A coil having resistance R and inductance L is first connected to a steady DC voltage source of V volts. The power dissipated in the coil is P₁ . The coil is then connected to an AC voltage source of RMS voltage V . If the inductive reactance of the coil at the AC frequency is equal to its resistance R , the power dissipated in this case is P₂ . The ratio P₁ : P₂ is :

Step-by-step solution

When connected to a steady DC source, the inductor acts as a short circuit. The impedance of the coil is simply its resistance R . Power dissipated in the DC circuit is P₁ = V^2 R When connected to the AC source, the impedance of the coil is Z = R^2 + X_L^2 Given that X_L = R , Z = R^2 + R^2 = R 2 Power dissipated in the AC circuit is P₂ = I_ rms ^2 R = ( V Z )^2 R P₂ = ( V R 2 )^2 R = V^2 2R^2 R = V^2 2R The ratio of the powers is P₁ P₂ = V^2 R V^2 2R = 2 1 Thus, P₁ : P₂ = 2 : 1 . Answer: 2:1