ABSTRACT
Problem 7. Determine the peak and mean values for a 240 V mains supply
For a sine wave, r.m.s. V = 0.707 × Vm A 240 V mains supply 240 V is the r.m.s. value,
hence Vm = V
0.707 =
0.707 V = peak value
Mean value VAV = 0.637Vm = 0.637 × 339.5 = 216.3 V
Problem 8. A supply voltage has a mean value of 150 V. 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 = 166.5 V
Now try the following Practice Exercises
1. An alternating current varies with time over half a cycle as follows:
Current (A) 0 0.7 2.0 4.2 8.4 8.2 time (ms) 0 1 2 3 4 5
Current (A) 2.5 1.0 0.4 0.2 0 time (ms) 6 7 8 9 10
The negative half cycle is similar. Plot the curve and determine: (a) the frequency, (b) the instantaneous values at 3.4 ms and 5.8 ms, (c) its mean value, (d) its r.m.s. value
2. For the waveforms shown in Figure 40.7 determine for each (i) the frequency, (ii) the average value, over half a cycle, (iii) the r.m.s. value, (iv) the form factor, (v) the peak factor
3. An alternating voltage is triangular in shape, rising at a constant rate to a maximum of 300 V in 8 ms and then falling to zero at a constant rate in 4 ms. The negative half cycle is identical in shape to the positive half cycle. Calculate (a) the mean voltage over half a cycle, (b) the r.m.s. voltage
4. Calculate the r.m.s. value of a sinusoidal curve of maximum value 300 V
5. Find the peak and mean values for a 200 V mains supply
6. A sinusoidal voltage has a maximum value of 120 V. Calculate its r.m.s. and average values
7. A sinusoidal current has a mean value of 15.0 A. Determine its maximum and r.m.s. values
1. Briefly explain the principle of operation of the simple alternator
2. What is meant by (a) waveform, (b) cycle? 3. What is the difference between an alternating
and a unidirectional waveform?
