Question

A capacitor of capacitance $C$ has initial charge $Q_0$ and connected to an inductance $L$. At $t=0$ switch $S$ is closed. The current through the inductor when energy in the capacitor is three times the energy of the inductor is :

• A. $\frac{Q_0}{2\sqrt{LC}}$
• B. $\frac{Q_0}{\sqrt{LC}}$
• C. $\frac{2Q_0}{\sqrt{LC}}$
• D. $\frac{4Q_0}{\sqrt{LC}}$

Answer 1

The correct option is A $\frac{Q_0}{2\sqrt{LC}}$

Initial charge in capacitor before switch $S$ is closed $=Q$

$\therefore$ Initial energy $U=\frac{Q_o^2}{2C}$

Now, after switch $S$ is closed

$U=U_2+U_C$ where $U_2$& $U_C$ are energies getting distributed & stored into the inductor & capacitor respectively

So, then $U_C=3U_2$

$U=U_2+3U_2=4U_2$

$\Rightarrow \frac{Q_o^2}{2C}=4\left(\frac{1}{2}L{i}^2\right)$

$\Rightarrow i^2=\frac{Q_o^2}{4<C}\Rightarrow i=\frac{Q_o}{2\sqrt{LC}}$