Page 348 - Basic Electrical Engineering

or,

= 687.8 μF

Example 3.15 An alternating voltage, V = (160 + j170)V is connected
across an L–R series circuit. A current of I = (12 – j5) A flows through the

circuit. Calculate impedance, power factor, and power consumed. Draw the
phasor diagram.

Solution:

Impedance,

= 12.1346.8°+ 22.6°

= 12.13 ∠46.8° + 22.6°

Z = 12.13(cos 69.4° + j sin 69.4°)

= 12.13 × 0.35 + j 12.13 × 0.93

= 4.24 + j 11.28 = R + jX L

The series circuit consists of a resistance of 4.24 Ω and an inductive

reactance of 11.28 Ω. The phasor diagram is drawn by considering a
reference axis. Let x-axis be the reference axis. The voltage applied has a

magnitude of 233 V and is making 46.8° with the reference axis in the
positive direction, i.e., the anticlockwise direction. Current flowing is 19.2 A