Question 31 of 180

A galvanometer $\left(G\right)$ of $2\Omega$ resistance is connected in the given circuit. The ratio of charge stored in ${C}_{1}$ and ${C}_{2}$ is:

A$\dfrac{2}{3}$B$\dfrac{3}{2}$C$1$D$\dfrac{1}{2}$

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Answer: (D) $\dfrac{1}{2}$

In steady state:
${R}_{eq}=12\Omega$
Current through the branches with resistance, $I=\dfrac{6}{12}=0.5A$
Now, potential drop across ${C}_{1}=3V$
and potential drop ${C}_{2}=4V$
Charge on the first capacitor, ${q}_{1}={C}_{1}{V}_{1}=12\mu C$
Charge on the second capacitor, ${q}_{2}={C}_{2}{V}_{2}=24\mu C$
Therefore, $\dfrac{{q}_{1}}{{q}_{2}}=\dfrac{1}{2}$

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