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(A) Given: Capacitance $C = 10 \mu F = 10 	imes 10^{-6} 	ext{ F}$.Source voltage $V = 50 \sqrt{2} \sin 100 t$.Comparing this with $V = V_{\max} \sin

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$50$

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(A) Given: Capacitance $C = 10 \mu F = 10 	imes 10^{-6} 	ext{ F}$.Source voltage $V = 50 \sqrt{2} \sin 100 t$.Comparing this with $V = V_{\max} \sin \omega t$, we get $V_{\max} = 50 \sqrt{2} 	ext{ V}$ and $\omega = 100 	ext{ rad/s}$.The capacitive reactance is $X_C = \frac{1}{\omega C} = \frac{1}{100 	imes 10 	imes 10^{-6}} = \frac{1}{10^{-3}} = 1000 \Omega$.The $RMS$ voltage is $V_{	ext{rms}} = \frac{V_{\max}}{\sqrt{2}} = \frac{50 \sqrt{2}}{\sqrt{2}} = 50 	ext{ V}$.The reading of the $AC$ ammeter gives the $RMS$ current $I_{	ext{rms}}$, which is $I_{	ext{rms}} = \frac{V_{	ext{rms}}}{X_C} = \frac{50}{1000} = 0.05 	ext{ A}$.Converting to milliamperes, $I_{	ext{rms}} = 0.05 	imes 1000 	ext{ mA} = 50 	ext{ mA}$.

$A$ capacitor of capacitance $10 \mu F$ is connected to an $AC$ source and an $AC$ Ammeter. If the source voltage varies as $V = 50 \sqrt{2} \sin 100 t$, the reading of the ammeter is (in $\text{ mA}$)

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