An alternating voltage v = 200 sqrt{2} sin(10 | vedclass

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(B) The given alternating voltage is $v = 200 \sqrt{2} \sin(100 t)$.Comparing this with the standard form $v = V_0 \sin(\omega t)$, we get the peak vo

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$20$

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(B) The given alternating voltage is $v = 200 \sqrt{2} \sin(100 t)$.Comparing this with the standard form $v = V_0 \sin(\omega t)$, we get the peak voltage $V_0 = 200 \sqrt{2} 	ext{ V}$ and angular frequency $\omega = 100 	ext{ rad/s}$.The capacitive reactance $X_C$ is given by $X_C = \frac{1}{\omega C}$.Given $C = 1 \mu F = 10^{-6} 	ext{ F}$, we have $X_C = \frac{1}{100 	imes 10^{-6}} = 10^4 \Omega$.The peak current $I_0$ is $I_0 = \frac{V_0}{X_C} = \frac{200 \sqrt{2}}{10^4} = 2 \sqrt{2} 	imes 10^{-2} 	ext{ A}$.The $A.C.$ ammeter measures the root mean square $(RMS)$ value of the current, $I_{rms}$.$I_{rms} = \frac{I_0}{\sqrt{2}} = \frac{2 \sqrt{2} 	imes 10^{-2}}{\sqrt{2}} = 2 	imes 10^{-2} 	ext{ A} = 20 	ext{ mA}$.

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

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