Search any question & find its solution
Question:
Answered & Verified by Expert
$\frac{50}{\pi} \mu \mathrm{F}$ capacitor is connected to a $250 \mathrm{~V}, 50 \mathrm{~Hz}$
AC supply. Then the rms current of the circuit is
Options:
AC supply. Then the rms current of the circuit is
Solution:
1216 Upvotes
Verified Answer
The correct answer is:
$1.25 \mathrm{~A}$
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,
$$
\begin{aligned}
& X_C=\frac{1}{2 \pi f C}=\frac{1}{2 \pi \times 50 \times \frac{50}{\pi} \times 10^{-6}}=200 \Omega \\
& I_{\mathrm{rms}}=\frac{V_{\mathrm{mms}}}{X_C}=\frac{250}{200}=1.25 \mathrm{~A}
\end{aligned}
$$
Source voltage,
$$
\begin{aligned}
& V=250 \mathrm{~V} \\
& f=50 \mathrm{~Hz}
\end{aligned}
$$
Capacitive reactance,
$$
\begin{aligned}
& X_C=\frac{1}{2 \pi f C}=\frac{1}{2 \pi \times 50 \times \frac{50}{\pi} \times 10^{-6}}=200 \Omega \\
& I_{\mathrm{rms}}=\frac{V_{\mathrm{mms}}}{X_C}=\frac{250}{200}=1.25 \mathrm{~A}
\end{aligned}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.