$\frac{50}{\pi} \mu \mathrm{F}$ capacitor is connected to a $250 \mathrm{~V}, 50 \mathrm{~Hz}$ AC supply.…
$\frac{50}{\pi} \mu \mathrm{F}$ capacitor is connected to a $250 \mathrm{~V}, 50 \mathrm{~Hz}$
AC supply. Then the rms current of the circuit is
- $1.25 \mathrm{~A}$
- $4.9 \mathrm{~A}$
- $5 \mathrm{~A}$
- $6 \mathrm{~A}$
Solution
Given, $C=\frac{50}{\pi} \mu \mathrm{F}=\frac{50}{\pi} \times 10^{-6} \mathrm{~F}$
Source voltage,
$
\begin{aligned}
& V=250 \mathrm{~V} \\
& f=50 \mathrm{~Hz}
\end{aligned}
$
Capacitive reactance,
$
\begin{aligned}
& X_C=\frac{1}{2 \pi f C}=\frac{1}{2 \pi \times 50 \times \frac{50}{\pi} \times 10^{-6}}=200 \Omega \\
& I_{\mathrm{rms}}=\frac{V_{\mathrm{mms}}}{X_C}=\frac{250}{200}=1.25 \mathrm{~A}
\end{aligned}
$
Asked in: AP EAMCET 2022 (06 Jul Shift 2)
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