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 .
(a) Find the total energy stored in the two capacitors before they are connected.
(b) Find the total energy stored in the parallel combination of the two capacitors.
(c ) Explain the reason for the difference of energy in parallel combination in comparison to the total energy before they are connected.

Explanation

Charge on `C_(1)` capacitor is `=q_(1)=C_(1)V_(1)`
charge on `C_(2)` capacitor is `q_(2)=C_(2)V_(2)`
(a) Total energy before they are connected `=(1)/(2)C_(1)V_(1)^(2)+(1)/(2)C_(2)V_(2)^(2)`

(b) After connecting with their common potential
`V=("Total Charge")/("Total Capacitance")=V=(q_(1)+q_(2))/(C_(1)+C_(2))=(C_(1)V_(1)+C_(2)V_(2))/(C_(1)+C_(2))`
Then , total energy `U=(1)/(2)(C_(1)+C_(2))V^(2)=(1)/(2)xx((C_(1)V_(1)+C_(2)V_(2))^(2))/((C_(1)+C_(2)))`
`V_(f)=(1)/(2)((C_(1)V_(1)+C_(2)V_(2))^(2))/(C_(1)+C^(2))`
(c ) After connecting in || charge flows through the connecting wires and due to that current heat is generated in the wires and some energy is lost . That is why energy difference is created.