An alternating voltage V = 300sqrt{2} sin(100 | vedclass

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(B) The given voltage is $V = V_{0} \sin(\omega t)$,where $V_{0} = 300\sqrt{2} 	ext{ V}$ and $\omega = 100 	ext{ rad/s}$.The capacitive reactance is

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$30$

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(B) The given voltage is $V = V_{0} \sin(\omega t)$,where $V_{0} = 300\sqrt{2} 	ext{ V}$ 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 peak current is $i_{0} = \frac{V_{0}}{X_{C}} = \frac{300\sqrt{2}}{10^{4}} = 300\sqrt{2} 	imes 10^{-4} 	ext{ A} = 30\sqrt{2} 	imes 10^{-3} 	ext{ A} = 30\sqrt{2} 	ext{ mA}$.The $AC$ ammeter measures the $rms$ value of the current.$i_{rms} = \frac{i_{0}}{\sqrt{2}} = \frac{30\sqrt{2}}{\sqrt{2}} = 30 	ext{ mA}$.

An alternating voltage $V = 300\sqrt{2} \sin(100t)$ is connected to a $1 \ \mu F$ capacitor through an $AC$ ammeter. The reading of the ammeter will be.....$mA$.

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