Two capacitors and a resistor in a circuit.

[Hunter1]

Have you tried changing the capacitor values and observing the steady state voltage?

Calculate the charge on C1:
C = C1
q = CV

Now rearrange for V but this time C = C1 + C2

Yes, i've simulated in LTspice changing resistance, capacitance and voltage alternately maintaining the others fixed.

Firstlaw of nature : ...

You are wrongly assuming ...

Well, i must say i didn't yet study maxwell's equations and stuff to be sure about what's happening. But i can say that as long as the electrons don't just change into something else like energy through photons or dissapear, the voltage will be EXACTLY the same in the final state(of course the voltage will be divided, so it's the algebric sum that will be the same) because the electrons generate the electric field, thus the potential difference (or voltage).

[Dave]

Firstlaw of nature : you cannot create energy , you cannot destroy energy. Only convert it.

As current is flowing from cap 1 to cap 2 ( as long as they don;t both have the same voltage) some of that energy is converted in heat. So after charge equalisation that energy has to be removed from the equation. this will result in a net drop of the voltage across the caps.

Otherwise you could create a perpeteuum mobile using three capacitors and a resistor. two in series , resistor , third. when balance has been achieved swap right one with one of the left ones and keep going...

Seriously, read what I wrote in one of my previous posts.

With ideal components, voltage will be exactly 10 volts. The energy loss is accounted for. The stored energy in a capacitor is W = 1/2 * C * V^2. If the voltage is halved, the energy is reduced to a quarter. You have two capacitors, so you have two quarters. Half of the total energy is transformed into heat.

Half of the initial energy is dissipated on the resistor, half of it is distributed between the capacitors. By having half the initial voltage on both caps, you have a quarter of the initial energy in each cap.
1(initial C1) + 0 (initial C2) = 1/2 (heat) + 1/4 (C1) + 1/4 (C2)

By this point I am starting to think you are just pulling my leg.

[IanB]

I'd like to know which equation can be used to calculate the current flowing in the circuit below. The voltage in both capacitors are measured with a DMM. Initially the capacitor 1 is charged, and has 20 volts; the second has 0 volts.

Just for completeness, here is a worked solution to the problem from first principles. (It's actually only a partial solution, because another step is needed to obtain the individual voltages V₁ and V₂. But the other information in this thread is enough to fill that gap.)

[free_electron]

Half of the initial energy is dissipated on the resistor, half of it is distributed between the capacitors. By having half the initial voltage on both caps, you have a quarter of the initial energy in each cap.
1(initial C1) + 0 (initial C2) = 1/2 (heat) + 1/4 (C1) + 1/4 (C2)

oh crap. i made a reasoning error. you are right. the power loss is accounted for. phew. no laws of nature were broken. ( heavy fines on that one ... )