In this figure the resistance of the coil of galvanometer G is 2 $$\Omega$$. The emf of the cell is 4 V. The ratio of potential difference across C$$_1$$ and C$$_2$$ is

$$\text{In steady state, branches } AC \text{ and } BD \text{ block DC } \implies I_{AC} = 0, \ I_{BD} = 0$$

$$\text{The remaining active DC path forms a single series loop: } A \rightarrow B \rightarrow C \rightarrow D \rightarrow \text{source}$$

$$R_{\text{total}} = R_{AB} + R_G + R_CD = 6 + 2 + 8 = 16\,\Omega \implies I = \frac{V}{R_{\text{total}}} = \frac{4}{16} = 0.25\text{ A}$$

$$V_{C1} = V_A - V_C = (V_A - V_B) + (V_B - V_C) = I \cdot R_{AB} + I \cdot R_G = 0.25(6) + 0.25(2) = 1.5 + 0.5 = 2\text{ V}$$

$$V_{C2} = V_B - V_D = (V_B - V_C) + (V_C - V_D) = I \cdot R_G + I \cdot R_CD = 0.25(2) + 0.25(8) = 0.5 + 2 = 2.5\text{ V}$$

$$\frac{V_{C1}}{V_{C2}} = \frac{2}{2.5} = \frac{4}{5}$$