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  4. In the circuit shown, switch S2 is closed first and is kept closed for a long time. Now S1 is closed. Just after that instant, the current through S1 is

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Q. In the circuit shown, switch $S_2$ is closed first and is kept closed for a long time. Now $S_1$ is closed. Just after that instant, the current through $S_1$ isPhysics Question Image

NEETNEET 2022

Solution:

Just before $S_1$ is closed, the potential difference across capacitor 2 is $2 \varepsilon$.
image
Just after $S_1$ is closed, the potential differences across capacitors 1 and 2 are 0 and $2 \varepsilon$, respectively.
If we take potential of $C$ and $D$ to be zero, then $V_A=\varepsilon, V_B=2 \varepsilon$
Since $V_B>V_A$, so current will flow
$i=\frac{2 \varepsilon-\varepsilon}{R_1}=\frac{\varepsilon}{R_1}$ towards left