Question

2. A coil connected across a 120 V, 60 Hz line draws a current of 5A.A Wattmeter placed in the circuit indicates 525 Watts. Calculate (a) Ps, (b) PqL

A (Ps) means apparent power and B (PqL) means reactive power.

Answer 1

Let coil have resistance R and inductive resistance is x_L line current I = A (magnitude) power loss p = 525 watt calculated or measured by watt meter Now power loss is coil only occurs in resistance p = 525 = I^2 R = (5)^2 R 525 = R 25 R = 525/25 therefore R = 215 Impedance of coil 7 = R + J x_L Vertical-bar z Vertical-bar = squareroot R^2 + x^2_i but Vertical-bar z Vertical-bar = v_rms/I = 120/5 = 24 ohm put in 1 apparent power or complex power or total power by source s = v_rms I * s = (120 angle 0 degree) (5 angle - 28.95 degree) s = (120 angle 0 degree) (5 angles 28.95) s = 600 angle 28.95 k v = (5600 cos 28.95 = 3600 sin 28.95 kv s = (525.025 + j 290 .42) kv active power: p = 525.025 watts & reactive power: R = + 290 .42 v A R at lagging power factor