An inductor of self-inductance 1 H is connected in series with a resistor of 100π Ω across an AC supply of 100π V (rms) and frequency 50 Hz. Find the maximum current in ampere.

Question

G1 Physics · Accepted Answer

Angular frequency: ω = 2πf = 2π(50) = 100π rad/s. Inductive reactance: X L X_L X L ​ = ωL = 100π×1 = 100π Ω. Series impedance: Z = √(R²+ X L 2 ) X_L²) X L 2 ​ ) = √[(100π)²+(100π)²] = 100π√2 Ω. RMS current: I r m s I_rms I r ​ m s = V r m s / Z V_rms/Z V r ​ m s / Z = 100π/(100π√2) = 1/√2 A. Maximum current: I m a x I_max I m ​ a x = √2 I r m s I_rms I r ​ m s = 1 A. Answer: 1.