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

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(B) 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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(B) 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 $X_C = \frac{1}{\omega C} = \frac{1}{100 	imes 1 	imes 10^{-6}} = 10^4 \Omega$.The peak current is $I_0 = \frac{E_0}{X_C} = \frac{200 \sqrt{2}}{10^4} = 2 \sqrt{2} 	imes 10^{-2} 	ext{ A}$.The $A.C.$ ammeter measures the root mean square $(RMS)$ current, $I_{	ext{rms}} = \frac{I_0}{\sqrt{2}}$.$I_{	ext{rms}} = \frac{2 \sqrt{2} 	imes 10^{-2}}{\sqrt{2}} = 2 	imes 10^{-2} 	ext{ A} = 20 	ext{ mA}$.

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

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