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In an ac circuit an alternating voltage e = 200 $\sqrt{2} \sin 100 t volts$ is connected to a capacitor of capacity $1 μF .$ . The r.m.s. value of the current in the circuit is:

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10 mA

100 mA

200 mA

20 mA

answer is D.

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Detailed Solution

$V_{r ms} = \frac{200 \sqrt{2}}{\sqrt{2}} = 200 V$

$I_{rms} = \frac{V_{rms}}{X_{C}} = \frac{200}{(\frac{1}{100 \times 10^{- 6}})} = 200 \times 10^{- 4}$

$= 20 \times 10^{- 3} = 20 mA$

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