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
Answer & explanation
Correct answer: option 2
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:
$I_0 = \sqrt{2} \, I_{rms} = \sqrt{2} \times 1 = 1.414 \, A$