A `10 muF` capacitor is charged to `15 V`. It is next connected in seties with an uncharged `5 muf` capacitor. The series cambination is finally connected across a `30 V` battery, as shown in. Find the new potential differnce across the `5 muF` and `10 muF` capacitors.

Explanation

The initial charge on the larger capacitor is

`Q=CDeltaV=10 muF(15 V) =150 muC` An additional charge is pushed through the `50 V` battery, giving the smaller capacitor `q` and the large charge `(150+q)`. In close loop (1)
`30-q/(C_(1))=((50+q))/(10)=0` or `q=50 muC`
Potential difference across `C_(1)`
`V_(1)=q/(C_(1)=(50)/5=10 V` and `V_(2)=(200)/(10)=20 V`.