In the L–C circuit, the capacitor C₁ is charged to a potential of 1 volt, and the capacitor C₂ and the inductor were initially uncharged. Now switch S₁ is closed at t = 0, and at $t=\frac{1.5\pi }{100}\sec$. switch S₂ is closed and S1 is opened. After $\frac{0.5\pi }{100}\sec$. , S₂ is also opened. Find final potential difference across the capacitor C₂ in volt.

 For L–C₁ system
$\begin{array}{l}{\omega }_1=\frac{1}{\sqrt{L{C}_1}}=\frac{100}{3}\\ \text{ and }{T}_1=2\pi \sqrt{L{C}_1}=\frac{6\pi }{100}\sec .\\ \text{ For }L-C_2\text{ system }\\ {\omega }_2=\frac{1}{\sqrt{L{C}_2}}=100\text{ and }\\ T_2=2\pi \sqrt{L{C}_2}=\frac{2\pi }{100}\sec .\end{array}$

$\text{ Circuit }\mathrm{L}-{\mathrm{C}}_1\text{ is on for }\frac{1.5\pi }{100}=\frac{{\mathrm{T}}_1}{4}\text{ sec }$

the charge on capacitor C₁ will be zero, so the entire energy $\left(\frac{1}{2}{\mathrm{C}}_1{\mathrm{V}}_1^2\right)\text{ will go in inductor. }$

Now after sec. $\frac{0.5\pi }{100}=\frac{{\mathrm{T}}_2}{4}\sec .$ 'i' in the inductor will be zero, so now all the energy $\left(\frac{1}{2}{\mathrm{C}}_1{\mathrm{V}}_1^2\right)$ will go in the capacitor C₂.

$\begin{array}{l}\Rightarrow \frac{1}{2}{C}_1{V}_1^2=\frac{1}{2}{C}_2{V_2}^2\\ \Rightarrow \frac{1}{2}\left(900\mu \right)(1{)}^2=\frac{1}{2}(100\mu ){V_2}^2\\ \Rightarrow V_2=3\text{ volt. }\end{array}$