A capacitor of capacity C_1 is charged to a p | vedclass

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(A) The initial energy stored in the capacitor $C_1$ is given by:$U_1 = \frac{1}{2} C_1 V_0^2$When the switch $S$ is closed,the charge $Q = C_1 V_0$ i

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$\frac{C_1 + C_2}{C_1}$

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(A) The initial energy stored in the capacitor $C_1$ is given by:$U_1 = \frac{1}{2} C_1 V_0^2$When the switch $S$ is closed,the charge $Q = C_1 V_0$ is redistributed between the two capacitors until they reach a common potential $V$. By the principle of conservation of charge:$C_1 V_0 = (C_1 + C_2) V$$V = \frac{C_1 V_0}{C_1 + C_2}$The final energy stored in the system is:$U_2 = \frac{1}{2} (C_1 + C_2) V^2 = \frac{1}{2} (C_1 + C_2) \left( \frac{C_1 V_0}{C_1 + C_2} \right)^2 = \frac{1}{2} \frac{C_1^2 V_0^2}{C_1 + C_2}$The ratio of the energy before and after the connection is:$\frac{U_1}{U_2} = \frac{\frac{1}{2} C_1 V_0^2}{\frac{1}{2} \frac{C_1^2 V_0^2}{C_1 + C_2}} = \frac{C_1 (C_1 + C_2)}{C_1^2} = \frac{C_1 + C_2}{C_1}$

$A$ capacitor of capacity $C_1$ is charged to a potential of $V_0$. After disconnecting it from the battery,it is connected to a neutral capacitor of capacity $C_2$ as shown in the adjoining figure. The ratio of the energy of the system before and after the connection of switch $S$ will be

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