Physics - Current Electricity Question with Solution | TestHub

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 $6\Omega$ and the galvanometer resistance. Mathematically,

$V_{C_1}=i\left[6\Omega +R_G\right]...\left(1\right)$

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

$V_{C_2}=i\left[R_G+8\Omega \right]...\left(2\right)$

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

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