As shown in the figure an inductor of inductance 200 mH is connected to an AC source of emf 220 V and frequency 50 Hz. The instantaneous voltage of the source is 0 V when the peak value of current is √a/π A. The value of a is ___________.

Solution

Physics sir · 3 steps

In a pure inductor the current peaks exactly when the voltage is zero, so just find the peak current.

1. Phase in an inductor. In a pure inductor the current lags the voltage by 90°. So when the source voltage is 0, the current is at its peak value I₀.
2. Inductive reactance. X_L = 2π f L = 2π × 50 × 0.2 = 20π Ω. The given 220 V is the rms value, so the peak voltage is V₀ = 220√2 V.

   I₀ = V₀/X_L = 220√2/20π = 11√2/π = √242/π A

3. So. Comparing with √a/π gives a = 121 × 2 = 242.

The common slip: 121. Treating 220 V as the peak value gives I₀ = 11/π, so a = 121. Use V₀ = 220√2 for the peak current.

The idea to keep. In a pure L or pure C circuit, current and voltage are 90° apart: one is zero when the other peaks. And a quoted AC voltage is rms unless stated, so multiply by √2 for the peak.

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