A pure inductor of 70 mH is connected to a source of 220 V. The frequency of the source is 50 Hz. The current in the circuit is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 10.0 A
A pure inductor opposes alternating current only through its inductive reactance \( X_L = \omega L = 2\pi\nu L \). The rms current is the rms voltage divided by this reactance.
\( L = 70\,\mathrm{mH} = 70\times10^{-3}\,\mathrm{H} \)
\( X_L = 2\pi\times50\times70\times10^{-3} \approx 22.0\,\Omega \)
\( \displaystyle I_{\mathrm{rms}} = \frac{V_{\mathrm{rms}}}{X_L} = \frac{220\,\mathrm{V}}{22.0\,\Omega} = 10.0\,\mathrm{A} \)
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 – …
