Charge Equilibrium

This is interesting, but whats now?

Hmm... What do you mean, Hurz?

Your last sentences is the higlight but you unfortunately stop then. Ok, reading it again "the charges will flow for a short time" with zero ohm between both caps the time must be very short, better said its 0. But the current must be infinite high as well. So this will cause issues, right? What I miss here are some facts and formulas and what is the expected result and what is everycircuit presenting as result e.g. of voltage at end of charge. And is this all correct or not. You see what I mean?

Or other said, EC says 8.33V after charge exchange for both capacitors, is this correct, or does EC fail? I can tell you it is correct, but wont it be nice to predict and verfiy be hand 8.333V as expected result?

Thank you for your feedback. Yes. When both of these capacitors are connected in this way, the charges should take a very short time to achieve the equilibrium state - it may take microseconds to achieve that. The principle used here is that charge is conserved in a circuit (principle of conservation of charge). Let us say that the initial potential differences across each capacitor are V_1 and V_2 respectively. That happens when you fully charge both capacitors, before you discharge them by closing the switch between the two capacitors. Here, the total charge stored in both capacitors is (V_1)(C_1) + (V_2)(C_2). When you disconnect the fully-charged capacitors from their respective power supplies, and close the switch between the capacitors, the charge will start flowing due to the difference in the potential differences across each capacitor. Charge equilibrium is achieved when the sum of the voltages in the loop is zero (Kirchhoff's Voltage Law). When the sum of the voltages across each capacitor is zero, the voltages across them must be the same, right? Here, the total charge stored in the circuit is (V_f)(C_1 + C_2). V_f will be the final voltage across each capacitor. To calculate the final result, all you need to do is to apply the principle of conservation of charge: Initial total charge = Final total charge (V_1)(C_1) + (V_2)(C_2) = (V_f)(C_1 + C_2) V_f = [(V_1)(C_1) + (V_2)(C_2)]/(C_1 + C_2). In this example, V_1 = 5 V, V_2 = 10 V, C_1 = 0.000005 F and C_2 = 0.00001 F. Hence, V_f = 8.333 V. Does that answer your question?

In this simulation, EveryCircuit assumes no resistance in the connecting wires. Of course - when there is no resistance, the initial current flow is infinitely high (I = (V/R)(exp(-t/(RC)))), but in this example EveryCircuit fails to show it. Notice that when you observe the graph of current flowing through the capacitor against time, the maximum current varies with the simulation speed. In reality, this will not happen because all wires have their own resistivity.

Right correct. Q=CU and this is fortunately even with zero ohm in between both caps reproduceable within EC. The thing with Kirchhoff is wrong, cuz he said, the sum of all voltagess in a loop is zero, but this does not mean that all single compoments in a loop do have the same voltage. They can have any different but the sum is zero. Anyway, thats not the point. Do you think it would make sense to extend the description to what you have written now in comments?

Good questioner. Thank you for pointing out my mistake. The truth is that when charge equilibrium is achieved, the capacitors will be arranged in parallel, i. e. the positive terminal of a capacitor will be connected to the other positive terminal of the other capacitor. In other words, the positive charges will be at one side of the circuit, and the negative charges will be at the other side. Since both of the capacitors are in parallel, the potential differences across each capacitor must be the same. If you use Kirchhoff's Voltage Law to prove it, you will eventually see that V_1 - V_2 = 0, which gives you V_1 = V_2 = V_f.

Why do you repeat yourself again and again, something you do not understand? Just ask for!

Didn't I answer your question? May I seek for guidance?