But that's an active circuit. You didn't say anything about control before. Did you forget to mention control before? Then who's controlling the superconductor?
Tim
Shall I quote your numerous claims that capacitors can just be "paralleled directly"?
OK so the DC-DC is just an alternative implementation?
I want to do it without a DC-DC. Can you show me an experimental setup (using superconductors if necessary) to prove the effect?
Tim
Do you have a setup I can test?
Two capacitors cant exist like this. This circuit needs 2 switches, not 1 switch.
If you assume the bottom 'wire' between the two capacitors has zero impedance, then it's essentially just one capacitor. Isn't it?
Two capacitors cant exist like this. This circuit needs 2 switches, not 1 switch.what?
Why will you need a second switch ? redundancy
What is exactly that you want to test ?
Is that moving energy from one charged capacitor to an identical discharged capacitor using a DC-DC converter will result in final voltage being significantly more than 1/2 and close to 0.707 ?
If energy is conserved
Two capacitors cant exist like this. This circuit needs 2 switches, not 1 switch.what?
Why will you need a second switch ? redundancyIf u had 2 open switches, with a charged capacitor on leftside & non-charged capacitor on righthandside, then if u closed one switch i think that the 2 capacitors would not end up with the same (say half each) charge. They would have i think a very mixed arrangement of 4 different charges.
As you said earlier, and as I said earlier: without a DC-DC converter.
Can you do it?
QuoteIf energy is conserved
Is a very, very big 'if'.
So can you do it? With just wires, superconducting or otherwise? As you said to do it earlier? I'm really curious to see what you propose. I haven't seen a 'no', no indication that it be impossible.
Granted, it's been a challenge getting anything of substance out of you at all... I'm beginning to think you don't actually know what you're talking about.
Tim
I do. I performed the test.
I observed an oscillating waveform, with the average voltage being 0.5 Vi, and the peak sine amplitude being 0.5 Vi. Put another way, the capacitors slosh alternately between 0 and 1 Vi. At no point are they simultaneously 0.71 Vi; and the current in the wire connecting them (inductor, really), when one is at 0.71 Vi, is nonzero, so opening the switch at that instant would result in energy loss.
Can you explain my findings?
Tim
If you consider the switch ideal (no capacitance) then if you close just one of two switches absolutely nothing will happen.
Discussing the case where you have switches in both the top and bottom wires...If you consider the switch ideal (no capacitance) then if you close just one of two switches absolutely nothing will happen.
Well, not absolutely nothing. If only one switch is closed, there will always be 10V measured over the other switch, so the act of closing one of the switches changes the voltages on the right hand side.
With both switches open the voltages on the right hand side are not at all clearly defined, as the two circuits are not connected.
(I'm waiting for some "the quantum wave function collapsed because you need to a meter to make the measurements" handwaving... from the person who adds DC-DC converters)
Also your do not provide any feedback on where my math is wrong. Math that shows the amount of energy in a capacitor.
Also your do not provide any feedback on where my math is wrong. Math that shows the amount of energy in a capacitor.
I have done that many times.... but one more time...
Can we agree that the total charge on the top half of the system is:
Qtotal = Cc1 Vc1 + Cc2 Vc2
So here is the initial state of the system:
Qtotal = 1F * 3V + 1F * 0V = 3 Coulomb.
I hope both agree that no charge can cross either capacitor regardless of the position of the switch. So the total charges on the top and bottom halves must remain the same before and after the switch is closed. If not, please let me know what mechanism is taking electrons out of the top half and moving them to the bottom half, as I am sure we both agree that no positively charged atoms are moving around either.
The switch is closed, and allowing for the incorrect assumption that everything settles down to a stable state when using ideal components:
Qtotal = Cc1 Vc1 + Cc2 Vc2
Qtotal = 1F * 1.5V + 1F * 1.5 = 3 Coulomb.
This is the same as you would have if you consider the two 1F capacitors now a single 2F capacitor:
Qtotal = 2F * 1.5V = 3 Coulomb.
We have the same amount of charge, but are left with only half the voltage over each capacitor. The incorrect assumption that the system becomes stable has let us down.
A simpler example that shows how flawed your position is
What about a single charged ideal capacitor (no ESR), ideal wires (zero ohms) and a single ideal switch (zero ohms when closed, perfect isolation when open), that shorts the capacitor. If, as you say, energy must be conserved, what happens if the capacitor has been charged to 3V and the switch is closed?
What happens with "close to ideal wires", that have zero ohms resistance, but a small amount of inductance?
The inductance is important, as it gives a solution where the total energy in the system is conserved.
A simpler example that shows how flawed your position is
What about a single charged ideal capacitor (no ESR), ideal wires (zero ohms) and a single ideal switch (zero ohms when closed, perfect isolation when open), that shorts the capacitor. If, as you say, energy must be conserved, what happens if the capacitor has been charged to 3V and the switch is closed?
What happens with "close to ideal wires", that have zero ohms resistance, but a small amount of inductance?
The inductance is important, as it gives a solution where the total energy in the system is conserved.
As I mentioned you are confusing charge with energy.
The important law is called Conservation of energy and not conservation of charge.
Please do a energy balance not a charge balance and you will understand what I'm saying.
When you do the transfer from one capacitor to the other you will be losing half of the energy as heat.
Once you understand the difference between charge and energy it should be easier to also understand what I'm saying.
3 Coulomb is like saying 0.833mAh so you say nothing about energy.
If one ignores the inductance in the circuit, and just treats it as having a finite resistance R in series with the two capacitors:
The final distribution of charge (1/2 on each capacitor) is independent of R, but the time required to equilibrate decreases as R decreases.
Since the final distribution is independent of R, it will still be true in the limit as R --> 0.
If one ignores the inductance in the circuit, and just treats it as having a finite resistance R in series with the two capacitors:
The final distribution of charge (1/2 on each capacitor) is independent of R, but the time required to equilibrate decreases as R decreases.
Since the final distribution is independent of R, it will still be true in the limit as R --> 0.
Yes 1/2 of the charge but 1/4 of energy. Half of the total energy was dissipate as heat on the series resistance no matter the value of the series resistance.
Please understand that discussion is about transfer of energy.