Introduction to Electric Circuits

4.1 Alternating quantities 71

For a sinusoidal current i = I m sin ~ot. Squaring gives i 2 -- Im 2 sin 2 o~t and the
27r, o)
mean of this over a complete cycle is (1/27r) f Im 2 sin 2 ~ot d(~ot). The rms value
is then the square root of this. 0
Using the identity sin 2 0 - (1 - cos 20)//2,
2rr/'oJ 2-rr o~
f Im 2 sin 2 ~ot d(oJt) - Im 2 f (l -- COS 2wt)/2 d(wt)
o o
277"'o)
-- Im 2/2[oot- sin o~t]o - 7rlm 2
The mean of this (obtained by dividing by 2 rr) is Im2/2. The rms value (obtained
by taking the square root) is
I = Im/X/2 (4.5)

The average value of a sinusoidai quantity

The average value of a sine wave over a complete cycle is zero, which is rather

meaningless, so the average value is taken to be the average over a half cycle.

This value is denoted by a capital letter with a subscript (e.g. Iav). For a current

represented by i = Im sin o~t,

2~,,/w 27r/'w

/av = (1/Tr) f Im sin wtd (wt) - (Im/rr)[-cos wt]0

0

= (Im/77")[ ( -- COS 77" -- ( -- COS 77" -- ( -- COS 0)]

9 ". Iav= 2Im/Tr (4.6)

The form factor of an a.c. waveform

This is defined to be the rms value divided by the average value, so that for a

sine wave the form factor is (Im/V'2)/(2Im/rr) - rr/(2V'2).

Form factor of a sine wave = 1.11 (4.7)

Example 4.3

An alternating voltage has an average value of 4 V and a form factor of 1.25.
Calculate (1) its rms value, (2) the peak value of a sinusoidal voltage having the
same rms value.

Solution

1 The form factor = rms value/average value. Thus the rms value = the form

factor x the average value"

V = 1.25 • 4 = 5 V

2 From Equation (4.5) we see that for a sinusoidal voltage the peak
value - ~/2 • the rms value. Thus Vm = ~/2 X 5 -- 7.07 V.