In an ac circuit,an alternating voltage e=200 | vedclass

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(B) Given,the alternating voltage is $e = 200 \sqrt{2} \sin(100 t) \; V$.Comparing this with the standard equation $e = E_0 \sin(\omega t)$,we get pea

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$20$

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(B) Given,the alternating voltage is $e = 200 \sqrt{2} \sin(100 t) \; V$.Comparing this with the standard equation $e = E_0 \sin(\omega t)$,we get peak voltage $E_0 = 200 \sqrt{2} \; V$ and angular frequency $\omega = 100 \; rad/s$.The rms voltage is $E_{	ext{rms}} = \frac{E_0}{\sqrt{2}} = \frac{200 \sqrt{2}}{\sqrt{2}} = 200 \; V$.The capacitance is $C = 1 \; \mu F = 1 	imes 10^{-6} \; F$.The capacitive reactance $X_C$ is given by $X_C = \frac{1}{\omega C} = \frac{1}{100 	imes 10^{-6}} = \frac{1}{10^{-4}} = 10^4 \; \Omega$.The rms current $I_{	ext{rms}}$ is given by $I_{	ext{rms}} = \frac{E_{	ext{rms}}}{X_C} = \frac{200}{10^4} = 2 	imes 10^{-2} \; A$.Converting to milliamperes $(mA)$,$I_{	ext{rms}} = 2 	imes 10^{-2} 	imes 10^3 \; mA = 20 \; mA$.

In an $ac$ circuit,an alternating voltage $e=200 \sqrt{2} \sin 100 t$ volts is connected to a capacitor of capacitance $1 \; \mu F$. The rms value of the current in the circuit is ..... $mA$.

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