Question

A capacitor of Capacitance $C$ is charged to potential $V_0$ and is connected in circuit as shown in the figure, Switch $S_1$ is closed at $t=0.$ After time $t=\frac{\pi \sqrt{LC}}{6},$ switch $S_1$ is opened , while $S_2$ is closed. If initially both the switches were open and capacitor of capacitance $2C$ was uncharged, the maximum charge stored on $2C$ would be

• A. $\frac{C{V}_0}{2}$
• B. $\frac{C{V}_0}{4}$
• C. $\frac{C{V}_0}{\sqrt{2}}$
• D. $\frac{C{V}_0}{\sqrt{3}}$

Answer 1

The correct option is C $\frac{C{V}_0}{\sqrt{2}}$

When switch $S_1$ is closed:

$Q=Q_0\cos \omega t=C{V}_0\cos \left(\frac{1}{\sqrt{LC}}\right)t......\left(1\right)$

The current in the circuit , $I=-\frac{dQ}{dt}$

$I=-\frac{d}{dt}[C{V}_0\cos \left(\frac{1}{\sqrt{LC}}\right)t]$

$I=\frac{C{V}_O}{\sqrt{LC}}\sin [\frac{1}{\sqrt{LC}}\times \frac{\pi \sqrt{LC}}{6}][\because t=\frac{\pi \sqrt{LC}}{6}]$
$\therefore I=\frac{C{V}_0}{2\sqrt{LC}}$

Energy stored in $L$ is , $E_L=\frac{1}{2}L{I}^2=\frac{1}{2}L(\frac{C^2{V}_0^2}{4LC})=\frac{C{V}_o^2}{8}$

Energy stored in the capacitor , $E_C=\frac{Q_{max}^2}{4C}=\frac{C{V}_0^2}{8}$

$\Rightarrow Q_{max}^2=\frac{C^2{V}_0^2}{2}$

$\Rightarrow Q_{max}=\frac{C{V}_0}{\sqrt{2}}$

Hence, option (c) is the correct answer.