For the attached circuit (which I've drawn rather poorly in GIMP), I need to figure out the charge on each capacitor when the switch has (a) been closed for a long time and (b) been open for a long time.
Loop and Junction rules, [itex]\Delta[/itex]V=IR, C=Q/[itex]\Delta[/itex]V, 1/Ceq = 1/C1+1/C2
For charging a capacitor, I = (Vemf/R)e^(-t/RC).
For discharging a capacitor, I = -(Qmax/RC)e^(-t/RC)
The Attempt at a Solution
I think I have a conceptual problem in thinking about where the current goes. I'm inclined to think that there is some sort of equilibrium, where the capacitors can discharge through the smaller circuits while the battery charges them through the larger circuits (both when the switch is open and closed). Can anyone help me think about this problem? Preferably sooner...