An inductor of self inductance 1 H is connected in series with a resistor of 100 π ohm and an ac supply of 100 π volt, 50 Hz . Maximum current flowing in the circuit is _________ A.

Solution

Physics sir · 3 steps

Find the impedance first, then remember 100π V is the rms value, so multiply the current by √2.

1. Inductive reactance. X_L = ω L = 2π f L = 2π × 50 × 1 = 100π Ω. This is the same size as the resistance R = 100π Ω.

   Z = √(R² + X_L²) = √((100π)² + (100π)²) = 100√2 π Ω

2. Rms current. A supply voltage quoted without a label is its rms value. So Iᵣₘₛ = Vᵣₘₛ/Z = 100π/(100√2 π) = 1/√2 A.
3. So. Iₘₐₓ = √2 Iᵣₘₛ = √2 × 1/√2 = 1 A, so the answer is 1.

The common slip: 0.71. Treating 100π V as the peak voltage gives Iₘₐₓ = 1/√2 ≈ 0.71 A, not 1.

The idea to keep. AC supply voltages are rms values unless the question says peak. Use Z = √(R² + X_L²) for the current, then multiply by √2 for the maximum.

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