An a.c. voltage is applied to a pure inductor. The current in the inductor would be
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → lagging the voltage by \( \displaystyle\frac{\pi}{2} \)
In a pure inductor the applied voltage is balanced by the self-induced emf, \( v = L\,di/dt \).
\( \displaystyle v = v_m \sin\omega t \)
\( \displaystyle \frac{di}{dt} = \frac{v_m}{L}\sin\omega t \)
\( \displaystyle i = -\frac{v_m}{\omega L}\cos\omega t = i_m \sin\left(\omega t - \frac{\pi}{2}\right) \)
So the current reaches its maximum a quarter cycle after the voltage: it lags the voltage by π/2.
Why the other options are incorrect
Option (1) leading the voltage by \( \displaystyle\frac{\pi}{2} \) — Incorrect. The current leads the voltage by π/2 in a pure capacitor, not in an inductor.
Option (3) leading the voltage by \( \displaystyle\frac{\pi}{4} \) — Incorrect. A phase difference of π/4 occurs in a series LR circuit with \( X_L = R \), not in a pure inductor.
Option (4) in phase with the voltage. — Incorrect. Current and voltage are in phase in a pure resistor.
NCERT (2026-27):
AC Voltage Applied to an Inductor – “… A comparison of Eqs. (7.1) and (7.12) for the source voltage and the current in an inductor shows that the current lags the voltage by \( \pi/2 \) or one-quarter (1/4) cycle.”