An alternating voltage e = 150sqrt{2} sin 100 | vedclass

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(C) The given alternating voltage is $e = 150\sqrt{2} \sin 100t 	ext{ V}$.Comparing this with the standard equation $e = E_0 \sin \omega t$, we get t

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$30$

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(C) The given alternating voltage is $e = 150\sqrt{2} \sin 100t 	ext{ V}$.Comparing this with the standard equation $e = E_0 \sin \omega t$, we get the peak voltage $E_0 = 150\sqrt{2} 	ext{ V}$ and angular frequency $\omega = 100 	ext{ rad/s}$.The root mean square $(RMS)$ voltage is $V_{rms} = E_0 / \sqrt{2} = 150 	ext{ V}$.The capacitive reactance $X_C$ is given by $X_C = 1 / (\omega C)$.Given $C = 2 \mu F = 2 	imes 10^{-6} 	ext{ F}$, we have $X_C = 1 / (100 	imes 2 	imes 10^{-6}) = 1 / (2 	imes 10^{-4}) = 5000 \Omega$.The $RMS$ current $I_{rms}$ is $I_{rms} = V_{rms} / X_C = 150 / 5000 = 0.03 	ext{ A}$.Converting to milliamperes, $I_{rms} = 0.03 	imes 1000 	ext{ mA} = 30 	ext{ mA}$.

An alternating voltage $e = 150\sqrt{2} \sin 100t \text{ V}$ is applied to a capacitor of capacity $2 \mu F$. The root mean square value of current in the circuit is (in $mA$)

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