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A $10 \mu \mathrm{F}$ capacitor is connected to a $210 \mathrm{~V}, 50 \mathrm{~Hz}$ source as shown in figure. The peak current in the circuit is nearly $(\pi=3.14)$ :
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The correct answer is:
0.93 A
Capacitive Reactance
$\begin{aligned}
X_C & =\frac{1}{\omega C}=\frac{1}{2 \pi f C}=\frac{1}{2 \times 3.14 \times 50 \times 10 \times 10^{-6}} \\
& =\frac{1000}{3.14}
\end{aligned}$
$\begin{aligned}
& V_{\mathrm{rms}}=210 \mathrm{~V} \\
& i_{\mathrm{rms}}=\frac{V_{\mathrm{rms}}}{X_C}=\frac{210}{X_C}
\end{aligned}$
$\begin{aligned}
\text { Peak current } & =\sqrt{2} i_{\mathrm{ms}}=\sqrt{2} \times \frac{210}{1000} \times 3.14=0.932 \\
& \simeq 0.93 \mathrm{~A}
\end{aligned}$
$\begin{aligned}
X_C & =\frac{1}{\omega C}=\frac{1}{2 \pi f C}=\frac{1}{2 \times 3.14 \times 50 \times 10 \times 10^{-6}} \\
& =\frac{1000}{3.14}
\end{aligned}$
$\begin{aligned}
& V_{\mathrm{rms}}=210 \mathrm{~V} \\
& i_{\mathrm{rms}}=\frac{V_{\mathrm{rms}}}{X_C}=\frac{210}{X_C}
\end{aligned}$
$\begin{aligned}
\text { Peak current } & =\sqrt{2} i_{\mathrm{ms}}=\sqrt{2} \times \frac{210}{1000} \times 3.14=0.932 \\
& \simeq 0.93 \mathrm{~A}
\end{aligned}$
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