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)
$\Rightarrow I^2=\frac{C}{L}{V}^2$
$\Rightarrow I=V\sqrt{C/L}$