In the circuit shown in the figure, if the galvanometer resistance is $6\Omega ,$ then in the steady state

 

Bridge is balanced, no current flows through galvanometer

$C_{eff}=\frac{10\times 20}{10+20}=\frac{20}{3}\mu F$

Capacitor in series, charge remains same.

$Q=C_{eff}.v=\frac{20}{3}\times 12=80\mu c.$

$R_{eff}=24\Omega ,$

Current through  $R_{2}is,i=\frac{V}{R_{eff}}=\frac{12}{24}=0.5A.$