Question

In the circuit given below, the capacitor $C$ is charged by closing the switch $S_{1}$ and opening the switch $S_{2}$. After charging, the switch $S_{1}$ is opened and $S_{2}$ is closed, then the maximum current in the circuit

• A. $V \sqrt{\frac{L}{C}}$
• B. $V \sqrt{\frac{C}{L}}$
• C. $\frac{V}{2 \pi} \sqrt{\frac{L}{C}}$
• D. $2 \pi V \sqrt{\frac{L}{C}}$

Answer 1

Answer: (B)

When switch $S_{1}$ is closed and $S_{2}$ is open, capacitor $C$ will be fully charged. After charging, when $S_{1}$ is opened and $S_{2}$ is closed, this $L-C$ circuit will act as tank circuit. In tank circuit,
$ \frac{1}{2} C V^{2} =\frac{1}{2} L I^{2}$ (Total energy)
$\Rightarrow I^{2} =\frac{C}{L} V^{2} $
$\Rightarrow I =V \sqrt{C / L}$