Find the current in each branch f the given circuit in steady state. Find the charge on each capacitor.

Explanation

In steady state circuit will be as shown
`KVL` for loop `1` gives
`10-i-5i_(1)=0`
`KVL` for loop `2` gives
`4-6i+5i_(1)=0`
From (`1`) & (`2`)
`i=2 Amp` & `i_(1)=1.6 Amp`

So `i-i_(1)=0.4Amp`
So, Current in right of `4V` battery `=i-i_(1)=0.4Amp`
Current in left of 4V battery `= i_(1) = 1.6 Amp`
Change on `C_(1)=C(i-i_(1))1=10xx0.4muC`
`rArrq_(1)=4muC`
Let charge on `C_(2)=q_(2)`
`-(q_(2))/(C_(2))-2xx(2)/(5)+(8)/(5)xx2=0`
`rArr q_(2)=C_(2)((12)/(5))=12muC`.