A capacitor of capacitance C, which is initially charged up to a potential differnce `epsilon`, is connected with a battery of emf `epsilon//2` such that the positive terminal of the battery is connected with the positive plate of the capacitor. After a long time.
i. find the total charge flow through the battery
ii. find the total work done by the battery
iii. Find the heat dissipated in the circuit during the process of charging.

Explanation

(i) Let potential of A is 0 so at B it is `(epsilon)/(2)`. So final charge on capacitor `=C epsilon //2`
Charge flow through the capacitor `=(C epsilon//2-C epsilon)=-C epsilon//2`
So charge is entering into battery.
(ii) finally, Change in energy of capacitor `=U_("final")-U_("initial")`
`=(1)/(2) C ((epsilon)/(2))^(2)-(epsilon^(2)C)/(2)=(1)/(8)epsilon^(2)C-(1)/(2)epsilon^(2)C=-(3 epsilon^(2)C)/(8)`
Work done by battery `=(epsilon)/(2)xx(-(epsilon C)/(2))=-(epsilon^(2) C)/(4)`
(iii) Work done by battery = Change in energy of capacitor + Heat produced
Heat produced `=(3 epsilon^(2)C)/(8)-(epsilon^(2)C)/(4)=(epsilon^(2)C)/(8)`