When the capacitor is charged, the voltage across it is equal to the voltage across the battery ( $V$ ). The voltage across a capacitor cannote change discontinuously, therefore the voltage across the capacitor immediately after the switch is thrown to contact $b$ is still $V$ . We can apply Kirchhoff's voltage loop around the right loop to find that
$\begin{align}V_C(t) - I(t)R = 0 \label{eqn1:1}\end{align}$
where $V$ is the voltage difference between the top and the bottom of the capacitor, and $I$ is the clockwise current. This equation must be true for all $t \geq 0$ . In particular, at $t = 0$ , $I(0) = V_C(0)/R$ . Since it was already argued that $V_C(0) = V$ , the voltage of the battery, we must have that $I(0) = V/R$ . This means that curves A and B are the only ones that could be correct. We can distinguish between them be solving for the current as a function of time. The current through a capacitor is given by
$\begin{align*}I(t) = C\frac{d V_C(t)}{dt}\end{align*}$
Because of the way we have defined the current $I(t)$ and the voltage $V_C(t)$ , the right equation is actually
$\begin{align*}I(t) = -C\frac{d V_C(t)}{dt}\end{align*}$
in this case.
So, if we differentiate equation (1) and use this relation, we find that
$\begin{align*}I(t)/C + R \frac{d I(t)}{dt} = 0\end{align*}$
The solution to this differential equation is (with the initial condition $I(0) = V/R$ )
$\begin{align*}I(t) = V/R \cdot e^{-t/(RC)}\end{align*}$
So, the current is $V/R$ at $t = 0$ and then it decays exponentially to $0$ . Thus, answer (B) is the correct.

Solution to 1996 Problem 1