I'm trying to determine as an exercise for myself the charge on a capacitor as a function of time when a resistor and a capacitor are parallel and connected to the battery. I know I have the wrong answer, but I'm not sure what I did wrong.
Through Kirchoff's loop rule, I can say that:
$$\epsilon – I*R = 0$$
Where epsilon is the emf of the battery. And
$$\epsilon – q/C=0$$
Therefore:
$$I*R = q/C$$
$$R * \frac{dq}{dt} = \frac{q}{C}$$
The solution to this differential equation I got was:
$$q(t) = Ce^\frac{t}{RC}$$
And I verified this through WolframAlpha.
But this would mean the capacitor will take on an arbitrarily large amount of charge. This does not have a $-t$ term like our RC circuit that's in series. So how could this be?
Best Answer
If they are connected in parallel, then $$I\ne\frac{dq}{dt}$$