A series LR circuit connected to an ac source dissipates an average power P₀ . The power factor of the circuit is 1 2 . If the frequency of the source is doubled while keeping the voltage amplitude constant, the new average power dissipated in the circuit will be :

Step-by-step solution

The initial power factor is = 1 2 . R R^2 + X_L^2 = 1 2 X_L = R The initial impedance squared is Z₁^2 = R^2 + X_L^2 = 2R^2 . The average power dissipated initially is P₀ = I_ rms ^2 R = V_ rms ^2 Z₁^2 R = V_ rms ^2 R 2R^2 = V_ rms ^2 2R . When the frequency is doubled ( ' = 2 ), the new inductive reactance becomes X_L' = 2X_L = 2R . The new impedance squared is Z₂^2 = R^2 + (X_L')^2 = R^2 + (2R)^2 = 5R^2 . The new average power dissipated is P' = V_ rms ^2 Z₂^2 R = V_ rms ^2 R 5R^2 = V_ rms ^2 5R . Taking the ratio