Rankers Physics

Alternating Current: Practice Problem & Solution

7. A small signal voltage $V(t) = V_0 \sin \omega t$ is applied across an ideal capacitor C : (2016 - I)
Current $I(t)$ lags voltage $V(t)$ by $90^\circ$
Over a full cycle the capacitor C does not consume any energy from the voltage source
Current $I(t)$ is in phase with voltage $V(t)$
Current $I(t)$ leads voltage $V(t)$ by $180^\circ$

Solution Explained:

To solve this problem, we apply the core principles of Alternating Current. Understanding the underlying formula is key to arriving at the correct answer below:

For an ideal capacitor, the phase difference between current and voltage is $\phi = 90^\circ$ (current leads voltage). The average power consumed over a full cycle is $P_{avg} = V_{rms} I_{rms} \cos(90^\circ) = 0$. Since average power is zero, the capacitor does not consume any energy from the voltage source over a complete cycle.

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