Analyse the circuit shown in the figure in steady state

Explanation

`8i_(1)+13i_(1)+1(i_(1)+i_(2))-24=0`
`22i_(1)+i_(2)=24`ldots (1)
For right mesh,`5I_(2) +2i_(2) + 1 + i_(2) + 2i_(2) - 13I_(1)-8i_(1) = 0`
or `10I_(2) - 2i_(1) + 1 = 0ldots(2)`
From equations (1) and (2)
`i_(1) = 1 amp` and, `i_(2) = 2 amp` Now, considering current in different parts when capacitors were charging (e. when Lwwwisti 20 steady state was not reached) Current charging 2 muf capacitor= current charging 1 muF capacitor + current charging 3 muF capacitor If steady state charges on 1 muF and 3 muF capacitors are a `q_(1)muC` and `q_(2)` `muC` respectively, then charge on` 2 muF` capacitor wil be`q_(1)+q_(2)`
Hence, in steady state the circuit will be as shown in the figure
. Applying Kirchoff.s voltage law on mesh BCKJB
`5i_(2)+q_(2)/(3)-q_(1)/(1)-8i_(1)=0` `ldots(3) 512 +9 9 -81, = 0(3)`
For mesh JKFGJ`q_(1)+(q_(2))/(3)-q_(1)/(1)-8i=0ldots(4)`
In equations (3) and (4), substituting. `1_(1) = 1 amp. i_(2) = 2 amp`
`q_(1)=4muC`
`q_(2)=6muC`