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Unit 1: DC Electricity 1-1

DC ELECTRICITY

1

AIMS

The aims of this unit are:

1. To explain the fundamental concepts of electricity, electric charge,

electric current, electromotive force, potential difference (voltage),

and resistance.

2. To demonstrate how to calculate DC voltages, currents and power

dissipation in simple DC resistor networks.

3. To introduce the concept of electric fields.

4. To introduce capacitance and capacitors.

5. To demonstrate how to calculate voltage, charge and energy in a

capacitor.

6. To introduce the main types of capacitor and higlight some of

their applications.

THE NATURE OF

ELECTRICITY

After studying this sub-unit, you should be able to:

1. Define or explain the meaning of the following terms:

. ampere

. conductor

. conventional current

. coulomb

. electricity

. electric charge

. electric current

. electromotive force (emf)

. free electronsdrift velocity

. ion

. insulator

. positive and negative charge

. potential difference (PD)

. schematic diagram

. volt

. voltage

2. Understand how positive and negative electric charges interact.

3. Explain electrical phenomena in terms of the atomic theory of

matter.

4. Perform calculations involving the units of coulombs and amperes.

OBJECTIVES

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1-2 Communication Technology 1

What is electricity? Where does it come from? How can it be produced?

Is there more than one type of electricity? How do positive and negative

electric charges interact? These are the sorts of questions that we will

shortly try to answer.

It was through rubbing certain materials together that mankind first

produced and investigated the mysteries of electricity. The first recordedexperiments took place around 600 BC, in the era of Pythagoras,

Confucius and Buddha. The pioneering physicist was Thales, who lived

in Miletus, a Greek-run territory located in the south-eastern corner of

what is now modern Turkey. Thales demonstrated that if amber ( a

yellowish substance used for ornamental purposes ) was rubbed with

animal fur, the amber would attract to it pieces of dry straw or feathers.

In England in 1600, William Gilbert, personal physician to Queen

Elizabeth 1, published a study which listed many other substances

capable of producing the same effect after rubbing. Gilbert called these

substances electrics, after elektron, the Greek word for amber. It was

French scientist Charles Dufay who produced the next step forward. Inthe 1730s he found that there was not just one type of electricity - but two.

Furthermore, Dufay found that like types of electricity repelled while

opposite types attracted. We can illustrate this point with the following

experiment.

Take a piece of amber and a piece of glass, two substances which can be

electrified by rubbing with fur and silk respectively. Electrify the amber

and glass, and suspend each by a thread so that it is free to move. Now

bring them close together. Result: the amber and the glass attract one

another. However, if two pieces of amber are brought close, they repel

one another. Similarly, two pieces of glass repel if they are suspended in

close proximity.

After numerous experiments with many substances, no type of electric

was ever discovered which was not attracted either to the amber or the

glass. In other words, there were two types of electricity and only two:

the amber-type and the glass-type.

In 1747 the American inventor and statesman Benjamin Franklin

introduced the terms positive charge and negative charge into the

language of electricity. Franklin described substances which act like

electrified amber as having a negative charge and those which act like

electrified glass as having a positive charge. We continue to use

Franklins terms today and we can now state Dufays discovery, which

is a fundamental law of electricity, in these terms:

Unlike charges attract one another, while like charges repel.

It is important to note that the words positive and negative are purely

arbitrary terms; positive electricity is not in any way better than

negative electricity.

We can summarise as follows the basic facts about electricity, all

verifiable from simple experiments:

. An object can be made electrified by friction. Some materials

are easier to electrify than others.

. There are two types of electric charge - positive and negative.

INTRODUCTION

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. Like charges repel; opposite ones attract.

. The strength of the attractive or repulsive electric force decreases

with distance.

. The electric force is far more powerful than gravity. And like

gravity, the electric force can act through empty space.

. The electric force which an object is capable of exerting is

independent of its mechanical properties - its shape, size, density,

etc.

THE ATOMIC

DESCRIPTION OF

ELECTRICITY

Benjamin Franklins terms of positive and negative charge were rooted

in his idea that electricity was some form of subtle fluid, of which

electrified materials had either too much (when positively charged) or

too little (when negatively charged). A material which was not charged

he called electrically neutral, as it contained its proper amount of this

mysterious fluid.

Franklin believed that all charged bodies sought to become electrically

neutral - to discharge their surplus or deficiency of electric charge. This

explained why oppositely charged bodies attract one another: the positive

and negative charges attempt to cancel one another out, as the body with

positive charge tries to surrender its excess of charge to the negative

body with a deficiency of charge. This theory also explained why bodies

with a similar charge repel one another. A positive body could not

discharge to another positive one; a negatively charged body could not

satisfy its deficiency of charge through another body with a similar

deficiency.

Franklin furthermore suggested that if the charge on two oppositely

charged bodies was large enough, the excess on one would leap or sparkacross to the other without the two bodies needing to be in physical

contact. It was from this theory that he developed the idea that lightning

in the atmosphere was simply the discharge of electricity from clouds to

the earth, and for lightning to take place, the clouds had to carry a large

electric charge opposite in sign to that of the area of earth beneath them.

It was known at the time that spark discharges were best attracted by

metal rods with sharp points. Franklin recommended placing such rods

(now called lightning conductors) on top of tall buildings, and connecting

each one by a length of wire to another metal rod, driven into the ground.

In this way, the electricity in the clouds would be discharged safely to the

earth. The lighting conductor provides a low-resistance path to earth forthe electricty, therebydiverting the electricity away from the fabric of

the building in the event of a lightning strike.

But what is the subtle fluid of which Franklin reckoned electricity was

composed? In the 1830s, English physicist Michael Faraday suggested

that there was within all matter two particular types of tiny particles, one

carrying a basic amount of positive charge and the other a basic amount

of negative charge. All observed charges were simply multiples of these

basic charges. As the atomic theory of matter took root in the nineteenth

century, scientists concluded that if matter was atomic, then so too was

electric charge. It is from these beginnings that our present theories have

developed.

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All matter consists of large numbers of tiny, spherical particles called

atoms, each with a radius of about 10-12 m. The structure of a typical

atom is illustrated in Figure 1.1. It consists of a central nucleus (radius

10-14 m), containing a number of sub-atomic particles called protons

and neutrons. Around the nucleus even smaller particles, the electrons,

orbit like planets around the sun. Each of the over 100 naturally-

occurring or artificially-produced elements is characterised by a uniquecombination of protons, neutrons and electrons in its atoms.

Electrons(negative charge)

Nucleus

Neutron(no charge) Proton

(positive charge)

-

-

--

+ +

+ +

Figure 1.1

Illustration of atomic structure

Neutrons, as their name suggests, carry no electric charge. Protons carry

a positive charge. Electrons carry a negative charge, equal in magnitude

but opposite in sign to the positive charge of the protons.

It is the charge on the protons and electrons in every atom of matter that

accounts for the phenomenon of electricity.

The atom is held together by the force of attraction between the protons

and the electrons. Normally, an atom contains the same number of each

and is therefore electrically neutral; that is, its net charge is zero. If an

atom gains or loses one or more electrons, it becomes negatively or

positively charged respectively. (An electrically charged atom is called

an ion.) In either situation, the atom is unstable and tries to return to the

neutral state, by repelling the gained electron into orbit around an

adjacent atom or by attracting an electron from an adjacent atom to make

up its deficiency of negative charge. We will return to this subject in

more detail in Unit 3.

PRINCIPLE OF

CHARGE

CONSERVATION

According to the principle of charge conservation, the charge of an

isolated system cannot change. If an additional positively charged

particle appears within a system, a particle with a negative charge of the

same magnitude will be created at the same time; thus, the principle of

conservation of charge is maintained.The unit of electric charge is called

the coulomb (C). The smallest sub-division of the amount of charge that

a particle can have is the charge of one proton, +1.602 10-19 coulomb.

The electron has a charge of the same magnitude but opposite sign, ie -

1.602 10-19 coulomb.

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In Italy in 1800 Alessandro Volta invented the electric battery, which for

the first time enabled scientists to experiment with a steady flow of

electric charge through a circuit. Previously, the only form of electric

charge movement which could be produced and investigated was the

sparking discharge that leapt from one electrified body to another. Such

was the fame brought to Volta by his invention that Emperor Napolean

called him to France in 1801 for a command performance of hisexperiments. Voltas battery truly revolutionized the study and application

of electricity; his invention was an essential component in the early

telegram and telephone communication systems which appeared in the

second half of the nineteenth century.

The flow of electric charge generated by Voltas battery was called an

electric current. Like their modern-day counterparts, the original

batteries had two terminals, one of which possessed a negative charge

and the other a positive one.

Which way does the electric current flow - from the positive end of the

battery or from the negative? Clearly, it must be one or the other. Theearly experimenters with electricity knew nothing of electrons, so they

had to choose one terminal or the other; in effect, they had to make an

even-money guess. And as it happened, they got it wrong. They decided

that current originates at the positive terminal of the battery and terminates

on the negative one. This convention, that current flows from positive

to negative, is still retained today, even though we now know that current

is carried by electrons and that electrons move from negative to positive.

In short:

Conventional current flow is from positive to negative; the actual

movement of electrons is in the opposite direction, from negative to

positive.

For practical purposes, it does not really matter in which direction we

believe the current to flow - as long as we are consistent in any particular

case.

Different materials exhibit different abilities to conduct electricity. For

example, copper conducts electricity very well and is therefore a good

conductor; plastic hardly conducts electricity at all and is accordingly

a good insulator.

This wide range of electrical conductivity shown by different materials

can be explained in terms of their atomic structure. Metals and other

good conductors have electrons in their outer orbital paths which are notstrongly attracted by the protons in the nucleus. These electrons are

called free electrons and normally drift randomly from atom to atom

throughout the conductor. However, when a battery or other electrical

source is applied across both ends of the conductor, these free electrons

all move in the same direction towards the positive terminal of the

source, so giving rise to an electric current. In insulators, all the electrons

are tightly bound to the nucleus, so that it is difficult to remove them from

their orbital paths. There are therefore no, or very few, electrons free for

conduction.

Consider a movement of electrons along a length of conducting metal

wire, similar to the flow of water through a pipe. See Figure 1.2.

ELECTRIC CURRENT

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Figure 1.2

Electron flow

Electron

Flow

Direction of Conventional Current

Area A

If the electrons move at a uniform rate and a total charge, denoted by the

letter Q, passes through a cross-section of the wire in a time of t seconds,

then this is equivalent to an electric current, I, where:

electric current =amount of moving charge

time taken

IQ

t=

As indicated previously, the units of electric charge are coulombs (C).

Current is measured in amperes (A). A current of 1A flows when 1C of

charge passes a point every second.

How many electrons are involved in carrying a current of 1A; that is, in

moving 1C of charge per second?

As previously indicated, the magnitude of the charge on each electron

has been measured to be 1.6 x 10 -19 C. Current is the rate of flow of

charge, ie the amount of charge passing a given point in a conductor

every second. Therefore for a current of 1A, we can say that the rate of

flow is 1 coulomb every second. Thereore the number of electronsrequired to make 1 coulomb of charge is

N =1

charge on each electron=

=

1

1 6 106 25 10

1918

..

As you can see, quite a large number of electrons is involved!

Another question is, how fast do the electons move?

In order to answer this question, imagine a length of conducting wire

with cross-sectional area A carrying a currentI, as shown in Figure 1.3,

and suppose there are n free electrons per unit volume of the wire, each

carrying the basic unit of negative charge, e, and moving at an average

velocity v. In a time of t seconds the electrons will travel a distance

l v t=

Thus in tseconds all those electrons Nin a volumeAl=Avtwill pass

through a plane P of the wire. Thus

N = nAvt

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and the total charge Q in motion is

Q Ne neAvt = =

Therefore, the electric current is given by the expression:

IQ

t

Ne

t

neAvt

tneAv= = = =

Electron Area A

electron flow

conventional current

Conductivity

Wire

Figure 1.3

Calculation of drift velocity in

a conductor

In order to estimate the drift velocity v of the electrons, supposeI=1.6

A,A=1 mm2 = 10-6 m2, e = 1.6 10-19 C, and n =1028 electrons / m3. Then

from the equation above

vI

nAem s mm s= =

= =

1 6

10 10 1 6 1010 1

28 6 19

3.

./ /

This is a surprisingly slow drift speed compared with the average

thermal speeds of the electrons which can be of the order of several

hundred metres per second. It should be noted therefore that theelectrons which leave the negative terminal of the electrical energy

source are not necessarily the same ones which enter the positive

terminal. Individual electrons progress only very slowly along a

conducting wire. As calculated above, their typical speed is about 1mm/

s or, in other words, in the above case, it takes nearly 17minutes for the

electrons to travel along one metre of conducting wire. The current,

however, travels at a speed close to that of the speed of light (3 108 m/

s). The situation may be compared to that of having a long tube filled

with marbles; push one marble at one end of the tube, and another marble

pops out at the other end. Each marble may be likened to a packet of

electrons carrying an amount of charge. While the marbles may pop out

quite slowly, the amount of charge emerging each second can be quite

high. The rate of flow of charge in coulombs per second gives the current

in amperes (A).

ELECTRIC CIRCUITS

An electric circuit is formed when electrical energy is applied, via an

uninterrupted conducting path, to one or more devices. Consider the

simple circuit shown in Figure 1.4, which consists of a battery, a length

of wire and a bulb.

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Figure 1.4

Battery and bulb circuit

Battery

Bulb

Connecting

Wire

Light andHeat

+ -

The bulb lights (and heats up) due to the flow of electricity from one

terminal of the battery to the other, along the wire and through the bulb

itself. The source of electrical energy for a circuit may take many forms:

battery, dynamo, alternator, mains supply, etc. The device to which the

electrical energy is transferred may be a bulb, water-heater, television

set, washing machine, computer system or whatever. It must be stressed

that for a current to flow in any circuit, there must be a continuousconducting path from one terminal of the source, through the device, and

back to the other terminal of the source. Most circuits contain a switch;

if open, the switch interrupts the conducting path and so no current can

flow; if in the closed position, the switch enables the current to flow.

To simplify the drawing of electric circuits, a standard set of symbols is

used, each one representing a typical circuit component. Figure 1.5

shows the schematic diagram of the battery and bulb arrangement

illustrated in Figure 1.4 and includes a switch. Additional circuit

component symbols will be introduced as we encounter them.

ELECTROMOTIVE

FORCE AND

POTENTIAL

DIFFERENCE

Voltas battery and its modern-day counterparts are examples of asource of electromotive force (emf). An emf source provides the

electrical energy which is necessary for an electric current to flow.

The concepts of electromotive force and potential difference (PD) may

be more readily understood by considering the simple analogy shown in

Figure 1.6, in which a pump continuously circulates water around a

closed-pipe system.

PositiveTerminal

NegativeTerminal

Battery Symbol

Bulb

Terminal

Bulb Symbol Bulb

Terminal

Switch

+ -

Figure 1.5

Schematic diagram of a battery

and bulb circuit

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Figure 1.6

The water-system analogy

At this point water

(current) can take one

of two different routes

pressuredifference

=

potential

difference

pump = emf

water flow

= current

To keep the water flowing in the system, the pump must be in continuous

operation. Similarly, in an electric circuit the continuous output from an

electrical energy source (a battery, generator or whatever) is required tokeep the electric current flowing in the circuit. The driving force

provided by the electrical energy source is called the electromotive force

or emf.

In the water system, water flows between any two points only when there

is a pressure difference between them. Similarly, in an electric circuit,

current can only flow between any two points when there is a potential

difference between them.

The unit of potential difference is called the volt, after Alessandro Volta.

The volt is defined as the energy in joules required to carry one coulomb

of charge from one point to another.

Thus, the emf of a source is the potential difference (PD) across its

terminals, expressed in volts.

SAQ 1For an electric current to flow between any two points in a

conductor, state which of the following conditions are true and

which are false:

a. A potential difference must exist between the points.

b. The same potential must exist at each of the points.

c. There must be a movement of electrons between the

points.

d. The conductor must be free of electrons.

e. A source of emf must be present.

f. The circuit may have a break in it.

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SAQ 2A torch battery is capable of delivering a current of 300mA

continuously for a period of 5 hours before running flat.

Calculate the initial charge stored in the battery.

SUMMARY

1. An atom is a particle of matter consisting of a central nucleus of

protons and neutrons, around which electrons orbit.

2. Protonscarry apositive charge;electronsanegativeone.Neutrons

carry no charge and are electrically neutral.

3. Forces exist between electrically charged particles and bodies, such

that like charges repel and opposite ones attract.

4. Electric charge, denoted by the symbol Q, is measured in units of

coulombs (C).

5. An electric current, symbol I, is a flow of negative charge, carried

by electrons.

IQ

t=

Electric current is measured in amperes (A), where 1A represents

1C of charge flowing past a point for one second.

6. Good conductors have many free electrons. Good insulators have

none or very few.

7. An electromotive force (emf) is a source of electrical energy

required to produce an electric current. Current flows between any

two points in a circuit between which there is a potential difference

(PD).

8. The unit of both PD and emf is the volt (V).

9. For an electric current to flow in a circuit, there must be a continuous,

uninterrupted path between the positive and negative terminals of

the source.

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ANSWER TO SAQSSAQ 1

a. True.

b. False.

c. True.

d. False.

e. True.

f. False.

SAQ 2

charge = current time:

Q = It => Q = 300 10-3 60 60 5 = 5400 C

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DC CIRCUITS

OBJECTIVES



. direct current (DC)

. electric power dissipation

. Kirchoffs laws

. Ohms law

. resistance

. internal resistance

. resistivity

. superposition theorem

2 Calculate the resistance of a conductor given resistivity and the

appropriate data.

3. Perform simple circuit calculations involving Ohms and Kirchoffs

laws.

4. Calculate the equivalent resistance of series, parallel and series-

parallel resistor networks.

5. Calculate the voltage, current and power dissipation in series,

parallel and series-parallel resistor networks.

INTRODUCTIONWe begin our analysis of electric circuits by considering a particular typeof circuit, called a direct current or DC circuit. In a DC circuit, current

always flows in the same direction through the circuit. For such circuits,

the voltage output of the source of emf stays constant with time as shown

in Figure 1.7.

emf

(volts)

V

Time t

Figure 1.7

Voltage output of a DC source

The other main source of electricity is the AC (alternating current)

type. As its name implies, the direction of the current in AC electricity

is constantly changing. The most common example of an AC source of

emf is the mains electricity supply. Many of the principles of DC circuits

can also be applied to AC ones. We will be studying AC electricity in

the next unit.

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RESISTANCE AND

OHMS LAW

So far in our study of electric circuits we have learnt of potential

difference and current, and that the first produces the second. We have

also learnt that some materials are better at carrying a flow of current

than others; or, to put it the other way around, some materials offer less

of a resistance to current flow than others.

Are these three quantities - potential difference, current and resistance- related in some simple way? It was this task of finding such a

relationship that the German physicist George Ohm set himself in the

years immediately following Voltas invention of the battery.

In 1827, Ohm showed that the flow of current in a conductor (at constant

temperature) is directly proportional to the potential difference across

the conductor and inversely proportional to the resistance of the material

carrying the current. Today, this statement is known as Ohms law.

Letting V represent potential difference (voltage), I for current and R for

resistance, we can write Ohms law mathematically as follows:

I VR

V RI R VI

= ; = ; =

Ohm was a humble school teacher at the time of his discovery. Instead

of bringing him fame and fortune, however, Ohms law embroiled him

in controversy with other scientists of the day, as a result of which he lost

his school teaching position and spent the next six years in poverty and

bitter disappointment. Eventually, recognition came. Five years before

his death at the age of 52, Ohm was made Professor at the University of

Munich. When he died in 1854, a statue was raised to him and a Munich

street named in honour of the man who discovered that V = RI.

Today, the unit of resistance is the Ohm, represented by the Greeksymbol omega (). A material has a resistance of 1 if a current of 1A

flows through it when an emf of 1V is applied across it.

SAQ 3a. A PD of 20V is applied between the ends of a conductor

and a current of 2A flows. Calculate the resistance of the

conductor.

b. What PD must be connected across a 100 resistor to

obtain a current of 10mA flowing in it?

c. A car light bulb has a resistance of 6. Calculate the

current, which flows through it when it is connected

across the cars 12V battery.

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RESISTIVITY

The resistance of a conductor depends on both its size and shape, and on

the material from which it is made. The resistance, R, of a regularly

shaped piece of material is proportional to its length, L, and inversely

proportional to its cross-sectional area, A. Mathematically, we may

write

RL

A=

Where is a constant of the material called the resistivity. The unit of

resistivity is the Ohm-metre (m).

SAQ 4Calculate the resistance of a cylindrical piece of copper wire

of length 80cm and diameter 1mm. The resistivity of copper

is 1.72 10-8m.

So far, we have considered resistance as being the property of a single

conductor to resist current flow. In addition, electronic components are

manufactured with specific resistive properties: these are calledresistors.

A typical resistor is shown in Figure 1.8, together with its circuit symbol.

A wide range of resistors are manufactured, with the resistance value

indicated in coded form by coloured band markings.

R

Colour Code

Bands

Tolerance Band

Figure 1.8Illustration of a resistor

component

RESISTORS IN SERIES

When two or more resistors are connected end-to-end, as shown in

Figure 1.8, they are said to be connected in series. If an emf is applied

across the ends of the network, a current I will flow. Since there is only

one path for current to travel, it must be the same in all parts of the circuit.

That is:

I = I1 = I2 = I3 (1)

Figure 1.9

Resistors in series across a

battery

R1 R2 R3

+ - + - + -

I1 I2 I3

V1 V2 V3

V

I

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For the current I to flow through the resistors, there must be a potential

difference (voltage) across each. These are labelled V1, V2 and V3respectively in Figure 1.9, with the plus sign (+) indicating the point of

highest potential and the minus sign (-) indicating the point of lowest

potential. The sum of the voltages V1, V2 and V3 must equal the supply

voltage, V. That is:

V = V1 + V2 + V3 (2)

Using Ohms law for each resistor, we can say that:

V1 = I1R1 ; V2 = I2R2; V3 = I3R3

If Rs represents the equivalent resistance of R1, R2 and R3, then we can

write:

V = V1 + V2 + V3IRs = IR1 + IR2 + IR3Rs = R1 + R2 + R3 (3)

Thus, for resistors connected in series, the equivalent resistance is

equal to the sum of the individual resistances. This rule holds for any

number of resistances connected in series.

EXAMPLE A 30V DC power supply unit (PSU) is applied across 3 resistors of value10, 20 and 30 connected in series. Calculate the current supplied

by the power supply and the voltage across each resistor.

SOLUTION The circuit is illustrated in Figure 1.10.

Let I denote the current flowing in the circuit, and let V1, V2 and V3represent the potential differences (PDs) across the three resistors.

Total resistance, Rs = R1 + R2 + R3= 10 + 20 + 30

= 60

By Ohms law, I = V/R

= 30/60

= 0.5A

Therefore, the current supplied by the PSU is 0.5A. Now let us find thevoltages, V1, V2 and V3, across each resistor.

V1 = IR1= 0.5 10

= 5V

V2 = IR2= 0.5 20

= 10V

V3 = IR3

= 0.5 30= 15V

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As a check, let us see if these three voltages add up to the supply voltage,

V = 30V.

V = V1 + V2 + V3= 5 + 10 + 15

= 30V

Figure 1.10

Series resistor network

SAQ 5a. A DC electric motor, with a resistance of 3, is operated

from a 24V supply. What value of resistor must be

connected in series with the motor if the current through

it is to be limited to 500mA?

b. When 3 identical resistors are connected in series across

a 15V DC supply, a current of 10mA flows. Find the

potential difference across each resistor and its resistance

value in Ohms.

RESISTORS IN

PARALLEL

When two or more resistors are connected side-by-side, as shown in

Figure 1.11, they are said to be connected in parallel. When a source of

emf is applied across a parallel connection of resistors, there is more than

one path through which current can flow.

Figure 1.11

Resistors connected in parallel

across a battery

R1 = 10

+ _ + _ + _

V1 V2 V3

30V

I

R2 = 20 R3 = 30

+ _

V

DC Power

Supply Unit

R1

R2

R3

+

+

+ _

_

_

V1I1

I2

I3

+ _

30V

V

I

V3

V2

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In addition, as each individual resistor is connected across the supply, the

PD across each resistor is equal to that of the supply voltage. That is:

V = V1 = V2 = V3

The total current, I, flowing into the parallel circuit divides amongst the

various branches in inverse proportion to the individual resistance ineach branch. The sum of the individual branch currents must equal the

current flowing into the parallel circuit. That is:

I = I1 + I2 + I3 (4)

Let Rp denote the equivalent resistance of the parallel connection. Then,

using Ohms law and equation (4), we can write

V

R

V

R

V

R

V

Rp= + +

1

1

2

2

3

3

But V = V1 = V2 = V3. Therefore:

V

R

V

R

V

R

V

Rp= + +

1 2 3

1 1 1 1

1 2 3R R R Rp= + + (5)

Therefore, for a network of resistors connected in parallel, the reciprocal

of the equivalent resistance is equal to the sum of the reciprocals of each

of the individual resistances.

A special situation arises when there are just two resistances, R1 and R2,

connected in parallel, as shown in Figure 1.12.

R1

is equivalent to

R2

Rp

R1R2

R1+ R2Rp=

Figure 1.12

Special case of two resistors

in parallel

Using equation (5) to obtain the equivalent resistance gives:

1 1 1

1 2

1 2

1 2R R R

R R

R Rp= + =

+

RR R

R Rp =

+=

1 2

1 2

product of resistances

sum of resistances

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The expression provides a useful short-cut to finding the equivalent

resistance of two resistors connected in parallel. It is important to

remember, however, that it does not hold true for more than two

resistors.

EXAMPLE Three resistors with resistance 5, 10 and 30 are connected inparallel across a 30V battery. Calculate:

a. The current flowing in each resistor.

b. The total current supplied by the battery.

c. The equivalent resistance of the network.

SOLUTION The circuit diagram is given in Figure 1.13

a. Using Ohms law, I1 = V/R1= 30/5

= 6A

I2 = V/R2= 30/10

= 3A

I3 = V/R3= 30/30

= 1A

b. Total current, I = I1 + I2 + I3= 6 + 3 + 1

= 10A

c. Let Rp be the equivalent resistance. Then:

1/Rp = 1/R1 + 1/R2 + 1/R3= 1/5 + 1/10 + 1/30

= 6/30 + 3/30 + 1/30

= 10/30

Rp = 30/10

= 3

As a check, the total current supplied by the battery can be calculated

using the expression:

I = V/R

= 30/3

= 10A

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+

+

+

+ _30V

V

I

R1 = 5

I1

I2

I3

R2 = 10

R3 = 30_

_

_

Figure 1.13

Parallel resistor network

SAQ 6Three resistors of 10, 15 and 30 are connected in parallel

across a supply of V volts. The total current drawn from the

supply is 2A. Calculate the supply voltage V and the current

through each resistor.

SAQ 7What value of resistor must be connected in parallel with a

47k resistor to obtain an equivalent resistance of 42k?

Very often resistor networks contain both series and parallel circuits.Such networks can be analysed by dealing with each series section and

each parallel section separately. This procedure is demonstrated in the

following example.

EXAMPLE For the circuit shown in Figure 1.14, calculate:

a. The total current supplied by the battery.

b. The PD across each resistor.

c. The current flowing in each resistor.

Figure 1.14

Network of series and parallel

resistors

18V

I

+ +

+

_ _

_

V1 V2 V3

R1 = 4 R2 = 6

R3 = 40

R4 = 10

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a. To calculate the total current supplied by the battery, we need to

calculate the total equivalent resistance of the circuit. Consider

firstly the equivalent resistance, Rp, of R3 and R4 in parallel.

Rp = R3R4/(R3 + R4)

= 40 10/(40 + 10)

= 400/50= 8

Thus, the circuit can now be redrawn as in Figure 1.15.

Total resistance, R = 4 + 6 + 8

= 18

From Ohms law, I = V/R

= 18/18

= 1A

Therefore, the current supplied by the battery is 1A.

b. Let V1, V2 and V3 denote the PDs across R1, R2 and Rprespectively.

Again from Ohms law, V1 = IR1= 1 4

= 4V

V2 = IR2= 1 6

= 6V

V3 = IRp

= 1 8

= 8V

Since Rp represents the effective resistance of R3 and R4 in parallel,

we can also say that the voltage V3 = 8V is the voltage across R3 and

R4 in Figure 1.15.

c. The current flowing in R1 and R2 is the battery current, I = 1A.

Let I1 and I2 denote the currents flowing through R3 and R4

respectively.

From Ohms law, I1 = V3/R3= 8/40

= 0.2A

I2 = V3 /R4= 8/10

= 0.8A

As a check, determine if I = I1 + I2 = 1A

I = I1

+ I2= 0.2 + 0.8

= 1A

SOLUTION

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Figure 1.15

Equivalent circuit

SAQ 8For the circuit shown in Figure 1.16, determine the current

through the 20 resistor.

40V

R1 = 9

R2=12

R5=15

R4=20

R3 = 6

Figure 1.16

Circuit for SAQ 8

KIRCHOFFS LAWS

So far we have analysed electric circuits by simplifying them, step-by-

step, into a form to which we can apply Ohms law. For relatively simple

circuits this procedure is quite adequate. However, it is not always easy

to reduce more complex circuits to a form in which Ohms law can be

directly applied. In such situations, Kirchoffs two laws for electric

networks are extremely useful.

First, let us define two terms. Ajunction is a point in a circuit at which

three or more conducting paths are joined. A loop is any closed

conducting path in an electric circuit.

We can now state the two laws of Gustav Kirchoff (1824-1887), the

eminent German physicist, developed by him when he was only twenty

years old.

Kirchoffs first law (the current law) states that:

The total current flowing into any junction equals the total current

flowing out of it.

If we adopt the convention that current flowing into a junction is positive

and that current flowing out is negative, we can re-state Kirchoffs

current law as follows:

The algebraic sum of the currents at any junction in a network is zero.

18V

I

+ + +_ _ _

V1 V2 V3

R1 = 4 R2 = 6 Rp = 8

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Consider Figure 1.17.

Junction(node)

I1

I5

I4

I3

I2

I1 + I4 = I2 + I3 + I5Figure 1.17

Illustration of Kirchoffs

current law

The current entering junction = current leaving junction

I1 + I4 = I2 + I3 + I5Or,

I1 - I2 - I3 + I4 - I5 = 0

Kirchoffs second law (the voltage law) states that:

In any closed loop in a network, the algebraic sum of the voltage drops

(PDs) around the loop is equal to the emf acting in that loop.

Consider Figure 1.18.

Figure 1.18

Illustration of Kirchoffs

voltage law

In this case, there is only one loop.

emf = sum of voltage drops

= V1 + V2 + V3= IR1 + IR2 + IR3

In applying Kirchoffs voltage law, we must carefully assign the correct

polarity (positive or negative value) to the voltage drops. The general

convention used is as follows. When progressing around a loop, a PD

going from high to low in the direction of progress is taken as positive;

conversely, a PD going from low to high is taken as negative. Therefore,

if we progress around a loop and the current is in the same direction as

our progress, the PD is taken as positive. If the current direction is

opposite to our direction of progress, the PD is taken as negative.

EV2

V3

V1

+

_

_

+

+

E = V1 + V2 + V3

_

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EXAMPLE As an example of the application of Ohms and Kirchoffs lawsconsider the circuit shown in Figure 1.19. Given the information

indicated on the diagram, calculate the values of I1

and R2

.

SOLUTION From Ohms law - : V3 = I3R3 = 3 4 =12V, which is also equal to

V2 since the resistors R2 and R3 are in parallel.

From Kirchoffs Voltage law - : V1+V

2=36V. Therefore V

1= 36 -

V2

= 36 - 12 = 24V .

From Ohms law - : I1

= V1/ R

1= 24 / 6 = 4A.

From Kirchoffs current law - : I1

= I2

+ I3. Therfore I

2= I

1- I

3= 4 -

3 = 1A.

From Ohms law - : R2

= V2

/ I2

= 12 / 1 = 12.

Figure 1.19

Illustration of Ohms and

Kirchoffs Laws

SAQ 9Given the information indicated on the diagram in Figure

1.20, calculate the values of I1

, I2

, I3

, V2

and V3.

Figure 1.20

Circuit for SAQ 9

36V

I1

+ -

V1

R = 6 + -V

2

R2

R3 = 4

I2

I3= 3A

+ -V3

1

I1

I3

I2

R3 = 2R1 = 2

R2 = 6

+ -

4V

+-

V3

+

-

V210V 4V

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INTERNAL

RESISTANCE

So far the voltage sources in our circuits have been ideal in that they have

had zero internal resistance. Practical voltage sources, including power

supplies and batteries possess some internal resistance. When a

voltage source delivers current to a circuit a potential difference is

developed across this internal resistance and so not all of the e.m.f of the

source is produced at its terminals; some potential (voltage) is lost across

the internal resistance of the source. Practically, the internal resistanceof a source can be represented as a resistance, r, in series with the e.m.f.

as shown in Figure 1.21 (a). Now consider this source connected to an

external resistance R as shown in Figure 1.21(b). This circuit is just a

source of e.m.f which causes a current to flow in a resistive circuit

consisting of two resistances, r and R, connected in series, which is a

simple circuit and can be analysed using the methods we have been

studying. The main point is that because there is a voltage drop across

the internal reistance r, the voltage developed across the external

resistance R is less than the e.m.f. value E. Thus in circuit analysis

problems where it is required to take into account the internal resistance

of the source, the technique is to place the internal resistance value in

series with the source of e.m.f and then treat it like any other resistivecomponent.

A

B

E VAB = E

+

+

_

_

r

(a)

A

B

E VAB = .E

Vr

+

+

+

_

_

_r

(b)

R R

R+r

I=E

R+r

Figure 1.21

Internal Resistance of a Voltage

Source

SUPERPOSITION

THEOREM

The Superposition Theorem states that:

In any (linear) circuit,the voltage or current in any part of the network

is the sum of the voltages or currents caused by each generator acting

separately, with the other generators replaced by their internalresistances.

To use the Superposition Theorem, one of the sources of e.m.f. is

selected and the circuit is re-drawn with only this source of e.m.f.

present, and all other sources of e.m.f. are replaced with their internal

resistances only. The voltages or currents at the required parts of the

circuit are then determined using Ohms law and the relationships for

resistors in series and in parallel. This is then repeated in turn for each

of the other sources of e.m.f., and finally all of the voltage or current

values are summed. The sum is the actual voltage or current due to alll

of the sources of e.m.f. in the circuit acting together.

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ELECTRIC POWER

Suppose we have a current I flowing through a resistance R, as illustrated

in Figure 1.22, where the PD across the resistance is V.

Figure 1.22

Illustration of power

dissipation in a resistor

If the current, I, flows for a time t seconds, then a charge of Q coulombs

will be transferred from B to A, where Q = It. It may be shown that, the

work done, W, in transferring this charge Q from B to A, against a

potential V, is given by:

W = QV = VIt joules

This work done by the voltage source is expended in the form of thermalenergy (heat). The energy dissipated per second in the resistor is defined

as the power dissipation in the resistor. It is measured in watts. Hence:

P = W/t

= VI t/t

= VI watts

From Ohms law:

V = RI or I = V/R

P = VI

= (IR)I= I2 R

Also:

P = VI

= V(V/R)

= V2/R

Therefore, there are three, equivalent expressions for electric power

dissipation in a resistance. These are:

P = VI; P = I2R; P = V2/R

EXAMPLE For the circuit shown in Figure 1.23, calculate:

a. The power delivered by the battery.

b. The power dissipated in each resistor.

c. Verify that the total power dissipation is equal to the power

delivered by the battery.

10VI

R = 31 R = 22

Figure 1.23Two resistor circuit

Heat

A B

V

R

I+ _

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SOLUTION a. I = 10/(3 + 2)= 2A

The power, P, delivered by the battery is:

P = VI

= 10 2= 20W

b. Let P1 and P2 denote the power dissipation in R1 and R2 respectively:

P = I2R1= (2)2 3

= 12W

P = I2R2= (2)2 2

= 8W

c. Total power dissipation = P1 + P2= 12 + 8

= 20W

This equals the total power delivered by the battery, as calculated in part

a. of this example.

SAQ 10a. A car light bulb is rated at 24W, 12V. What current flows

in the bulb when it is switched on and what is its resistance?

b. An electric soldering iron has an element of resistance

100 and carries a steady current of 0.5A for 1 hour. How

much electrical energy is converted into heat energy

during this period? What is the power rating of the iron?

SUMMARY

1. A DC source of emf is one whose voltage output remains constant

with time.

2. The resistance R of a conductor of length L and cross-sectional

area A is given by:

R = L/A where is a constant of the material known as theresistivity.

3. Ohms law states that resistance, voltage and current are all inter-

related as follows:

R = V/I; V = RI; I = V/R

4. In a series resistor circuit:

Rs = R1 + R2 + R3 + ........ + RnI = I1 = I2 = I3 = ............. = In

V = V1 + V2 + V3 + ........ + Vn

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5. In a parallel resistor circuit:

1/Rp = 1/R1 + 1/R2 + 1/R3 + ...... + 1/RnI = I1 + I2 + I3 + ................... + InV = V1 = V2 = V3 = ............. = Vn

In the special case of two parallel resistors:

Rp = R1R2/(R1 + R2) (= product/sum)

6. For series-parallel resistor networks, each series and parallel section

has to be treated separately and their equivalent resistances calculated.

These are progressively substituted in the combined circuit until the

original circuit is reduced to a simple series or parallel circuit.

7. Kirchoffs first law (the current law) states:

The algebraic sum of the currents flowing towards a junction in an

electric circuit is zero. That is, the current flowing into the junction

is equal to the current flowing out of it.

8. Kirchoffs second law (the voltage law) states:

In any closed loop in an electric circuit, the algebraic sum of the

voltage drops (PDs) around the loop is equal to the emf acting in

that loop.

9. The Superposition Theorem states:

In any (linear) circuit,the voltage or current in any part of the

network is the sum of the voltages or currents caused by each

generator acting separately, with the other generators replaced by

their internal resistances.

10. Electric circuits can be analysed to obtain the currents and voltages

in each part of the circuit using a combination of Ohms law,

Kirchoffs laws and the Superposition Theorem.

11. When an electric current flows in a resistance R, electrical energy

from the source of emf is converted into heat. The power dissipation,

P, in watts, which defines the rate at which energy is converted, is

given by the three equivalent expressions:

P = VI; P = I2R; P = V2/R.

12. The power, P, produced by a source of emf, V, which delivers a

current, I, to a circuit is given by:

P = VI.

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ANSWERS TO SAQSSAQ 3

From Ohms Law: R = V/I; V = IR; I = V/R

a. R = V/I

= 20/2= 10

b. V = IR

= 10 10-3 100

= 1V

c. I = V/R

= 12/6

= 2A

SAQ 4

The cross-sectional area, A, of the wire is given by the

expression:

A = r2

= (0.0005)2 m2

= 7.85 10-7 m2

The resistance of the wire is given by the expression:

R = L/A

where:

= 1.72 10- 8 Ohm-metres

L = 0.8m

A = 7.85 10-7 m2

Therefore:

R = (1.72 10-8 0.8)/(7.85 10-7)

= 0.0175

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SAQ 5

a. The circuit is illustrated in Figure 1.24

Let R = the resistance of series resistor

Rm = resistance of motor

I = V/(R + Rm)

0.5 = 24/(R + 3)

0.5(R + 3) = 24

0.5R + 1.5 = 24

0.5R = 22.5

R = 45

b. The circuit is illustrated in Figure 1.25

Let R represent the resistance of each resistor. The total

resistance, RT, is therefore:

RT = R + R + R = 3R

RT = V/I

3R = 15/(10 10-3)

= 1500

R = 1500/3

= 500

V1 = V2 = V3 = IR

= 10 10-3 500

= 5V

V

24V

R

I = 500mA

= 0.5A

Motor

+_

R = 3

Figure 1.24Circuit for SAQ 5a

Figure 1.25

Circuit for SAQ 5b

I = 10mA

V = 15V

V1

R R

+ _ V2+_ V3+

_R

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SAQ 6

The circuit is illustrated in Figure 1.26

Let Rp represent the equivalent resistance of R1 , R2 and R3

in parallel.

1/Rp = 1/R1 + 1/R2 + 1/R3= 1/10 + 1/15 + 1/30

= 3/30 + 2/30 + 1/30

Rp = 30/6

= 5

V = IRp= 2 5

= 10V

Let I1, I2 and I3 denote the currents through R1, R2 and R3respectively.

I1 = V/R1= 10/10

= 1A

I2 = V/R2= 10/15

= 0.67A

I1 = V/R3= 10/30= 0.33A

V

I1 I2 I3

I = 2A

R1=10

R2=15

R3=30

Figure 1.26

Circuit for SAQ 6

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SAQ 7

For two resistors R1 and R2 in parallel, their equivalent

resistance Rp is given by the expression:

Rp = R1 R2 /(R1 + R2)

In the case, Rp = 42k and R1 = 47k

The problem is to calculate R2 :

47 R2 /(47 + R2) = 42

47R2 = 42(47 + R2)

47R2 = 1974 + 42R25R2 = 1974

R2 = 394.8k

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SAQ 8

To find the current through the 20 resistor R4, we need to

know the voltage across it. Since R2, R4 and R5 are in

parallel, it is possible to reduce these to a single series circuit,

as shown in Figure 1.27. It is now easy to find the voltageacross Rp and hence R2 .

1/Rp = 1/R2 + 1/R5 + 1/R4= 1/12 + 1/15 + 1/20

= 5/60 + 4/60 + 3/60

= 12/60

Rp = 60/12

= 5

Total resistance, R = R1 + Rp + R3

= 9 + 5 + 6= 20

Current, I = V/R

= 40/20

= 2A

Voltage across Rp = IRp= 2 5

= 10V

Current through R4 = I4I4 = 10/20

= 0.5A

V 40V

IR1= 9

R3= 6

Rp

Figure 1.27

Circuit for SAQ 8

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SAQ 9

Using Ohms law - : I1

= V1

/ R1

=4 /2 = 2A.

Using Kirchoffs voltage law- : V2

= 10 - 4 = 6V.

Using Ohms law - : I2

= V2

/ R2

= 6 / 6 = 1A.

Using Kirchoffs current law - : I1 + I3 = I2 .Therefore I

3= I

2- I

1= 1 - 2 = -1A.

The minus sign above indicates that the current is in fact

opposite in direction to that indicated, ie a value of 1A in the

left to right direction through R3.

From Ohms law - : V3

= I3R

3= -12 = - 2V

Note again that the negative sign indicates that the polarity of

the voltage across R3is opposite to that originaaly marked on

Figure 1.20, ie the left side of R3

is at a potential which is 2V

higher than the right side.

This shows that current directions and voltage polarities must

be borne in mind when dealing with circuits of this type.

SAQ 10

a. Power = voltage current

P = VI

24 = 12I

I = 2A

Resistance, R = V/I

= 12/2

= 6

b. Electrical energy transferred = power time

= (VI) t

= (I2R) t

In the case:

I = 0.5AR = 100

t = 1 hour = 60 60 seconds = 3600 seconds

Energy converted to heat = (0.5)2 100 3600

= 90,000 joules

= 90kJ

Power, P = I2R

Power rating of soldering-iron = (0.5)2 100

= 25W

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ELECTRIC FIELDS

OBJECTIVES



electric field

electric field strength

equipotential lines

lines of electric force

permittivity

2. Calculate the force on electric charges situated in an electric field,

given the appropriate data.

3. Calculate the work done in moving a charge in an electric field.

GRAPHICAL

REPRESENTATION OF

ELECTRIC FIELDS

We have already seen that charged bodies in close proximity exert forces

upon one another. For like charges, the forces are repulsive; for opposite

charges, the forces are attractive. We will now try to explain how these

forces arise. One of the first people to try and do this was the English

physicist Michael Farady, who, in the 1830s, suggested that the space

around charged objects was filled with something he called an electric

field. This field served as the medium for transmitting the electrical

forces, just as water transmits waves or air carries sound through it.

Consider two parallel plates a small distance d apart, having a potential

difference of V volts between them as shown in Figure 1.28. Assume that

this potential difference causes plate A to have a positive charge +Q andplate B to have a negative charge -Q. An electric field exists in the area

between the two plates. If the distance between the plates is very small

compared to their area, virtually all of the field will be confined to the

region between them. There will be a slight fringing effect, ie the

electric field also exists in a small region outside that exactly between the

two plates. (See Figure 1.28.)

As an aid to visualising this electric field, Faraday developed a pictorial

scheme called lines of force. These imaginary lines represent electrical

force in the following way:

. The direction of the force at any point in space is along the lines.

. The strength of the force is greatest where the lines are closely

spaced.

. The direction of the lines is that in which a positively charged body

would move if it were placed in the field.

. The lines always repel and never cross one another.

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Figure 1.28

Field between two parallel

plates

If a small positive charge is inserted anywhere in the field between the

two plates, it will experience the same net force. Although the force of

attraction towards plate B and the force of repulsion from plate A varies

with the distance of the charge from the plates, the resultant of the forces

is always constant. Since the force on a given charge is the same at all

points, the strength of the electric field must be the same at all points: the

field is said to be uniform.

Another useful device in visualising the electric field is the idea of

equipotential lines. These lines are drawn at right angles to the lines of

force and link points in an electric field which are at the same potential.

If the potential on the plates is V1 and V2 respectively, and the field

between them is uniform, the potential of points in the field must

progress linearly from V1 at plate A to V2 at plate B. This means thatlines of equipotential can be drawn through the field parallel to the plates

as shown in Figure 1.29(a). There exists a potential gradient which

varies linearly with distance between the plates as shown in Figure

1.29(b).

d

V2

(a) Equipotential

Lines

(b) Potential

Gradient

Equipotential

Lines

V1

V1

V2

d

vFigure 1.29

Equipotential lines and

potential gradient

d

A

V1 V2

V1 - V2

d

V1 > V2

'fringing'

E =

+Q -Q

B

+ - V = V1 - V2v

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The potential difference between these lines compared to the distance

between them is a measure of the electric field strength E:

E =

V

dvolts per metre

The direction of field strength is positive in the direction of lower

potential, which is also the direction of the force on a positive charge

placed between the two points.

Electric field strength is sometimes called voltage gradient.

SAQ 11(a) A potential of 200V is applied between two parallel plates

spaced 2mm apart. Calculate the field strength between

the plates.

(b) Two parallel plates are situated 1mm apart. What potentialmust be applied between the plates to produce a field

strength of 5000V/m?

An alternative definition of electric field strength is as follows:

The electric field strength at any point in a field is defined as the force

which would be exerted on one unit of positive charge if it were placed

at that point.

Thus if a charge Q is placed in an electric field E, it experiences a force

F given by:

F = QE

SAQ 12A charge of 10C is situated in an electric field of strength

2000V/m. Determine the force acting on the charge.

The work done, W, in moving a charge, Q, a distance, d, in an electric

field of strength, E, is equal to the product of the force on the charge and

the distance moved in the direction of the force. That is:

W = Fd

= QEd

But E = V/d. So:

W = Q(V/d)d

= QV

Or, V = W/Q

This expression leads to a formal definition of potential difference:

The potential difference between two points in an electric field is the

work done in moving a unit of positive charge between the two points.

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SUMMARY

1. Electric fields can be represented graphically by lines of force.

Lines of force are directed from positive to negative charge (potential)

and never cross one another. Their closeness represents field strength.

2 . The electric field strength, E, is defined in magnitude and direction

as the force exerted by the field on a unit of positive charge. For two

conductors a distance d apart having a potential difference Vbetween them, the electric field strength, E, is given by:

E = V/d

3. An equipotential line links points in an electric field which are at

the same potential. Their direction is always at right-angles to the

lines of force.

4. A charge, Q, placed in an electric field of strength, E, experiences

a force F given by the expression:

F = QE

5. The potential difference between two points in an electric field is the

work done in moving a unit of positive charge between the two

points.

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ANSWERS TO SAQSSAQ 11

(a) Electric field strength = voltage between plates/distance

between plates

E = V/d= 200/0.002

= 100 000V/m

= 105 V/m

(b)

E = V/d

5000 = V/0.001

V = 5000 0.001

= 5V

SAQ 12

Force = charge electric field strength

F = Q E

= 10 10-6 2000

= 0.02N

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CAPACITORS AND

CAPACITANCE

OBJECTIVES



. capacitance

. capacitor

. dielectric

. dielectric field strength

. equivalent capacitance

. Farad

. maximum working voltage

. parallel-plate capacitor

. relative permittivity

. time-constant

. parasitic capacitance

2. Calculate the amount of charge stored on a capacitor, given the

appropriate data.

3. Calculate the capacitance of a parallel-plate capacitor, given the

appropriate data.

4. Calculate the electric field strength in the dielectric of a capacitor,

given the appropriate data.

5. Calculate the energy stored in a capacitor, given the appropriate data.

6. Calculate the time-constant of simple resistor-capacitor networks.

7. List the main types and applications of capacitors.

CAPACITANCE

One of the most important circuit components of our electronic age is the

capacitor.A capacitor is a device for storing charge. They are used to

minimise voltage variations in power supplies, to increase the efficiency

of electric power transmission, to measure time, and to detect and

generate radio waves.

We can illustrate the operation of a capacitor by considering the circuit

shown in Figure 1.30. The two terminals of a battery are each connectedto a length of conducting wire. Each length of wire goes to a small, flat

plate, also made of conducting material. The two plates are separated

from each other by a small air gap. What happens in such a circuit?

The emf of the battery causes current to flow until the plate connected

to the positive battery terminal receives a charge Q and achieves the

same positive potential as the battery terminal itself. Similarly, the plate

connected to the negative terminal of the battery receives a negative

charge, - Q. In other words, the flow of current stops when the potential

difference across the capacitor equals the potential difference across the

battery. Note that the charges on the two plates are equal in magnitude

but opposite in sign.

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Figure 1.30

Operation of a capacitor

Suppose we remove the battery. What happens? Answer: The positive

and negative charges remain on the two plates. Hence, this arrangement

of two flat plates separated by an air gap has the ability to store an electric

charge.

The capacity of a device to store electric charge, called its capacitance,

is defined as the ratio of charge on either plate to the potential difference

between the two plates. If Q (in coulombs) is the charge on either plate

and V (in volts) represents the potential difference between the plates,

then C (the capacitance) is equal to:

Capacitance = Charge on either plate/Potential difference

C = Q/V coulombs/volts

Rather than measure capacitance in coulomb/volts, physicists have

given it a special unit of its own: the farad (F), named after MichaelFaraday. A device has a capacitance of one farad if it holds a charge of

one coulomb when a potential of one volt is applied across it.

In practice, the farad is an extremely large unit. For example, a capacitor

of 1 farad capacity, with two parallel plates 1mm apart in air, would

require each plate to have an area of over 40 square miles!

Practical capacitors typically have values in microfarads (F) nanofarads

(nF) and picofarads (pF):

1 microfarad (F) = 10-6 F

1 nanofarad (nF) = 10-9 F

1 picofarad (pF) = 10-12 F

We may summarise the properties of a capacitor thus:

. A capacitor is not like a resistor: it does not allow DC current to pass

through it.

. A capacitor is not like a battery or other emf generator: it must be

charged before it can supply a current. It can also be charged to many

different voltages.

. A capacitor stores electric charge.

V

Conducting

Plates V

-Q

Charging current flows

into capacitor

+Q

+

_

I

Air-gap V

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THE PARALLEL-

PLATE CAPACITOR

Any configuration of conductors separated by an insulating material is

capable of storing charge and, consequently, of possessing capacitance.

For example, the earth and the ionosphere, a layer of charged particles

surrounding our planet at a height of roughly 100km, may be together

considered as a huge capacitor.

The simplest form of capacitor is the type considered above - two flat,metal plates, separated by a thin layer of insulating material (air). (Figure

1.31). This is termed a parallel-plate capacitor. Assuming the

potential of the battery or other emf source is constant, how can we vary

the capacitance of our parallel-plate device?

Because the two plates are oppositely charged, there must be an electric

field between them; if the plates are close together, we can assume that

the electric field lines are straight and parallel. In the area within the

plates, the electric field (E) is simply the potential difference (V) divided

by the distance (d) between them.

electric field = potential difference/distance apart

E = V/d

One value which cannot change in this equation is V, the potential

difference across the plates; as V is generated by the constant emf source,

V remains fixed in magnitude.

If we reduce the distance (d) between the two plates, the electric field (E)

must increase, as the potential difference (V) cannot change. As E

increases, then so too must the charge (Q) on each plate. This in turn

increases the capacitance. Hence, the capacitance is inversely proportional

to the distance separating the two plates. Mathematically:

C 1/d

Another option is to increase the area (A) of the two plates. Suppose two

larger plates are held the same distance apart and the same potential

applied across them. The charge per unit area must remain the same,

since the electric field is the same as before. As the area is greater, so too

must be the amount of charge stored. Hence, the capacitance is directly

proportional to the area of the plates:

C A

Combining this with the previous equation gives us:

C A/d

In summary, the capacitance of a parallel-plate capacitor (Figure 1.31)

is directly proportional to the area of its plates and inversely proportional

to their distance apart.

For any given insulating material between the plates, it is found that:

C = A/d

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where is a constant known as the permittivity of the insulator. By

permittivity we mean the degree to which the insulator permits the

storage of charge on the two plates; the greater the permittivity, the

greater the charge stored and the greater the capacitance. The units of

permittivity are farads per metre (F/m).

DIELECTRIC

Spacing d

Area AFigure 1.31

Parallel-plate capacitor

Usually, the permittivity is not expressed directly, but in relation to the

permittivity of free space (a vacuum), where 0= 8.854 x 10-12 F/m. The

relative permittivity of an insulator, denoted by the symbol r, is its

own, unique permittivity divided by the permittivity of free space.

Relative permittivity has no units.

relative permittivity = permittivity of insulator/permittivity of free space

r = /0

Insulators used to separate the two plates of a capacitor are calleddielectrics. r is therefore termed the relative permittivity of a dielectric.

In conclusion, we can now answer the question with which we began this

section: for a constant emf source, how can capacitance be varied?

. by moving the plates closer together or further apart

. by increasing or decreasing the area of the plates

. by using a dielectric with a greater or lesser permittivity.

These three factors are related to capacitance by the following equation:

C = r0A/d

EXAMPLE A parallel-plate capacitor consists of two plates of dimension 10cm 5cm, separated by 1mm layer of mica which has a relative permittivity

of 5. Calculate the capacitance of the device and the charge stored when

the PD between the plates is 100V.

(0 = 8.85 10-12 F/m)

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SOLUTION area of plates = 10 5cm= 50cm2

= 0.005m2

capacitance C = 0r A/d

= (8.85 10-12

5 0.005)/10-3

= 0.22 10-9 F

= 220pF

charge stored = capacitance voltage

Q = CV

= 0.22 10-9 100

= 0.22 10-7 C

= 0.022C

SAQ 13List three possible ways of changing the capacitance of a

parallel-plate capacitor.

SAQ 14Calculate the amount of charge stored in a 0.1F capacitor

which is charged to 25V; that is, with a potential difference

of 25V across it.

SAQ 15 A 2700pF capacitor is to be constructed using a paper dielectricof thickness 0.5mm. If the relative permittivity of paper is 2.5,

calculate the required area of the plates.

(0= 8.854 10-12 F/m)

An important quantity in capacitor design is dielectric field strength.

This is the magnitude of the applied electric field which the material of

the dielectric is capable of withstanding without breaking down. Because

of the very close proximity of the plates in a parallel-plate capacitor, the

electric field in the dielectric can be very large. (E = V/d; E is large if d

is small). If high enough, the electric field can cause the dielectric tobreak down and conduct current. This happens when the force of the

electrons on the atoms of the dielectric (normally an insulator) is so great

that these electrons become detached from their nuclei and cause

conduction, often damaging the dielectric.

The maximum amount of field strength which a given dielectric can

withstand is called the dielectric field strength of that dielectric. As the

distance d between the plates is fixed during manufacture, an upper limit

must be set on the voltage that can be applied across the capacitor. Since:

dielectric strength, E = V/d

Vmax = dE

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In practice, a maximum working voltage, which is less than the value

capable of causing dielectric breakdown, is quoted for the capacitor, so

that the dielectric strength limit is never reached in normal use.

SAQ 16

A capacitor uses a dielectric of thickness 0.1mm and operatesat a voltage of 50V. Calculate the amount of dielectric field

strength across the dielectric.

CAPACITORS IN

SERIES AND

PARALLEL

Like resistors, capacitors may be connected together into series or

parallel networks, or in some combination of each. In every case, these

networks may be represented by a single, equivalent capacitance.

Figure 1.32Parallel Capacitor Network

Figure 1.32 shows three capacitors connected in parallel. Note here the

standard, circuit symbol for a capacitor: two parallel lines, separated by

a small space. It may easily be shown that the equivalent capacitance Cp

of this parallel connection is

Cp

= C1 + C2 + C3

Thus, for any number of capacitances connected in parallel, the equivalent

capacitance is equal to the sum of the individual capacitances.

Figure 1.33 shows three capacitors connected in a series network. It

maybe shown that the equivalent capacitance Csfor this series connection

is given by

Figure 1.33

Series Capacitor Network

1/Cs = 1/C1 + 1/C2 + 1/C3

Thus, for any number of capacitors in series, the reciprocal of the

equivalent capacitance is equal to the sum of the reciprocals of the

individual capacitances.

You will notice that the above cases are exactly opposite to those for

resistors in series and parallel.

VC1 C2 C3

Q1 Q2 Q3

S

S

Charging

Current I

C1 C2 C3

Q1 Q2 Q3

+ _ ++V1 V2 V3

V

_ _

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A capacitor, a powerhouse of stored charge, is also a powerhouse of

stored electrical energy. When a capacitor receives charge from an emf

source, energy is transferred from the source to the capacitor and stored

in the form of an electric field between the plates. The capacitor will

continue to hold this energy even when the supply is removed; it

therefore becomes a small energy source in its own right. The energy

stored in a capacitor can be regained by allowing the capacitor todischarge through a resistor or other circuit component, for example an

electronic flash for a camera.

It may be shown that the amount energy E stored in a capacitor C charged

to a voltage V may be quantified as

E = CV2 /2

The energy stored in a capacitor is therefore equal to half the product of

the capacitance and the square of the voltage across the two plates.

Since capacitors can store electrical energy, could they be used toreplace batteries, such as the standard car battery for example?

The typical 12V car battery can transfer a total of about 3.6 10-4 C of

charge. Using the equation C = Q/V, the capacitance of the battery is 3

10-3 F. Assuming that capacitor plates are 1mm apart and the dielectric

is free space, the area of the plates required is given by the equation:

A = Cd/0

which gives the result of 3.39 105 m2. This corresponds to a square of

about 360 miles to a side! We can therefore conclude that, in this

example at least, a capacitor cannot be used as a large source of electrical

energy.

As energy storers, however, capacitors can perform functions that

batteries cannot. For example, camera flash-bulbs are powered by a

capacitor which is in turn charged by a battery. The advantage of using

the capacitor is that it can provide a short duration flash as it is

discharged. In addition, capacitors are not damaged by such sudden

discharge of all their electrical energy; most types of battery would be.

ENERGY STORAGE IN

A CAPACITOR

CHARGE AND

DISCHARGE OF A

CAPACITOR

THROUGH A

RESISTANCE

Consider the circuit of Figure 1.34(a) consisting of an initially uncharged

capacitor C, a resistor R and a switch S that are connected in series to a

dc power supply E. How will the voltage VC

across the capacitor vary

when the switch is instantaneously closed? Closing the switch suddenlyconnects the positive terminal of the supply to the resistor forming a

complete circuit and current begins to flow. This current begins to charge

the capacitor, so that a positive charge accumulates on the plate A of the

capacitor ( and a negative charge accumulates on the other plate B). As

a result, the potential (voltage) on plate A increases from zero to a

positive value. As more charge accumulates on the capacitor, this

positive potential continues to increase. As it does so, the value of the

potential across the resistor is reduced; consequently, the current decreases

with time, approaching the value of zero as the capacitor potential

reaches its final maximum value of E volts.

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Figure 1.34

Charging a Capacitor from a

Voltage Source through a

Resistance

Figure 1.34(b) graphs this voltage VC

as a function of the time and Figure

1.34 (c) shows how the current i varies with time. Theoretically full

charging of the capacitor takes an infinite amount of time but in practice

it becomes virtually fully charged after a finite amount of time. The

quantity = RC is known as the time-constant of the circuit. After a time

equal to one time-constant , the capacitor is charged to 63.2% of its fully

charged value and after a time equal to 5 (5 x time-constant), the voltage

has reached 99.3% of its fully charged value.

The charge and discharge of a capacitor has many important applications

in modern electronics, particularly in relation to timing circuits.

TYPES OF CAPACITOR

AND MAIN

APPLICATIONS

There are many capacitor types and they are utilised in a wide variety of

application areas across the broad spectrum of electronic engineering.

For example they are used as the basic storage element in certain types

of digital computer memories a charged capacitor represents the

storage of a binary value of 1 and an uncharged capacitor represents

the storage of a binary value of 0. They are used as part of a

smoothing circuit in dc power supplies in which a time-varying

alternating current (ac) input voltage is converted to a constant dc output

voltage. As indicated above they are are utilised in timer circuits which

operate on the time for the capacitor to be charged to a certain voltage.

In conjunction with resistors and other electronic components they areused to design electronic filter circuits which are capable of passing

signals in a certain frequency range and rejecting frequencies outside

this range. This can be particularly useful in separting wanted signals

which are contaminated by electronic noise. These are just some

examples of the application of capacitors and throughout the remainder

of this unit and the course in general you will study these and other

applications of capacitors in more detail.

Capacitors are classified according to the type of dielectric. Figure 1.35

provides summary information on a variety of capacitors classified

according to the type of dielectric. Table 1.1 summarises the main

features of each capacitor type and also lists some broad application

areas.

S

(a) Circuit

(b) Variation of capacitor

voltage with time

Ri

1k

C

1F

VcE(10V)

E

+

_

+

_

Vc(t)

Vc(t) = E(1 - e-t/RC)

(c) Variation of charging

current with time

E/R

t t

i(t)

i(t) = e-t/RCE

R

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Figure 1.35

Classification of Capacitor

Types

Ceramics

Film

Temperaturecompensation

Highdielectricconstanttype

Semiconductortype

Polyester(Myler)

Polypropylene

Polystyrene

Aluminum

electrolytic

Tantalum

electrolytic

Mica

Multi layered

Single layered

Multi layered

Single layered

Single layered

Capacitors

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Type Feature Application circuit

Multi-layered ceramic

capacitor

This type of capacitor has been developed to meet

demands for high-density ceramic capacitors. Multi-

layered ceramic capacitors incorporate multiple printedlayers of electrode plates made of 20 to 50-f thick

ceramic sheets. These capacitors are more compact

and have better temperature characteristics than single-

layered ceramic capacitors. Multi-layered ceramic

capacitors are, however, rather expensive because their

electrode plates use precious metals. With the further

development of materials for electrode plates, these

capacitors are expected to take the lead in the main

stream of ceramic capacitors. Like a single-layered

ceramic capacitor, a multi-layered ceramic capacitor is

either a product of high dielectric constant construction

or a product that has excellent temperature

characteristics ideal for temperature compensation.

Circuits of general electronic

equipment

Single-layered ceramic

capacitor

This type of capacitor that incorporates a ceramic

dielectric is either a product ideal for temperature

compensation, a product that has a high dielectric

constant, or a product that is of a semiconductor type,

which depends on the kind of ceramic dielectric. The

single-layered ceramic capacitor for temperature

compensation does not have a high electrostatic capacity

but it covers a wide temperature characteristic range

including temperatures below 0C. The capacitor of

high dielectric constant construction is compact yet ithas a high electrostatic capacity. The semiconductor

type capacitor is far more compact yet it has the highest

electrostatic capacity of all single-layered ceramic

capacitors.

Circuits that require excellent

frequency characteristics, such

as high-frequency circuits and

digital circuits

Film capacitor The high-frequency and temperature characteristics of

film capacitors excel those of ceramic capacitors.

Furthermore, high-capacity film capacitors are

available, which are, however, more expensive and

larger than ceramic capacitors that are the same in

capacity. Polyester (Myler), polypropylene, orpolystyrene can be used for the film of this type of

capacitors.

High-frequency circuits

and analog circuits

Metal-glazed

film capacitor

This type of capacitor incorporates electrode plates

made of film vacuum-evaporated with metal such as

aluminum. Metal-glazed film capacitors can be more

compact yet higher in electrostatic capacity than film

capacitors. Polyester, polypropylene, or polycarbonate

can be used for the film of this type of capacitors.

High-frequency circuits,

switching circuits, and analog

circuits

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Aluminum

electrolytic capacitor

This type of capacitor incorporates a metal oxide film

dielectric produced from electrolysis and its anode is

made of aluminum. That is the reason this type of

capacitor is called aluminum electrolytic capacitor. It

is possible to produce high-capacity aluminum

electrolytic capacitors. Their frequency and temperature

characteristics are, however, bad.

Power supply circuits, audio

circuits, timer circuits, and

backup circuits

Tantalum electrolytic

capacitor

This type of capacitor is similar to aluminum electrolytic

capacitors, but its anode is made of tantalum instead of

aluminum. Tantalum electrolytic capacitors are a little

inferior in electrostatic capacity to aluminum

electrolytic capacitors. The frequency and temperature

characteristics of tantalum electrolytic capacitors,

however, excel those of aluminum electrolytic

capacitors.

Noise limiters, coupling circuits,

and filter circuits

Others Mica, glass, and paper are used for dielectric elements

as well as the materials described above. Mica is the

best dielectric but it is expensive. Glass ensures a

stable temperature coefficient over a wide range. Paper

is used for high-voltage capacitors.

Precision equipment and high-

voltage equipment

Table 1.1

Summary of CapacitorFeatures and General

Application areas

PARASITIC

CAPACITANCE

Unfortunately capacitance arises naturally in many situations in electronic

circuits when it is unwanted since it can be detrimental to their

performance. For example, two parallel copper tracks on a printed

circuit board have a small capacitance between them. The copper tracks

act as the parallel plates and the insulating material of the board substrate

acts as a dielectric. This type of naturally occurring capacitance is

known as parasitic capacitance and it can greatly reduce the speeed at

which electronic signals can be transmitted on the printed board in thisexample. Because of the materials used and the way in which electronic

components such as transistors, diodes, inductors, and even resistors are

constructed, parasitic capacitance exists thro

ct1 unit 1

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