An alternating voltage e = 200 sqrt{2} sin(10 | vedclass

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(D) The given alternating voltage is $e = e_0 \sin(\omega t)$, where $e_0 = 200 \sqrt{2} 	ext{ V}$ and $\omega = 100 	ext{ rad/s}$.The capacitive re

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$20$

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(D) The given alternating voltage is $e = e_0 \sin(\omega t)$, where $e_0 = 200 \sqrt{2} 	ext{ V}$ and $\omega = 100 	ext{ rad/s}$.The capacitive reactance is $X_C = \frac{1}{\omega C} = \frac{1}{100 	imes 1 	imes 10^{-6}} = \frac{1}{10^{-4}} = 10^4 \Omega$.The $RMS$ voltage is $V_{rms} = \frac{e_0}{\sqrt{2}} = \frac{200 \sqrt{2}}{\sqrt{2}} = 200 	ext{ V}$.The reading of the $AC$ ammeter gives the $RMS$ current $I_{rms} = \frac{V_{rms}}{X_C}$.Substituting the values, $I_{rms} = \frac{200}{10^4} = 2 	imes 10^{-2} 	ext{ A} = 20 	ext{ mA}$.

An alternating voltage $e = 200 \sqrt{2} \sin(100 t) \text{ V}$ is connected to a $1 \mu F$ capacitor through an $AC$ ammeter. The reading of the ammeter is: (in $\text{ mA}$)

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