The value of current through an inductor of inductance 2.0 H and negligible resistance when connected to an AC source of 157 V, 50 Hz will be:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( \displaystyle\frac{1}{4}\,\mathrm{A} \)
An ideal inductor in an a.c. circuit offers the inductive reactance \( X_L = \omega L = 2\pi fL \), and the current follows \( I = V/X_L \) for the same type of value (rms or peak) of voltage and current.
\( X_L = 2\pi fL = 2\pi\times50\,\mathrm{Hz}\times2.0\,\mathrm{H} = 200\pi\,\Omega \approx 628\,\Omega \)
\( \displaystyle I = \frac{V}{X_L} = \frac{157\,\mathrm{V}}{628\,\Omega} = 0.25\,\mathrm{A} = \frac{1}{4}\,\mathrm{A} \)
Shortcut: Notice that \( 157 = 50\pi \). Then \( I = \frac{50\pi}{200\pi} = \frac{1}{4}\,\mathrm{A} \) exactly, with no need to put in the value of \( \pi \).
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 – …
