The average power dissipated in a pure capacitor in an AC circuit is:

1. A $V_{rms}I_{rms}$
2. B $V_{rms}I_{rms}/2$
3. C zero Correct answer
4. D $I_{rms}^2 X_C$

Solution

Phase difference 90° ⇒ $P_{avg} = V_{rms}I_{rms}\cos 90° = 0$.

About this question

Subject: Physics · Chapter: Alternating Current · Topic: Power in AC Circuits

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