The peak value of current in an ideal inductor of (2/π) H inductance connected to a 200 V, 50 Hz ac supply will be

1.414 A

The correct answer is Option (2) → 1.414 A

Given:

Inductance $L = \frac{2}{\pi}$ H

Voltage $V = 200$ V (rms)

Frequency $f = 50$ Hz

Inductive reactance:

$X_L = 2 \pi f L = 2 \pi (50) \left(\frac{2}{\pi}\right) = 200 \, \Omega$

RMS current:

$I_{rms} = \frac{V}{X_L} = \frac{200}{200} = 1 \, A$

Peak current:

$I_0 = \sqrt{2} \, I_{rms} = \sqrt{2} \times 1 = 1.414 \, A$