The AC 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 Figure 6.1 .

An e.m.f. is generated in the coil (from Faraday’s laws) which varies in magnitude and reverses its direction at regular intervals. The reason for this is shown in Figure 6.2 . In positions (a), (e) and (i) the conductors of the loop are effectively moving along the magnetic fi eld, no fl ux is cut and hence, no e.m.f is induced. In position (c) maximum fl ux is cut and maximum e.m.f is induced. In position (g), maximum fl ux is cut and 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 shown as E . In positions (b), (d), (f) and (h) some fl ux is cut and some e.m.f is induced. If all such positions

C H A P T E R 6

Figure 6.1 : Coil rotates at constant angular velocity

124 Chapter 6

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 AC generator (i.e., the alternator).

6.2 Waveforms

If values of quantities that vary with time t are plotted to a base of time, the resulting graph is called awaveform . Some typical waveforms are shown in Figure 6.3 . Waveforms (a) and (b) areunidirectional waveforms , for, although they vary considerably with time, they fl ow 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 Figure 6.3(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 Figure 6.3(g) ).

The time taken for an alternating quantity to complete one cycle is called the period or theperiodic time, T , of the waveform.

Alternating Voltages and Currents 125

The number of cycles completed in one second is called the frequency , f , of the supply and is measured inhertz , Hz. (The standard frequency of the electricity supply in the U.S.

is 60 Hz and in Great Britain is 50 Hz.)

T 1 1

f f

or T Example 6.1

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

Solution

(a) Periodic time T

1f 1

50 0.02 s or 20 msv (b) Periodic time T

1f 1

20 000 0.000 05 s or 50 sμ Example 6.2

Determine the frequencies for periodic times of (a) 4 ms, (b) 4 μ s.

Figure 6.3 : Typical waveforms

126 Chapter 6

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

They are represented by small letters, i , υ , e , etc. (See Figures 6.3(f) and (g). )

The largest value reached in a half cycle is called the peak value or the maximum value or theamplitude of the waveform. Such values are represented by V_m , I_m etc. (See

Figures 6.3(f) and (g). ) A peak-to-peak value of e.m.f is shown in Figure 6.3(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

Alternating Voltages and Currents 127

The area under the curve is found by approximate methods such as the trapezoidal rule, the mid-ordinate rule or Simpson’s rule. Average values are represented by V_AV , I _AV , 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 (rms) value and whenever an alternating quantity is given, it is assumed to be the rms value. The symbols used for rms values are I , V , E , etc. For a nonsinusoidal waveform as shown in Figure 6.4 the rms value is given by:

I i i i

n ⎛ ₁² ₂² n²

⎝⎜⎜⎜⎜ ⎞

⎠⎟⎟⎟

⎟

where n is the number of intervals used.

For a sine wave,

rms value 0.707 maximum value (i.e., 1/ 2maximum value)

Form factor rms value

average value

For a sine wave, form factor 1.11

Figure 6.4 : Nonsinusoidal waveform

128 Chapter 6

w w w. n e w n e s p r e s s . c o m

rms 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 6.4

For the periodic waveforms shown in Figure 6.5 determine for each: (i) frequency, (ii) average value over half a cycle, (iii) rms value, (iv) form factor, and (v) peak factor.

Solution

(a) Triangular waveform ( Figure 6.5(a) )

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

T

1 1

20 10

1000

3 20 50 Hz (ii) Area under the triangular waveform for a half cycle

¹₂ baseheight ¹₂ (1010³)200 1 volt second

Figure 6.5 : Waveforms for Example 6.4

Alternating Voltages and Currents 129

(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 rms value is found to be 115.6 V)

130 Chapter 6

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

(iv) Form factor rms value

The following table gives the corresponding values of current and time for a half cycle of alternating current.

time t (ms) 0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 current i (A) 0 7 14 23 40 56 68 76 60 5 0

Assuming the negative half cycle is identical in shape to the positive half cycle, plot the waveform and fi nd (a) the frequency of the supply, (b) the instantaneous values of current after 1.25 ms and 3.8 ms, (c) the peak or maximum value, (d) the mean or average value, and (e) the rms value of the waveform.

Solution

Alternating Voltages and Currents 131

(b) Instantaneous value of current after 1.25 ms is 19 A , from Figure 6.6.

Instantaneous value of current after 3.8 ms is 70 A , from Figure 6.6.

(c) Peak or maximum value 76 A.

(d) Mean or average value area under curve length of base

Using the mid-ordinate rule with 10 intervals, each of width 0.5 ms gives:

area under curve.

(0.5 10 ³)[3 10 19 30 49 63 73 72 30 2] (see Figure 6.6 )

(0.5 10 ³)(351)

Figure 6.6 : Half cycle of alternating current for Example 6.5

132 Chapter 6

Calculate the rms value of a sinusoidal current of maximum value 20 A.

Solution