An inductor of self inductance 1 H 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.

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Impedance of circuit ​ Z = R 2 + ( X L ​ ) 2 ​ = R 2 + ( ω L ​ ) 2 ​ = ( 100 π ) 2 + ( 2 π × 50 × 1 ) 2 ​ = ( 100 π ) 2 + ( 100 π ) 2 ​ = 2 ​ × 100 π I rms ​ = 2 V ​ = 2 ​ × 100 π 100 π ​ = 2 ​ 1 ​ I m a x ​ = 2 ​ I rms ​ = 2 ​ × 2 ​ 1 ​ = 1 Ampere ​ Correct Answer : 1

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