A small signal voltage \( e=e_0\sin\omega t \) is applied across an ideal capacitor of capacitance C. The correct statement is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → Capacitor does not consume any energy from the source of emf over a complete cycle.
For the voltage \( e = e_0\sin\omega t \) across the capacitor, the charge is \( q = Ce \), and the current is its rate of change:
\( i = \frac{dq}{dt} = \omega C e_0\cos\omega t = i_0\sin\left(\omega t + \frac{\pi}{2}\right) \)
The current leads the voltage by \( \pi/2 \). The instantaneous power taken from the source is
\( p = ie = \omega C e_0^{2}\sin\omega t\cos\omega t = \frac{\omega C e_0^{2}}{2}\,\sin 2\omega t \)
and its average over a complete cycle is zero: the capacitor stores energy while it charges and returns it while it discharges, so it consumes no energy.
Why the other options are incorrect
Option (1) Phase difference between Current I(t) and Voltage V(t) is zero — Incorrect. A phase difference of zero is the case of a pure resistor; here it is \( \pi/2 \).
Option (3) Current I(t) lags behind the voltage V(t) by \( \frac{\pi}{2} \) — Incorrect. The current leads the voltage in a capacitor; it lags in a pure inductor.
Option (4) Current I(t) leads the voltage V(t) by \( \pi \) — Incorrect. The lead is \( \pi/2 \), not \( \pi \).
NCERT (2026-27):
Summary – “… The current through the capacitor is π/2 ahead of the applied voltage. As in the case of inductor, the average power supplied to a capacitor over one complete cycle is zero.”