Question

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 $S_1$ is

• A. $\frac{\epsilon }{R_1}$ towards right
• B. $\frac{\epsilon }{R_1}$ towards left
• C. zero
• D. $\frac{2\epsilon }{R_1}$

Answer 1

Answer: (B)

Just before $S_1$ is closed, the potential difference across capacitor 2 is $2\epsilon$.

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