When a d c voltage of 100 V is applied to an inductor, a d c current of 5 A flows through it. When an ac voltage of 200 V peak value is connected to inductor, its inductive reactance is found to be 20 √3 Ω. The power dissipated in the circuit is _________ W.

Solution

Physics sir · 4 steps

DC tells you the coil's resistance, AC tells you the current, and only the resistance burns power.

1. Resistance from DC. With steady DC the inductor offers no reactance, so only its wire resistance limits the current. R = 100/5 = 20 Ω.
2. Impedance in AC. With AC the same R is in series with X_L = 20√3 Ω. So Z = √(20² + (20√3)²) = √(400 + 1200) = 40 Ω.
3. Find the current. Peak current is I₀ = 200/40 = 5 A, so the rms current is Iᵣₘₛ = 5/√2 A.

   P = Iᵣₘₛ² R = 25/2 × 20 = 250 W

4. So. Only the resistance dissipates energy; the pure inductance stores and returns it. The power dissipated is 250 W.

The common slip: 500. Using the peak current in I² R gives 5² × 20 = 500 W. Average power needs the rms current, which is I₀/√2.

The idea to keep. Average AC power is always Iᵣₘₛ² R, never with peak values. Reactance changes the current but never burns energy itself.

Answer from the NTA final answer key. Solution by PrepWiser.