In a pure capacitive ac circuit

The correct answer is Option (4) → The current leads the voltage in phase by π/2.

In a capacitor the current is the rate of flow of charge onto the plates, and the charge is proportional to the voltage. With \( v = v_m\sin\omega t \):

    \( \displaystyle q = Cv = Cv_m\sin\omega t \)

    \( \displaystyle i = \frac{dq}{dt} = \omega Cv_m\cos\omega t \)

    \( \displaystyle i = i_m\sin\left(\omega t + \frac{\pi}{2}\right) \)

The current reaches its maximum a quarter cycle before the voltage, so the current leads the voltage by \( \pi/2 \).

Why the other options are incorrect

• Option (1) The current and voltage differ in phase by π. — Incorrect. A phase difference of \( \pi \) (opposite phase) does not occur for a single pure element.

• Option (2) The current lags the voltage in phase by π/2 — Incorrect. The current lags the voltage by \( \pi/2 \) in a pure inductive circuit, not a capacitive one.

• Option (3) The current and voltage are in phase. — Incorrect. Current and voltage are in phase in a pure resistive circuit.

NCERT (2026-27):
Summary – “… An ac voltage \( v = v_m\sin\omega t \) applied to a capacitor drives a current in the capacitor: \( i = i_m\sin(\omega t + \pi/2) \). Here,

    \( \displaystyle i_m = \frac{v_m}{X_C},\quad X_C = \frac{1}{\omega C} \) is called capacitive reactance.

The current through the capacitor is \( \pi/2 \) ahead of the applied voltage.”