In the given circuit find out charge on `6mu F` and `1mu F` capacitor.

Explanation

It can be simplfied as `C_(eq)=(18)/(9)=2 mu F`
charge flow through the cell `=30xx2 mu C`
`Q=60 mu C`
Now charge on `3 mu F`= Charge on `6mu F=60 mu C`
Potential difference across `3mu F=60//3=20 V`
`:. ` Charge on `1mu F=20 mu C`.