In a purely capacitive ac circuit, the current

The correct answer is Option (2) → leads the voltage by \( 90^\circ \)

For a source \( v=v_m\sin\omega t \) across a capacitor, the charge on the plates is \( q=Cv \) and the current is the rate of flow of charge, \( i=dq/dt \).

    \( \begin{aligned} i &= \frac{d}{dt}\left(Cv_m\sin\omega t\right) \\ &= \omega Cv_m\cos\omega t \\ &= i_m\sin\left(\omega t+\frac{\pi}{2}\right) \end{aligned} \)

From the above, the current has the phase \( \omega t+\pi/2 \) while the voltage has the phase \( \omega t \).

Option (1) leads the voltage by \( 180^\circ \) — Incorrect. A shift of \( 180^\circ \) would make the current always opposite in sign to the voltage. The capacitor current is shifted by only a quarter cycle.

Option (2) leads the voltage by \( 90^\circ \) — Correct. The current \( i_m\sin(\omega t+\pi/2) \) is ahead of the voltage by \( \pi/2 \). It is maximum when the capacitor has just begun to charge and zero when the plates are fully charged.

Option (3) lags the voltage by \( 180^\circ \) — Incorrect. A lag of \( 180^\circ \) again means current and voltage in opposite phase, which does not happen in a purely capacitive circuit.

Option (4) lags the voltage by \( 90^\circ \) — Incorrect. A current that lags the voltage by \( \pi/2 \) belongs to a pure inductor, where the self-induced emf delays the current.

NCERT (2026-27):
AC Voltage Applied to a Capacitor – “A comparison of Eq. (7.16) with the equation of source voltage, Eq. (7.1) shows that the current is π/2 ahead of voltage.”

    \( v=v_m\sin\omega t \qquad (7.1) \)

    \( i=i_m\sin\left(\omega t+\frac{\pi}{2}\right) \qquad (7.16) \)