An AC alternating emf E = 110 sqrt{2} sin(100 | vedclass

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(A) Given: $C = 2 \mu F = 2 	imes 10^{-6} 	ext{ F}$,$E = 110 \sqrt{2} \sin(100t) 	ext{ V}$.Comparing with $E = E_0 \sin(\omega t)$,we get peak volt

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$22$

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(A) Given: $C = 2 \mu F = 2 	imes 10^{-6} 	ext{ F}$,$E = 110 \sqrt{2} \sin(100t) 	ext{ V}$.Comparing with $E = E_0 \sin(\omega t)$,we get peak voltage $E_0 = 110 \sqrt{2} 	ext{ V}$ and angular frequency $\omega = 100 	ext{ rad/s}$.The capacitive reactance is $X_C = \frac{1}{\omega C} = \frac{1}{100 	imes 2 	imes 10^{-6}} = \frac{1}{2 	imes 10^{-4}} = 5000 \Omega$.The peak current is $I_0 = \frac{E_0}{X_C} = \frac{110 \sqrt{2}}{5000} 	ext{ A}$.The $rms$ current is $I_{rms} = \frac{I_0}{\sqrt{2}} = \frac{110 \sqrt{2}}{5000 \sqrt{2}} = \frac{110}{5000} 	ext{ A}$.$I_{rms} = \frac{11}{500} 	ext{ A} = 0.022 	ext{ A}$.Converting to $mA$,$I_{rms} = 0.022 	imes 1000 	ext{ mA} = 22 	ext{ mA}$.

An $AC$ alternating $emf$ $E = 110 \sqrt{2} \sin(100t) \text{ V}$ is applied to a capacitor of $2 \mu F$. The $rms$ value of current in the circuit is $...... \text{ mA}$.

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