Page 346 - Basic Electrical Engineering

Power consumed, P = VI cos ϕ

= 230 × 10.75 × 0.703 W

= 1738.16 W

Example 3.14 Two coils having impedance Z and Z are connected in
1
2
series across a 230 V, 50 Hz power supply as shown in Fig. 3.42 (b).

Figure 3.42 (b)

The voltage drop across Z is equal to 120 ∠ 30° V. Calculate the value of
1
Z .
2

Solution:

We have, V = V + V 2
1
or, V = V − V = 230 ∠0 − 120∠ 30°
1
2
= 230(cos 0° + j sin 0°) − 120(cos 30° + j sin 30°)

= 230 − 120 × 0.866 − j 120 × 0.5

= 139.6 ∠ − 25.4°

Since this is a series circuit, the current flowing through the circuit is the

same.
The circuit current can be calculated by using any of the following

relations