A coil is connected to an AC source of RMS voltage 100 V. It draws an RMS current of 5 A and consumes 300 W of power. If a capacitor of capacitive reactance 21 , is then connected in series with the coil, the new power factor of the circuit will be:

Step-by-step solution

First, we determine the resistance and inductive reactance of the coil. The power consumed by the coil is due to its resistance R : P = I_ rms ^2 R 300 = (5)^2 R R = 300 25 = 12 , The initial impedance of the coil is: Z = V_ rms I_ rms = 100 5 = 20 , Using the impedance formula for the coil Z = R^2 + X_L^2 : 20^2 = 12^2 + X_L^2 400 = 144 + X_L^2 X_L^2 = 256 X_L = 16 , Now, a capacitor with X_C = 21 , is added in series. The new impedance Z' of the circuit is: Z' = R^2 + (X_L - X_C)^2 Z' = 12^2 + (16 - 21)^2 Z' = 14