Question

Analyse the circuit shown in the figure in steady state.

Answer 1

In steady state, no current flows through capacitors. Therefore, to calculate current in resistors, the circuit can be analysed after removing capacitors. Suppose that in the steady state, a discharging current $i_2+i_2$ flows through the $24$ volt battery. Current distribution (according to Kirchoff's current law) will be shown in the figure.
Applyiing Kirchoff's voltage law on left mesh
$8i_1+13i_1+1(i_1+i_2)-24=022i_1+i_2=\mathrm{24....}\left(1\right)$
For right mesh
$5i_2+2i_2+1+2i_2-13i_1-8i_1=0or10i_2-21i_1+1=\mathrm{0....}\left(2\right)$
From equations (1) and (2)
$i_1=1amp$ and $i_2=2amp$
Now. considering current in different parts when capacitors were charging (ie when steady state was not reached)
Current charging $2\mu F$ capacitor
= current charging $1\mu F$ capacitor + current charging $3\mu F$ capacitor
If steady state charges on $1\mu F$ and $3\mu F$ capacitors are $q_1\mu C$ and $q_2\mu C$ respectively then charge on $2\mu F$ capacitr will 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
$5i_2+\frac{q_2}{3}-\frac{q_1}{1}-8i_1=\mathrm{0...}\left(3\right)$
For mesh JKFGJ
$\frac{q_1}{1}+\frac{(q_1+q_2)}{2}+2i_2-13i_1=\mathrm{0...}\left(4\right)$
In equations (3) and (4) substituting $i_1=1amp;i_2=2amp$
$q_1=4\mu C$

$q_2=6\mu C$