How does a capacitor work?

Discussion in 'General Electronics Chat' started by NichA, Jun 21, 2007.

1. NichA Thread Starter Member

Jun 10, 2007
13
0
Here's what I know:

When a capacitor is place in parallel with a voltage source, electrons will get pushed onto one plate and removed from the other. This will cause one plate to have a charge of +Q and the other to have a charge of -Q.

Here's what I don't understand:

I keep reading that electrons will continue to flow until the voltage across the capacitor is equal to the voltage across the voltage source. This doesn't make sense to me. Isn't the voltage across the two plates the same as the voltage source initially since they are connected in parallel?

I need some help understanding the relationship between the voltage across the plates and the voltage source and the flowing current and all that...

2. Dave Retired Moderator

Nov 17, 2003
6,960
145
You need to think about this differently. For a real application you best thinking of this in terms of a first-order RC circuit with a stepped input volatge - ref. http://en.wikipedia.org/wiki/Image:Series-RC.svg

The resistance is present because the wire connecting the voltage source and capacitor will have some finite resistance. The current through capacitor is i(t) = C(dv/dt), therefore current only flows as the voltage across the capacitor is changing -this is to be expected because of the redistribution of charge.

The sum of the voltage across the resistor and capacitor are equal to the voltage at the source (KVL in a series circuit).

The voltage across the resistor is Vr(t) = Vexp(-t/RC) and the voltage across the capacitor is Vc(t) = V(1-exp(-t/RC)), where RC is the time constant and describes the convergence rate, and V is the voltage source magnitude. As t tends to infinity the voltage across the capacitor approaches that of the voltage asymptotically resulting in i(t) tending to zeros and consequently the voltage across the resistor approaches zero.

So the answer is that the voltage across the capacitor is not instantly equal to the voltage at the source because there is an associated resistance which means that the voltage across the capacitor builds up to the source voltage in accordance with the RC time constant. For the theoretical case of R = 0 plug the numbers into the above equations and what do you get?

Dave

3. NichA Thread Starter Member

Jun 10, 2007
13
0
Thanks Dave, that really helped me. I can totally visualize what's happening now.

4. Dave Retired Moderator

Nov 17, 2003
6,960
145
No problems. Any other questions, feel free to ask.

Dave