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

Explanation

` (##MOT_CON_JEE_PHY_C25_SLV_018_S01.png" width="80%">
It can be simplified as
` C_( eq) = ( 18)/( 9) = 2mu F `
Charge flow through the cell ` = 30xx 2mu C `
` (##MOT_CON_JEE_PHY_C25_SLV_018_S02##) `
` Q= 60 mu C `
Now charge on ` 3mu F = ` Charge on ` 6 mu F = 60 mu C `
protential diffeence across ` 3mu F `
`= ( 60 )/(3) = 20 V `
` therefore ` Charge on ` 1mu F = 20 mu C `