Q126 / 200 · NEET 2011 (Screening) · Physics
In an $\mathrm{AC}$ circuit an alternating voltage $e=200 \sqrt{2} \sin 100 t$ volt is connected to a capacitor of capacity $1 \mu \mathrm{F}$. The rms value of the current in the circuit is
- A$100 \mathrm{~mA}$
- B$200 \mathrm{~mA}$
- Correct answerC$20 \mathrm{~mA}$
- D$10 \mathrm{~mA}$
Solution
Given, $e=200 \sqrt{2} \sin 100 t$
and
$$
\begin{aligned}
C & =1 \mu \mathrm{F} \\
E_{\mathrm{rms}} & =200 \mathrm{~V} \\
X_C & =\frac{1}{\omega C}=\frac{1}{1 \times 10^{-6} \times 100}=10^4 \Omega \\
i_{\mathrm{rms}} & =\frac{E_{\mathrm{rms}}}{X_C} \\
i_{\mathrm{rms}} & =\frac{200}{10^4}=2 \times 10^{-2} \mathrm{~A}=20 \mathrm{~mA}
\end{aligned}
$$
Hence, the correct answer is option: 'C'.