DOC UM EN T R ES UM E

ED 021 739 24 SE 004 642

By- Balabanian, Norman Root, Augustin A.CAPACITORS AND RESISTANCE-CAPACITANCE NETWORKS.Syracuse Univ., N.Y. Dept. of Electrical Engineering.Spons Agency- Office of Education (DHEW), Washington D.C. Bureau of ResearchReport No- 5Bureau No-BR-5-0796Pub Date 64Contract- OEC- 4- 10- 102Note- 162p.EDRS Price MF-$0.75 HC-$6,56Descrip_tors-*COLLEGE SCIENCE ELECTRICITY, *ELECTRONICS *ENGINEERING EDUCATION, *INSTRUCTIONAL

MATRIALS PHYSICAL SCIENCES *PROGRAMED INSTRUCTION, UNDERGRADUATE STUDY

Identifiers-Syracuse University, United States Office of EducationThis programed textbook was developed under a contract with the United States

Office of Education as Number 5 in a series of materials for use in an electricalengineering sequence. It is divided into three par ts--(1) capacitors, (2)voltage-current relationships, and (3) simple resistance-capacitance networks. (DH)

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CAPAC ITORS AND

RESISTANCE -CAPAC ITANCE NETWORKS

by

Norman Balabanian

Augustin A. Root

Electrical Engineering Department

Syracuse University

C ontract No

OE 4-10-2.02

U.S

.Off ice of Education

Copyright 1964 ©

nd

2Edition

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Table of Contents

Page

Introduction

1

Section 1

Capacitors

3

Section 2 Voltage-Current Relationships

11

Section 3

Simple R-C Networks

41

a

t

CAEACITORS AND RC NETWORKS

This unit of programmed text is concerned witha determination of voltages or

currents (network variables) ina network containing resistors, sources and capac-

itors.

What we want to do is to be able to apply Kvl, Kcl

and variations of Ohm's

law to networks like this:

I.Such networks are very much like the

ones with which we have been working; except

for branches containing capacitors.

As soon as we can write expressions for thecurrents and voltages associated

with capacitors, we

can treat this network in the same way that we handled purely

resistive networks.

When you finish this

program you will be able:

2 1.

to determine

the values ofvoltage and current

in network branches

containing

capacitors when thevoltages andcurrents in other

appropriate parts of

the

network are knownat a

specillictime, by applying

the basic

relationships:

Kv11 Kcl andthe v-i

relationships'of thebranch.

2.

to do number1, above, whenthe known

currents and. voltages

are functions of

time.

3.

to write the

differential

epaations that result

from an applicationof the basic

relationships to

a network of moderate

complexity which

contains resistors,

capacitors and

sources.

4.

to solve theabove differential

eqUations when the

network contains

one capac-

itor, one

source and. oneor more resistors.

The voltage

or current sourcemay

be (a)

constant, (b)

an exponential function, (c)

a polynomial in time,

or (d)

a sinusoid.

5.

to describe,

mathematically andgraphically, the

response of single RC

networks

to switching

changes of current

or voltage.

1.

Capacitors

A capacitor is nothing more than a device

consisting of two conducting bodies

which carry equal and opposite electrical

charges.

The tuning capacitor used ina

radio for selecting stations, and shown inFig. 1, is a familiar example of a

capacitor.

Considered as a whole, the net charge on the whole

capacitor equals

Fig. 1

Answer:

zero.

(h charge on one

of.two bodies, orplates, is

always balancedby an equal and"opposite"

charge on the otherplate, i.e.,

their

algebraic sum is

always zero.)

Even though there is no net charge on a

capacitor, the fact that there is a

posItive charge on one part of the

capacitor and a negative charge on the other,

and these two parts are separatedfrom each other, means that therewill be an

electric field between the two parts andtherefore a potential difference, or

voltage across the capacitor.

Let q refer to the positive charge on one

part of the capacitor and let v be

the voltage between the two parts, withreference plus at the positively charged

part.

We call q the charge on the capacitor, or

the capacitor charge, even though

it is the charge on only one part.

It is empirically found that these

two quan-

tities

,the voltage and the charge

are proportional to

each other.

Hence, we

can write

q = Cv

where C is the constant of proportionality;

it is called the capacitance.

As

evident from this relationship, capacitancehas the units of

Answer:

'capacitance has the units of coulombs/mkt.

7

In honor of Michael Faraday, the English scientist, this unit

(coulombs/

volt) is called the farad.

The voltage across the two parts of a capacitor is 10 volts and on one of

.

the parts there are a trillion, 10121 excess

electrons.

Find the capacitance

in microfarads, abbreviated gf.

(The mass of an electron is 9 x

l0-31

kg and

its charge is -1.6 x

10-19cou1ombs.)

Answer:

; ...6.

14.1

.1

1012

0t.6

x10')

f arad s='

'10

(Nbte that q is the positivecharge w4c4

is equal,

bu opposite,

tO the negatiVe

charge of

theelectrOili:)

,

v

Remark In general, a capacitor maybe considered to be nytwo conducting bodies

0

carrying equal and opposite

charges, as shown in Fig.

2(a).

In almost any elec-

trical device one can thinkof, there are conductingparts which carry charges

and are at different

potentials.

These parts, then, canbe considered tO be

capacitors and the

influence of their capacitances shouldbe taken into account in

analyzing the characteristics

of the electrical device.

Sometimes an engineerwill

take considerable pains to eliminate

"stray" capacitance effects.

Other times,

the engineer will deliberatelyintroduce capacitors into a

device or system in

order to obtain desired results.

As you learn about the

functions which capacitors

perform, you will beable

to analyze the performance

of networks involvingcapacitors, and design

circuits to

obtain the characteristics you want.

Just as resistors canbe manufactured in many

sizes (not just Physical sizes

but values of R), so also

capacitors can be manu-

factured in many sizes, coveringthe range of values

from about 100micromicrofarads

(10-12gf) to 100 microfarads

(100 p.f), and in many sizes and

shaped..

(You will

learn more about the evaluationof the capacitance of

capaciioks having different

geometries when you studyelectric fields, later.)

IOne Of the most practical

shapes is the parallel-plate

capacitor, Fig.

2(b),

in which the two conducting parts

are plane surfaces separatedby a small distance.

'(The radio tuning capacitor

has such a structure with anumber of parallel plates.)

.In an effort to

increase the area of the plates,

which is a factor in determining

;the capacitanceiralue, many capacitors aremade of two sheets of conductingfoil

separated by an insulatingsheet, the whole thingbeing rolled into a compact

cylindrical shape.

Fig. 2(a)

external

network

with

sources

,.

i$

.....40,

/,

/

+ charges/

ii

I/

4....

1- charges //

.--

... a

ro.

.111

010w

OW

.

PARALLEL PLATE

CAPACITOR

Fig. 2(b)

11

2.

Voltage-Current Relationships in Capacitors

Consider the situation in Fig. 2(h) where two conducting bodies are connected

to a network containing sources with time-varying voltages or currents.

The charges

on the two "plates" will vary with time.

Charge will be conducted across from one

plate to the other (assuming an ideal nonconductor between them).

If we don't concern ourselves with anything internal to the dashedlines in

the diagram and look at the device only from its terminals, it will appear as

if

there is a current entering at one terminal and being conducted through to

the

other terminal.

There will also appear a voltage across these terminals.

The symbol for a capacitor is shawn in

Fig. 3.

A single current is shown as if it

symbol for

a capacitor

were being conducted straight through.

Remembering that q = Cv for the capacitor,

write an expression relating the current i

to the voltage.

i =

ww

w

Fig.

3

12

Answer:

=67

17(since i =

dt)

laN

(Tb emphasize that these

are functions of time,

we sometimes write

dv(t)

i(t)

dt

").t

rig.

,bt

13

This expression, i = C

cgt.

is the current-voltage relationship

for a

capacitor.

In importance, it ranks wlth Ohm's lawwhich gives the analogous

ralationship for a resistor.

Figure 4 shows some possible capacitor

voltage waveforms.

Sketch on

the axes below each waveform, the

corresponding current waveforms andlabel

them appropriately.

IA Ans

wer

:

15

Suppose the voltage v on a 2 tlf capacitor is, successively, each ofthe

functions given below.

Write the expressions for the corresponding currents,

with the following references:

-

2p. f

(a)

v1(t) = 100

e-t/5

volts; il(t) =

(b)

v2(t) =

45 volts; i2(t) =

ma.

(c)

v3(t) = 10 sin 1000t volts;

i3(t)

ma.

1

11111111111111111111111111111L

9_0TX Ot " mg

..../01.

C-eta

000T soo 03 = (;) T

(o)

0 ti (1-)°T

(q)

a OOT (T

OT x

"Ma

0 tO

-(-)T1

(V)

_:aattsuv

9T

.0111111.'

17

If the current through a capacitor is.proportional to the derivative

of

the voltage across it, then,

conversely, the voltage is

" ""

''

s,,

1"-'"

18

Answer:

proportional to the Lamm],

(or antiderivative) Of the

current.

imiti

mm

illiN

INIM

III/1

1111

1111

1111

19

Specifically, the expression

for the capacitor voltage in terns

of its

current is

20

Answer:

v(1.) =

fi(t)dt

K (a constant)

(You may have left out the constant.

If you did, go back and add it; it

is essential.)

2.

21

Consider a 1 pf capacitance with the following voltage and

current references:

e-171(--_.

.Suppose there is initially a voltage of + 100volts across the capac-

itor (having the same polarity as the

reference).

Then, a constant current of 0.1 ma

is caused to flow in the reference direction.

(You might think of an ideal constant

current source being connected to the capacitor to

make this happen.)

Find the value of the voltage after 0.2 second,

assuming the current started

at t = 0.

(a)

v =

Also, find the value of the voltage after 0.2 second,

assuming the polarity of

the initial voltage is reversed, everything else being the same.

(b)

v =

T

22

Answer:

(a)

v = 120 volts

(b)

v = -80 volts

This is obtained as follows:

"IS

.4,

a,V

=1 T.

vdt

K =106 f10-4

dt

K = 100t

K

To find K, note that when t = 0, v = 100; hence, K = 100 and so

v = 100t

100.

Putting t = .2 here leads to the result.

If the polarity of the initial voltage is reversed, then when

evaluating K we would say:

when t = 0, v

- 100; hence,

K = - 100

and

v =

100t - 100.

Putting t = .2 leads to the

result.

i.,23

Finally, suppose the current in a

.5 tlf capacitor has the waveform

given below, starting at time t . 0.

Assuming the same references as

before,

find the expression for the correspondingvoltage as a function of time.

i(t)

=1

e-2t

ma.

(Before the current starts, the capacitor

10

voltage is 50 volts.)

v(t).

volts.

Answer:

1

(a)

v(t) = 150, - 100

e-2t

volts.

.,

The solia;:don goes like thiz:

V =

1

.5 x 10-6

= 2 x 103

e-2t(10-3)dt

at t = 0, v = 50 volts

*

50 = -100 + K ; K = 150

Therefore, v = 150 - 100 e.-2t

volts.

+ K = - 100 e-2t

+ K

25

Another, slightly different

form, used in expressing the capacitorvoltage

in terms of the current, canbe found by integratingboth sides of the expression

dy

1.

cTE=

1(t)

from some time t1-,

considered to be a starting time,

to an arbitrary time t.

The resulting expression as an

indicated integration will be:

.

26

Answer:

it

dt

1

%Ay

1ft:

i(t)dt

t

Remark:

Note that the uRRELLUELL of integration

is t and so is the variable of

integration -- what is usually called the

*dummy" variable.

It is-a dumpy

because it disappears upon integration and substitutionof limits.

TO illustrates

(8t+6t2)dt=

4t2-1.2t31ab

4b2+20-4a2-2a3.

In the final result the variable t,

which was the dummy, does not appear.

Thus, instead of t any other symbol can be

used. as a dummy variable.

This is especially appealing if the same

symbol appears

in the limits of integration.

/a such eases, another symbol is

sometimes used for

the dumogy variable to avoid confusion.

Thus, the above result could be written

/1t

dV

c/X1 d'Edx

ti(x)dx Where X replaces t as the dummy variable.

However, a

different symbol does not carry the connotation of"time.

Hence, at the risk of

sone confusion, we will continue to use

t as the dummy variable even if t appears

in the limits of integration.

The integration withrespect to t on theleft-hand side can

an integrationwith respect to v by noting

that (dv/dt) dt = dv,

integration is now in terms

of v, the limits ofintegration must

The upper limit on

the left will thenbe what v is at time tand

will be Write the resulting

expression.

27

be changed to

Since the

also be changed..

the lower limit

28 Ans

wer

:

e;I

t,

OW

,,

4t, +

,fl

29

The left-hand side can now be integrated and so the expression for the

capacitor voltage in terms of the currentbecomes:

v(t)

1

Answer:

value of the capacitor

voltage at the initial

time tl.

t

Vote that i i(t)dt is simply the charge conducted to the capacitor by the

t

31

current during

1the time interval from t

to t.

Thus, the expression for the

1voltage can be interpreted in words as follows:

The capacitor voltage at any time

after the initiation of current through it equals ih4 sum 3f two quantities:

(1)

the vOltage resulting from the flow of charge over an interval of time plus (2) the

value of the capacitor voltage at the time of initiation of the current.

The expression

t i(t)dt

is a definite integral in contrast to the indefinite

integral previousiy obiained relating v and i.

Write both expressions.

v(t)

v(t)

32

Answer:

v(t) = 26-* fi(t)dt

K

1f

v(t) = 7,

rt

ikt)dt

v(t1)

(Examine these expressions carefully and note

their similarities and differences.)

,

33

Very often the time at which the current is initiated is conveniently taken

to be zero time.

Write the appropriate expression for the capacitor voltage in

the definite integral form for this case.

Ans

wer

:

v(t)

ft i(

.6)d

.t +

v(0

).C

0

,

35

Suppose a alf capacitor has an initial charge of 0.1 millicoulomb when at

n0

timeta0acurrentimoe-t/i

microamps. is caused to flow.

Find an expres-

sion for the capacitor voltage at any time thereafter.

v( t )=

Also state the value of the capacitor voltage after the passage of a very

long time.

36

Answer:

v(6)

as

=40et/1°VOUS

90 volts

t-+.

Solution

V =

.0001

= 50 volts at t=0

2x10

Jrdt

+ K

1.

0-t

=8x10

e15

dt +-K

1-6

2x10

f(Verify by inserting.t

ss 0 that the expression for

v(t) does reduce to the value of the capacitor

voltage at the initial time.)

= -

40 e-t/10

4,e

e

To summarize:

a new devide has been introduced calleda

37

It is characterized bya parameter C called the

and measured in

ti

J8'

An"

swer

:

cb.p

acit6

.nce

,.

faiit

tab

(91.

illic

rofa

radS

)

39

The charge on one plate is

proportional to the voltage across

the capacitor.

From this, it follows that

the voltage-current relationship

for the following

:777t

references (

) is either

-7.71.71

or

i ( t )

v ( t )

,v1,

'q.v

.'''.

,'`n.

..

lto Answer:

or

i(t) = C

alv(t) i=

-26.

fi d

t + K

t

v(t) pi

2c--

f, i

(tA

t + v

(o).

`1.

" (Now,

you are equipped.towrite the

current and voltage relationships

in simple circuits.).

I !

3.

Simple RC Networks

You are now prepared to analyze the characteristics of

networks which include

capacitors, as well as sources and resistors.

In such networks there is only one

new factor which was not present when you analyzed

resistive nebworks, namely the

v-i relationship of the capacitor.

All other principles still apply

Indeed,

Kirchhoff's two laws do not depend on what the branches are made up of, but only

on the existence of branches forming closed paths and

nodes.

Hence, they will

apply in networks containing capacitors, among other things.

Consider the network in Fig. 5.

Write Kcl at the upper node (a) in terms of

iR and ic.

.001

1.

11.2

Ans

wer

:

igiR

C

ig=

10

cos

200t

Fig.

5

MID

-

Suppose that at t = 0 we know

the current

ic

is to be 6 amps.

Find the

value of v at this time.

V =

44 Answer:

v = 4.O volts.

(At t.0 and

i,(0).6 amps.)

(This is obtained by findingthe value of

iR

from the previous Kcl

equation after

putting t = 0 in 10 cos

200t4, then

using Ohm's

law.)

10 cos 200 t

Fig.

5(Repeated)

i. 10 cos

200t

Fig. 5

(repew,,ed)

,

......

..sra

mor

;3

Next, give the rate at which the voltage v is changing with time.

That

is, write the value of dv/dt at this time.

dv

dt

at t = 0

.

.Answer:

L....6._

ts sec

at t

0.

1

dt (This follows simply from the.v-i

relationship of the

capacitor.)

i= 10 cos 200t

Fig. 5

.41

47

The pattern that appears in the preceding

development is similar to that which

was used inresistiire circuits in which the basic laws were appliedalternately in

computing voltages

Lnd currents.

Here, also, the same pattern is

followed with the

added factor that in addition to Kell KVl and Ohm's

law, there is now avelilable the

v-i relationship of the capacitor as well:

i = C dv/dt.

When dealingwkth

resistive networks, the relationships involved areall

algebraic.

But sin e 'the v-i relationship of a capacitor

involves a derivative,

when carrying out the analogous procedure in

networks containing capacitors, it may

sometimes be necessary to differentiate in order to

arrive at solutions.

As an illustration, let's again consider Fig.

5.

This time suppose that di

/dt

is specified to be

103 amps/sec. at t

= 0.

It is required to find the value

of i

at this time.

As before, applying Kel gives:

ig(t) = iR(t)

i (t).

(The func-

tional notation is deliberately introduced to emphasize

that this is valid at any

time.

But di /dt is the specified quantity.

How can di

/dt be obtained from this

Kcl equation.

Siate what ta do and do it.

48

Answer:

Differentiate

the

1C1

equation.

di

a(0

diR(t)

alc(t)

."1""""""'"".".

dt

dt

+1

dt

1 I I i

Now, at t = 0, di /dt has been given numerically as 103amps/sec.

Further-

more, i

is given as 10 cos 200t,. and its derivative can be found.

With this

information

diR can be evaluated at t = 0.

It will be:

dt

at t = O.

Ans

wer

:

diR

=103amps./sec. at t = 0.

dt

Remember that what we reallywant is the value

of i

at t m 0.

What we have

obtained is dydt. But iR

is proportional to v byOhm's law, and

is related

ip

to v.

Using these facts,

ic

at t = 0 is foundto be

icat t = O.

52 Ans

wer

:1

a -

.02

amps

. ago

t0.

(ic

= c

av/

at =

c a

(RiR

)/at

= R

Ccu

R/a

t)

v=

100

cos

100r

t/3

Fig.

6

go.a

*,00

/14,

1.1.

.*.^

.

53

To illustrate further

the alternate use of thebasic relationships

(Kcl, Kvl

and the v-i relationships

for resistors andcapacitors), consider the network shown

in Fig. 6.

When t = 10 milliseconds,

it is found that the capacitor

voltage is

v = 30 volts

and its rate of change is

dv/dt = - 107

volts/sec.

It is desired to

find the value of C for which

these values will hold.

Note that from the v-i

relationship i = C

dv/dt, C can be found if both i

and dv/dt are known.

The latter is given

numerically.

An expression for i, valid

for any time, can be obtained

in terms of other branch

currents by

When Ohm's law is subsequentlyused to express these

currents in terms of appropriate

branch voltages, this p_allearlfor i which 1 Ads at any t

becomes:

54 Answer:

applying Kcl.

DATA

At t = 10 milliseconds,

v --v

200

100

v = 30 volts

v= 100 cos 100gt/3

g

100 cos 100g.eLLZ. ...ILA .

dv

(or i

= -107

volts/sec.

200

-100i

dt

55

Inserting numerical values, this becomes i =

amps.

(iou

should get i =

amps.)

Finally, the value of C becomes:

56 Answer:

C0.02 4.

So1ution:

100cosit./

30

200

100

. .10 - .30

-.20 amps.

.0 = (.2(101)

C = .02 x 10

.02 pf

20K

Fig.

7

ilMa.

1...0

,110

MM

0

57

Up to this point emphasis has

been on numerical values at a

given point in time.

However, the same principles apply

to general solutions as

functions of time.

Thus,

in Fig. 7 suppose that current

i is specified to be

1-t/10

)i(t)

= 7.5.05(1

- e

and it is desired to find v as a

function of time.

As a first step vg can beobtained i terms of the

voltages across the two

resistors by XVI, and these, in turn, canbe expressed in terms of

the currents by

Ohm's law.

The result is

58 .

Ans

wer

:v

22 2

0100

0 (i

a+i)

g20

K

Fig

7

59

But iI

can be expressed in

terms of i and the capacitor

current,loy Kel.

The capacitor current, in turn, canbe expressed in terms of thevoltage v by

means of the v-i

relationship of the capacitor.

Finally, v is related to

i

by Ohm's law.

As a result of all these,

the previous expression

for v

can be

written:

Vg=

(This should be in. terms of' ionly.)

6o Answer:

12.11) volts.

=(i

-4.

dt

Reminder

1

To

-t/10)

i(t) =

iT)

..

Fina

lly, i

nser

ting

the

spec

ifie

d.va

lue

of i(

t) le

ads

to

V11

2

g

62.

Answer:

vg

Iwo (1

.99e-t/10)

signal

source

Iilter or

coupling

etworx or

amplifier

Fig. 8(a)

receiving

device

i4111

11.

=D

~alp

Ico...L.IANN

I

Ii-4,0

signal

__...4.1

i.:.

1

:

output

signal

input

IT II

I

C

0..0

JFig. 8(b)

'63

Up to this point, in consideringnetworks in_which capacitors appear,

it has

been assumed that a voltage or current in a

capacitor or resistor is known

either

at some instant of time or as a

function of time.

However, the more usual

problem

is to determine how the voltages

,nd currents In such a network varywith time after

some initial instant atwhich a source is switched on or

off.

In such a case some

additional factors eater into the

problem.

We shall next devote attentionto this

class of problems.

The situation illustrated in Fig.

8 is renresentative of a large

number of

applications.

A signal mhich starts at some

time is to be transmitted

through some

kind of transmitting network to areceiving device of some

kind.

It is necessary

to know how the signal

(which is simply

current or a voltage) is

modified by the

transmitting system.

A particularly simple case is

shown in Fig.

8(b).

In the following pages, we

shall analyze the characteristics

of this circuit in considerable

detail.

In order

to consider a number of

possible variations all at the same

time

we shall deal

withl

the situation illustrated in Fig.

9.

..

Vp11111R

ir .6111111MO

MO

MO

ovaaltaa,

6 *2-ra

IT

2A

:.

, .

fr9

asigismil

65

Assume that the

capacitor CI in Fig.

9, has been chargedat some time in the

past to a voltageVo At a particular

time, which canbe designated t =0, the

switch is closed,

thereby applying the

voltage source v ,

which is the source

of

the signal, to theR and C network.

(It is assumed that v

is known.)

It is

desired to find

expressions for the

capacitor voltage v

and the current

i as a

function of time

following the closingof the switch.

As a first step,

write Kvl aroundthe loop, at the same

time eliminating the

resistor voltage

by Ohm's law.

66

Answer :

v = Ri + v

g (whe

re th

e ca

paci

tor

vis

som

e fu

nctio

nof

'the

.in

itial

voltage V0

and.

the

source voltage ' vg

. )

67

In this expressionboth v and. i appear

and, both areunknowns.

Either one

of them can be

eliminated by using a

relationship

(W., Kbl, Ohm's law and

the

v-i relationshipof the

capacitor) that has not yetbeen used in arrivingat

this equation.

State therelationship whichshould now be used.

IPK

+',Z

4Ns.

68

Answer:

The v-i relationship of the capacitor.

69

In writing an equation for this v-i relationship, we

have a choice.

In the

expression v

= Ri + v, i and. v are the capacitor

current and. voltage.

We can insert an appropriate expression eitherfor i or for v, and thus

obtain two different equations.

One of these expressions would lead to a solution

more easily than the other.

Write out both of these, equations.

a)

V=

b)

v=

/

71

Let us temporarily concentrate on

the first of these:

dv

RC -- + v = v

dt

Here we find something new;

this is not a simple algebraic

eqUation.

Not only

does the unknown, v, appear inthe equation, its derivative appears

also.

An

equation like this in which one or more

derivatives of an unknown appear

is called

a differential

equation.

In the list below check

those equations which are

differential equations)

just to be sure that the above

definition of a differential

equation is clear.

a)

2y + 5rt

b)

4y + 3x +

z-dc (5sin x) = 0

c)

2v + f

e-2tdt

=6i

d)

a+

= ci

dt2

72 Answer:

a) and d) are

differential

equations.

b) is not a

differential equation,

even

though an indicated

differentiation appears,because there is no

derivative

of an unknown,

i.e., performingthe

indi

cate

d op

erat

ions

would not leadto an expression

containing

derivatives.

73

Note that in one

of the equations

in the last

frame, there WRS afirst

derivative and in the

other there was a

second derivative.

In general,

derivatives

of any order mayavoear.

We define theorder of the

differential equation as

the order of thehighest

derivative appearingin the equation.

Thus, if thehighest derivativethat appears

is the thirdderivative, the

differential equation

is of thirdorder.

The follow-

ing differential

equation:

d2v

dv

5+

23 - 2v = 0

dt

dt

is of

order.

Ans

wer

:

seco

nd

Fig.

10

75

We are going to be

concerned with such

differential equationswhich arise when

the basic electrical

laws are applied toRC networks.

To gain same experience

in

writing the equations

from a given network,

let's consider a few more

cases.

In Fig. 10 is shown anetwork in which a switch

is closed at aparticular

time taken to be t =

0.

It is desired to find a

single equation

containing only

the capacitor voltage as a

variable, like the one

found for the previous

network.

Redraw the network after

the switch is closed.

State three different

procedures

by which the desired

equation can be obtained.

There are several ways

that you could attackthis probleml to

obtain an

expression for the capacitor

voltage.

You have used these

different methods many

times.

State three procedures

by which the desired

equation can be

obtained.

1.

2.

3.

76 Answer:

1.

Convert the current source and. R into avoltage source equivalent,

thus arrivingat the same structure as

the previous example.

2.

Write a nil equation around the loop

formed by R and C.

3.

Write a node equation.

77

Let's consider the last two possibilities

in some more detail, taking

first

the loop equation.

Let i be the capacitor currentwith such a reference that

Ls= + C dv/dt.

1)

Write the loop equation and. useOhm's law to replace the resistor voltage.

(Watch out for the resistor current!)

2)

Ube an appropriate relationshipto substitute for i in

terms of v.

78

Answ

er:

or.

Fig.

9.1iepeated)

79

This expression is much like the one

obtained for the previous network

of Fig.

9

dv

RC

+ v =

vdt

The only difference is the appearance

of Ri

instead of v

.

Now let's write a node equation

for Fig. 10.

(When writing the algebraic

sum of the currents

leaving a node, express the branch

currents in terms of the

node voltages.)

The node equation is

Answer:

or

dv

vC

+= ig

dv

RC

-E-t+ v

sa R

i g

(You should feel good that you got the same

answer by unin,

twb 3.1 ':::. rent approaches

to the problem. )

0kg. 11

81

As a final example, let's

consider the network in Fig.

11.

(NUmerical values

are used for the

resistances to avoid undue

algebraic complications../ At t = 0,

the switch is closed and it is

desired to find an equation for

the voltage across

the capacitor.

One possible method of procedure

is to replace everything but

the

capacitor by a Thsvenin equivalent

and thereby arrive at a series

R and CI just

like the first network we considered.

(Do this as a check on what you will do

in

the following.)

Two other procedures are:

a)

b)

82 Answer:

a)

Write a set of loop equations.

b)

*Write a set of node equations.

(The Thevinin equivalent circuit would look

like this:

3

dv

320C+v=i-v

dt

g

C)

Fig..11

(after switch is Closed)

83

If node 0 is chosen as a datum andnode equations are written at nodes a

(node-to-datum voltage = va) and b

node-to-datum voltage =v), the result will be:

node a:

va-v6

,

va

,

va -v

20

5-

-r

51v

-iv

=v

a1

g--

or

v-va

dv

node b:

C= 0

dt

dv

vv + 5.0

0a

dt

This is a set of two equations in

two unknowns, va and v.

Since we are interested

in v we eliminate vaby solving for va

from the second equation and

substituting into

the first.

The result of these steps is:.

(5 C

v)

-v =

vg

or

20 u

v =

vdt

4g

dv

2

Note the similarity between this

equation and the corresponding ones

for the previous

networks.

The final possibility is towrite a set of loop equations

and to carry out a

procedure similar to the one just

completed.

(You'll have to eliminate the capacitor

current appropriately.

Carry out this procedure;

let's see if you can arrive at

the

.

corpo

.r.c.m113+

84

Answer:

Ilrst choose loop currents

(see'diagram).

Then write 'Oil equations

around the two

*windows'.

20i1 + 60(il-i)

v8oi1

-6oi = v

ssow

swas

410.ii

-60(i1 -i) + 5i + v as o

-60i1

+ 65i + v sa o

2.

ien solve fori1 gram say, the second

equat on, and substitute

into the

first:

814

i-60

or

20i+vuiv

3.

Finally, replace i byC dv/dt and get

20

dtdv

(The last two steps can beinterchanged.)

85

What you might suspect

from these examples

is that in a network

having a

single capacitor

and source, but anynumber of resistors,

the equation for

the capacitor voltage

is a differential equation

of first order.

And. your

suspicions

would be right.

Thus, by a

detailed study of thenetwork of Fig. 91

we will at

the same time be coveringa large

number r)f other cases as

well.

41 lb

Kvl;

Ri + v= v

e *CD

dv

+C

onor

OP

v = v

dt

Fig. 9

86

imir

ise1

1111

1111

1111

1111

1111

10i

Let's now return to the equation

for the capacitorvoltage in Fig.

9.

dv

RC -c-ct + v

v

In order to carry out a solution,

it will be necessary to

knaw specifically

what the source voltage v

is.

Let us first assume that the source

is a battery so that v

is a

constant:

v= V1

so that the equationbecomes

/

dv

RC

+ v = V

dt

1

in order-to solve for v, this equation

should somehow be

integrated.

.4

87

Two students discussed thisproblem, each proposing a

different approach:

dv

IPlan 1:

Write the equation

asa-t- = -

(v

-V1

)and integrate both

sides with

RC

respect to t.

Plan 2:

Write the equation as inPlan 1, then divide both

sides by v - V1'

and

then integrate with respect

to t.

That is, write

1dv

v-V1

dt

-then integrate.

Which plan would you choose?

Plan 1:

go to page

88.

Plan 2:

go to page

89.

1RC

This one won't work because v

is really v(t) which is anunknown fUnction

of time.

If we try to integrate -(v -N1)/W) we can write

1/.

1v1)

at = -

1

RC

RC

fvdt +

RC

f dt

t.

The second term can be integrated;

but the integrand of the first

term is vl and v

is not known as a functionof time; so it can't be integrated.

Go back and choose

the other plan.

89

This plan will work.

But as we prepare to integrate, we must first decide

whether to find an antiderivative (indefinite integral)or to find the definite

integral from the initial time t 3= 0 to a later time t.

Remembering that

1dv

1v4/1

dt

RC

let's do both in order to

compare their results.

In each case carry out the

integration on the right and show the indicated integration on the left.

1.

Indefinite integral

2.

Definite integral

Indefinite integral: ri

dt

K = -

tdt

RC

Definite integral:

101

dv

v-Vl dt

(The constant of

integration must

appear.

If you

did not have it,

return and insert

it.)

91

Again note that (dv/dOdt

dv, and so the integration with respect to t can

be changed to one with respect to v.

In the case of the definite integral, the

limits of integration must also' be changed appropriately.

Do this and carry out

the integration for both cases.

The result will be:

1.

Indefinite integral

2.

Definite integral

Answer:

1.

Indefinite integral: ,en

[v(t)

-V1]

KRC

v(t)

-

2.

L.:..2inite integral:

Ln

(Note, the integrands become

dv/(v-41) which is of the form

du/u, whose

antiderivative is Ln u.

In the case of the

definite integral,

the limits

are V0

(lower) and. v (upper).

The functional notation,

v(t), is used for

emphasis.)

93

The

fir

st o

f'th

ese

can

be r

educ

ed.to the second. by evaluating

the constant K.

Thus, in. the first expressionlet t = 0.

The value of v(t) at this

time is

1and. so K becomes

_,

94

Answer:

v(t) st t = 0 is v(0) =Vo.

K = - pi (110 -V1)

Vow substitute this value of K back into the

original expression

/n (v(t) - V1) -I- K = - t/RC

and, remembering that /n x - /n y = /n x/y, convince yourself that

both integrals lead to the

same result; namely,

xn

V'0 - V1

v(t)

-v1

ON

O

t-RC

..

95

In this form, v is not given explicitly in terms

of t.

As a final step

.take the antilogarithm of both sides and solve for

v(t).

v(t) =

for t > 0

96 Answer:

or

v(t)

= v1

4-(VO

NT1) e-t/RC

for t > 0

= v

e-t/RC

It

"t/RC)

v(t)

+(1

01

e

Note:

V1 is the applied battery voltage and V0_is the

initial capacitor voltage.

rarr

aMilt

-97

This expression

give

s th

ecapacitor voltage explicitly as a

function of time .

It remains to examine

the nature of this

relationship or the shape

of the plot of

the output voltage as time passes

after the closing of the

switch.

First, consider the

exponential

e-t/Rc.

Since the exponent of anexponential

should be dimensionless,

what is the dimension of

the product RC?

98 Answer;

RC has the dimension of time.

Alin

tr-

99

For this reason RC is called the

time constant.

Figure 12 is a plot of the

exponential f(t) = e

tEvaluate the slope of this

function at t = 0.

Slope of e-t/RCat t = 0 is

Draw a straight line having this

slope tangent to the curve at t =Op and

extend it until it intersects the t-axis.

From the geometry, state thevalue

of t at which this line intersects

the axis.

100

Answer:

1slope = -RC

The straight line intersects the

axis at t = RC.

RC

2RC

1.

Fig. 12(a)

6..

K,

4.91

Thus, the slope of the line tangent to the exponential at t = 0, is

the

negative reciprocal of the

102

Answer:

time constant, RC (Say

it out loud.

Remethber it.

We'll be using it many

times.)

4.

103

We shall use T

(capital tee) as a general symbol for the time

constant.

The

time constant can be taken as a unit

of time.

Thus, the time axis in theplot of

the exponential was labeled in terms

of units of RC.

Suppose the time constant is

4 milliseconds; then an interval of 10

milliseconds can be said to last

for 2 and

1/2 time constants.

In Fig. .12(a) how many time

constants elapse for the line

drawn tangent to

the exponential curve at t'= 0

to drop to zero?

time const.

,

Ans

wer

:

One time const.

1

A f

(t )

=e-

tin

1

RC

II

2RC

3RC

t

Lai

ci t

105

Let's examine the nature of this exponential function.

-When t = T, that is when t = RC, thenv(t)

= e

1.

Since e-1

=lie

= 0.37,

the exponential curve has,dropped from 1.0 to 0.37 in the

elapsed time of one

time constant (when t = T = RC).

This drop amounts to 63 per cent of the

total

drop which will occur as time continues to infinity.

As t -*op, the exponential

function approaches zero.

What percentage of the total change from t = 0 to m

is completed by the

expanential in (a) 3 time constants and

(b) 4 time constants.

a)

per cent

b)

per cent

Draw the approximate shape of this curve on the axes provided on page

104.

io6

Answer: (a)

9 per cent

(b)

91 per cent

(Not

e th

ate-

83 i;

*0.05 ande-4

sz. .02)

Remark

For many practical purposes 2 per

cent

can often be considered

negligible.

In fact, in

making most electrical measurements,

the normal

accuracy of ourmeasurements is often of the order

of 1 or 2 per cent.

Hence, it can oftenbe assumed

that an exponential has completedits change in

4 time constants.

This is a useful number to

remember.

4.11

1.ae

r

107

Now let's return to the expression for

the capacitor voltage in Fig. 9 which

we found on page

96: v(t)

= vo e-t/T+ V (1 - e

)for t > 0

1-t/T,

where T = RC is the time constant.

Here the result is grouped in two terms.

The

first term has as a multiplier the initial voltage

Vo of the capacitor; the second

has a multiplier which is

.(Write it in words.)

It is possible to view the whole solution intwo parts using the supe-rposition

principle:

first, consider that the capacitor has a nonzero

initial voltage Vo

but that the source voltage is zero

(de-activated); then, consider that

108

Answer:

the source voltageV1

the source voltage has a nonzero valueV1

but that the

initial capacitor voltage is zero.

Reminder

The expression is:

v(t) =

0e-t/T

+ V1(1 -

e-t/T)

109)

Write the separate expressions

for the capacitor

voltage for

each of these

two cases:

1.

Initially charged capacitor

being discharetedthrough a r

>0

..-F-.V0 at t = 0

esistor:

0for t

v(t) =

2.

Initially unchargedcapacitor being

charged througha resistor by a

suddenly appliedconstant voltage source.

v(t) =

for t >

= 0 at t; =0

Also sketch waveforms

of v(t) for these

two cases.

Label the pertinent

points properly.

(For definiteness)

letV1 >V0.)

v(t)

Av(t)

2T

3T

4T

I

T2T

3T

4T

110

Ans

wer

:1.

vt)

=V

oe-

t/T

I.

111

In the case of the

discharging capacitor, the curve

clearly shows that the

capacitor vo14-age starts

at its initial value Voand drops exponentially

to 2

per cent of its

initial value in

(how many) time constants.

In its

final state, the capacitor

voltage reaches

On the other hand, the

voltage of the charging

capacitor starts from zero

and rises

(in what manner?) toward its

final value, which equals

112

Answer:

4 timesconstants

zero

expo

nent

ially

Vi

113

Let's considerthe initiallycharged capacitor,

which is discharging

through

a resistor)

in more detail.

The expression

v(t) =

e-t /I/

= V0

e-t/T

for t > 0

Vo

is an exponential.

There are two

constants in this

expression which

determine the

curve of v,....ltageagainst time:,

1.

the quantity Vo

(which is the

),

determines the point onthe voltage axis

at which thri curve

starts at

t = 0, and

2.

the

,which determines

the rate atwhich the curve

falls to its final

lialue, zero.

o

I i

Answer:

the initial value of the capacitor

voltage

time constant T

I

115

Once these two quantities

are known,

together with the

fact that the

final value

of the capacitor

voltage is zero, the curve

of the variation

of v against t can

be

sketched.

Suppose a 2 pf

capacitor is chargedto an initial

voltage of 10 N,

a long

time.

At t = 0 a switch

is closed connecting a

10 kilohm resistor

acrosz.

capac-

itor.

Sketch the plot ofthe capacitorv.Itage

approxiMately to scale fromt = - 20

to

100 milliseconds.

Lable appropriate

points; in particular

the points cor-

responding to 1 and 3

time constants.

10 5

-20

100

in millisec.)

u6

Answer:

v (v

olts

)

10

3 7

(curve starts

tangent)

1 time

const.

3time

const.

20

140

6o

eolo

u

t = RC

Thus, tomake a sketchof the capacitor

voltage it is onlynecessaryto know the

initial voltage, Vo,

the final voltage,

and the time

constant.

117

Let's turnnext to:the

initially unchargedcapacitor whichischarged from a

41source

through a resistor.

The expression

.17or the voltagefollowing.the

switch-

ing is

v(t) = Vi (1 -e-t/T)

Here also

there are just

two constants

which will

determine the

shape of the curve

of thefunction: V

the source

voltage and T)

the time

constant.

In the present

case, the

initial value

of the capacitor

voltage, and its

final

(ultimate) value

after the passage

of a' long time

are:

a)

initial valla

of v

b)

final value

of v =

Answer;

initial value of v

0

final value of v

V

,101

,111

11M

a,ag

uros

ream

man

imIll

IMIM

MIN

ON

NIII

SM

Ila

119

Knowing the initial value

and the final value, a

sketch of the voltage can

be made as a function

of time once the time

constant is known, since

the time

constant determines the

rate at which the curve goes

from its initial to

its

final value.

Thus, suppose an initiallyuncharged .5 p.f capacitor

is charged from a

20-volt

battery through a 200 kilohm

resistor.

Sketch the plot of the

capacitor voltage

approximately to scale for

t =

- 200

to + 500 milliseconds.

Label appropriate

points.

millisec.)

120

Answer:

20 19 12.6

(volts)

MM

...1

10 0

=0

IMIM

0NO

MIN

D M

IND

MIM

OM

NI,

.=11

, ON

O=

MP

410

111

OW

N 6

1=1,

MM

INID

Ml=

b M

INN

S

AO

M =

IM 4

=0

MD

MO

M M

E,

OM

MW

WW

I .11

1011

.1 4

. 0=

0 M

MIN

D

1 time const.

3 time const.

-200

100

200

300

ntsec.)

Finally, let us returnto the general

situation

in Fig. 9 in whichthere is both an initialvoltage

and a charging source.

The capacitor voltage

after

the closing of the switch was

earlier found to be

v(t) = V0e-t/T

(1

e-t/T)

or

v(t)

V0 + (V1 -V0)

(1

- e-t/T

)

(You should. verify that these

removing the

parentheses.)

In this case the

initial

a)

initial value =

b)

final value =

121

Fig. 9

are differentforms of the same

expression by

value and final value

of the Capacitorvoltage are:

122

Ans

wer

:

a)in

itial

val

ue =

Vo

b)fi

nal v

alue

=V

1

The expression

V(t) = Vo

(Vi

Vo)(1 -e-tn)

is perhaps

the simplestwhich can be

used to sketch

the voltage as a

function of

time.

In this form,

there are two

additive terms.

The first is a

constant Vo; the

second is like

the case wehave just

completed.

Sketch each of

the two terms

separately as

dotted lines on a

set of axes;

then sketch

their sum as a

full line.

Assume that V =2.5 V

10'

124.

Ans

wer

:v

T0.

.....

V1

T...

... M

P10

MO

N M

ON

40 O

M a

m.

ma

ma

..a o

we

am o

w a

wm

.ao

vo

l

.....

. 4111

110

1 rI

O

(1/ 1

-V0

) (1

-e-

-t/

owl..

Ia...

...."

.w

ow.

ON

IMM

NI O

ON

NO

Va/

e.rw

o

--

WIW

I* M

EM

O

V0

RC

= 1

00p.

sec.

125

Here again we

find that the

voltage rises

from its initial

value to its

final

value exponentially.

The rate atwhich it rises

will depend onthe time

constant.

If a sketch

of the voltage

is required,

it is only necessaryto know the

initial and

final values

and the time

constant; it is

not necessaryto solve a

differential

equation.

In Fig. 13,

the capacitorIns an initial

voltage of I- 50volts when, at

time

= 0,

the

switch is closed.

a)

Write anexpression for the

capacitor

voltage as a

function of time

follow-

ing the closing

of the switch.

b)

Sketch the waveformof the capacitor

voltage) labeling

all appropriate

points.

)On the same axes

sketch in a dotted

line the voltage

waveform if the

initial voltage is -

50 volts instead

of + 50 volts.

126

Ans

wer

:4

v(t)

= 5

o +

70(

1e-

aot)

Av

120

94-

57 50

.10

mm

alo

mm

e=

maw

mow

am

mo

..wm

...m

me

ag.

oNor

as.m

.wra

...om

wan

..

-50

A

200

300

4-00

500

>t (

p.se

c. )

wow

127

When theproblem of Pig. 9 'was firit started, it was the

intention to .examine

both the capacitor voltage and its

current.

Having concentrated on the

voltage,

let us now turn to.the current.

The voltage in the general case was

found to be

(nut/T)

'o'

''1 '

v(Fig. 9)

Using the voltage-currentrelationship of the Capacitor:and remembering what the

time constant is in terms.of It and C, write 'anexpression for the current i(t).

128

Answer:

v =V

i(t) =

1R

0e-t / T

(Manipulate your answer

until it agrees

with this.)

`111

1111

31

..701

11JI

MIIM

EN

b,

129

For the two distinct cases:

(a) V1

ita 0

1.V

of.4 0 and. (b)

Vo

ea 0

, V1

y4 0

1

the form of the expression is

just the same, a decaying exponential.

Case (a)

i(t)

is -

Vo

eVi = 0

Case (b)

i(t) s*V1

e-t/T

ts 0

Vo

This is in contrast with thevoltage, which is a dem= exponential

in case

but a

in the other case.

130

Answer :

case ial

rising (or increasing, or growing) exponential

Note thitt the actual direction of current in case (a)

will depend on the polarity of the initial voltage.

If

V 0.is positivel.meaning the'initia1 voltage actually has

.the same polarity as the reference, then the current

will be negative, meaning the actual current will be

opposite to the reference.

Fig.

9

131

In the general case when

neither V0nor V1is zero, the expression

for the

current is still an

exponential.

Its sign will depend.on th-

relative values of

V1

and V0.

Let's consider the cirauit in

Fig.-9 again.

Suppose the initial capacitor

voltage is 25 volts and the

dc source voltage

is 10 volts, with R = 1 kilohm

and C = 114f.

Sketch the curve of the

current

against time following the

switching.

Label the time axis inmilliseconds and

the current in milliamps.

What is the actual directionof current relative to

the reference direction?

^ry

132

The actual current is opposite to

.

the reference.

v

(a)

ti

(b)

Figi*:-.111.

Fig. 15

(c)

t

133

Another item of unTinishedbusiness is the consideration of the casewhen the

source voltage in Fig.

9 is somethl.ng other than a

constant.

Some signal waveforms

that are of great importance in

communications and in computers are shown inFig. 14.

The first one is called a step

-Punction. .14 until sone time,

the voltage

(assuming the signal is a voltage) is zerowhen it suddenly jumps to a non-zero

value and stays there.

This descrirtion fits the caseof a battery suddenly switched

into a network and indeed we can sayin the case of Fig. 9 that avoltage steD

function is aDDlied to the RC

network.

Let's consider the step function

in somewhat more detail.

In order to describe

the step function shown in Fig.

150 we would need to state two

things:

(k) when

the discontinuous step takes placeand (b) what the magnitude of the step

is.

In

Fig. 15 these two quantities are:

a)

the time at which the step occurs =

b)

the magnitude of the step m

Ans

wer

:

a )

t 1

b )

V1

135

When the magnitude of a

step is unity, it is

called a unit step function.

A unit step functionoccurring at time t = t1 is

denoted by p(t-t1).

ThusI

mathematically,, the unit step

is expressed as

ol(t-ti)

1

= 0

Draw a graph of the

functions,

v1

=5u(t-.02)

and

v2

=-10u(t-.01)

for

t > ti

fort < t1

Label the points at whichthe steps occur.

,.t

137

Now write the mathematical expressions for

the step functions shown graphically.1

We will take these to be steps of current

rather than voltage so that you won't get

the idea that there are only voltage steps.

i1

0.5

1(t) =

3

parr

.

.13

8

Ans

wer

:

1(t) = -2u(t -.5)

.

i2(t)

3u(t-40)

3u(t)

(Nbte that the unitstep tunction occurring

at t

0 is simply u(t).)

10

t(sec.)

0.3

Fig. 16

139

Now consider the Dulse having a10-volt magnitude shown in Fig.

16.

On the

set of axes below show how two

step functions can be combinedgraphically to

yield the given pulse.

Also write a mathematical expression.

Answer:

vA

10

0.3

0 .1

-10--

10

t(sec.)

Add. the two step functions.

Everywhere beyondt = 0.3

they cancel eachother and leave only the pulsebetween

Fig. 17

0.1 and 0.3.

The mathematical expression

is

..°

.2

v = 10 u(t-0.1) -

10 u(t-0.3)

Thus, if the source voltage in Fig. 9 is a pulse, we can find the capacitor

voltage and current by using, twice, the previously obtained solution

for a constant

source voltage, once with a positive value, and once

with a negative value but

occurring at a later time.

In Fig. 17 suppose v (t) is the pulse shown and suppose the

capacitor is

uncharged before.the initiation of the pulse.

It is desired to find the capacitor

voltage v and its current i from t.= 0 to 1 second.

There are two periods of time to consider; the time immediately following

each of the two step functions.

For each time period the curve of v or i against

time can be.sketched when three quantities are known.

These three quantities are:

a)

b)

c)

Answer:

a)

the initial value

b)

the ultimate, or final, value

c)

the time constant

Remark The ultimate value,is the value which the variable

in question (v or i)

would reach if the state of the network did not change.

The network cannot

anticipate a change which will take place at a later time.

Thus, after the

pulse is initiated and before it is terminated) thenetwork will behave as if

the step which initiated the pulse will remain on.

143

Let's take the first time period

(when the pulse is on).

Specify, in the

table below, the appropriate values of time constant,

initial value and final

value of the capacitor voltage and current

voltage

current

time constant

sec.

sec.

initial value

volts

ma.

ultimate value

volts

ma.

Hint:

Verify that Kvl is satisfied by the initial values and

final values in

your answer.

Answer:

FOR''TBE FIRST TIME PERIOD

voltage

current

time constant

initial value

ultimate value

.01 sec.

0 volts

10 volts

.01 sec.

.1 ma.

0 ma.

Note that

105i

v must eqUal 10 volts at all times

after the pulse is initiated.

The ultimate values given here arethe values that would exist if the pulse were to

remain on for a very long time compared to the time constant.

l0

51a

RE

IM M

IS M

r sl

ow

gsec)

,

41,

145

Let's continue tnis analysis of the

first time period,

Now write the general

expressions'for the capacitor voltage

andcurrent as

.they apply to this time period,

using the vaiues just

computed to evaluate all

unknown constants.

v(t)

,2=

volts

0 < t < 0.2

i(t)

amps.

0 < t < 0.2

-enr

rIl.W

WW

.19,

7

146

Ans

wer

:

v(t)

= l0

(l-e

-100

t) v

olts

i(t) =10-4e-loot

amps.

Solution for v(t):

dv

dv

1051

+ v = 10 ;

10 (.

106

+ v = 10

= 1000 - 100v

dt

dt

dv

1000-100v-

dt

1-

10-5

Zn(

1000

-100

v) =

t +

-100

t;

1000 - 100v = k2.

at t = 0, v = 0

;K2 =

1000

;v = 10(1 - e-100t)

011.

411a

NIM

IIMN

INM

EIN

IIIIIN

SIO

N11

1111

14,,

114-

7

The durationlor length,

of the pulse is 0.2 secoudwhich is a factor of

times the time constant.

What can you then concludeconcerning the

degree to which the capacitorvoltage and current havereached their ultimate

values by the end of the

pulse?

Compute the value of v andi just at the endof the pulse to verify your

.conclusion.

.

148

Answer:

20

Since the capacitorvoltage andcurrent reach a value

within 2J3er cent

of their finalvalue in

4time con-

stants, in 20

time constants.theywill have reachedtheir

ultimate value

almost completely.

v(0.2)

1(0.2)

= =

10(1-e-2°)

--

10

4e

20=

= 1010(1-2)00-8) 1 10

?2)a0

8) A 0

-4-

10

5

Fig. 17

(repeated)

ato

t

149

Now let's turn to the second time period

(after the termination of the pulse).

The initial time for this period is the endof the pulse.

Complete the table

below for the capacitor voltage and current.

voltage

current

time constant

initial value

ultimate value

sec.

Volts

volts

sec.

ma.

ma.

(Be careful of references) and make sure your answers are

consistent with Kvl.)

150

Answer:

voltage

current

time constant

initial value

ultimate value

.01 sec.

10 volts

0 volts

.01 sec.

-.1 ma.

0 ma.

Note that the current has now reversed.

105

Nig

10

.2

Atif...Omt(see)

151

It is desired now- to write expressions forthe voltage and current as a func-

tion of time during the period fol4owing the

end of the pulse.

We know the initial

values of v and i at the start of this interval.

The situation can be considered

as tha start of a new

problem.

It is convenient to consider thestarting point as

the zero of time.

For this reason we can define a new time

variable to which is

0.2 sec. less than the old time variable t;

that is, t'

t0.2.

Thus, at the

end of the pulse, when t = 0.2, the newtime is t' se 0.

Wtite expressions for v and i in terms

of the time variable to using thevalues

from the previous page.

Then substitute for tl

in*terms of t.

v(t') so

volts.

t'

i(t') =

amps.

t' > 0

or

v(t) =

volts

t > 0.2

i(t) =

amps.

t > 0.2

152

Answer:

-300

t1 "

v(v)

m m

e-

irat

h

i(t')

- 0

e-4

-m

ots

amps.

v(t)

=1o

e-10

0(t-

0.2)

i(t) m-m

e-10

0(t-

4).2

)

t > 0

.2

t > 0

.2

These are*to be

com

pare

d w

iththe previously found expressions

(repeated below)

for the period when the pulse is on.

v(t)

=1o

(1-e

-100

t)

-4e-100t

i(t)

= 1

0

O < t < 0.2

O <

t< 0.2

153

make a careful sketch on the axes below of the voltage and current

waveshapes as a function of time) from slightly before the initiation of the pulse

to a time equal to twice the pulse duration.

(The outline of the applied pulse

is shown dotted.)

A v

10 0

0 .2

t(sec.)

1.1.

11.1

"."8

.401

6

t(sec.)

0 .2

t(sec.)

10-4

155

Nbtice 'tbe shape of these curves.

The capacitor voltage still looks like 4

pulse, but with somewhat rounded corners.

The current consists of two spikes,

the maximum current occurring when there is a change in the input voltage.

As the

capacitor voltage increases toward the pulse voltage, the voltage across the

resistor (and hence also the current) decreases, in accordance with Kirchhoff's

voltage law.

The degree of rounding of the corners of.the pulse will depend on the

.For the case under consideration, the time constant is

(how many?) times the pulse width.

In this case we find that the applied voltage

pulse is transmitted to the output with a small amount of distortion.

Answer:

time constant

1/20, or 0.05

100

MIN

NO

W

112f

Fig.

18

c

lbw

Let's consider changing the

time constant to

observe what influence it has on

the waveshape of tile

capacitor voltage and current.

Suppose that the

capacitance in Fig. 17 is changedto lpf: everything

else remaining the same.

Sketch the resulting curves

of v(t) and i(t).

Also, write the corresponding

expressions for v(t) andi(t) in the twoAime inter-

:

vals.

(You may have some difficulty withthe second

interval.

Note that it starts at t = .2

and so your

express;'e n should accountfor this.)

> 0.2

v(t) =

i(t)

0 < t < 0.2

v(t)

i(t) as

157

0.02

158

10A i

.02

.02

A.

-.u07x10

0 < t <0.2

v(t) = 10(1-e-10t)

4 -10t

i(t)

10 --e

t ). 0.2

v(t) = 8.7e-10(t -.2)

i(t) = .87)40-4e-10(t -.2)

Mbtice how, much more distorted the capacitor voltage is nowcompared to what it

was when the time constant was smaller.

1111

1111

1111

1111

111

Now consider the network shown in

Fig. 19.

The switch has been closed for

a long time so thatthe network is in

equilibrium.

At a time which we can

take to be t = 0, the switch is opened.

Ten tisec. later it is again closed.

Sketch the waveshapes of v(t),

il(t)

and 12 (t) approximately to scale from

t = -10 to t = 40 ilsec.

Label the axes.

50C

gi5o

cg

159