Don't let the pretty pictures distract you.
You are a hard person to goad something out of, but eventually it works.
Is that your replay ? How about you understand now that if you transfer energy more efficiently from one capacitor to the other that is identical you get close to 0.707 the voltage of the charged capacitor in both.
It should not be me that is doing the work.
Just test with a DC-DC converter and you will see very close to ideal is possible. If the DC-DC converter was 100% efficient (so ideal) then you get 0.707 Vi but you do not need a DC-DC converter if there is no resistance to get the same result.
(And of course if you have an ultra efficiency DC-DC convertor to make the transfer then you might get close to 71%, but we don't. We have two ideal caps, some ideal wire and an ideal switch).
You keep on bringing up DC-DC converters
...
So finally we have the proof that everybody else missed. That if you transfer energy to and from a magnetic field, using an inductor, (thus taking the energy "out of the wire") that energy only flows in the wires.
You keep on bringing up DC-DC converters
...
So finally we have the proof that everybody else missed. That if you transfer energy to and from a magnetic field, using an inductor, (thus taking the energy "out of the wire") that energy only flows in the wires.
At first you and others mentioned that you can not have 0.707 * Vi because that will violate the conservation of charge witch likely you confused with energy.
Then after pointing out the formula for energy stored in a capacitor it was still argued that 0.707 * Vi is not possible in an circuit made with superconductors so zero resistance.
Then I mentioned something that is way easier to test (a DC-DC converter) and claim remained the same that voltage at the end of the experiment can not be higher than 0.5 Vi.
The energy seen through the bulb in the first 65ns in Derek's experiment is related to line capacity so no energy travels through air.
Any transmission line has both inductance and capacitance (both are energy storage devices) so the transmission line model that any electrical engineer uses is a realistic and accurate way to predict what happens.
I simplified the two capacitors spice model even more and below is the schematic and the graph
In the schematic you have a 3V supply to charge one of the 1000uF capacitors then the supply is disconnected by the switch S1 after 20ms and then starting at 40ms the S2 switch is closed for just 6ms then open.
The charged capacitor C1 will discharged from 3V down to around 2V while the C2 that started fully discharged is charged using that energy at 2V so you start with C1 at 3V and you end up with both identical capacitors charged at 2V using only energy available in C1
The only other two components are the 47mH inductor and a diode. The inductor is an intermediary energy storage helping reduce the losses due to circuit resistance. There is no outside energy coming in to the circuit and some is still lost as heat but much less than just paralleling the two capacitors directly.
The red graph is the control for the S2 switch showing the period switch was closed. The blue is the voltage across C1 and green voltage across C2.
(Attachment Link)
(Attachment Link)
You cannot add inductors and diodes or DC-to-DC converters. The problem is about connecting two capacitors in parallel without any other components.
How do you think energy is transferred through a capacitor? Let's take a parallel plate capacitor with vacuum between the plates.
What about the airgap of a transformer?
It is also 'cheating'. I too could get any answer I want if I am free to add to the system. If I said "let me put an inductor in there, a diode, a switch and a trained imp that can push the switch really quickly" would you not agree that that is not the same problem any more?
Is it possible to have an electric field without an associated particle? Is it possible to have an electric field without a charged particle?
If the area of the floor of the swimming pool is increased from A to 2A (analogous to adding the initially-discharged capacitor) then the height of the water (analogous to the voltage across the capacitors) must be halved from h to h/2. The gravitational potential energy of the water (analogous to the electrical potential energy of the capacitors) must be halved, too. The weight of the water (analogous to the charge on the capacitors) does not change when the floor area (analogous to the total capacitance) is doubled.
The pathological nature of the Two Capacitor Paradox problem leads to the observation that only the Law of Conservation of Charge is satisfied, but the Law of Conservation of Energy is violated.
The pathological nature of the Two Capacitor Paradox problem leads to the observation that only the Law of Conservation of Charge is satisfied, but the Law of Conservation of Energy is violated.
Law of conservation of energy has never been broken/violated. There is no paradox related to the two parallel capacitors so that page on Wikipedia is misinformation as anyone can write a wiki page and write whatever it wants.
As I clearly mentioned a few times all energy is accounted for. Due to resistance in the two identical capacitors circuit (circuit perfectly symmetrical) after closing the switch half of the energy is wasted as heat in the conductors (that includes wires, switch and capacitor plates) the other half remains in the two capacitors quarter of the energy in each capacitor.
There is absolutely no mystery in the two parallel capacitor problems unless you do not understand what capacitors are.
Adding an inductor (another energy storage device that people do not understand) and a diode will not bring any extra energy as system is still isolated and the initial total energy at the beginning of the test is the same and contained all in the charged capacitor.
The end result with the inductor and diode added is closer to ideal case where circuit will have had no resistance at all case where no energy will be wasted as heat.
Did you even looked closely at the equations you posted from wikipedia to see how absurd and stupid they are.
Your point is moot. The whole point is that all the conductors are assumed to be perfect. There is no resistance.
Check if your solution satisfies the law of preservation of charge.
Also, check what the charge and voltage ratios as a function of the series resistance and what the limits as R-->0 are.
Exactly.
Since the final result is independent of the finite resistance in the circuit, it will also be the result in the limit as R-->0.
I mean that in the context of mathematical limit calculations.
(I was once criticized here for suggesting "epsilon-delta" (q.v.) calculations of limits, but that's how it works.)
Even if the resistance is 1x10-6 ohms (unphysical), the loss of total energy is still 1/2.
Therefore, for any specified resistance value the loss is unchanged.
The factor (1/2), of course, is only valid for two equal capacitors.
In the Spice model you posted above (reply 477), there is zero resistance in series with your finite inductor.
The switches have finite resistance, but the second one is not on for a long time.
However, the diode model included in the circuit can bleed charge from the system, assuming it includes losses.
In a resistor-only circuit with two capacitors, only resistors in parallel with the capacitors (e.g., capacitor leakage) will remove charge from the system as the capacitors slowly discharge with the power switch open.
(PS: when I distinguished active and passive circuits, I mean that an active circuit has an external power source. So a diode in the dark is passive, but it has losses.)
What happens if you leave S2 on for a long time in your simulation?