Q122 / 125 · BITSAT 2015 · Physics
In the circuit shown below, the ac source has voltage $V=20 \cos (\omega t)$ volt with $\omega=2000 \mathrm{rad} / \mathrm{s}$
The amplitude of the current will be nearest to
- Correct answerA$2 \mathrm{~A}$
- B$3.3 \mathrm{~A}$
- C$2 / \sqrt{5} \mathrm{~A}$
- D$\sqrt{5} \mathrm{~A}$
Solution
$$
\begin{aligned}
Z=& \sqrt{R^{2}+\left(\omega L-\frac{1}{\omega C}\right)^{2}} \\
&=\sqrt{10^{2}+\left(2000 \times 5 \times 10^{3}-\frac{1}{2000 \times 50 \times 10^{-6}}\right)^{2}} \\
=& 10 \Omega \\
\quad i=\frac{V_{0}}{Z}=\frac{20}{10}=2 A .
\end{aligned}
$$
Hence, the correct answer is option: 'A'.