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Veritasium "How Electricity Actually Works"

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electrodacus:

--- Quote from: T3sl4co1l on May 03, 2022, 03:25:17 pm ---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

--- End quote ---

You understand that a DC-DC converter has no stored energy. So he will be supplied from the existing energy in that stored capacitor.
Even if this active circuit uses some of the energy that we want to transfer it is way more efficient than just paralleling the two capacitors directly.
I do not get your confusion.

T3sl4co1l:
Shall I quote your numerous claims that capacitors can just be "paralleled directly"?

electrodacus:

--- Quote from: T3sl4co1l on May 03, 2022, 04:02:28 pm ---Shall I quote your numerous claims that capacitors can just be "paralleled directly"?

--- End quote ---

Of course they can be paralleled directly.
What I was saying (main point of the entire discussion) is that energy is transferred trough the wire and not outside the wires.
Since normal wires have resistance you end up with half the initial energy as the other half was lost as heat in the wire thus you end up with just half the Vi in both capacitors.
If you had an ideal setup with no resistance so all conductors are superconductors then you have no loss and end up with 0.707 * Vi in both capacitors.
Then some of you claimed that is not true that with ideal circuit you end up with that high voltage as you confuse voltage with energy.

Since is hard and expensive to setup an experiment with superconductors I offered a much easier solution and that is a DC-DC converter.
Using a 80% efficient DC-DC converter CC-CV working in CC mode will result in almost 90% of the energy still being present in the two capacitors after the energy is split equally between them so close enough to 0.707 * Vi to prove my point.

If you agree that moving the energy from charged capacitor to discharged capacitor will result in a higher final voltage in both capacitors than using wires then my point is made. If you do not then all I can say is you need to do the experiment.
By adding a DC-DC converter to the circuit you do not bring any external energy. You just more efficiently move the existing energy from one place to another.

T3sl4co1l:
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

electrodacus:

--- Quote from: T3sl4co1l on May 03, 2022, 04:51:43 pm ---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

--- End quote ---

Yes DC-DC is an alternative way to transport energy from the charged capacitor to the discharged one.

The proof is already there with the normal capacitors.  After you parallel the capacitors you end up with only half the energy the other half escaped as heat due to circuit resistance.
There is no electric field in the discharged capacitor and there will only be a field as soon as an electron gets to one of the plates and simultaneously an electron will leave the opposite plate.
So energy transfer is done trough wires.

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