A capacitor of 50 mu F is connected to an a.c | vedclass

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(D) Given the voltage equation $e = 220 \sin(50t)$.Comparing this with the standard form $e = E_0 \sin(\omega t)$,we get the peak voltage $E_0 = 220 \

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$0.55 	ext{ A}$

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(D) Given the voltage equation $e = 220 \sin(50t)$.Comparing this with the standard form $e = E_0 \sin(\omega t)$,we get the peak voltage $E_0 = 220 	ext{ V}$ and angular frequency $\omega = 50 	ext{ rad/s}$.The capacitive reactance $X_C$ is given by $X_C = \frac{1}{\omega C}$.Substituting the values: $X_C = \frac{1}{50 	imes 50 	imes 10^{-6}} = \frac{1}{2500 	imes 10^{-6}} = \frac{10^6}{2500} = 400 \Omega$.The peak current $I_0$ is given by $I_0 = \frac{E_0}{X_C}$.$I_0 = \frac{220}{400} = \frac{22}{40} = 0.55 	ext{ A}$.

$A$ capacitor of $50 \mu F$ is connected to an a.c. source $e = 220 \sin(50t)$ (where $e$ is in volts and $t$ is in seconds). The value of the peak current is:

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