An inductive coil of resistance R = 0 Ω, L = 50 mH is connected to a source of 220 V, 50 Hz. The r.m.s. current is:-
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 14 A
With R = 0 the coil is a pure inductor, and the rms current is the rms voltage (220 V) divided by the inductive reactance.
\( \displaystyle X_L = 2\pi f L = 2 \times 3.14 \times 50\,\mathrm{Hz} \times 50\times10^{-3}\,\mathrm{H} \approx 15.7\,\Omega \)
\( \displaystyle I_{\mathrm{rms}} = \frac{V_{\mathrm{rms}}}{X_L} = \frac{220\,\mathrm{V}}{15.7\,\Omega} \approx 14\,\mathrm{A} \)
Note: NTA picked this question from an NCERT (2026-27) example and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27):
AC Voltage Applied to an Inductor – …
