A capacitor `(C=50 muF)` is charged to a potential difference of `20V`. The charging battery is disconnected and the capacitor is connected to another cell of emf `10V` with the positive plate of capacitor joined with the positive terminal of cell.
(a) find charge flown through `10V` cell.
(b) is work done by the cell ot is it done on the cell? find its magnitude.
(c) find the heat developed in connecting wires.

Explanation

Charge on capacitor
`Q_(1)=50xx20=1000muC`
Charge on capacitor
`Q_(2)=50xx10=500muC`
`Q_(2) lt Q_(1)`
(a) charge flown through cell of emf `10V`
`DeltaQ=Q_(1)-Q_(2)=500muC`
(b) Since `Q_(2) lt Q_(1)`. Work is done on the cell
`W_(b)=DeltaQV=500xx100xx10^(-6)`
`=5xx10^(-3)J`
`W_(b)` is`-ve`.
(c) `DeltaU=U_(f)-U_(i)`
`=(1)/(2)C(V_(2)^(2)-V_(1)^(2))=(1)/(2)xx50xx10^(-6)(10^(2)-20^(2))`
`=-7.5xx10^(-3)J`
`W_(b)=DeltaU+H`
`-5xx10^(-3)=-7.5xx10^(-3)+H`
`H=2.5xx10^(-3)J`
where `H`: heat developed in connecting wires.