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. $\frac{\varepsilon}{R_{1}}$ towards right
• B. $\frac{\varepsilon}{R_{1}}$ towards left
• C. zero
• D. $\frac{2 \varepsilon}{R_{1}}$

Answer 1

Answer: (B)

Just before $S_{1}$ is closed, the potential difference across capacitor $2$ is $2 \varepsilon$.
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, the $V_{A}=\varepsilon, V_{B}=2 \varepsilon$
Since $V_{B}>V_{A}$, so current will flow
$i=\frac{2 \varepsilon-\varepsilon}{R_{1}}=\frac{\varepsilon}{R_{1}}$ towards left