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Chapter 8

Alternating voltages and currents

8.1. Introduction

Electricity is produced by generators at power stations and then distributed by a vast network

of transmission lines (called the National Grid system) to industry and for domestic use. It is

easier and cheaper to gener- ate alternating current (a.c.) than direct current (d.c.) and a.c. is

more conveniently distributed than d.c. since its voltage can be readily altered using trans-

formers. Whenever d.c. is needed in preference to a.c., devices called rectifiers are used for

conversion.

8.2 The a.c. generator

Let a single turn coil be free to rotate at constant angular velocity symmetrically between the

poles of a magnet system as shown in Fig.8.1

Fig.8.1

An e.m.f. is generated in the coil (from Faraday’s laws) which varies in magnitude and reverses

its direc- tion at regular intervals. The reason for this is shown in the following Fig.8.2. In

positions (a), (e) and (i) the conductors of the loop are effectively moving along the magnetic

field, no flux is cut and hence no e.m.f. is induced. In position (c) maximum flux is cut and

hence maximum e.m.f. is induced. In position (g), maximum flux is cut and hence maximum

e.m.f. is again induced. However, using Fleming’s right-hand rule, the induced e.m.f. is in the

opposite direction to that in position (c) and is thus shown as −E. In positions (b), (d), (f) and

(h) some flux is cut and hence some e.m.f. is induced. If all such positions of the coil are

considered, in one revolution of the coil, one cycle of alternating e.m.f. is produced as shown.

This is the principle of operation of the a.c. generator (i.e. the alternator).

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Fig.8.2

8.3. Waveforms

If values of quantities which vary with time t are plotted to a base of time, the resulting graph

is called a wave- form. Some typical waveforms are shown in the Fig.8.3 Waveforms (a) and

(b) are unidirectional waveforms, for, although they vary considerably with time, they flow in

one direction only (i.e. they do not cross the time axis and become negative). Waveforms (c)

to (g) are called alternating waveforms since their quantities are continually changing in

direction (i.e. alternately positive and negative).

A waveform of the type shown in Fig. 8.3(g) is called a sine wave. It is the shape of the

waveform of e.m.f. produced by an alternator and thus the mains electricity supply is of

‘sinusoidal’ form.

One complete series of values is called a cycle (i.e. from O to P in Fig. 8.3(g)). The time taken

for an alternating quantity to complete one cycle is called the period or the periodic time, T, of

the waveform.

The number of cycles completed in one second is called the frequency, f , of the supply and is

measured in hertz, Hz. The standard frequency of the electricity supply in Great Britain is 50

Hz

T=1/f or f=1/T

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Fig.8.3

Example: Determine the periodic time for frequencies of (a) 50 Hz and (b) 20 kHz.

(a) Periodic time T =1/f=1/50= 0.02 s or 20 ms

(b) (b) Periodic time T =1/20000= = 0.00005 s or 50 μs

Example: Determine the frequencies for periodic times of (a) 4 ms (b) 4 μs.

(a) Frequency f =1/T= = 250 Hz

(b) (b) Frequency f = = 250 000 Hz or =250 kHz or 0.25 MHz

Example: An alternating current completes 5 cycles in 8 ms. What is its frequency?

Time for 1 cycle = (8/5) ms = 1.6 ms = periodic time T . Frequency f =1/T= = 625 Hz

8.4. A.C. values

Instantaneous values are the values of the alternating quantities at any instant of time. They are

represented by small letters, i, v, e, etc., (see Fig.(f) and (g)).

The largest value reached in a half cycle is called the peak value or the maximum value or the

amplitude of the waveform. Such values are represented by Vm, Im , etc. (see Fig. (f) and (g)).

A peak-to-peak value of e.m.f. is shown in Fig. (g) and is the difference between the maximum

and minimum values in a cycle. The average or mean value of a symmetrical alternating

quantity, (such as a sine wave), is the average value measured over a half cycle, (since over a

complete cycle the average value is zero).

Average or mean value = area under the curve / length of base

The area under the curve is found by approximate meth- ods such as the trapezoidal rule, the

mid-ordinate rule or Simpson’s rule. Average values are represented by VAV, IAV, EAV, etc.

For a sine wave:

average value = 0.637 × maximum value (i.e. 2/π × maximum value)

The effective value of an alternating current is that current which will produce the same heating

effect as an equivalent direct current. The effective value is called the root mean square (r.m.s.)

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value and whenever an alternating quantity is given, it is assumed to be the rms value. For

example, the domestic mains supply in Great Britain is 240 V and is assumed to mean ‘240 V

rms’. The symbols used for r.m.s. values are I, V, E, etc. For a non-sinusoidal waveform as

shown in Fig.8.4 the r.m.s. value is given by:

where n is the number of intervals used.

Fig.8.4

For a sine wave:

rms value = 0.707 × maximum value (i.e.1/√2 × maximum value)

Form factor =r.m.s. value/ average value

For a sine wave, form factor = 1.11

Peak factor =maximum value/ r.m.s. value

For a sine wave, peak factor = 1.41.

The values of form and peak factors give an indication of the shape of waveforms.

Example: For the periodic waveforms shown in Fig.8.5 determine for each: (i) frequency; (ii)

average value over half a cycle (iii) r.m.s. value; (iv) form factor and (v) peak factor.

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Fig.8.5

(a) Triangular waveform (Fig. 8.5(a)).

(i) Time for 1 complete cycle = 20 ms = periodic time, T. Hence

frequency f =1/T=1/20*e-3= = 50 Hz

(ii) Area under the triangular waveform for a half cycle = 1/2× base × height

= 1/2× (10 × 10−3) × 200 = 1 volt second

Average value of waveform= area under curve/ length of base

= 1 volt second/10 × 10−3=100 V

(iii) In Fig. 8.5(a), the first 1/4 cycle is divided into 4 intervals. Thus

= 114.6V

(Note that the greater the number of intervals chosen, the greater the accuracy of the result. For

example, if twice the number of ordinates as that chosen above are used, the r.m.s. value is

found to be 115.6V)

(iv) Form factor = r.m.s. value/average value=114.6/100=1.15

(v) Peak factor =maximum value/r.m.s. value=200/114.6=1.75

(b) Rectangular waveform (Fig. 8.5(b)).

(i) Time for 1 complete cycle = 16 ms = periodic time, T. Hence

frequency, f =1/T=1/16 × 10−3=62.5 Hz

(ii) Average value over half a cycle=area under curve/length of base

=10 × (8 × 10−3)/= (8 × 10−3)=10 A

(iii)

however many intervals are chosen, since the waveform is rectangular.

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(iv) Form factor =r.m.s. value/average value=10/10= 1

(v) Peak factor =maximum value/r.m.s. value=10/10= 1

Example: Calculate the r.m.s. value of a sinusoidal current of maximum value 20A.

For a sine wave, r.m.s. value = 0.707 × maximum value = 0.707 × 20 = 14.14 A

Example: Determine the peak and mean values for a 240V mains supply.

For a sine wave, r.m.s. value of voltage V = 0.707 × Vm.A 240V mains supply means that

240V is the r.m.s. value, hence

Vm =V/0.707=240/0.707== 339.5V= peak value

Mean value

VAV = 0.637Vm = 0.637 × 339.5 = 216.3V

Example: A supply voltage has a mean value of 150V. Determine its maximum value and its

r.m.s. value. For a sine wave, mean value = 0.637 × maximum value.

Hence

maximum value =mean value/0.637=150/0.637= 235.5 V

R.m.s. value = 0.707 × maximum value= 0.707 × 235.5 = 166.5V

8.5. The equation of a sinusoidal waveform

In Fig. 8.6, 0 A represents a vector that is free to rotate anticlockwise about 0 at an angular

velocity of ω rad/s. A rotating vector is known as a phasor.

Fig. 8.6

After time t seconds the vector 0A has turned through an angle ωt. If the line BC is constructed

perpendicular to 0A as shown, then

sin ωt =BC/0B i.e. BC = 0B sin ωt

If all such vertical components are projected on to a graph of y against angle ωt (in radians), a

sine curve results of maximum value 0A. Any quantity which varies sinusoidally can thus be

represented as a phasor.

A sine curve may not always start at 0◦. To show this a periodic function is represented by y =

sin(ωt ± φ), where φ is the phase (or angle) difference compared with y = sin ωt. In Fig. 8.7(a),

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y2 = sin(ωt + φ) starts φ radians earlier than y1 = sin ωt and is thus said to lead y1 by φ radians.

Phasors y1 and y2 are shown in Fig. 8.7(b) at the time when t = 0.

Fig. 8.7

In Fig. 8.7(c), y4 = sin(ωt − φ) starts φ radians later than y3 = sin ωt and is thus said to lag y3

by φ radians. Phasors y3 and y4 are shown in Fig. 8.7(d) at the time when t = 0.

Given the general sinusoidal voltage, v = Vm sin(ωt ± φ), then

(i) Amplitude or maximum value = Vm

(ii) Peak to peak value = 2Vm

(iii) Angular velocity = ω rad/s

(iv) Periodic time, T = 2π/ω seconds

(v) Frequency, f = ω/2π Hz (since ω = 2πf )

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(vi) φ = angle of lag or lead (compared with v = Vm sin ωt)

Example: An alternating voltage is given by v = 282.8 sin 314 t volts. Find (a) the r.m.s.

voltage, (b) the frequency and (c) the instantaneous value of voltage when t = 4 ms.

(a) The general expression for an alternating voltage is v = Vm sin(ωt ± φ). Comparing v =

282.8 sin 314 t with this general expression gives the peak voltage as 282.8V.

Hence the r.m.s. voltage = 0.707 × maximum value = 0.707 × 282.8 = 200 V

(b) Angular velocity, ω = 314 rad/s, i.e. 2πf = 314. Hence frequency,

f =314/2π= 50 Hz

(c) When t = 4 ms, v = 282.8 sin(314 × 4 × 10− 3) = 282.8 sin(1.256) = 268.9V

Note that 1.256 radians =1.256×180/π=71.96

Hence v = 282.8 sin 71.96◦ = 268.9V, as above.

Example: An alternating voltage is given by v = 75 sin(200πt − 0.25) volts. Find (a) the

amplitude, (b) the peak-to-peak value, (c) the r.m.s. value, (d) the periodic time, (e) the

frequency, and (f) the phase angle (in degrees and minutes) relative to 75 sin 200πt.

Comparing v = 75 sin(200πt − 0.25) with the general expression v = Vm sin(ωt ± φ) gives:

(a) Amplitude, or peak value = 75 V

(b) Peak-to-peak value = 2 × 75 = 150 V

(c) The r.m.s. value = 0.707 × maximum value = 0.707 × 75 = 53 V

(d) Angular velocity, ω = 200π rad/s. Hence periodic time,

T =2π/ω=2π/200π=0.01 s or 10 ms

(e) Frequency, f = 1/T=1/0.01= 100 Hz

(f) Phase angle, φ = 0.25 radians lagging 75 sin 200πt

0.25 rads = 0.25 ×180◦/π= 14.32◦

Hence phase angle = 14.32◦ lagging

8.6 Combination of waveforms

The resultant of the addition (or subtraction) of two sinusoidal quantities may be determined

either:

(a) by plotting the periodic functions graphically, or

(b) by resolution of phasors by drawing or calculation

Example: The instantaneous values of two alternating currents are given by i1 = 20 sin ωt

amperes and i2 = 10 sin(ωt + π/3) amperes. By plotting i1 and i2 on the same axes, using the

same scale, over one cycle, and adding ordinates at intervals, obtain a sinusoidal expression for

i1 + i2.

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i1 = 20 sin ωt and i2 = 10 sin(ωt + π/3) are shown plotted in Fig.8.8. Ordinates of i1 and i2

are added at, say, 15◦ intervals (a pair of dividers are useful for this).

For example, at 30◦ , i1 + i2 = 10 + 10 = 20A at 60◦ , i1 + i2 = 17.3 + 8.7 = 26A at 150◦ , i1 +

i2 = 10 + ( − 5) = 5A, and so on.

Fig. 8.8

The resultant waveform for i1 + i2 is shown by the broken line in Fig. It has the same period,

and hence frequency, as i1 and i2. The amplitude or peak value is 26.5A

The resultant waveform leads the curve i1 = 20 sin ωt by 19◦ i.e. (19 × π/180) rads = 0.332 rads

Hence the sinusoidal expression for the resultant i1 + i2 is given by:

iR = i1 + i2 = 26.5 sin(ωt + 0.332) A

Example: Two alternating voltages are represented by v1 = 50 sin ωt volts and v2 = 100 sin(ωt

− π/6)V. Draw the phasor diagram and find, by calculation, a sinusoidal expression to represent

v1 + v2.

Phasors are usually drawn at the instant when time t = 0. Thus v1 is drawn horizontally 50 units

long and v2 is drawn 100 units long lagging v1 by π/6 rads, i.e. 30◦. This is shown in Fig. 8.9(a)

where 0 is the point of rotation of the phasors.

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Fig. 8.9

Procedure to draw phasor diagram to represent v1 + v2:

(i) Draw v1 horizontal 50 units long, i.e. oa of the Fig 8.9(a)

(ii) Join v2 to the end of v1 at the appropriate angle, i.e. ab of the Fig. 8.9(b)

(iii) The resultant vR = v1 + v2 is given by the length ob and its phase angle may be measured

with respect to v1

Alternatively, when two phasors are being added the resultant is always the diagonal of the

parallelogram, as shown in Fig. 8.9(c).

From the drawing, by measurement, vR = 145V and angle φ = 20◦ lagging v1.

A more accurate solution is obtained by calculation, using the cosine and sine rules. Using the

cosine rule on triangle 0ab of Fig. 8.9(b) gives:

and φ = sin−1(0.3436 )= 20.096◦ = 0.35 radians, and lags v1. Hence

vR = v1 + v2 = 145.5 sin(ωt − 0.35)V

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8.7. Rectification

The process of obtaining unidirectional currents and voltages from alternating currents and

voltages is called rectification. Automatic switching in circuits is achieved using diodes.

Half-wave rectification Using a single diode, D, as shown in the Fig., half-wave rectification is

obtained. When P is sufficiently positive with respect to Q, diode D is switched on and current

i flows. When P is negative with respect to Q, diode D is switched off. Transformer T isolates

the equipment from direct connection with the mains supply and enables the mains voltage to

be changed.

Thus, an alternating, sinusoidal waveform applied to the transformer primary is rectified into a

unidirectional waveform. Unfortunately, the output waveform shown in Fig.8.10 is not

constant (i.e. steady), and as such, would be unsuitable as a d.c. power supply for electronic

equipment. It would, however, be satisfactory as a battery charger.

Fig. 8.10

Full-wave rectification using a centre-tapped transformer

Two diodes may be used as shown in Fig.8.11 to obtain full-wave rectification where a centre-

tapped transformer T is used. When P is sufficiently positive with respect to Q, diode D1

conducts and current flows (shown by the broken line in Fig.8.11). When S is positive with

respect to Q, diode D2 conducts and current flows (shown by the continuous line in Fig.8.11).

Fig. 8.11

The current flowing in the load R is in the same direction for both half-cycles of the input. The

output waveform is thus as shown in Fig.8.11. The output is unidirectional, but is not constant;

however, it is better than the output waveform produced with a half-wave rectifier. Section 8

explains how the waveform may be improved so as to be of more use.

A disadvantage of this type of rectifier is that centre-tapped transformers are expensive.

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Full-wave bridge rectification

Four diodes may be used in a bridge rectifier circuit, as shown in Fig.8.12 to obtain full-wave

rectification. (Note, the term ‘bridge’ means a network of four elements connected to form a

square, the input being applied to two opposite corners and the output being taken from the

remaining two corners). As for the rectifier shown in Fig.8.11, the current flowing in load R is

in the same direction for both half cycles of the input giving the output waveform shown.

Following the broken line in Fig.: When P is positive with respect to Q, current flows from the

transformer to point E, through diode D2 to point G, and back to the transformer.

Following the full line in Fig.8.12:

When Q is positive with respect to P, current flows from the transformer to point G, through

diode D3 to point F, then through load R to point H, through D1 to point E, and back to the

transformer. The output waveform is not steady and needs improving; a method of smoothing

is explained in the next section.

Fig. 8.12

8.8 Smoothing of the rectified output waveform

The pulsating outputs obtained from the half- and full- wave rectifier circuits are not suitable

for the operation of equipment that requires a steady d.c. output, such as would be obtained

from batteries. For example, for applications such as audio equipment, a supply with a large

variation is unacceptable since it produces ‘hum’ in the output. Smoothing is the process of

removing the worst of the output waveform variations.

To smooth out the pulsations a large capacitor, C, is connected across the output of the rectifier,

as shown in Fig.8.13; the effect of this is to maintain the output voltage at a level which is very

near to the peak of the output waveform. The improved waveforms for half- wave and full-

wave rectifiers are shown in more detail in Fig.8.14(a)

During each pulse of output voltage, the capacitor C charges to the same potential as the peak

of the waveform, as shown as point X in Fig.8.14. As the waveform dies away, the capacitor

discharges across the load, as shown by XY. The output voltage is then restored to the peak

value the next time the rectifier con- ducts, as shown by YZ. This process continues as shown

in Fig.8.14.

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Capacitor C is called a reservoir capacitor since it stores and releases charge between the peaks

of the rectified waveform.

The variation in potential between points X and Y is called ripple, as shown in Fig.8.14; the

object is to reduce ripple to a minimum. Ripple may be reduced even further by the addition of

inductance and another capacitor in a ‘filter’ circuit arrangement, as shown in Fig.8.15.

The output voltage from the rectifier is applied to capacitor C1. and the voltage across points

AA is shown in Fig.8.15, similar to the waveforms of Fig.8.14. The load current flows through

the inductance L; when current is changing, e.m.f.’s are induced. By Lenz’s law, the induced

voltages will oppose those causing the current changes.

Fig.8.13

Fig. 8.14

Fig. 8.15

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As the ripple voltage increases and the load current increases, the induced e.m.f. in the inductor

will oppose the increase. As the ripple voltage falls and the load current falls, the induced e.m.f.

will try to maintain the current flow.

The voltage across points BB in Fig.8.15. and the current in the inductance are almost ripple-

free. A further capacitor, C2, completes the process.

Problems

1. The value of an alternating current at any given instant is:

(a) a maximum value (b) a peak value

(c) an instantaneous value (d) an r.m.s. value

2. An alternating current completes 100 cycles in 0.1 s. Its frequency is:

(a) 20 Hz (b) 100 Hz (c) 0.002 Hz (d) 1 kHz

3. In the following Fig., at the instant shown, the generated e.m.f. will be:

(a) zero (b) an r.m.s. value (c) an average value (d) a maximum value

4. The supply of electrical energy for a consumer is usually by a.c. because:

(a) transmission and distribution are more easily effected

(b) it is most suitable for variable speed motors

(c) the volt drop in cables is minimal

(d) cable power losses are negligible

5. Which of the following statements is false?

(a) It is cheaper to use a.c. than d.c.

(b) Distribution of a.c. is more convenient than with d.c. since voltages may be readily altered

using transformers

(c) An alternator is an a.c. generator

(d) A rectifier changes d.c. to a.c.

6. An alternating voltage of maximum value 100V is applied to a lamp. Which of the

following direct voltages, if applied to the lamp, would cause the lamp to light with the

same brilliance?

(a) 100V (b) 63.7V (c) 70.7V (d) 141.4V

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7. The value normally stated when referring to alternating currents and voltages is the:

(a) instantaneous value (b) r.m.s. value (c) average value (d) peak value

8. State which of the following is false. For a sine wave:

(a) the peak factor is 1.414

(b) the r.m.s. value is 0.707 × peak value

(c) the average value is 0.637 × r.m.s. value

(d) the form factor is 1.11

9. An a.c. supply is 70.7V, 50 Hz. Which of the following statements is false?

(a) The periodic time is 20 ms

(b) The peak value of the voltage is 70.7V

(c) The r.m.s. value of the voltage is 70.7V

(d) The peak value of the voltage is 100V

10. An alternating voltage is given by v = 100 sin(50πt − 0.30)V. Which of the following

statements is true?

(a) The r.m.s. voltage is 100V

(b) The periodic time is 20 ms

(c) The frequency is 25 Hz

(d) The voltage is leading v = 100 sin 50πt by 0.30 radians

11. The number of complete cycles of an alternating current occurring in one second is

known as:

(a) the maximum value of the alternating current

(b) the frequency of the alternating current

(c) the peak value of the alternating current

(d) the r.m.s. or effective value

12. A rectifier conducts:

(a) direct currents in one direction

(b) alternating currents in one direction

(c) direct currents in both directions

(d) alternating currents in one direction

14. A sinusoidal voltage has a mean value of 3.0A. Determine it’s maximum and r.m.s.

values.

15. The instantaneous value of current in an a.c.circuit at any time t seconds is given by:

i = 50 sin(100πt − 0.45) mA. Determine

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(a) the peak to peak current, the frequency, the periodic time and the phase angle (in degrees)

(b) the current when t = 0

(c) the current when t = 8 ms

(d) the first time when the voltage is a maximum. Sketch the current for one cycle showing

relevant points.