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A capacitor of capacity $C_1$ is charged to the potential of $V_0$. After disconnecting with the battery, it is connected with a neutral capacitor of capacity $C_2$ as shown in the adjoining figure. The ratio of energy of system before and after the connection of switch $S$ will be
$\frac{{{C_1} + {C_2}}}{{{C_1}}}$
$\frac{{{C_1}}}{{{C_1} + {C_2}}}$
$C_1C_2$
$\frac{{{C_1}}}{{{C_2}}}$
Solution
$U_{1}=\frac{1}{2} C_{1} V_{0}^{2}$
$C_{1} V_{0}=\left(C_{1}+C_{2}\right) V$
$V=\frac{C_{1} V_{0}}{C_{1}+C_{2}}$
$\mathrm{U}_{2}=\frac{1}{2}\left(\mathrm{C}_{1}+\mathrm{C}_{2}\right) \frac{\mathrm{C}_{1}^{2} \mathrm{V}_{0}^{2}}{\left(\mathrm{C}_{1}+\mathrm{C}_{2}\right)^{2}}=\frac{1}{2} \frac{\mathrm{C}_{1}^{2} \mathrm{V}_{0}^{2}}{\left(\mathrm{C}_{1}+\mathrm{C}_{2}\right)}$
$\frac{U_{1}}{U_{2}}=\frac{1 / 2 C_{1} V_{0}^{2}}{1 / 2 \frac{C_{1}^{2} V_{0}^{2}}{\left(C_{1}+C_{2}\right)}}$
$\frac{U_{1}}{U_{2}}=\frac{C_{1}+C_{2}}{C_{1}}$
