In the circuit shown in figure S₁ is open and, S₂ and S₃ are closed. The circuit is in steady state. At time t = 0, switch  1 is closed and S₂ and S₃ are opened simultaneously. V = 100 volt, R = 10 $\mathrm{\Omega }$, C = 100 $\mathrm{\mu }$F, L = 0.03 H
 If the maximum charge that will appear on the capacitor at any time is $q_0$ and 
 the time at which the charge on the capacitor will become zero for the first time is $t_0$, then

(a) At time t = 0⁺, situation is as shown below
$\mathrm{q}=\mathrm{CV}=100\times {10}^{-6}\times 100=0.01\mathrm{C}$
Energy stored in the capacitor
${\mathrm{U}}_1=\frac{1}{2}{\mathrm{CV}}^2=\frac{1}{2}\times 100\times {10}^{-6}\times {100}^2=0.5\mathrm{J}$
When charge on capacitor become maximum energy stored in is $={\mathrm{U}}_1+{\mathrm{U}}_2=2\mathrm{J}$
$\begin{array}{l}\frac{{\mathrm{q}}_0^2}{2\mathrm{C}}=2\\ \Rightarrow {\mathrm{q}}_0^2=4\times 100\times {10}^{-6}\\ \Rightarrow {\mathrm{q}}_0=2\times {10}^{-2}=0.02\mathrm{C}\end{array}$
(b) Angular frequency of oscillation $\mathrm{\omega }=\frac{1}{\sqrt{\mathrm{LC}}}=\frac{{10}^3}{\sqrt{3}}\mathrm{rad}/\mathrm{s}$
Charge on capacitor changes with time according to q = q₀ sin$(\mathrm{\omega t}+\mathrm{\delta })$
At t = 0, q = 0.01
$\begin{array}{r}\therefore 0.01=0.02\sin \mathrm{\delta }\\ \Rightarrow \mathrm{\delta }=\mathrm{\pi }/6\text{ or }5\mathrm{\pi }/6\end{array}$
We take$\mathrm{\delta }=\frac{5\mathrm{\pi }}{6}$as charge on the capacitor is initially decreasing (look at the direction of current initially)
$\therefore \mathrm{q}=0.02\sin \left(\mathrm{\omega t}+\frac{5\mathrm{\pi }}{6}\right)$
q is zero when$\mathrm{\omega t}+\frac{5\mathrm{\pi }}{6}=\mathrm{\pi }$
$\begin{array}{l}\Rightarrow \mathrm{\omega t}=\mathrm{\pi }/6\\ \frac{{10}^3}{\sqrt{3}}\mathrm{t}=\frac{\mathrm{\pi }}{6}\Rightarrow \mathrm{t}=\frac{3.14\times 1.73}{6\times {10}^3}=0.9\mathrm{ms}\end{array}$