A 10 mu F capacitor is connected to a 210 tex | vedclass

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$(A)$ The capacitive reactance $X_C$ is given by $X_C = \frac{1}{\omega C} = \frac{1}{2 \pi f C}$.Substituting the given values: $X_C = \frac{1}{2 	i

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$0.93$

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$(A)$ The capacitive reactance $X_C$ is given by $X_C = \frac{1}{\omega C} = \frac{1}{2 \pi f C}$.Substituting the given values: $X_C = \frac{1}{2 	imes 3.14 	imes 50 	imes 10 	imes 10^{-6}} = \frac{1}{3140 	imes 10^{-6}} = \frac{1000}{3.14} \approx 318.47  \Omega$.The $RMS$ voltage is $V_{	ext{rms}} = 210 	ext{ V}$.The $RMS$ current is $I_{	ext{rms}} = \frac{V_{	ext{rms}}}{X_C} = \frac{210}{1000 / 3.14} = \frac{210 	imes 3.14}{1000} = 0.6594 	ext{ A}$.The peak current $I_0$ is given by $I_0 = \sqrt{2} 	imes I_{	ext{rms}}$.$I_0 = 1.414 	imes 0.6594 \approx 0.932 	ext{ A}$.Thus, the peak current is approximately $0.93 	ext{ A}$.

$A$ $10 \mu F$ capacitor is connected to a $210 \text{ V}, 50 \text{ Hz}$ source as shown in the figure. The peak current in the circuit is nearly $(\pi = 3.14)$ : (in $\text{ A}$)

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