 " { t

L

V ( loo Q) '+

(50

Q) '

|

\

5u 12

Thts,

Z

=

44.7

O and

0

=

-63.4".

For

the circuit in

Figure l6-20(b),

the

total impedance

s

t

(1.0

kflx2

ko)

-l

/ I o k()\

Z = l

"

R . l z - t a n ' { + # l = 8 9 4 t - 2 6 . 0 .

Q

I

v

t

t . o kOr ' +

(2

kQ) ' l

\

- Ns .

/

Thlas,Z=

894 O

and I

=

-26.6".

Related Prohlem

Determine

Z

in Figure

l6-20(a) if

the frequency

s doubled.


16/64

IMPEDANCE

ND PHASE

NCLE

OF PARALLELC

CIRCUITS

613

Conductance/

usceptance/

nd Adm

ttance

Recall that

conductance,

G

is

the reciprocal

of resistance.

The

phasor

expression or

conductance

s expressed

s

Two

new terms

are now introduced

for use n parallel

RC

circuits.

Capacitive sus-

ceptance

(86)

is the reciprocal

of capacitive reactance

and

the

phasor

expression for

capacitive usceptance

s

G=#

=

Gt|o

Br=

nh=Bclgoo=+iBc

"

=zh=Y/'+o

(16-17)

(16-18)

(16-19)

Admittance

()z)

is the

reciprocal of impedance

and the

phasor

expression for

admit-

tance s

The

unit of each

of these erms s

the siemens

S),

which is the reciprocal

of

the ohm.

In working

with

parallel

circuits, it is often

easier to use

G, 86, and I rather

than

R, Xc, andZ. In

a

parallel

RC

circuit, as

shown n Figure

16-21, the

total admittance

s

simply

the

phasor

sum

of the conductance

nd he capacitive

usceptance.

Y = G + j B c

(16-20\

(a )

FIGURE 6_21

Admittance

n a

parallel

RC

circuit.

EXAMPLE 16-6

Determine

he total admittance

Y)

and

total impedance

Z)

in Figure 16-22.

Sketch

the

admittance

phasor

diagram.

C

0.2

pF

FIGURE

6-22


17/64

614

r

RC

CIRCUITS

Solution

From

Figure16-22,

R

=

330 Q;

thusG

=

llR= l/330

fl

=

3.03

mS.

The

capacitive

eactances

x r

=

: L

=

- - . . - ]

= ' 796

2n[C

2n(1000

z1Q.2

F1

The

capacitive usceptance

agnitudes

The total

admittance s

16_5

r

ANATYSIS

F

PARALLET

C

CIRCUITS

B . = L =

1

= 1 . 2 6 m S

"

xc

196a

Y,o,=

G +

Bc

=

3.03mS+

1.26

mS

which

can be expressed n

polar

form as

Y

o,

f c\ Blztan

(4\

'

\ G /

=

m

zrun

1.26S

\

=

t.zxzzz.6"

\ : .0 :ms - " ' * ' * "

Total admittance s converled o total impedanceas follows:

zr,

=

J-= - . - - l - .

-

=

3osZ-22.6"

L

Y,o,

(3.28222.6'mS1

The

admittance

phasor

diagram is shown in

Figure 16-23.

F ICURE

6-23

Br '=

1 .26mS

I

=

3.28 nS

G

=

3.03mS

Related Problem Calculate

the total admittance

n Fisure 16*22

if

f

is

to 2.5 kHZ

SECTION 16-4

1. Define

conductance,

capacitive

susceptance,and admittance.

REVf

EW

2. It Z= 100

e, whar s

the

vatue

of I?

3. In a certain

parallel

RC circuit,

R

=

47 Q

and

X.

=

75 {t. Determine

Y.

4. In

Question

, what s Z?

mS

In

the

previous

section,

you

learned

how to express

he impedance

of a

parallel

RC

circuit.

Now, Ohm's law and Kirchhoff\

current

law ure ased n

the analysis of RC

circuits.

Current and voltage relationships

in

a

parallel

RC circuit

are examined.

After

completing

this section,

you

should be able to

I

Analyze

a

parallel

RC

circuit

.

Apply

Ohm's law and Kirchhoff's

current law

to

parallel

RC

circuits

.

Express

he voltages and

currents as

phasor quantities


18/64

ANALYSISF

PARALLELC CIRCUITS 6.15

Show

how impedance and

phase

angle vary with

frequency

Convert

from a

parallel

circuit

to an equivalent seriescircuit

For convenience n the analysis

of

parallel

circuits, the Ohm's

law formulas using

imped-

ance,

previously

stated,

can be rewritten

for

admittance

using the

relation Y

=

l/2.

Remember, he use of boldface nonitalic letters indicatesphasorquantities.

v = I

Y

(16-2"t)

(16-22)

(16-23)

I = V Y

Y = f

v

EXAMPLE

6-7

Determine the total cuffent and

phase

angle in Figure

showing the

relationship of V" andl,or.

16-24. Draw a

phasor

diagram

FtcuRE 6-24

Solution

The capacitive

reactance s

Xr=;4 ;=

2nf

C

Zn(1.5

The capacitive

usceptanceagnitude

s

C

0.02

pF

=

5.31 o

kHz)(0.02

pF)

1 l

B c = 4 =

5 * 3 t

o

=

1 8 8

S

The conductancemagnitude s

1 1

o =

o = r ; * = 4 5 5 p S

The total admittance

s

Y,o,= G

+

Bc=

455

pS

+7188

pS

Converting to

polar

forrn

yields


19/64

616

I

RC

CIRCUITS

FtcuRE

6-26

Currents in a

parallel

RC

circuit.

The phase

angle s

22.5".

Use

Ohm's law to determine

he total

current.

l ,o,=Y,Y,o,=

(1020"

V)(492222.5'

pS)

=

4.92222,5'

mA

The magnitude

of the

total current is 4.92

mA, and it

leads the

applied voltage

by

22.5',

as he

phasor

diagram n

Figure 16-25

indicates.

FfGURE

6-25

-/

' , '=

-r 'e l

nA

, \ 2 2 . 5 .

,

I

=

' n v

Related

Prohlem

What is the total

cuffent

(in

polar

form) iflis

doubled?

Relationships

f the Currents

and Voltagesn

a Parallel

RC

Circuit

Figure

l6-26(a) shows

all the currents

n a basic

parallel

RC circuit.

The total

curterrt, ,o,,

divides at the

junction

into the two branch currents, Ia and 16,.The applied voltage,V,,

appearsacross

both the resistive

and the capacitive

branches,

so V",

Va, and Vg are all in

phase

and of the

samemagnitude.

L

I ' R

V

The current

through the resistor

is in

phase

with the voltage.

The current

through

the

capacitor eads

he voltage, and

thus the resistive

current, by

90'. By Kirchhoff's

cur-

rent law,

the total current

is the

phasor

sum of the

two branch currents,

as shown

by the

phasor

diagram in Figure

l6-26(b). The

total cuffent

is expressed

as

Ic

I,o,

l / i

l , / i

(b)

I,or=

Io +

jI6

This

equation

can be expressed in polar

form

as

l , o ,=Y I ' n+

I f l t an

where

the magnitude

of the total

current is

I 'o'

=f

I 'o

I"

and the

phase

angle

between he resistor

cunent

and the total

current is

-'l1"\

\ro

(16-24)

(1

6-2s)

(16-26)

-r l1c\

t - l

\1o

V,

"T

0 = t a n

('t6-27)


20/64

ANALYSISF PARALLEL

CCIRCUITS

617

Since

the resistor

cuffent and the applied voltage

are in

phase,

d

also represents

he

phase

angle between

he total current and the applied voltage.

Figure 16-27 shows

a com-

plete

current and voltage phasor

diagram.

FIGURE

I6-27

Current and voltage

hasor

liagram

or

a

parallel

RC circuit

(amplitudes

re arbitrary).

EXAMPTE 16-8

Determine he

value of eachculrent n Figure 16-28,

and describe he

phase

elation-

ship of each

with the applied voltage. Draw

the current

phasor

diagram.

FIGURE 6-28

Solution

The resistor

current, the capacitor

current, and the total current

are

expressed

s

ollows:

to=*

=ffi+=sl.sto'mA

I-

=

-L

-

l2zo'Y

=

8oz9o"mA

-

Xc r5oz-go' t)

l,o,

=

Ia +

jIc

=

54.5 mA +780 mA

Converting Iro, o

polar

form

yields

l ,o,

f

I 'o*

lztun-t( \

\1^/

=

@

ztun,( f lJn{ .

\=96.Btss.7o

A

\

54.5

mA

/

As the results

show, he resistor current s

54.5

mA

and is in

phase

with the volt-

age.The capacitor

urent is 80 mA and eads

he

voltage

by

90".

The total current s

96.8

mA and leads

he voltage by 55.7". The phasor

diagram n Figure 16-29 illus-

trates hese elationships.


21/64

618

r

RC

CIRCUITS

FIGURE 16-29

16

=

80

mA

1,o,

=

96.8 mA

54.5

mA

Related Prohlem

Determinehe otal

In a

parallel

cuffent.

In

=

10010' mA and lc

=

6029Oo tnA.

Conversionrom Parallel

o Series

orm

For every

parallel

RC

circuit, there is an equivalent series RC circuit. Two

circuits are

consideredequivalent when they both presentan equal impedance at their terminals; hat

is,

the

magnitude

of

impedance

and the

phase

angle are identical.

To obtain the

equivalent series circuit for a

given parallel

RC circuit, first find

the

impedanceand

phase

angle of the

parallel

circuit. Then use the values of Z and

d

to con-

struct

an

impedance

riangle shown n Figure 16-30. The vertical and horizontal

sides f

the triangle represent he equivalent series resistance and capacitive

reactance as indi-

cated. These values can be found using the following

trigonometric relationships:

R"q= Z cos0

Xc("q)= s in0

FIGURE16-30

Impedance triangle

for

the

series equivalent of a

parallel

RC circuit. Z anil

0 are

the known

values

or

the

parallel

circuit. R"o and Xs1"r:) re the series equivalent values.

EXAMPLE6-9

Convert the

parallel

circuit in Figure 16-31

to a series orm.

(16_28)

(16-29)

X6601= Z sin?

L c

2'7

Rrq= Z cos

e

FIGURE 6-31


22/64

Solution First, find the admittance of the

parallel

circuit as follows:

t l

O =

* = , r r = 5 5 . 6 m S

1 t

B - = L

= 3 7 . 0 m S

L

x c 2 7 A

Y = G + jBc = 55.6mS +737.0mS

Converting

to

polar

form

yields

Y

=f

c\

B\zrun-'(*\

\ G /

=

ffiz,un-' (..rj$

\

=

oo.zst.o.

\

) ) .o m)

/

Then, the total impedance

s

7

- l -

u t o t -

Y

6 6 . 8 2 3 3 . 6 ' m S

ANALYSIS F PARALLELC

CIRCUITS 619

=

15.01-33.6"

mS

Converting to rectangular orm yields

Z , o , = Z c o s 0 -

Z

s i n 0 = R . q -

X c G q )

=

15.0

cos(-33.6")

j15.0

sin(33.6')

=

12.5Q

-

j8.31

O

The

equivalent series RC circuit is a 12.5 f2 resistor in

series with a capacitive

tanceof 8.31 O. This is shown n Fisure 16-32.

FIGURE 6-32

^c(eq)

Related Problem The impedance

of a

parallel

RC circuit is Z

=

101,-26' kf).

Con-

vert to an equivalent

seriescircuit.

SECTION 16-5

1. The admittance f an RC circuit is

3.50 mS, and he appliedvoltage s

6 V. What is

REVIEW the total current?

2. In a certain

parallel

RC

circuit, the resistor current s 10 mA, and the

capacitor cur-

rent is 15

mA. Determine the magnitude and

phase

angle of the

total cuuent.

This

phase

angle

s

measuredwith respect o what?

3. What is the

phase

angle between

the capacitor cuffent and the applied voltage in a

parallel

RC

circuit?

I

Coverage

f

parallel

reactive circuits

continues in Chapter 17, Part 2, on

page

672.


23/64

16_6 SERIES.PARALLET

C

CIRCUITS

In this section, the concepts studied

with respect to series and

parallel

circuits ure

used to analyze circuits with combinutions of both series

and

parallel

R und C ele-

ments.

After completing this section,

you

should be able to

I

Analyze series-parallel RC circuits

.

Determine total impedance

.

Calculate currents and voltages

.

Measure mpedance and

phase

angle

Series-parallelcircuits consist of arrangementsof both

series and

parallel

elements.The

following two examples demonstrate

how

to approach

he analysis of series-parallelRC

networks.

EXAMPLE6-10

FIGURE 6_33

In the circuit of Figure

(a)

total impedance

16-33, determine he

following:

(b)

total current

(c)

phase

angle

by

which l,o,leads V"

c2

0.05,uF


24/64

SERIES-PARALLEL

C

CIRCUITS

621

Solution

(a)

First,

calculate

hemagnitudes

f capacitive

eactance.

x r , = - L = - , '

1

- = 3 1 8 f )

2n fC

2r r5kHzr r0 . l F l

xr,

=

-f=

=

6?'l a

2n[C 2n(5kHzlr0.05 Fl

Oneapproach

s to lind the mpedance

f the series

ortion

and

he mpedance

f

the

parallelportion

andcombine

hem

o

get

he total

mpedance.

he mpedance

of the series

ombination

f R1and

C1 s

Zt

=

Rt

-

jXct

=

1.0 O

-

j318

O

To determine

he mpedance

f the

parallel

portion,

irst

determine

he admittance

of the

parallel

combination

f R2 and

C2.

t . = I = -

= 1 . - l 7 m S

R2

680O

u. , ,=

*= #

"=

1 .57

S

Yz=

Gz.+

Bcz

=

1.47mS +

1.57

mS

Converting to polar

form

yields

Y

=

f c i ' . , - ' -z tan

( ls t1

'

\ G ,

=

fi

zrun_,( l ]\=2.t5t46.e"

s

\ r . 4 /

m s /

Then, the

impedance

of the

parallel

portion

is

Zr=

l=

=

4652-46.9"

)

Y2 2. I5146.9"

mS

Converting to rectangular

orm

yields

Zz= Zzcos

-. iZ2sin

0

=

(465

O)cos(-46.9')

j(465

Q)sin(-46.9')

=

318 A

-

fi39

A

The series

portion

and the

parallel

portion

are in series

with each

other. Combine

Z1

and 22 to

get

the total impedance.

Z , o , = Z r + 2 2

=

(1 .0kO

j318

O ) +

(318

A

-

j 3 3 9

O )

=

1318O-

j651

O

Expressing

Z,o, n

polar

form

yields

z,u,= z'3 f z- tan

(?\

'

\ z , l

=

@

t-tan-'

9\

=

1.47-26.5.

kO

\

l J 1 6

2

(b)

Use Ohm's law

to determine

he total current.

I,,=5

=== 4:-

=6.80226.5"

A

z,o,

1.41 -26.5"

kQ

(c)

The

total cunent

leads he

applied voltage

by 26.5".

Related

Problem

Determine

the voltages

across 21

and 22 in Figure 16-33

and

express n

polar

form.


25/64

622

I

RC CIRCUITS

EXAMPLE6_11

Determine

all currents n Figure 16-34. Sketch a current

phasor

diagram.

V,

220"V

f = 2 M H z

z1

R1

? ? ' o

L l

0.001

z2

K1

4 7 A

C"

OilOZ

P

P F I

r

FIGURE 6-34

Solution First, calculateXg1 ard Xs2.

Xr,

=

-L

=

79'6 e)

2nfC 2trQMHz)t0.001 pF)

X ? . = L

= 3 9 . 8 f )

2nfC 2n(2MHz)(0.002 p.F)

Next,

determine he

impedance

of each of the two

parallel

branches.

zt

=

Rr

-

jxct

=

33 a

-

j19.6

A

Zz= Rz-

jXcz=

47 Q

-

j39.8

O

Convert these mpedances o

polar

form.

Zr=.VR'r+xLz-tan- ' ( \1

\ R ' l

=

\,{x

ol +

oe.o

t z-tun-'(7?.6

\

=

86.2t-6j.5"

\ 3 3 o

" " - - " ' "

z,

=

f R', xL z-nn-

t

452\

\ R z l

=

V{47 e)2+

(3ei

af z-tan- ' (}2}9\=6t.6t-40.3. tL

\

4 7 f 2

)

Calculate achbranch uffent.

t'

=

$

=

- ?lo.-]- ^

=

23.2t67.5' A

z1

86.22-67.5"

r,=5

=

- ?lolu-.

=32.5t40.3'ma

z2 61.61-40.3" Q

To

get

the total current, expresseach branch cuffent in rectangular orm so that

they

can be added.

I r

=

8 .89mA+

j2 l .4mA

lz=24.8 mA +721.0mA

The total cuffent is

l r o r=11 12

=

(8.89

mA +

j21.4

mA) +

(24.8

mA +

j21.0

mA)

=

33.7mA +

j42.4

mA


26/64

Converting

I,o,

to

polar

form yields

1.,

=

@

za",(H*)

=

st.z:sr.6.

A

The

current

phasor

diagram

s

shown n

Figure

l6-35.

SERIES-PARALLEL

C

CIRCUITS

623

Circuit

ground

FlcuRE

6-3s

Related

Prohlem

Determine

the

voltages

across

each component

in Fieure

16-34

and sketch

a voltage phasor

diagram.

Measurement

of Zrorand

0

Now, let's

seehow

the values

of 2,,,,

and

0 for the

circuit

in Example

16-10

can be

deter-

mined

by

measurement.

First,

the total

impedance

s measured

as outlined

in

the follow-

ing

steps

and as

llustrated

n

Figure

l6-36

(other

ways

are also possible):

step 1.

Using

a sine

wave generator,

et he

source oltage

o a known

value

(10

v)

and

the frequency

to

5 kHz.

If

your

generator

s

not accurate,

hen it

is advisable

o

check

the voltage

with

an

ac voltmeter

and the

frequency

with a

frequency

counter

rather

than

relying

on

the marked

values

on the

generator

controls.

Frequency

counter

.-

Cz

0.05

F

R2

680 f,)

z , ^ , = v '

I o v

= t . + z t o

I,n,

6.79 mA

FIGURE

6-36

Determining

Z,o,

by measurement

f

V, and lro,.

1r

=

23.2

mA

40.3"

51.6 '


27/64

624

r

RC CIRCUITS

Step

2. Connect an ac ammeter as shown in

Figure 16-36,

and

measure he total cur-

rent.

Step 3. Calculate he

total impedanceby using Ohm's law.

To measure he

phase

angle, an

analog oscilloscope s used n this illustration. The

basic

method of measuring a

phase

angle on an analog oscilloscope was introduced n the

TECH TIP in Chapter 12. We will use that method here in an RC circuit.

To measure he

phase

angle, the source voltage and the total current must be dis-

played

on the screen n the

proper

time

relationship. Two basic types of scope

probes

are

available o measure he

quantities

with an oscilloscope:

he voltage

probe

and the cunent

probe.

Although the cunent

probe

s a convenientdevice,

t is often not as readily available

or

as

practical

to use as a voltage

probe.

For this reason,we will confine our

phase

mea-

surement

echnique to the use of voltage

probes

n conjunction with the oscilloscope.

A

typical oscilloscope voltage

probe

has two

points,

the

probe

tip and the

ground

lead, hat

are connected o the circuit. Thus, all

voltage measurementsmust be referenced o

ground.

Since

only voltage

probes

are to be used, the total cur:rent

cannot

be

measured

directly.

However, or

phase

measurement, he voltage acrossR1 s in

phase

with the total

cuffent and can be used to

establish he

phase

angle.

Before

proceeding

with the

actual

phase

measurement,note that there is a

problem

with displayin1 Vn.

If

the scope

probe

is connected across he resistor, as indicated n

Figure l6-31(a), the

ground

lead of the scope will short

point

B to

ground,

thus bypass-

ing the rest of the components

and effectively removing them from the circuit electrically,

as llustrated in

Figure 16-3'7(b)

(assuming

hat the scope s not isolated from

power

ine

ground).

To avoid this

problem, you

can switch the

generator

output

terminals so that one end

ofRl

is connectedo

ground,

as shown n Figure 16-38(a).

This

connectiondoes

not alter

This

pafi

of

the circuit is

shorted out by ground

connection through scope.

B is to short out the

rest

of

the circuit.

FIGURE 6-37

Unilesirable ffects

of measuring lirect$ acrossa componentwhen he instrumentand

the

circuit ure

grounded.

F-

Ground

(a)

Ground lead on scope

probe grounds point B.

(b)

The effect of

grounding point


28/64

SERIES-PARALLEL

C

CIRCUITS

625

the

circuit electrically

because

Rt still has

the same

series elationship

with

the rest of

the

circuit.

Now the

scopecan

be connected

across t to

display

Va1,as ndicated

n

part

(b)

of

the

figure. The

other

probe

s

connected

across he voltage

source

o display

V" as ndicated.

Now channel

1 of the scope

has vp1

as an input,

and channel 2

has v". The

trigger source

switch on the

scopeshould

be on internal

so that

each race on

the screen

will be triggered

by

one of the inputs

and the

other will

then be shown

n the

proper

time relationship

to it.

Sinceamplitudesare not important, the volts/div settingsare arbitrary.The sec/divsettings

should be adjusted

so that one

half-cycle

of the waveforms

appears

on the screen.

/ar\

f ^ 1

r> r6t rd

\ - / Y Y

C

R

'A-m--litll

R z 3

(a)

Ground epositioned so that

one end of R1 s grounded

FIGURE6-39

Measurement

f the

phase

angle on the

oscilloscope,

(b)

The

scopedisplays

a half-cycle

of Val and I/r.

7p1 ep resents he phase

of the

total current.

Decalibrate

both channels o

make both voltages

appear o have

the same amplitude.

Onehalf-cycle

=

I80'

c2

itioning

ground

so that

a direct voltage

measurement

can be

made with respect

to

without

shorting out

part

of the circuit.

Beforeconnecting

he

probes

o the

circuit n Figure

16-38,

you

must

align he

wo hor-

izontal

lines

(traces)

so hat

they appear

as a single ine

across he center

of the screen.

To do

so,

ground

he probe ips and adjust heverticalposition knobs to move he traces oward the

center ine

ofthe screen

until they are

superimposed.

his

procedure

ensures

hat both wave-

forrns have

he samezero

crossing

so hat an accurate hase

measurement

an

be made.

The

resulting

oscilloscope

isplay s

shown n Figure

16-39.

Since here are 180'

in one

half-cycle,

each of the

ten horizontal

divisions across

he screen

epresents

8o.

Thus, the

horizontal

distance between

the correspondingpoints

of the

two waveforms is

the

phase

angle n degrees

as ndicated.

| ^ |

*

a l

I ^:^ I

L

Each iv is ion

t8 .

A

L

Phase

angle

V/div V/div

s/div

@ 0 0

CHl

CHz


29/64

626

r

RC CIRCUITS

sEcTtoN 6-6

REVIEW

1. What is the equivalent

eriesRC circuit for the series-parallel

ircuit in Figure

l6-33?

2.

What

s

the otal mpedancen

polar

orm of the circuit n

Figure 16-34?

I

Coverage of series-parallel reactive

circuits continues in Chapter 17, Pafi

3, on

page

679.

16_7

.

POWER N RC

CIRCUITS

In a

purely

resistive

ac circuit, all of the energy delivered by the source

is dissipated

in the

form

of heat by the resistance. n a

purely

capacitive

ac circuit, all of the

energy

delivercd by the source is stored by the

capacitor during a

portion

of the voll.

age cycle

qnd

then returned to the source

d.uring another

portion

of the cycle so that

there is no

net conversion to heat.

When

there is

both

resistsnce

and capacitance,

some

of

the

energy is alternately stored snd returned by the

capacitunce and some s

dissipated

by the

resistance.

The emount of energy

converted o heut is determineil by

the relative

values of the resistance and the

capacitive

reactance.

\fter

completing this section,

you

should be able to

I

Determine

power

in RC

circuits

.

Explain true and reactive

power

.

Draw the

power

triangle

.

Define

power

factor

.

Explain apparent

power

.

Calculate

power

in

an

RC

circuit


30/64

POWERN RC

CIRCUITS 627

It is reasonable

o assume hat when the resistance

s

greater

han the capacitive eactance,

more ofthe

total energy delivered

by the source s converted

o heat by the resistance

han

is stored

by the capacitance.Likewise,

when the reactance

s

greater

han the resistance,

more

of the total energy s

stored and returned than is

converted o heat.

The formulas for

power

in a resistor, sometimes

called true

power (P*"),

and the

power

in

a capacitor, called reactive

power (P,),

are estated

here. The unit of

true

power

is the watt, and the unit of reactivepower is the VAR (volt-amperereactive).

Pou.= IzR

Pr=

IzXc

(1

6-30)

(1

-31)

The PowerTriangle

or RC

Circuits

The

generalized

mpedance

phasor

diagram is shown in Figure

1640(a). A

phasor

rela-

tionship

for the

powers

can

also be represented y

a similar diagram because

he respec-

tive magnitudes

of the

powers,

P-" and P,, differ from

R and Xs by a factor

of 12. This

is shown in Figure 16-40(b).

Watts

W)

D

I,

X,

N-

(reactive)

Volt-amperes

reactive

(VAR)

(b)

Power

phasors (c)

Power triangle

a)

mpedance

hasors

RE

t

6-40

of the

power

triangle

for

an RC circuit.

The

resultant

power phasor,

2Z,

rcpresents he

apparent

power,

P". At any nstant

in time Po s

the total

power

that appears

o be transferred

between he source and the RC

circuit. The

unit of apparent

power

is

the volt-ampere,

VA. The expression or apparent

power

s

Po= I2Z

(16-32)

The power phasor diagram in Figure 16-40(b) can be rearranged n the form of a

right

triangle, as shown in

Figure 16-40(c). This is

called the

power

triangle.

Using the

rules of trigonometry, P,,u"

can be expressed

as

Poo"

=

Pocos

d

(1

6-33)

Since Po

equals 2Z or

VI,

the

equation or the true

power

dissipation in an RC cir-

cuit can be

written as

P*"

=

VI cos0

where

V

is

the applied oltage

and

1

s the otal current.

(16-34)


31/64

628

T

RC

CIRCUITS

For the case

of a

purely

resistive

current,

0

=

0o and cos

0o

=

1, so P*" equals

VL

For the case

of a

purely

capacitive

circuit,

0

=

90o and cos

90o

=

0, so P-" is zero. As

you

already know,

there s no

power

dissipation n

an ideal capacitor.

The Power Factor

The term cos 0 is called the power factor and is statedas

P F = c o s 0 (1

-3s)

As the

phase

angle between

applied voltage and

total current increases,

he

power

factor

decreases,ndicating

an increasingly

reactive

circuit. The smaller

the

power

factor,

the

smaller the

power

dissipation.

The

power

factor

can vary from

0

for

a

purely

reactive

circuit to 1 for

a

purely

resis-

tive

circuit. In an RC

circuit, the

power

factor is referred

to as a leading

power

factor

because he current

leads he

voltase.

EXAMPLE 6-12

Determine the

power

factor and the

true

power

in the

circuit of Figure

1641.

FIGURE 6-41

p

C

v,

1 5 V

/

=

10kHz

:

Solution

The capacitiveeactance

s

Xr=

=)-

=

r - ] -

=

3.18 o

2nfC 2n(10

Hz)(0.005

F)

The otal mpedance

f the circuit

n rectangularorm

is

Z

=

R

-

jXc=

1.0 Q

-j3.18

kO

Convertingo

polar

orm

yields

z = f

R 2 x l z - t u n - ' ( * \

\ R /

The angle associatedwith the impedance s 0, the angle between the applied voltage

and the

total current; therefore,

he

power

factor is

The current masnitude

is

PF= cos

0

=

cos(-'72.5')

0.301

t

=L=

, 1 :

y ,

=

4 .50

A

z 3.33 O

The rue

power

s

P^.=

V,I

cos

d

=

(15

VX4.50mAX0.301)

20.3mW

Related

Problem

What

s the

power

actor

flis reduced

y half in Figure 164I?


32/64

POWER

N

RC CIRCUITS

629

The

Significance

f Apparent

Power

As mentioned,

apparent

power

is the

power

that

appears

o be transfered

between

the

source

and the load,

and it

consists

of two components-a

true

power

component

and a

reactive power

component.

In

all electrical

and electronic

systems,

t is the

true

power

that does

the work. The

reactivepower is simply shuttledback and forth between he source and load. Ideally, in

terms

of

performing

useful

work, all

of the

power

transferred

o the load

should

be true

power

and none

of it reactive power.

However,

in most

practical

situations

the load

has

some

reactance

associated

with it, and

therefore you

must

deal with

both

power

compo-

nents.

In Chapter

15, we

discussed

he use of

apparent

ower

in relation

o transformers.

For

any reactive

oad, there

are two

components

of the total

current:

the resistive

compo-

nent

and the reactive

component.

If

you

consider

only the

true

power

(watts)

in

a load,

you

are dealing

with only

a

portion

of the total

current

that the load

demands

from a

source.

n order

to have

a

realistic picture

of the

actual

cuffent that

a load will

draw,

you

must consider

apparent

power (VA).

A

source such

as an ac

generator

can

provide

current to

a load up

to some maxi-

mum value.

If the load

draws more

than

this maximum

value,

the sourcecan

be damaged.

Figure l6-42(a)

shows

a 120

V

generator

hat can

deliver a maximum

current

of

5

A

to a

load. Assume

that the

generator

s rated

at 600 W

and is connected

o a

purely

resistive

load of 24

Q

(power

factor

of 1).

The ammeter

shows

that the

current is

5 A, and the

wattmeter

indicates

that the

power

is 600

W. The

generator

has no problem

under these

conditions,

although

it is operating

at maximum

current

and

power.

Ammeter

indicates

that

current s excessive

Wattmeter

indicates

that power

s

below

rated value.

Z = 1 8 Q

PF

=0.6

\

(a)

Generatoroperating

at its limits

with a

resistive

oad.

(b)

Generator s

in danger

of intemal damage

due to

excess

culaent, even

though the

wattmeter

indicates

that the

power

is

below the maximum

wattage rating.

FICURE

6-42

Wattage ating

of a source s

inappropriate

when the

load is reactive.

The rating

should

be

in

VA

rather

thqn

in v,atts.

Now,

consider what

happens f

the load is

changed

o a reactive

one with an imped-

ance of 18 o

and a

power

factor of

0.6, as indicated

in Figure

1642(b).

The cur:rent

s

120

Y/r8 a

=

6.61 A, which

exceeds

he maximum.

Even though

the wattmeter

reads

480

W

which is

less than

the

power

rating

of the

generator,

the excessive

current

probably

will

cause damage.

This illustration

shows that

a true

power

rating can

be

deceiving and

s inappropriate

or

ac sources.The

ac

generator

should be

rated at

600 VA,

a rating that manufacturersgenerally use, rather than 600 W.

5Amax

600 W max


33/64

630

I

RC

CIRCUITS

EXAMPLE

6-13

For

the circuit

in Fisure

16-43, find

the true power,

the reactive power,

and

the appar-

ent power.

FICURE

6-43

C

0.15

pF

Solution From

Figure

caDacitiveeactance

s

16-43.R

=

4'70 ):

hus

G

=

IlR

=

I /410

Q

=

2.13

mS. The

X c =

1 _

=

1061

2rfC

2r(1000

HzX0.15

F)

The

true

power

is

, _ v , _

l O v

_ . r ,

l o =

R ' = 4 . , 0 A = 2 l . J m A

,

v " 10v

I r = i =

1 0 6 l O = 9 . - l 3 m A

pru.=

IIR

=

(21.3

mA)2(4:.0

)

=

213mW

The reactive ower

s

p,=

ISXc

=

(9.43

mA)2(1061

)

=

94.3mVAR

The

apparent ower

s

P"=f P'ou"a

l=

Related

Problem

What is

the rrue power

in

Figure

1643 if

the frequency

changed

o 2 kHz?

(213

mW)2

+

(94.3

mVAR)2

=

233mVA

sEcTtoN

6-7

REVIEW

16-8

r

BASIC

PPLICATIONS

1. To which

component

n an RC

circuit is

the

power

dissipation

due?

2. The phase

angle,

0,

is

45" What

is the

power

factor?

3. A

certain series

RC circuit

has the

following paramerer

values: R

=

330 {1,

X,

=

460

{r, and

=

2 A. Determine

the

true

power,

the reactive power,

and the

apparent

power.

RC

circuits are

found

in

a variety

of applications,

often

as

part

of a more

complex cir-

cuit. Two

major

applirations,

phase

shift networks

and

frequency-selective

networks

ffilters),

are covered

n this

section.

After

completing

this section,

you

should be

qble

to

I

Discuss

some

basic RC

applications

.

Discuss

and

analyze he

RC lag network

.

Discuss

and

anaTyze

he RC lead

network

.

Discuss how

the RC

circuit operates

as

a filter


34/64

BASICAPPLICATIONS

631

The RC LagNetwork

The RC lag network is

a

phase

shift circuit in which the output voltage ags the input

volt-

age by a specifiedamount. Figure 1644(a) shows a seriesRC circuit with

the output volt-

age taken across he capacitor. The source voltage is the input,

Vi,.

As

you

know,

0, the

phase

angle between he current

and the

input

voltage, s also the

phase

angle between he

resistor voltage and the input voltage becauseVp and are in phasewith each other.

FIGURE 6-44

RC lag network

(Vou,

=

Vd.

(a)

A basic RC lag network

(b)

Phasor voltage diagram

showing the

phase

ag

between Vi, andVou,

Since Vg lags Vp by

90o,

the

phase

angle between the

capacitor voltage and the

input voltage is the difference between

-90o

and

d,

as

shown

in Figure 16-44(b).

The

capacitor voltage is the output, and it lags the input,

thus creating a basic lag network.

When the input and

output voltage waveforms of the lag network are displayed on

an

oscilloscope,a

relationship

similar to that in Figure 16-45 is observed.The amount of

phase

difference, designated

@,

between he input and the output is dependenton the rel-

ative sizes of the capacitive reactanceand the resistance,as is the magnitude of the out-

put

voltage.

@ phase ag)

FIGURE 6-45

Oscillnscopedisplny of the input and output voltage waveformsof

a

lag network

(Vou,

ags

V). The angle shown s arbitrary.

Phase Difference Between

Input and Output As already established,

0

is

the

phase

angle between 1 and V;n. The angle between Vou,and

V;,

is designated

$

(phi)

and is

developed s

ollows.

The polar

expressions or the input voltage and the current are

V;,10o

ard I/-0,

respectively.The output voltage is

Q=

-90"

+

0

(phase

ag)

Y,*,= (IZ0)(Xsl-90') = IXcZ(-9}" + 0)


35/64

632

r

RC

CIRCUITS

The

preceding

equation

states hat the

output voltage

is at an

angle of

-90"

+

0 with

respect o the

input voltage.

Since

0

=

*tarf

'(xclR),

the angle

between he input

and out-

put

is

d

=

_eoo..""_,(f)

(1

6-36)

This

angle s always

negative,

ndicating that

the output voltage

lags the input

voltage,

as

shown n Figure

1646.

FIGURE

6-46

vor,

(a)

EXAMPLE

6-14

FIGURE 6_47

Determine

the

amount

of

phase

ag from

input to output

in each ag network

in Figure

1647.

u,

" f = 1 k H z

(a)

(b)

Solution For

the ag network

n Figure 1647(a),

d

=

-90'*

run-'(*)

=

-90o

+ tan-l/

j9

)

=

-90'+

r8.4"

=

-7r.6o

\ R / \ 1 5 k o /

Theoutput ags he nputby 71.6".

For

the ag network

n Figure 1647(b),

first

determinehe capacitive

eactance.

x r = = J : = ^ ; * 1 -

n = 1 . 5 9 k e

2nfC

2n(1k[1z)0.1 pr)

d

=

-90'

,un-'1*)

=

-e0o

,un-'/

'1?

o \

=

-23.2"

\ R / \

6 8 0 C 2

/

The

output lags the

inputby 23.2'.

Related

Prohlem In

a lag network, what

happens

o the

phase

ag

if the frequency

increases?


36/64

BASIC PPLICATIONS

633

Magnitude

of

the Output Voltage

To

evaluate he

output voltage

in terms

of its mag-

nitude,

visualize

the RC lag network

as

a voltage divider.

A

portion

of the

total input

volt-

age

s

dropped

across he resistor

and

a

portion

across he capacitor.

Because

he

output

voltage

is the voltage

across

he capacitor, t

can be

calculated

as

v ,= ( - \ v , ,

o ' 1 - \ f * * x Z ) ' "

Or it

can be calculated

sing

Ohm's law

as

Vory

=

IXs

The

total

phasor

expression or

the output

voltage

of an RC lag

network is

('t6-37)

(1

6-38)

Your= VourZQ

(1

6-39)

The

RC Lead

Network

The RC

lead network

s a

phase

shift circuit in

which the output

voltage eads

he nput

volt-

ageby a specified

amount.

When the output

of a seriesRC

circuit is

taken across he resistor

rather

han

across he capacitor,

as shown n

Figure r6a9@),

it becomes

a ead network.

EXAMPLE

16-15

For

the lag network

in Figure

1641(b)

(Example

16-14),

determine

the output

volt-

age n phasor

orm

when the input

voltage

has an rms

value of 10

V. Sketch

the input

and

output voltage

waveforms

showing

the

proper phase

relationship.

X6

(1.59

kO)

and

g

(-23.2o)

were found in Example

16-14.

Solution The

output voltage

in

phasor

orm is

/ x - \

You,=

--+= lv,,lE

\ Y R ' + X I /

_ /

l . s e k o

\ , ^ . ,

(m

)t0t-23.2'

Y

=

9.20/-23.2o

rms

The

waveforms

are shown n Fisure

16-48.

FICURE

6-48

V

Related

Prohlem In

a lag network,

what happens

o

the output voltage

if the fre-

quency

ncreases?

Vurt=9'20Y

rm s


37/64

634

r

RC CIRCUITS

(a)

A basic RC lead network

(b)

Phasor

voltage diagram showing

the

phase

ead between I{, and Vou,

@

phase

ead)

(c )

FIGURE

6-49

RC lead network

(Vo,t

=

Vd.

Phase Difference Between Input and Output

In a seriesRC circuit, the current

eads

the

input voltage. Also, as

you

know, the resistor voltage is in

phase

with the curent.

Since the output voltage

is

taken across

he resistor, the output leads the input, as ndi-

cated by the

phasor

diagram

in Figure 1649(b). A typical oscilloscope display of the

waveforms s shown n Figure 1649(c).

As in the lag network, the amount of

phase

difference between he

input

and

output

and also the magnitude of the output voltage in the lead

network is

dependenton

the rel-

ative values of the resistance and the capacitive reactance.When the input voltage s

assigneda reference angle of 0', the angle of the output

voltage is

the same

as

d

(the

angle between otal current and applied voltage) because he

resistor voltage

(output)

and

the current are in

phase

with each other. Therefore, since

Q

=

0

in

this case, he expres-

sion

s

(1

6-40)

This

angle

is

positive

because

he output leads the input. The following example llus-

trates he computation of

phase

angles

or lead networks.

. - ' lX . \

d = t a n ' l - l

\ R i


38/64

EXAMPLE

16-16

Calculate

he output

phase

angle or

eachcircuit in Figure

16-50.

vt,

500Hz

(a) (b)

FICURE 6-50

Solution For the lead network

in Figure

16-50(a),

b

=

tan-'(* l

=

tun-' l***)

=:+.: .

\ R / \ z z u s z l

The

output eads he input

by 34.3'.

For the lead network

in Figure 16-50(b),

first

determine he

capacitive eactance.

X c = - L = # = l - 4 5 k f )

2rc[C

2n(500Hztt0.22pFl

e

=

tan-'

\\

=,un-'/

"'o^s,9

)

=

5s.4"

\ R / \

t . 0 k o /

The output eads

he input by

55.4'.

Related Problem

In a lead network,

what happens

o the

phase

ead if the frequency

increases?

BASIC PPLICATIONS

635

Magnitude

of the

Output Voltage Since the

output voltage

of an RC lead network

is

taken across

he resistor, he magnitude

can

be calculatedusing

either the

voltage-divider

formula

or Ohm's law.

r /

- /

R

\ , ,

Y o a l - \ . V R \ x i ) " ' '

Vo6

=

IR

The

expression or

the output voltage in phasor

form is

('t6-4"t)

(16-42)

Y

or,

=

VourZ$

(1

6-43)


39/64

636

T

RC

CIRCUITS

EXAMPLE6_17

The input

voltage in Figure 16-50(b)

(Example

16-16) has

an

rms

value of 10 V.

Determine

the

phasor

expression or

the output voltage.

Sketch he waveform relation-

ships for

the

input

and output voltages

showing

peak

values. The phase

angle

(55.4")

andXs

(1.45

kO)

were found in Example 16-16.

Solution The phasorexpression or the output voltage is

r / |

R

\ , , / t . O k o \ ,

\ o u , = l

/ . . l V i , l Q = l _ | 1 0 2 5 5 . 4 ' V

= 5 . 6 8 1 5 5 . 4 o V r m s

\ Y

n ' + X l l

\

t ' l o K l 2 /

The

peak

value

of the input voltage is

Vi,@) l.4l4vin(*,)

=

|

.414(10 )

=

14. 4

V

The

peak

value of

the output voltage is

Vo,t(p) l .4l4Vout(*,)

=

|

.414(5.68 )

=

8.03V

The

waveforms are shown n Fieure 16-51.

FIGURE 6-51

Related Problem

In

a

lead network,

what happens

o the output voltage if

the

fre-

quency

s reduced?

TheRC

Circuitasa Filter

Filters

are

frequency-selective

ircuits

that

permit

signals of certain frequencies

o

pass

from the input

to the output while

blocking all others. That is,

all frequencies but the

selected

ones are filtered out. Filters

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PoEC 16 RC Circuits

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