The peak value of current in an ideal inductor of \( (2/\pi)\,\mathrm{H} \) 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
An ideal inductor offers only its inductive reactance \( X_L = \omega L = 2\pi f L \). The supply voltage quoted for an ac mains is its rms value, so the reactance gives the rms current first, and the peak current is \( \sqrt{2} \) times it.
\( \displaystyle X_L = 2\pi f L = 2\pi \times 50\,\mathrm{Hz} \times \frac{2}{\pi}\,\mathrm{H} = 200\,\Omega \)
\( \displaystyle I_{\mathrm{rms}} = \frac{V_{\mathrm{rms}}}{X_L} = \frac{200\,\mathrm{V}}{200\,\Omega} = 1\,\mathrm{A} \)
\( \displaystyle I_0 = \sqrt{2}\,I_{\mathrm{rms}} = 1.414\,\mathrm{A} \)