An alternating voltage E = 100 sqrt{2} sin(50 | vedclass

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(A) The given alternating voltage is $E = E_0 \sin(\omega t)$, where $E_0 = 100 \sqrt{2} 	ext{ V}$ and $\omega = 50 	ext{ rad/s}$.The root mean squa

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$10$

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(A) The given alternating voltage is $E = E_0 \sin(\omega t)$, where $E_0 = 100 \sqrt{2} 	ext{ V}$ and $\omega = 50 	ext{ rad/s}$.The root mean square (rms) voltage is $E_{	ext{rms}} = \frac{E_0}{\sqrt{2}} = \frac{100 \sqrt{2}}{\sqrt{2}} = 100 	ext{ V}$.The capacitive reactance is $X_C = \frac{1}{\omega C} = \frac{1}{50 	imes 2 	imes 10^{-6}} = \frac{1}{100 	imes 10^{-6}} = 10^4 \Omega$.The ammeter measures the rms current $I_{	ext{rms}}$, which is given by $I_{	ext{rms}} = \frac{E_{	ext{rms}}}{X_C}$.Substituting the values, $I_{	ext{rms}} = \frac{100}{10^4} = 10^{-2} 	ext{ A} = 10 	ext{ mA}$.Thus, the ammeter reading is $10 	ext{ mA}$.

An alternating voltage $E = 100 \sqrt{2} \sin(50 t)$ is connected to a $2 \mu F$ capacitor through an a.c. ammeter. The ammeter reading will be (in $\text{ mA}$)

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