Question

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

• A. $\frac{Q_{0}}{2 \sqrt{L C}}$
• B. $\frac{Q_{0}}{\sqrt{L C}}$
• C. $\frac{2 Q_{0}}{\sqrt{L C}}$
• D. $\frac{4 Q_{0}}{\sqrt{L C}}$

Answer 1

Answer: (A)

$E_{\text {total }}=E_{i}+E_{c}=\frac{Q_{0}^{2}}{2 C}$
$4 E_{i}=\frac{Q_{0}^{2}}{2 C}$
$E_{i}=\frac{Q_{0}^{2}}{8 C}=\frac{1}{2} L i^{2}$
$i^{2}=\frac{Q_{0}^{2}}{4 L C}$
$i=\frac{Q_{0}}{2 \sqrt{L C}}$