Solution to 1996 Problem 1

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.

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