A complete solution to wrap up this thread:
When the switch is closed resistor ##R_2## and the capacitor ##C## are in parallel. We can re-draw the circuit, swapping these two branches to make the voltage divider clear:
When the switch is first closed the capacitor is uncharged and looks like a short circuit for the first instant. That means ##R_2## is bypassed and the full potential of the battery will be across ##R_1##, yielding an initial battery current:
## I_o = \frac{E}{R_1}##
Eventually the capacitor will have charged to its final value and its current ##I_c## will be zero. Effectively it then looks like an open circuit to the rest of the circuit, leaving just the voltage divider comprised of ##R_1## and ##R_2##. The battery just sees the two resistors in series, so the final current will be:
##I_f = \frac{E}{R_1 + R_2}##
So the source current will start at ##I_o## at time t = 0 and drop to at ##I_f## as t → ∞.
Since its a first order circuit the transition will follow an exponential curve as shown. What is required to pin down the details is the time constant, ##\tau##.
In the circuit diagram above we can imagine replacing the battery and resistor divider with a Thevenin equivalent. Then we will have a simple RC circuit whose time constant will be ##\tau = R_th C##. We don't need the Thevenin voltage here, just the Thevenin resistance.
Opening the circuit at a-b we can see that the Thevenin resistance is given by:
##R_th = R_1 || R_2 = \frac{R_1 R_2}{R_1 + R_2}##
So the circuit's time constant will be
##\tau = \frac{R_1 R_2}{R_1 + R_2} C##
We can now write the expression for the current by inspection of the current vs time sketch:
##I(t) = I_f + (I_o - I_f) e^{- \frac{t}{\tau}}##
We can plug in the various expressions for the parameters from their derivations above and rearrange for a "pleasing" form for the equation. Perhaps:
$$ I(t) = \frac{E}{R_1 + R_2} \left( 1 + \frac{R_2}{R_1} e^{- t\frac{R_1 + R_2}{R_1 R_2 C}} \right) $$
