Analyze the circuit, find the charge in capactiors `C_1 C_2`, and `C_3` in steady state.

Explanation

In steady state, no current flows theoufh capacitor
therefore, to calculate current in resistors, the circuit can be analyzed after removing capacitors.
Let in steady state, a discharging , curerent `(i_1+I_2)` flow
thorugh 24V battery. current distribution (accosding to kirchhoff's current law) will be as shown in fing
Applying Kirchhoff's vlotage law on leftn meach, we get
If in steady state the charges on `1muF` and `3muF` capacitors are `q_1 and q_2,` respecitvly then the charge on `2muF` capactior will be `q_1 +q_2` hence in steady state, the curicit will be as shown in fig
Applying Kirchhoff's voltage law on mesh BCKJB
`5I_2 +(q_2)/(1) - 8I_1 = 0 (iii)`
For mesh JKFGJ,
`(q_1)/(1) + ((q_1 +q_2))/(2) +2I_2 - 13I_1 =0 (iv)`
In Eqs. (iii) and (iv) substituting `I_1 = 1A and I_2 = 2A,` we get
`q_1 = 4muC and q_2 =6muC`