An alternating voltage given as V = 100 sqrt{ | vedclass

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(A) The given alternating voltage is $V = 100 \sqrt{2} \sin(100t) 	ext{ V}$.Comparing this with the standard form $V = V_m \sin(\omega t)$,we get the

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(A) The given alternating voltage is $V = 100 \sqrt{2} \sin(100t) 	ext{ V}$.Comparing this with the standard form $V = V_m \sin(\omega t)$,we get the peak voltage $V_m = 100 \sqrt{2} 	ext{ V}$ and angular frequency $\omega = 100 	ext{ rad/s}$.The capacitive reactance is given by $X_C = \frac{1}{\omega C}$.Substituting the values,$X_C = \frac{1}{100 	imes 1 	imes 10^{-6}} = \frac{1}{10^{-4}} = 10^4 	ext{ } \Omega$.The root mean square $(RMS)$ voltage is $V_{	ext{rms}} = \frac{V_m}{\sqrt{2}} = \frac{100 \sqrt{2}}{\sqrt{2}} = 100 	ext{ V}$.The $RMS$ current $I_{	ext{rms}}$ is given by $I_{	ext{rms}} = \frac{V_{	ext{rms}}}{X_C}$.$I_{	ext{rms}} = \frac{100}{10^4} = 10^{-2} 	ext{ A}$.Converting to milliamperes,$I_{	ext{rms}} = 10^{-2} 	imes 10^3 	ext{ mA} = 10 	ext{ mA}$.

An alternating voltage given as $V = 100 \sqrt{2} \sin(100t) \text{ V}$ is applied to a capacitor of $1 \mu\text{F}$. The current reading of the ammeter will be equal to . . . . . . $\text{mA}$.

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