Question 338 of 380

In this figure the resistance of the coil of galvanometer $G$ is $2Ω$. The emf of the cell is $4V$. The ratio of potential difference across ${C}_{1}$ and ${C}_{2}$ is

A$1$B$\dfrac{4}{5}$C$\dfrac{5}{4}$D$\dfrac{3}{4}$

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Answer: (B) $\dfrac{4}{5}$

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,
${V}_{{C}_{1}}=i\left[6Ω+{R}_{G}\right]...\left(1\right)$
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,
${V}_{{C}_{2}}=i\left[{R}_{G}+8Ω\right]...\left(2\right)$
Divide equation (1) by equation (2) to obtain the required ratio.
$\begin{matrix}\dfrac{{V}_{{C}_{1}}}{{V}_{{C}_{2}}} & = & \dfrac{i\left[6Ω+{R}_{G}\right]}{i\left[{R}_{G}+8\Omega \right]} \\ & = & \dfrac{6Ω+2\Omega }{2\Omega +8\Omega } \\ & = & \dfrac{4}{5}\end{matrix}$

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