In an inductor connected to an ac source,

The correct answer is Option (3) → the current lags the voltage in phase by \( \pi/2 \)

For a pure inductor the applied voltage balances the self-induced emf.

    \( \displaystyle v = v_m\sin\omega t = L\frac{di}{dt} \)

Integrating,

    \( \displaystyle i = -\frac{v_m}{\omega L}\cos\omega t = i_m\sin\left(\omega t - \frac{\pi}{2}\right) \)

So the current lags the voltage by \( \pi/2 \) (a quarter cycle).

Why the other options are incorrect

• Option (1) the current and voltage are in same phase. — Incorrect. Current and voltage are in phase only in a pure resistor.

• Option (2) the current leads the voltage in phase by \( \pi \) — Incorrect. A phase difference of \( \pi \) does not occur in any single element; in a capacitor the current leads by \( \pi/2 \).

• Option (4) the current lags the voltage in phase by \( \pi \) — Incorrect. The lag in an inductor is \( \pi/2 \), not \( \pi \).

NCERT (2026-27):
AC Voltage Applied to an Inductor – …