In Wheat stone’s bridge shown in the adjoining figure galvanometer gives no deflection on pressing the key, the balance condition for the bridge is :

At balance, the potentials of point B and D are same and there will be no current in the arm BD. Thus,

${\mathrm{i}}_1{\mathrm{R}}_1=\frac{\mathrm{q}}{{\mathrm{C}}_1}...\left(1\right)$
where q is the charge on both the capacitor plates connected in series.
Quite similarly ${\mathrm{V}}_{\mathrm{B}}-{\mathrm{V}}_{\mathrm{C}}={\mathrm{V}}_{\mathrm{D}}-{\mathrm{V}}_{\mathrm{C}}$
or ${\mathrm{i}}_1{\mathrm{R}}_2=\frac{\mathrm{q}}{{\mathrm{C}}_2}...\left(2\right)$
Dividing eqs. (i) and (ii), we get
$\frac{{\mathrm{R}}_1}{{\mathrm{R}}_2}=\frac{{\mathrm{C}}_2}{{\mathrm{C}}_1}$