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An $8 μF$ capacitor is connected across 220 V 50 Hz line. What is peak value of charge through capacitor?

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$2.5 \times 10^{- 3} C$

$2.5 \times 10^{- 4} C$

$5 \times 10^{- 5} C$

$7.5 \times 10^{- 2} C$

answer is A.

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Detailed Solution

$q_{rm 3} = V_{rms} C = 1 .76 \times 10^{- 3} C$

$q_{peak} = \sqrt{2} q_{rms} = 1 .76 \times 10^{- 3} \times \sqrt{2} = 2 .5 \times 10^{- 3} C$

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