Question

In the circuit shown in the figure, the steady charge on $6\mu F$ capacitor is $q\mu C$. Find the value of q

Answer 1

In steady state, no current flow in the capacitor branches.
The current flow in the top loop is $I_t=\frac{10}{3+2}=2A$
The current flow in the bottom loop is $I_b=\frac{20}{4+6}=2A$
Potential across $3\Omega$ is $V_3=3I_t=6V$
Potential across $4\Omega$ is $V_4=3I_b=8V$
Now we can redraw the circuit as shown in figure.
here, the capacitors are in series so $C_{eq}=\frac{3\times 6}{3+6}=2\mu F$
and $Q_{eq}=C_{eq}(V_4-V_3)=2\times (8-6)=4\mu C$
As they are in series so charge on each capacitor is equal to $Q_{eq}$