Introduction to Electric Circuits

82 Single-phase a.c. circuits

Solution

1 The impedance Z = ~//(R 2 + [X g -- Xc]2). Now

XL = 2rrfL = 2rr 100 X 150 • 10 -3 -- 94.2 f/

and

Xc = 1/2rrfC = 1//2rr 100 x 10 x 10 -6= 159 a

SO

Z- X/(122 + [94.2- 159] 2) -65.9

2 The current drawn from the supply is given by I = V/Z where I and V are

rms values. The peak value of the supply voltage is 100 V so that the rms

value is 100/~/2 = 70.7 V, so

I- V/Z- 70.7/65.9- 1.07 a

3 The phase angle of the circuit is 4) = cos -1 (R/Z) = cos -1 (12/65.9) = 79.5 ~

Because the capacitive reactance is greater than the inductive reactance, the

circuit is predominantly capacitive and so the phase angle is a leading one

(i.e. the current leads the voltage).

The expression for the current is i = I m sin (27rft - 4)) with I m = V'2I and

4) = 79.5 ~ = 79.57r/180 rad. Thus i - 1.5 sin (2007rt - 0.4470 A.

4.4 COMPLEX NOTATION

We have seen that a phasor quantity is one for which both magnitude and

direction is important. These quantities may be represented by phasor diagrams

in the manner shown earlier in the chapter. In Fig. 4.21(a), the phasor V1 is

(a) Vi (b) I V~_cos (;th

Vl sinr Vl sin(lh

(reference) ~ I

V~ cosr

V2

Figure 4.21

(reference)

shown to be leading the reference phasor by 4)1 degrees, whereas the phasor V2
is shown as lagging the reference by ~J)2 degrees. It is conventional to take the
horizontal axis as the reference direction.
V~ may be represented by

V 1 -'- IVllL-~)I

In this notation IVll indicates the magnitude of the quantity and is represented