In an ac circuit, the current leads the voltage by π/2 . The circuit…

The correct answer is Option (4) → purely capacitive

The phase difference between current and voltage tells which element the circuit contains. For a capacitor, \( v = v_m\sin\omega t \) gives

    \( \displaystyle i = C\,\frac{dv}{dt} = \omega C v_m \cos\omega t = i_m \sin\left(\omega t + \frac{\pi}{2}\right) \)

So in a purely capacitive circuit the current leads the voltage by exactly \( \pi/2 \).

Option (1) purely resistive — Incorrect. In a purely resistive circuit the current and voltage are in phase.

Option (2) should have circuit elements with resistance equal to reactance. — Incorrect. With \( R = X \) the phase difference is \( \tan^{-1}(X/R) = \pi/4 \), not \( \pi/2 \).

Option (3) purely inductive — Incorrect. In a purely inductive circuit the current lags the voltage by \( \pi/2 \).

Option (4) purely capacitive — Correct. The current leads the voltage by \( \pi/2 \) only in a purely capacitive circuit.

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. … Thus, we see that in the case of an inductor, the current lags the voltage by π/2 and in the case of a capacitor, the current leads the voltage by π/2.”