A capacitor of capacitance C, which is initially uncharged , is connected with a battery of emf `epsilon`. Find the heat dissipated in the circuit during the process of charging .

Explanation

Final status
Let potential at point A is 0, so at B also 0 and at C and D it is `epsilon`.finally, charge on the capacitor
`Qc=epsilon C`
`U_(i)=0`
`U_(f)=(1)/(2) CV^(2)=(1)/(2) C epsilon^(2)`
work done by battery `=intPdt`
`W=int epsilon i dt=epsilon int i d t=epsilon. Q = epsilon=epsilon^(2)C`
(Now onwards remember that w.d. by battery `=epsilon Q ` if Q has flown out of the cell from high potential and w.d.on battery is `epsilon Q` if Q has flown into the cell through high potential)
Heat produced `=W-(U_(f)-U_(i))=epsilon^(2)C-(1)/(2) epsilon^(2)C=(C epsilon^(2))/(2)`.