A capacitor of capacitance `C_1` is charged to a potential `V_1` , while another capacitor of capacitance `C_2` is charged to a potential difference `V_2` . The capacitors are now disconnected from their respective charging batteries and connected in parallel to each other
Find the total energy stored in the two capacitors before they are connected .

Explanation

Initially capacitor of capacitance `C_1` is charged to a potential `V_1` , hence energy stored in it `U_(1) = (1)/(2) C_(1) V_(1)^(2)` .
Similarly for second capacitor of capacitance `C_2` charged to a potential `V_2` the stored energy `U_2 = (1)/(2) C_(2) V_(2)^(2)`
`therefore` Total electric energy stored in the combination before they are connected `U_(2) = U_(1) + U_(2) = (1)/(2) [C_(1) V_(1)^(2) + C_(2) V_(2)^(2)]`