In the circuit shown,the switch is shifted fr | vedclass

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(B) Initially,the switch is at position $1$ for a long time,so the capacitor is fully charged by the battery $\varepsilon$. The initial charge on the

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(B) Initially,the switch is at position $1$ for a long time,so the capacitor is fully charged by the battery $\varepsilon$. The initial charge on the capacitor is $q(0) = C\varepsilon$.At $t = 0$,the switch is shifted to position $2$. The capacitor is now connected to a battery of $2\varepsilon$ through a resistor $2R$. The capacitor will charge further until it reaches a steady state.The final steady-state charge on the capacitor will be $q(\infty) = C(2\varepsilon) = 2C\varepsilon$.Since the charge increases from $C\varepsilon$ to $2C\varepsilon$ exponentially,the correct graph is the one that starts at $C\varepsilon$ and approaches the asymptote $2C\varepsilon$. This corresponds to graph $B$.

In the circuit shown,the switch is shifted from position $1 \rightarrow 2$ at $t = 0$. The switch was initially in position $1$ for a long time. The graph between the charge on capacitor $C$ and time $t$ is:

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