Q106 / 120 · VITEEE 2013 · Physics
The natural frequency of the circuit shown in adjoining figure is
- Correct answerA$\frac{1}{2 \pi \sqrt{L C}}$
- B$\frac{1}{2 \pi \sqrt{2 L C}}$
- C$\frac{2}{2 \pi \sqrt{L C}}$
- Dzero
Solution
In the given circuit, two condensors and the inductor are in series.
$\therefore \mathrm{L}_{\mathrm{s}}=\mathrm{L}+\mathrm{L}=2 \mathrm{~L}$
and $\frac{1}{\mathrm{C}_{\mathrm{s}}}=\frac{1}{\mathrm{C}}+\frac{1}{\mathrm{C}}=\frac{2}{\mathrm{C}} \Rightarrow \mathrm{C}_{\mathrm{s}}=\frac{\mathrm{C}}{2}$
$\therefore$ Natural frequency of the circuit
$\begin{array}{l}
\mathrm{v}=\frac{1}{2 \pi \sqrt{\mathrm{L}_{\mathrm{s}} \mathrm{C}_{\mathrm{s}}}}=\frac{1}{2 \pi \sqrt{2 \mathrm{~L}} \times \sqrt{\mathrm{C} / 2}} \\
=\frac{1}{2 \pi \sqrt{\mathrm{LC}}}
\end{array}$
Hence, the correct answer is option: 'A'.