A small signal voltage V(t) = V_0 ω t is applied across an ideal capacitor C :

1. Current I(t), lags voltage V(t) by \(90^\circ\).

2. Over a full cycle the capacitor C does not consume any energy from the voltage source.

3. Current I(t) is in phase with voltage V(t).

4. Current I(t) leads voltage V(t) by \(180^\circ\).

Doubtly Asked 30 Oct 2026 NEET UG 2016 Phase 1 Practice test neet Save WhatsApp Report

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Answer: 2. Over a full cycle the capacitor C does not consume any energy from the voltage source.

Key idea: in an ideal capacitor, current leads voltage by \(90^\circ\), so the average power is zero.

Option 1 is wrong: current leads, not lags.
Option 2 is right: energy is stored and returned each quarter cycle, so net consumption over a cycle is zero.
Option 3 is wrong: they are not in phase.
Option 4 is wrong: the lead is \(90^\circ\), not \(180^\circ\).

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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