Unit 2: AC Electricity 2-1

AC ELECTRICITY

2

AIMS

The aims of this unit are to describe and quantify the behaviour ofresistors, inductors and capacitors in AC circuits.

INTRODUCTION

In Unit 1 we studied direct current or DC electricity. In DC circuits, thevoltage remains constant with respect to time and the current flows inone direction only through the circuit.

In this unit we will be studying another form of electricity, called

alternating current or AC. In AC circuits, the value of the voltage is

constantly changing with time, as is the direction of current flow through

the circuit. In particular, we will be looking at the behaviour of simple

AC circuits containing resistances, inductances and capacitances. For

the purpose of this course, this unit will complete our study of electricity.

Why use AC? Why build circuits in which the magnitude and directionof the voltage is constantly changing, and in which the current isoscillating back and forth, rather than flowing steadily in one direction?DC circuits seem so straightforward in comparison: all that are neededin analysing them are Ohm’s and Kirchoff’s laws. So why not use DC inall electric and electronic applications?

The answer is that AC is the only form in which electrical power can beeconomically transmitted over long distances. The following exampledemonstrates this point very clearly. Suppose we wish to supply a citywith ten million watts of electrical power. Our generating station islocated some distance away, and we wish to minimise the powerdissipation along the transmission lines.

Remember that electrical power (in watts) equals the line voltage (involts) multiplied by the current (in amps). To supply our city with power,we can use any combination of voltage and current, so long as theproduct of the two gives us the figure of ten million watts. Shall we sendthe power out from our generating station in the form of high-voltage andlow-current? Or as low-voltage and high-current?

If we set the voltage at 100V, then the current must be 100,000A tosatisfy the electricity consumers in the city. Alternatively, if we increasethe voltage to 100,000V, then 100A of current will suffice.

However, even the best conducting transmission lines will present someresistance to the flow of current, and accordingly, some power will bedissipated. Assuming line resistance is 20Ω, the power lost in transmissionis given by the equation, P = I2 R. If we generate 100,000A at 100V, thenthe power loss equals 200,000 million watts. This is some 20 thousandtimes the power required by the city!

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On the other hand, if the power is sent out along the lines at 100,000Vand only 100A, then the lost power is just 200,000 watts or 2% of the loadrequired by the city. The reason for these hugely varying results is thatpower dissipated in the transmission of electricity increases not just withthe current, but with the square of the current.

It is for this reason that electrical power is distributed at high-voltagesand low currents. But could we not transmit the high-voltage, low-current power in DC form? Yes, we could; but then we would be unableto step-down the line voltage with transformers before passing it on tothe power consumers. Only AC can be manipulated in this way. Today,most power stations produce electricity at 100,000-400,000V (100-400kV) AC, which transformer stations near the point of usage step-down to the mains line voltage of 240V AC.

Radio waves are of their nature also constantly changing in voltage andcurrent. Their transmission, detection, processing, amplification andtranslation into audible sound are all other important areas in which anunderstanding of AC is essential. In summary, AC is a vital element inour electric and electronic world.

SINUSOIDAL

WAVEFORMS

OBJECTIVES

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

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

. alternating current and voltage

. amplitude

. angular frequency

. cycle

. frequency

. hertz

. instantaneous current and voltage

. leading and lagging phase angles

. peak value

. periodic time

. phase angle and phase difference

. sinusoidal waveform

2. Write the mathematical equation for the instantaneous value of a

voltage or current sinusoidal waveform, given the appropriate data.

We learnt in Unit 1 that DC voltage and current remains constant withrespect to time, as shown in Figure 2.1. As indicated above, very oftenin electric and electronic devices voltages and currents are used whichare not constant with respect to time. The variation of these voltages orcurrents with respect to time are known as AC (alternating current )wavefoms. By far the most important type of AC waveform is thesinusoid as shown in Figure 2.2. Note that although the abbreviation ACliterally stands for alternating current, it is used generally to describeboth current and voltage waveforms.

Unit 2: AC Electricity 2-3

A sinusoidal waveform is continuously changing in direction frompositive to negative to positive and so on, ie the magnitude and polarity(direction) both change with time. It consists of a basic pattern, repeatedevery T seconds. We refer to a single and complete basic pattern as acycle of the waveform. The duration of the cycle is called the period,denoted by the capital letter T. The sinusoid is therefore said to beperiodic.

Voltage

Current

V

I

time t

time t

Figure 2.1Time variation of a DCvoltage or current

Figure 2.2Sinusoidal Waveforms

The frequency, f, is defined as the number of complete cycles it goes

through every second. Frequency was formerly expressed in cycles per

second, but the unit hertz (Hz) is now universally used. Thus, when the

frequency of the mains electricity supply is said to be 50Hz, this means

that 50 complete cycles of the sinusoidal voltage waveform take place

every second. The relationship between frequency, f, in hertz and period,

T, in seconds, is given by the expression:

f = 1/T

Mathematically, we can represent the sinusoid as a function of timeusing the equation

v t V ft V ft( ) ˆ sin ˆ sin ,= =2 2π θ θ π =

where V is the peak value of the voltage waveform and f is its frequencyin Hz..

voltageor

currentT

t

Sinusoidal Wave

0

+

-

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This is simply the sine function of an angle θ measured in radians andwhere the angle varies with time in accordance with the frequency of thevoltage ω, ie θ = 2πft.

If this voltage is connected across a resistor R the current flowing in theresistor will follow the same pattern of periodic variation as the emf, andits value at any instant is expressed as follows:

i t I ft( ) ˆ sin= 2π , where ˆˆ

IV

R= is the peak value of the current flowing in

the circuit.

It is important to note that this equation for instantaneous current appliesonly in what is termed a pure resistive circuit. A pure resistive circuit isone in which there is no capacitance or inductance, only resistance.

Another way of expressing how fast the sinusoidal waveform is changingis in relation to its angular frequency in radians per second or Hz. Sinceone complete cycle corresponds to an angle of 2π radians (360o), afrequency of f Hz (cycles per second) corresponds to an angularfrequency of 2πf radians per second. Angular frequency is denoted bythe symbol ω. Thus:

ω = 2πf or f = ω/2π

As f = 1/T, where T is the time for one period, we can write:

ω = 2π/T or T = 2π/ω

We now have two equivalent expressions for both the instantaneousvoltage and instantaneous current of an AC sinusoidal waveform:

v t V ft V t( ) ˆ sin ˆ sin= =2π ω

i t I ft I t( ) ˆ sin ˆ sin= =2π ω

where V and I are the peak voltage and peak current respectively, and

ω is the angular frequency in radians per second and f is the waveformfrequency in cycles per second (hertz).

Voltage and current in DC circuits are represented usually by the capitalletters, V and I. AC voltage and current are represented by the sameletters, but in lower case form, v and i. When instantaneous values of ACvoltage and current are discussed, the symbols v(t) and i(t) are mostcommonly used. The peak value (amplitude) of an AC voltage or currentwaveform is shown by the symbols

.

V or I

Unit 2: AC Electricity 2-5

An emf, e, is given by the expression e(t) = 20 sin (2000πt) VoltsDetermine:

(a) The amplitude of the waveform.

(b) The angular frequency of the waveform.

(c) The frequency of the waveform in Hz.

(d) The time for one complete period of the wave.

0.5 1 1.5

e (t)

+20V

-20V

T = 1ms

t (ms)

Figure 2.3Graph of the emf waveform

The waveform is illustrated in Figure 2.3.

Comparing the equation e(t) = 20 sin (2000πt) to the general equation:t

v t V ft V t( ) ˆ sin ˆ sin= =2π ω

we can deduce that:

(a) The amplitude of the waveform is 20V

(b) The angular frequency, ω = 2000π rad/s

(c) The frequency, f = ω/2π= 2000π/2π= 1000Hz

(d) The periodic time, T = 1/f= 1/1000 s= 1ms

EXAMPLE

SOLUTION

SAQ 1In a domestic light bulb connected to the mains electricitysupply, the instantaneous current is zero twice in each cycle ofthe current. Why doesn’t the light go out during these times ofzero current?

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SAQ 2A sinusoidal current, i(t), has a maximum value of 10A and afrequency of 100Hz. Calculate:

(a) The angular frequency in radians per second.

(b) The time for one complete cycle of the current.

(b) Write down an expression for the instantaneous current,

i(t), given that i(t) = 0 when t = 0 and sketch the waveform.

SAQ 3A sinusoidal voltage v(t) is described by the equation

v(t) = 100 sin (100πt) Volts.

Calculate:

(a) The instantaneous value of the voltage after 6ms.

(b) The first time after t = 0 that the instantaneous voltage is50V.

SUMMARY

1. A voltage or current which varies with some particular pattern iscalled a waveform. Waveforms which vary in magnitude and whichchange from positive to negative and vice versa are known asalternating or AC waveforms.

2. A periodic waveform is composed of the repetition of a single basicwaveform pattern.

3. The following parameters may be defined for any periodic waveform:

amplitude - the maximum value of the waveformcycle - a complete basic wave patternperiod - the time T for one cyclefrequency - the number of cycles in one second

The unit of frequency is the hertz (Hz), where f = 1/T and T is the

periodic time.

4. The instantaneous current and voltage of an AC sinusoidalwaveform are given by the expressions:

v t V ft V t( ) ˆ sin ˆ sin= =2π ω

i t I ft I t( ) ˆ sin ˆ sin= =2π ω

5. Two sinusoidal waveforms are said to be in phase when they havetheir maximum and minimum values at the same instant of time;otherwise, they are described as out of phase.

Unit 2: AC Electricity 2-7

ANSWERS TO SAQSSAQ 1

One of the reasons that the frequency of the mains electricitywas chosen at 50 Hz throughout Europe was that this frequencywas high enough to stop electric lights from flickering. Inother words, the current does not stay at zero long enough toallow the light to diminish.

Other electric devices for which the frequency of the mainssupply is important are those driven by synchronous motors,such as record players and clocks. 50Hz is also convenient forthem. In the United States, the mains frequency is slightlyhigher at 60 Hz.

SAQ 2

Peak value or amplitude of current, I =10Afrequency, f = 100Hz

(a) angular frequency, ω = 2πf= 2π × 100= 628 rads/s

(b) periodic time, T = 1/f= 1/100= 0.01s= 10ms

(c) instantaneous current, i(t) = I sin ωti(t) = 10 sin (200πt)

SAQ 3

(a) At t = 6ms = 6 × 10-3 s and measuring angles in radians,

v(t) = 100 sin (100π × 6 × 10-3)= 100 sin (1.88)

= 100 × 0.951= 95.1V

(b) Let t1 = time at which v(t) is 50V

50 = 100 sin (100πt1)

1

2100 1= sin( )πt

sin−

=1

11

2100πt

0.52 = 100πt1

giving t1

= 0.52/100π= 1.67 ms.

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POWER DISSIPATION

AND RMS VALUES

OBJECTIVES

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

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

. average power

. effective value of AC current or voltage

. pure resistance

. rms voltage and current

2. Calculate rms and peak values of sinusoidal AC waveforms, given

the appropriate data.

AVERAGE POWER

DISSIPATION IN AN AC

CIRCUIT

If a DC current, I, flows in a pure resistance, R, and is constant withrespect to time, as shown in Figure 2.4 then the power dissipation, P, isgiven by the expression P = I2R

CurrentI

I2R)

Power

(

t

t

(a) Current variation through a resistance R in a DC circuit

(b) Power dissipation in a resistance R in a DC circuit

Figure 2.4Power dissipation in a DC

circuit

However, consider a sinusoidal AC current, i(t), flowing through a pureresistance, R. The resulting instantaneous power dissipation, P(t), isgiven by the expression:

P(t) = i2(t)R

In the case of an AC current, the instantaneous power, P(t), is clearly notconstant with respect to time. Figure 2.5 shows how both the instantaneouscurrent, i(t), and the square of the instantaneous current i2(t), continuallyvary with respect to time.

Figure 2.5Power dissipation in an AC

circuit

Current

I

I

I

2

21/2

i (t)2

i2

i (t)

Time, t

^

^

^

πω

πω2 π

ω3

Unit 2: AC Electricity 2-9

Note that squaring the instantaneous current, i(t), has the effect ofremoving the negative sign and making the result permanently positive;that is, the graph of the square of the instantaneous current, i2(t), neverpasses beneath the horizontal axis. Thus, the instantaneous powerdissipation, P(t), is always positive, regardless of whether theinstantaneous current, i(t), is flowing in the positive or negative direction.

In this situation, it is more convenient to deal with average powerdissipation, Pav , rather than with the constantly changing instantaneouspower, P(t). The average power dissipation is defined as the averagevalue of the instantaneous power dissipation over one complete period.

Pav = average value of P(t) over one period, T = 2π/ω

To find Pav we perform the following, four-step operation:

1. We take the various positive and negative values of the instantaneouscurrent, i(t), over one complete cycle.

2. We square all these values, making them all positive.

3. We then calculate the average of the squared values, i 2 .

4. Finally, we multiply the average of the squared values over onecomplete cycle by the value of the resistance, R, thus finding theaverage power dissipated in that resistance.

Therefore Pav = i 2 R where i 2 is the average of the squares of theinstantaneous current values over one complete period.

For a sinusoidal AC current, such as that in Figure 2.5, we can draw astraight, horizontal line which is proportional to the constant value of theaverage power, given by the expression

Pav = i 2 R

It can be shown that: the average value of the square of the instantaneouscurrent is equal to half the square of the peak value of that current:

i I2 21

2= ˆ

Thus, the average power dissipation, Pav, can be written as:

P iI

Rav = R22

2=

ˆ

The average power dissipation of a pure resistance in an AC circuit istherefore equal to half the square of the peak current multiplied by theresistance.

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ROOT-MEAN SQUARE

(RMS) VALUES OF AC

CURRENTS AND

VOLTAGES

We now have an expression for the average power dissipation for asinusoidal AC current flowing in a pure resistance. Is this expressionrelated to the average current? Reference to Figure 2.5 reveals that theanswer to this question clearly is no. The area enclosed by the i(t) graphabove the t-axis is equal to the area under the t-axis enclosed by thenegative part of the i(t) graph. The average value of i(t) is therefore zero.Instead of average values for AC currents (or voltages), we use so-calledeffective values, where:

An effective value of an AC current or voltage is that value which, if inDC form, would produce the same average power dissipation.

The equation for average power dissipation in a purely resistive ACcircuit contained the expression for the average value of the square of the

instantaneous current, i 2 . If we find the square root of this, we then havethe effective current:

effective current = i I I2 2 2 0 707= =ˆ / . ˆ

Note how we obtained this effective value: we found the square root ofthe average of the square. Another word for average is mean. Theeffective value of an AC current is therefore the square root of the meanof the square; or, as it is commonly known, the root-mean square (rms)value.

We can follow the same reasoning and derive a similar expression for therms value of an AC voltage. These expressions for the rms values of ACcurrent and voltage are important ones and you should memorise them:

i I

v V

=

=

0 707

0 707

. ˆ

. ˆ

RMS values are almost universally used to specify AC voltages andcurrents. For example, the voltage of the mains supply is quoted as 240V.By this is meant that the rms value of the voltage is 240V. Since the ACmains voltage is sinusoidal, is peak value is:

240 2 339× = V

SAQ 4Calculate the rms values of a sinusoidal voltage with a peakvalue 10V.

SAQ 5Calculate the peak values of a sinusoidal voltage waveformwith an rms value of 20V.

Unit 2: AC Electricity 2-11

SUMMARY

1. When an AC current, i(t), is applied across a pure resistance, R, the

power dissipation at any instant, P(t), continually varies as the

instantaneous current varies. Mathematically:

P(t) = i2(t)R

2. The average power dissipation of an AC current in a pure resistance

is equal to half the square of the maximum value of that current,

multiplied by the resistance:

P I RAV = 1

22ˆ

3. The effective value or root-mean square (rms) value of a sinusoidalAC current or voltage is that value, which if in DC form, wouldproduce the same average power dissipation.

4. The rms values of a sinusoidal AC current and voltageare related to

its peak amplitude values by the expressions:

i I

v V

==

0 707

0 707

. ˆ

. ˆ

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ANSWERS TO SAQSSAQ 4

For a sinusoidal voltage, the rms value, v , is 0.707 times thepeak value. That is:

v = ˆ /v 2

v = 0.707 V= 0.707 × 10= 707V

SAQ 5

For a sinusoidal waveform:

ˆ .V v v= =2 1 14 = 1.414 × 20 = 28.3V

Unit 2: AC Electricity 2-13

COMPONENT

BEHAVIOUR IN AC

CIRCUITS

OBJECTIVES

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

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

. capacitive reactance

. impedance

. inductive reactance

. reactance

. phasor diagram

2. Calculate the rms values of current, voltage and average power insimple resistive AC circuits.

3. Describe and calculate the response of a pure resistance to an ACvoltage.

4. Describe and calculate the response of a pure capacitance to an ACvoltage.

5. Describe and calculate the response of a pure inductance to an ACvoltage.

RESISTANCE IN ACCIRCUITS

When an AC voltage, v, is applied across a resistance R, as shown inFigure 2.6, an alternating current, i, flows.

Figure 2.6AC current and voltagerelationship in a pureresistance

If the applied voltage is sinusoidal, then the voltage at any instant, v(t),is given by the expression:

v t V t( ) ˆ sin= ω

voltageor

current V

I

v

i t

(a) graphical relationship of voltage and current

(b) phasor relationship of voltage and current

^

^

i v

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From Ohm’s law, the instantaneous current will be:

i tv t

R

V

Rt I t( )

( ) ˆsin ˆ sin= =

=ω ω

where ˆˆ

IV

R=

Therefore the current and voltage are in phase as shown in Figure 2.6(a).Generally phase relationshipscan be conveniently indicated on what’scalled a phasor diagram In a phasor diagram, each sinusoidal waveformis represented by a single line called a phasor. The length of the phasoris proportional to the amplitude of the wave; the greater the amplitude,the longer the line. All lines are drawn from a single point, which maybe imagined to be the centre of a clock face. The phase difference inradians between each waveform is represented by a corresponding anglebetween the lines. Leading phase angles are drawn in an anti-clockwisesense from the reference phasor, while lagging phase angles are drawnin a clockwise direction. In this case the voltage and current are in phaseand therefore the phase angle between the phasors is zero as shown inFigure 2.6(b).

For one half-cycle, the current is flowing in one direction through theresistor and for the next half-cycle, it is flowing in the opposite directionand so on. Furthermore, at two instants in time during each cycle, thecurrent is zero.

Multiplying both sides of the equation V = IR by 0.707 converts thepeak values of voltage and current to their rms values:

which is the same basic form of Ohm’s law as used in DC circuits. Thus,if rms values of voltage and current are used, a resistor behaves the samefor AC as it does for DC. The techniques already encountered in ourstudy of DC circuits - Ohm’s law, Kirchoff’s laws - are equallyapplicable to AC resistor networks.

From the definition of the root-mean square (rms) value of AC waveforms,it follows that the average power, Pav , dissipated in a resistance, R, isgiven by the expression:

Pav = v i= i2R= v 2/R

where v is the rms value of the potential difference across the resistor

and i is the rms value of the current through it. So: power calculationsin AC resistive circuits are exactly the same as in DC circuits, when therms values of voltages and currents are used. For this reason we will drop

the bar over the symbols for voltage v and current i , and proceed on thebasis that when we use these lower case symbols in ac circuits, we areAs we shall see, AC circuits which contain capacitors or inductorsbehave quite differently assuming that they are rms values.

0 707 0 707. ˆ . ˆV IR

v iR

==

Unit 2: AC Electricity 2-15

In conclusion: in AC resistive circuits, the current and voltage are inphase. If rms values of current and voltage are used, Ohm’s andKirchoff’s laws can be applied, and the calculations of power dissipationare the same as those in DC circuits.

EXAMPLE For the circuit shown in Figure 2.7, calculate:

(a) The rms current supplied by the generator.(b) The rms voltage across the 2.2kΩ resistor.(c) The average power dissipated in the 6.8kΩ resistor.

18Vrms1kHz

V

R2

= 2

.2 k

ΩR

1 =

6.8

kΩ

Figure 2.7AC circuit with tworesistances

SOLUTION (a) The total circuit resistance, RT = 6.8 + 2.2 = 9kΩ

The rms current, i = v/R= 18/(9 × 103)= 2mA

(b) Let v2 = rms voltage across R2 = 2.2kΩ

v2 = iR2= 2 × 10- 3 × 2.2 × 103

= 4.4V

(c) Let P = average power dissipation in R1 = 6.8kΩ P = i2Ri

= (2 × 10-3)2 × 6.8 × 103

= 27.2 × 10-3

= 27.2mW

SAQ 6In the circuit shown in Figure 2.8, calculate:

(a) The rms current supplied by the generator.

(b) The average power dissipated in the 5.6 kΩ resistor.

(c) The frequency of the AC current in Hz through the 2.7 kΩresistor.

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Figure 2.8Circuit for SAQ 6

3.3kΩ

v(t) = 30 sin (400πt) volts

2.7kΩ

5.6kΩ

The effect of placing a capacitor in a DC circuit is to stop the flow ofelectric current. Electrons cannot travel across the gap between thecapacitor plates and so the continuous conducting path required for acurrent to flow is interrupted. For example, if a capacitor is connected inseries with an electric light bulb and a DC power supply, the bulb willnot illuminate as the circuit is open at the capacitor.

If we remove the DC power supply, however, and replace it with an ACone, then the electric bulb does light up! If a variable capacitor is used,it can also be seen that the brightness of the bulb is directly proportionalto the value of the capacitance and to the waveform frequency of theapplied voltage. In this section, we shall examine in some detail thebehaviour of a capacitance in an AC circuit.

In our study of capacitance in Unit 1, we learnt that a capacitor ofcapacitance C, charged to a DC voltage V, stores an amount of charge,Q, given by the expression:

Q = CV

If an alternating voltage v is applied across a pure capacitance C, asshown in Figure 2.9, the capacitor will be continually charged anddischarged.

CAPACITANCE IN ACCIRCUITS

Figure 2.9AC voltage applied across a

capacitor C

X

Y

CV sin t

i

^

vc

ω

Unit 2: AC Electricity 2-17

During the positive half-cycle of the voltage waveform, plate X of the

capacitor becomes positively charged and plate Y negatively charged.

During the negative half-cycle, X receives a negative charge and Y a

positive one. There is therefore an alternating flow of charge or alternating

current, i, through the capacitor Thus, unlike the case for a dc voltage

where the capacitor eventually charges up to the value of the dc supply

voltage and no more current flows, a capacitor continually conducts an

ac current. Thus we may say that a capacitor passes ac and blocks dc.

It may be shown that the alternating current is also sinusoidal but leadsthe supply voltage by a phase angle of π/2 radians (90o). The expressionfor the current i(t) maybe shown to be

i t I t I t I CV( ) ˆ sin( / ) ˆ cos , ˆ ˆ= + = =ω π ω ω2 where

Thus the graph of the instantaneous voltage across the capacitor is a sinefunction and the graph of the instaneous current through it is a cosinefunction.

Figure 2.10 shows the plots of v and i as functions of time, illustratingthe current/voltage relationship in a capacitor.

voltage orcurrent

iCvC

tFigure 2.10AC current and voltage

relationship in a pure

capacitance

The equations for instantaneous current and voltage are:

v t V t i(t) CV t( ) ˆ sin ˆ sin( / )= = +ω ω ω π and 2

These equations, together with their corresponding graphs, indicate veryclearly that the instantaneous voltage and current in a capacitor are outof phase by a quarter of a cycle - π/2 radians or 90o - with the currentleading the voltage. Thus, the phasor diagram is as shown in Figure 2.11.

The peak value of the current , I , is related to the peak value of the

voltage, V , by the expression.

ˆ ˆ

ˆ

ˆ

I CV

V

I C

=

=

ω

ω1

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Figure 2.11Phasor diagram showing the

relationship between the

instantaneous voltage and

current in a capacitor

90

VC

iC

°

Converting to rms values gives:

i Cv

v

i C fC

=

= =

ω

ω π1 1

2

where f is the waveform frequency in hertz.

Note that this expression is in the form of Ohm’s law, V/I = R. Thequantity 1/ωC or 1/2πfC is the factor resisting the flow of currentthrough the capacitor and is called capacitive reactance. The termresistance is not used to denote opposition in this case, for the followingreason: power is dissipated in a resistance, but not in a capacitor.However, the units of capacitive reactance are ohms.

The symbol for capacitive reactance is XC, where:

X V I v ic c c c c= =ˆ / ˆ /= 1/ωC= 1/2πfC ohms

Note that XC is inversely proportional to both the capacitance, C, and thewaveform frequency, f. Hence, if either (or both) f or C increase, thecapacitive reactance decreases and vice versa. In other words, the greaterthe waveform frequency or the capacitance, the smaller the reactanceand the larger will be the current flowing across the capacitor. Therelationship between capacitive reactance and waveform frequency isshown in Figure 2.12.

Figure 2.12Variation of capacitivereactance with waveformfrequency

capacitivereactance X c

X = c

frequency f(Hz)

12πfC

Unit 2: AC Electricity 2-19

EXAMPLE A 1µF capacitor has an rms current of 2mA flowing through it at afrequency of 1000Hz. Calculate the rms voltage across the capacitor.

SOLUTION Capacitive reactance, XC = 1/2πfC= 1/(2π × 1000 × 10-6)= 159Ω

rms voltage across capacitor, v= iXc= 2 × 10-3 × 159= 0.32V= 320mV

SAQ 7Why do capacitive reactances become small in high-frequencycircuits, such as those in FM radios or TV sets?

SAQ 8A sinusoidal source of emf of 10V rms and frequency 2kHz isapplied across a 0.1µF capacitor. Calculate the rms value ofthe current flowing through the capacitor.

In conclusion we can state the following points concerning the behaviourof a capacitance in an AC circuit:

If a sinusoidal waveform is applied across a capacitor, the current willlead the voltage by 90o. Capacitive reactance is the opposition whichevery capacitor presents to the flow of current. Its units are ohms.

INDUCTANCE IN ACCIRCUITS

A piece of wire wound in the form of a coil possesses an electricalproperty known as inductance. The property arises from the observablephenomenon that if a current flowing through the coil changes for somereason then an emf e is induced in the coil which tries to oppose thecurrent change. The magnitude of the induced emf is proportional to therate of change of current. The constant of proportionality is known as theinductance of the coil L and is measured in a unit called the henry.Mathematically we can summarise this with the equation

e Ldi

dt= −

Where di/dt denotes the charge of current with respect to time. Considera sinusoidal voltage, v, applied across a lossless (zero resistance)inductor of inductance L, as shown in Figure 2.13.

Figure 2.13Instantaneous voltage andcurrent relationship in a pureinductance

voltage orcurrent

vL

vL

iLiL

(t)(t)

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This results in a sinusoidal current flowing through the inductor, i , whichis taken as the reference waveform with equation

i t I t( ) ˆ sin= ω

It may be shown that the voltage across the inductor is given by

Figure 2.13 plots v and i as functions of time. Again, the equations forv and i indicate that the instantaneous voltage and current in an inductorare out of phase by an angle of 90o (π/2 radians), with the voltage leadingthe current. The phasor diagram will be as shown in Figure 2.14.

90

vL

iL

°

Figure 2.14

Phasor diagram showing the

relationship between the

instantaneous voltage and

current in an inductor

Since

=

ˆ ˆ

ˆ / ˆ /

V LI

V I v i L

fL

=

= =

ω

ωπ2

where f is the waveform frequency in hertz.

The quantity ωL = 2πfL is the factor resisting the flow of current and iscalled the inductive reactance, XL , measured in ohms.

XL = v/i= ωL

= 2πfL

XL is directly proportional to frequency; the higher the frequency, thegreater the inductive reactance. See Figure 2.15.

inductivereactance

XLXL = 2 fL

frequency (Hz)

πFigure 2.15

Variation of inductivereactance with waveformfrequency

v t V t V t LI( ) ˆ cos ˆ sin( / ), ˆ ˆ= = +ω ω π ω2 where V =

Unit 2: AC Electricity 2-21

Like capacitive reactance, XC, inductive reactance, XL, cannot be addeddirectly to resistance in an electric circuit. No power is dissipated in apure inductance; like a pure capacitance, it presents no resistance to theflow of current. Energy is repeatedly stored in the magnetic fieldsurrounding the inductor and released back to the supply. A device thatbehaves in this way is called a reactor. Lossless capacitors and inductorsare pure reactors. Practical capacitors and inductors are never quitelossless and contain some resistance, which dissipates some of thesupplied power.

SAQ 9Manufacturers of resistors, those circuit components in circuitswhich offer specified resistances to the flow of current,always show the resistance value of their component bymeans of colour-coded bands on the outer surface of eachresistor. In this way, the user can readily distinguish between- say - a 30kΩ resistor and a 2Ω one. Why don’t capacitor andinductor manufacturers similarly display the reactance oftheir components?

SAQ 10When an AC voltage is applied across a pure inductance of0.1H, a sinusoidal current, i = 300 sin (100πt) mA, flows.Calculate:

(a) the rms current

(b) the rms value of the applied voltage

(c) the peak value of the supply voltage

Write a mathematical expression for the instantaneous valueof the supply voltage.

PHASE DIFFERENCEBETWEEN SINUSOIDAL

WAVEFORMS

In circuits containing resistors and capacitors and/or inductors the phasedifference between the voltages and currents may vary in a general wayand do not necessarily have a phase difference of just 90°, or one quarterof a cycle, (lag or lead), as is the case for a pure inductance orcapacitance. Consider the two sinusoidal waveforms, v1 and v2, sketchedin Figure 2.16(a).

2-22 Communications Technology 1

Figure 2.16Illustration of phase difference

(a) in phase

(b) out of phase

πω

π2

2

π ππ

t

= θωωω

πω

π

ω

πφ

π3 θ=

ω2π 3π

ωθ=

ω

t

2 t

3 θ = ωtπ3

td

v2

v1

v2v1

We can say that v1 and v2 are in phase because v1 has its maximum andminimum values at exactly the same instants of time as v2. For thewaveform in Figure 2.16 (b), however, this is not the case. The waveformsare displaced relative to each other and are said to be out of phase. Inother words, there is a phase difference between v1 and v2.

The amount of phase difference is expressed in terms of the phase angleφ, measured in radians on the ωt-axis. If we define v1 as the referencewaveform, then v2 lags behind v1 (since v2 reaches it maximum valueafter v1) by a phase angle φ . Thus:

A phase lag φ on the θ-axis corresponds to a time lag td on the t-axis. Bothvoltages are sinusoidal waveforms with v1 = 0 when t = 0 and v2 = 0 whent = td. That is:

v2 (td ) = 0

ωtd - φ = 0and ωtd = φthus td = φ/ω

We can therefore conclude that: the time delay between two out-of-phasesinusoidal waveforms is equal to the phase angle between them dividedby their angular frequency.

v t V t

v t V t

1 1

2 2

( ) ˆ sin

( ) ˆ sin( )

=

= −

ω

ω φ

v t V t td2 2( ) ˆ sin( )= −ω ω

Unit 2: AC Electricity 2-23

EXAMPLE A voltage v1 in an electric circuit is sinusoidal, and has a peak value of4V and a frequency of 200Hz. Another voltage, v2, has the samefrequency as v1 and a peak value of 2V. Also it reaches its peak 1msbefore v1 does.

(a) Determine the phase angle between v1 and v2 .

(b) Write mathematical expressions for v1 and v2 using v1 as thereference waveform.

SOLUTION (a) Time lead of v2 over v1 = 1ms

angular frequency of v2 and v1, ω = 2πf= 2π × 200= 400π

phase lead of v2 over v1, φ = ωtd

= 400π × 10-3

= 0.4π radians

(b) If v1 is the reference voltage, then:

v t V t1 1( ) ˆ sin= ω

where V V1 4=

and ω = 400π rads/sthus v1(t) = 4 sin (400πt)V

Since v2 leads v1 by a phase angle:

v t V t2 2( ) ˆ sin( )= +ω φ

where V V2 2=

. ω = 400π rads/sφ = 0.4π rads

v2(t) = 2 sin (400πt + 0.4π)V= 2 sin π(400t + 0.4)V

SAQ 11A current in an AC circuit, i1, is sinusoidal and has a peakvalue of 100mA and a frequency of 500Hz. Another current inthe same circuit, i2 , has the same frequency but is twice aslarge as i1 and lags it by a phase angle of π/4 radians (45∞).

(a) Write the mathematical expressions for i1 and i2.

(b) Calculate the time interval between a peak in the waveformof i1 and a peak in the waveform of i2.

(c) Sketch roughly i1 and i2 on the same axis.

2-24 Communications Technology 1

SUMMARY

1. In a purely resistive AC circuit, the current and voltage are in phasewith one another. The rms values of voltage and current are relatedto resistance by Ohm’s law:

v = iR i = v/R R = v/i

2. The average power dissipation in a purely resistive circuit is givenby the expressions:

P = vi or P = i2R or P = v2/R

where v and i are rms values.

3. In a purely capacitive AC circuit, the current leads the voltage by90o or π/2 radians.

4. Capacitive reactance, XC , measured in ohms, is the oppositionthat a pure capacitance, C, presents to the flow of AC current. It isinversely proportional to the angular or waveform frequency of theapplied voltage.

XC = 1/ωC or XC = 1/2πfC

5. The voltage and current in a purely capacitive circuit are related bythe expression:

v = iXC

where v and i are rms. values.

6. In a purely inductive AC circuit, the current lags the voltage by 90o

or π/2 radians.

7. Inductive reactance, XL, measured in ohms, is the opposition thata pure inductance, L, presents to the flow of AC current. It is directlyproportional to the angular or waveform frequency of the appliedvoltage.

XL = ωL or XL = 2πfL

8. The voltage and current in a purely inductive circuit are related bythe expression:

v = iXL

where v and i are rms values.

9. Capacitive and inductive reactances cannot be added directly toresistance, as a component which offers reactance does not dissipatepower.

10. The phase difference between two out-of-phase sinusoidalwaveforms is expressed in terms of the phase angle φ between them.The phase angle is equal to the time difference, td, between the twowaveforms multiplied by their angular frequency:

φ = ωt d

Unit 2: AC Electricity 2-25

11. When comparing two out-of-phase waveforms, one is described asthe reference waveform and the other waveform as lagging orleading the reference one.

2-26 Communications Technology 1

ANSWERS TO SAQSSAQ 6

(a) The equivalent resistance, Rp, of 5.6kΩ and 2.7kΩ inparallel is:

Rp = (5.6× 2.7)/(5.6 + 2.7) = 1.82kΩ

Total circuit resistance, RT = 3.3 + 1.82= 5.12kΩ

RMS voltage of generator, C = 30 × 0.707= 21.2V

RMS current of generator, i = v/R= 21.2/(5.12 × 103)= 4.14 × 10-3

= 4.14mA

(b) Let v1 be the rms voltage across R1, the 3.3kΩ resistorand let v2 be the rms voltage across Rp, the parallelcombination of the 5.6k Ω and the 2.7kΩ resistor.

v1 = iR1= 4.14 × 10-3 × 3.3 × 103

= 13.7V

v2 = v - v1= 21.2 - 13.7= 7.5

The average power dissipation in the 5.6kΩ resistor isgiven by:

= (7.5)2 /(5.6 × 103) = 0.01W = 10mW

(c) The frequency of the AC current through the 2.7kΩresistor is the same as the frequency of the supplyvoltage.

angular frequency, ω = 400πtwaveform frequency, f = ω/2π

= 400π/2π= 200Hz

Unit 2: AC Electricity 2-27

SAQ 7

Because capacitive reactance decreases as the frequencyacross the capacitor increases; the higher the frequency, thelower the reactance.

SAQ 8

Capacitive reactance, XC = 1/2πfC= 1/(2π × 2000 × 0.1× 10 -6)= 796Ω

rms current through capacitor, i = v/Xc= 10/796= 0.0126A= 12.6mA

SAQ 9

The resistive value of a resistor remains constant, regardlessof the frequency of the AC current passing through it. However,the reactances of capacitors and inductors vary according tothe AC frequency applied across them. As a result,manufacturers of capacitors and inductors cannot possiblypredict their reactances in every circumstance; that dependson how the components are used.

2-28 Communications Technology 1

SAQ 10

(a) The rms current = 0.707 I = 0.707 × 300 = 212mA

(b) Frequency of supply, f = ω/2π= 100π /2π= 50Hz

inductive reactance, XL = 2πfL= 2π × 50 × 0.1= 31.4Ω

rms value of supply voltage, v= iXL= 212 × 10-3 × 31.4= 6.7V

(c) Peak value of supply voltage,

V v=×2

= 6.7 1.414

= 6.7V

The voltage across a pure inductance leads the currentthrough it by π/2 radians or by 90o.

v(t) = 9.5 sin (100πt + π/2) volts

Unit 2: AC Electricity 2-29

SAQ 11

(a) i t I t I ft1 1 1 2( ) ˆ sin ˆ sin= =ω π

= 100 sin (2π × 500 × t)= 100 sin (1000πt) mA

If i2 is twice as large as i1 , then:

ˆ ˆI I2 12== 2 × 100= 200mA

i2 lags i1 by a phase angle of φ = π/4 radians (45o).Therefore i2 is of the form:

i2 = 200 sin (1000πt - π/4) mA

(b) Time delay between waveforms, td = φ/ω= (π/4)/1000π= 2.5 × 10-4

= 0.25ms

(c) Your sketch of i 1 and i 2 should look like that shown inFigure 2.17.

1 2 3

200

100

0

100

0.25ms

t (ms)

current(mA)

200

-

-

i2

ii

Figure 2.17Waveforms in SAQ 11