A $40 \mu \mathrm{~F}$ capacitor is connected to a $200 \mathrm{~V}, 50 \mathrm{~Hz}$ ac supply. The

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A $40 \mu \mathrm{~F}$ capacitor is connected to a $200 \mathrm{~V}, 50 \mathrm{~Hz}$ ac supply. The r.m.s value of the current in the circuit is, nearly\

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(c) : Here, $C=40 \mu \mathrm{~F}=40 	imes 10^{-6} \mathrm{~F}$\ $V_{\mathrm{rms}}=200 \mathrm{~V} ; v=50 \mathrm{~Hz}$\ The value of the current, $I_{\mathrm{rms}}=\frac{\varepsilon_{r m s}}{1}=\varepsilon_{r m s} \omega \mathrm{C}$ $$ \omega C $$ or $\quad I_{\mathrm{rms}}=200(2 \pi 	imes 50) 	imes\left(40 	imes 10^{-6}ight)=2.51 \mathrm{~A} \quad(\because \omega=2 \pi v)$\