A 40 , mu F capacitor is connected to a 200 , | vedclass

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(D) Given: Capacitance $C = 40 \, \mu F = 40 	imes 10^{-6} \, F$, Voltage $V_{rms} = 200 \, V$, Frequency $f = 50 \, Hz$.The capacitive reactance is

Vedclass · Accepted Answer

$2.5$

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(D) Given: Capacitance $C = 40 \, \mu F = 40 	imes 10^{-6} \, F$, Voltage $V_{rms} = 200 \, V$, Frequency $f = 50 \, Hz$.The capacitive reactance is given by $X_C = \frac{1}{\omega C} = \frac{1}{2 \pi f C}$.The rms current is given by $I_{rms} = \frac{V_{rms}}{X_C} = V_{rms} 	imes 2 \pi f C$.Substituting the values:$I_{rms} = 200 	imes 2 	imes 3.1416 	imes 50 	imes 40 	imes 10^{-6}$.$I_{rms} = 200 	imes 314.16 	imes 40 	imes 10^{-6} = 2.513 \, A$.Rounding to the nearest value, we get $I_{rms} \approx 2.5 \, A$.

$A$ $40 \, \mu F$ capacitor is connected to a $200 \, V, 50 \, Hz$ $AC$ supply. The rms value of the current in the circuit is nearly $....... A$.

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