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

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(B) The given 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

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$20$

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(B) The given 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 $X_C = \frac{1}{\omega C}$.Given $C = 1 \mu F = 1 	imes 10^{-6} 	ext{ F}$.$X_C = \frac{1}{100 	imes 10^{-6}} = 10^4 \Omega$.The $RMS$ voltage is $E_{rms} = \frac{E_0}{\sqrt{2}} = \frac{200 \sqrt{2}}{\sqrt{2}} = 200 	ext{ V}$.The reading of the $AC$ ammeter is the $RMS$ current $I_{rms} = \frac{E_{rms}}{X_C}$.$I_{rms} = \frac{200}{10^4} = 2 	imes 10^{-2} 	ext{ A} = 20 	ext{ mA}$.

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

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