Introduction to Electric Circuits

4.2 Single-phase a.c. circuits in the steady state 73

by its arrowhead. After the completion of the positive half cycle, of course, both

arrowheads will reverse.

Let the voltage be represented by v = Vm sin wt. The value of the current i

flowing at any instant will be given by v/R=(Vm/R) sinoot, i.e.

i= (Vm/R)sin ~ot. Now (Vm/R) is the maximum value (lm) reached by the

current, so i = Im sin ~ot. Note that there is no phase difference between the

voltage and the current expressions. The waveforms and the phasor diagram

are as shown in Fig. 4.7(a) and (b).

v,i~ v

(^0) t
(a) Waveforms (b) Phasordiagram
Figure 4.7
The rms value (V) of the voltage is Vm/~/2 and that of the current (I) is
Im/V'2. Now Im/X/2 = (Vm/V'2)/R, so that I = V/R (which of course is Ohm's
law). For a purely resistive a.c. circuit then,
Vii- R (4.8)
Example 4.4
In the circuit of Fig. 4.6, R = 10 11 and v - 25 sin 314t. Determine (1) the rms
value of the current, (2) the phase angle of the circuit, (3) the frequency of the
supply.
Solution
1 The peak value of the voltage is 25 V so that the peak value of the current
is 25/R = 2.5 A. The rms value of the current is therefore 2.5/~/2 - 1.77 A.
2 For a purely resistive circuit the current and voltage phasors are in phase
with each other so that the phase angle is zero.
3 The angular frequency is ~o = 314 rad s -~. The frequency is
f= 60/2Ir- 314/6.28 = 50 Hz
Purely inductive circuits
The diagram of Fig. 4.8 overleaf shows a pure inductor L connected to a single-
phase voltage source V. Let the current be represented by i - Im sin o)t. Since
this changing current produces a changing flux which will link the inductor then,