A pure inductor of 0.25 H is connected to a source of 220 V. If the frequency of the source is 50 Hz, then the rms current in the circuit will be.
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 2.8 A
The a.c. supply voltage 220 V is an rms value. In a pure inductor the current is limited by the inductive reactance.
\( \displaystyle X_L = 2\pi fL = 2\pi \times 50\,\mathrm{Hz} \times 0.25\,\mathrm{H} \approx 78.5\,\Omega \)
\( \displaystyle I_{\mathrm{rms}} = \frac{V_{\mathrm{rms}}}{X_L} = \frac{220\,\mathrm{V}}{78.5\,\Omega} \approx 2.8\,\mathrm{A} \)
Note: NTA picked this question from an NCERT (2026-27) exercise and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27):
Exercises – …
