In the circuit shown in the figure $K_1$ is open. The charge on capacitor $C$ in steady state is $q_1$. Now key is closed and at steady state charge on $C$ is $q_2$. The ratio of charges $q_1/q_2$ is
A$1.67$
B$0.6$
C$1$
D$0.67$
Medium
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A$1.67$
a $q_1 = CE$
$q_2 = CE × \frac{3}{5}$
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