In the circuit, shown in fig. $‘K’$ is open. The charge on capacitor $C$ in steady state is $q_1$. Now key is closed and at steady state, the charge on $C$ is $q_2$. The ratio of charges $\left( {\frac{{{q_1}}}{{{q_2}}}} \right)$ is
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(a) When key is open, charge in steady state will be ${q_1} = CE.$
When key is closed, potential difference across capacitor will be $V = \frac{{2R}}{{R + 2R}}E = \frac{2}{3}R$
Charge in steady state will be ${q_2} = \frac{2}{3}CE$ $==>$ $\frac{{{q_1}}}{{{q_2}}} = \frac{3}{2}$.
When key is closed, potential difference across capacitor will be $V = \frac{{2R}}{{R + 2R}}E = \frac{2}{3}R$
Charge in steady state will be ${q_2} = \frac{2}{3}CE$ $==>$ $\frac{{{q_1}}}{{{q_2}}} = \frac{3}{2}$.
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