In the circuit shown in fig calculate chare on capacitors `C_1` and `C_2` in steady stete.

Explanation

In steady no current flows thorugh capacitors,
therefore, there are four unkonw quantites in the given
circuit .
I. curretn in left mesh ABGHA
ii. Curent in rieght mesh CDEFC

iii. Charge `q-1` on capcitor `C_2`
iv. Charge q-2 on capactor `C_2`
But by applyingh Kirchhoff's voltage law, three unique
equations can be formed. At steady to current passes
through capacitor branches. In an isolated system, net charge
should be zero, i.e the sum of charges on plates in zero. Hence,
charge on both capacitors should be equal. So
`q_1 = q_2 = a (say)`

Considering this fact, in steady state, circuit will be as
shown in Applying Kirchhoff's voltage law in
mesh ABGHA, we get
`I_1R_1 +I_1r_1 -E_1 = 0 or I_1 = 0.5A`
For mesh CDEFC,
`I_2r_2 +I_1R_2 = 0 or I_2 =3A`
Now applying Kirchhoff's voltage law in mesh BCFGB, we
get
`+(q)/(C_1) +E_3 _ I_2R_2 +(q)/(C_2) = I_1R_1 = 0`
or `q = 10muC`