In this figure the resistance of the coil of galvanometer G is 2   Ω . The emf of the cell is 4…

Current flowing through both capacitors will be zero at steady state.

The potential difference across capacitor $C_1$ will be the sum of the potential difference across the resistor $6 \mathrm{\Omega }$ and the galvanometer resistance. Mathematically,

${\mathrm{V}}_{{\mathrm{C}}_1}=i\left[6\Omega +{\mathrm{R}}_{\mathrm{G}}\right] ...\left(1\right)$

Similarly, the potential difference across capacitor $C_2$ will be the sum of the potential difference across the resistor $8 \mathrm{\Omega }$ and the galvanometer resistance. Mathematically,

${\mathrm{V}}_{{\mathrm{C}}_2}=i\left[{\mathrm{R}}_{\mathrm{G}}+8\Omega \right] ...\left(2\right)$

Divide equation (1) by equation (2) to obtain the required ratio.

$\begin{array}{rcl}\frac{V_{C_1}}{V_{C_2}}& =& \frac{i\left[6\Omega +R_G\right]}{i\left[R_G+8 \mathrm{\Omega }\right]}\\ & =& \frac{6\Omega +2 \mathrm{\Omega }}{2 \mathrm{\Omega }+8 \mathrm{\Omega }}\\ & =& \frac{4}{5}\end{array}$