A coil is connected to a 12 V DC source and draws a steady current of 4 A . When the same coil is connected to an AC source of 12 V (rms) and angular frequency 50 rad/s , the rms current in the circuit is 2.4 A . The inductance of the coil is:

Step-by-step solution

When the coil is connected to a DC source, the inductive reactance is zero. The resistance R of the coil is: R = V_ DC I_ DC = 12 4 = 3 When connected to an AC source, the net impedance Z of the coil is: Z = V_ rms I_ rms = 12 2.4 = 5 The impedance of the coil (which acts as a series LR circuit) is given by: Z = R^2 + X_L^2 5 = 3^2 + X_L^2 25 = 9 + X_L^2 X_L^2 = 16 X_L = 4 The inductive reactance is X_L = L . Substituting the given values: 4 = 50 L L = 4 50 H = 0.08 H = 80 mH Answer: 80 mH