In the circuit shown, the switch is shifted from position $1 \rightarrow 2$ at $t = 0$. The switch was initially in position $1$ since a long time. The graph between charge on capacitor $C$ and time $'t'$ is
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At $t = 0$, the charge on cap acitor is $Q_0 = CE$ After a long time, the charge on cap acitor will be $2Q_0 = 2CE$ :. graph is $(2)$
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