Question

At $t=0$ , switch $S_1$ is closed and remains closed for a long time and $S_2$ remains open. Now $S_1$ is opened and $S_2$ is closed. Find out the current through the capacitor immediately after that moment.

• A. $\frac{3E}{R}$
• B. $\frac{2E}{3R}$
• C. $\frac{E}{3R}$
• D. $\frac{2E}{R}$

Answer 1

The correct option is A $\frac{3E}{R}$

Let, the potential at point $A$ is zero. The potential at point $B$ and $C$ will be $E$.



$V_B-V_A=(E-0)$

$\therefore$ Charge on capacitor $Q_c=C(E-0)=CE$

Now, when $S_2$ is closed and $S_1$ is open.

Potential difference across capacitor and charge on it will not change suddenly.
Potential at $A$ is zero so at $D$ is $-2E$.

$\therefore$ Current through the capacitor

$=\frac{E-(-2E)}{R}=\frac{3E}{R}\left(BtoD\right)$

Hence, option (a) is the correct answer.