A capacitor of 50 mu F is connected to a powe | vedclass

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(D) Given: Capacitance $C = 50 \mu F = 50 	imes 10^{-6} 	ext{ F}$ and voltage $V = 220 \sin 50 t 	ext{ V}$.Comparing with the standard equation $V

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$\frac{0.55}{\sqrt{2}} 	ext{ A}$

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(D) Given: Capacitance $C = 50 \mu F = 50 	imes 10^{-6} 	ext{ F}$ and voltage $V = 220 \sin 50 t 	ext{ V}$.Comparing with the standard equation $V = V_0 \sin \omega t$,we get peak voltage $V_0 = 220 	ext{ V}$ and angular frequency $\omega = 50 	ext{ rad/s}$.The capacitive reactance $X_C$ is calculated as:$X_C = \frac{1}{\omega C} = \frac{1}{50 	imes 50 	imes 10^{-6}} = \frac{1}{2500 	imes 10^{-6}} = \frac{10^6}{2500} = 400 \Omega$.The peak current $i_0$ is given by:$i_0 = \frac{V_0}{X_C} = \frac{220}{400} = 0.55 	ext{ A}$.The rms current $i_{	ext{rms}}$ is given by:$i_{	ext{rms}} = \frac{i_0}{\sqrt{2}} = \frac{0.55}{\sqrt{2}} 	ext{ A}$.

$A$ capacitor of $50 \mu F$ is connected to a power source $V = 220 \sin 50 t$ (where $V$ is in volts and $t$ is in seconds). The value of the rms current (in amperes) is:

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