An alternating voltage E = 200sqrt{2} sin(100 | vedclass

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(B) The given alternating voltage is $E = 200\sqrt{2} \sin(100t)$.Comparing this with the standard equation $E = E_0 \sin(\omega t)$,we get the peak v

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$20$

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

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

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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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