Q133 / 160 · AP EAMCET 2014 · Physics
A capacitor $50 \mu \mathrm{F}$ is connected to a power source $V=220 \sin 50 t$ ( $V$ in volt, $t$ in second). The value of rms current (in ampere)
- A$\frac{\sqrt{2}}{0.55} \mathrm{~A}$
- B$0.55 \mathrm{~A}$
- C$\sqrt{2} \mathrm{~A}$
- Correct answerD$\frac{(0.55)}{\sqrt{2}} \mathrm{~A}$
Solution
The given
$$
\begin{aligned}
& C=50 \mu \mathrm{F}=50 \times 10^{-6} \mathrm{~F} \\
& V=220 \sin 50 t
\end{aligned}
$$
But we know that
$$
V=V_0 \sin \omega t
$$
Comparing both equations
$$
V_0=220 \mathrm{~V}, \omega=50 \mathrm{rad} / \mathrm{s}
$$
The capacitive reactance of the circuit is
$$
\begin{aligned}
X_C & =\frac{1}{\omega C} \\
& =\frac{1}{50 \times 50 \times 10^{-6}}=400 \Omega
\end{aligned}
$$
The peak and the rms values of current in the circuit are
$$
i_0=\frac{V_0}{X_C}=\frac{220}{400}=\frac{11}{20}
$$
and
$$
\begin{aligned}
i_{\mathrm{ms}} & =\frac{i_0}{\sqrt{2}} \\
& =\frac{11 / 20}{\sqrt{2}}=\frac{0.55}{\sqrt{2}} \mathrm{~A}
\end{aligned}
$$
Hence, the correct answer is option: 'D'.