Two capacitors of `25 mu F` and `100 mu F` are connected in series to a source of `120 V`. Keeping their charges uncharged, they are separated and connected in parallel to eachother. Find out
(i) pot. Diff. between the plates of each capacitor
(ii) energy loss in the process.

Answer

`38.4 V, 0.052 J`

Explanation

Here, `C_(1) = 25 muF, C_(2) = 100 muF, C_(s) = 20 muF`,
`q = C_(s) V_(s) = 2400 mu C` on each capacitor `C_(p) = 125 mu F, V_(p) = (2400+2400)/(125) = 38.4 V`
Loss of energy `= (1)/(2) C_(s) V_(s)^(2) - (1)/(2) C_(p) V_(p)^(2)`