# Charging a capacitor in parallel with a resistor?

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?

• It's worth drawing a diagram that shows all the currents. If you did so, the problem would jump out at you. The usual situation would have a capacitor in series with the resistor - then the sum of EMF, voltage drop across the resistor, and integral of current through the capacitor divided by capacitance would be zero. That's the setup that will give you the usual decaying exponential. Mar 14 '17 at 13:28

If they are connected in parallel, then $$I\ne\frac{dq}{dt}$$
• You can just ignore the resistor. Then you have $q/C=\epsilon$, $q=C\epsilon$. In a real circuit there will inevitably be some resistance in series with the capacitor. If this is very small, then that means the time constant is very small, and the capacitor becomes almost fully charged very quickly. Mar 14 '17 at 5:38
• @rb612, the point to note here is that you have three currents: the current $I_R$ through the resistor, current $I_C$ through the capacitor and the current $I_B$ through the battery. Of course $I_R + I_C = I_B$. The equations should work properly if you remember they are not the same. (Perhaps we should mark the caps' current as $i_C(t)$ since it's not constant over time, but whatever.) Mar 14 '17 at 11:12