When the capacitor is charged, the voltage across it is equal to the voltage across the battery (). The voltage across a capacitor cannote change discontinuously, therefore the voltage across the capacitor immediately after the switch is thrown to contact is still . We can apply Kirchhoff's voltage loop around the right loop to find that
where is the voltage difference between the top and the bottom of the capacitor, and is the clockwise current. This equation must be true for all . In particular, at , . Since it was already argued that , the voltage of the battery, we must have that . 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
Because of the way we have defined the current and the voltage , the right equation is actually
in this case.
So, if we differentiate equation (1) and use this relation, we find that
The solution to this differential equation is (with the initial condition )
So, the current is at and then it decays exponentially to . Thus, answer (B) is the correct.