A capacitor of capacity ${C_1}$ is charged to the potential of ${V_o}$. On disconnecting with the battery, it is connected with a capacitor of capacity ${C_2}$ as shown in the adjoining figure. The ratio of energies before and after the connection of switch $S$ will be
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(a) Energy $(U)$ $ = \frac{{{q^2}}}{{2C}}$. $q$ remains same so $U \propto \frac{1}{C}$
$==>$ $\frac{{{U_{Before}}}}{{{U_{After}}}} = $$\frac{{{C_1} + {C_2}}}{{{C_1}}}$
$==>$ $\frac{{{U_{Before}}}}{{{U_{After}}}} = $$\frac{{{C_1} + {C_2}}}{{{C_1}}}$
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