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Veritasium "How Electricity Actually Works"

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T3sl4co1l:
You can just admit you don't know how superconductors behave, like, it's okay to not know things sometimes??  I know it's awfully late in the thread to do so [sunk cost fallacy], but late is still better than never?

The whole reason superconductors were brought up is you claim a conservation of energy, when none exists in any possible conceivable experiment that can be constructed, using such materials and no active circuitry (e.g. DC-DC).

You've had many "outs" presented to you throughout this thread:
- Energy goes into AC mode
- "Oops, I meant RMS voltage"
- Admit the trivial circuit (two capacitances and ""ideal"" wires) cannot be constructed
- Admit the trivial circuit (two capacitances and real wires, superconducting or otherwise) behaves as expected

After the first few pages, the main curiosity in this thread, I think, revolves around the peculiar psychology causing this reluctance; the technical topics are all well settled, after all.

Tim

SandyCox:

--- Quote from: electrodacus on May 13, 2022, 06:54:29 pm ---
--- Quote from: SandyCox on May 13, 2022, 06:25:48 pm ---Check if your solution satisfies the law of preservation of charge.

--- End quote ---

You do not understand what conservation of charge means and how it is applied.
The clue is in the fact that adding another charged particle to a capacitor that is at 1V and same capacitor that is at 2V requires different amounts of energy.

--- End quote ---
:palm: What you are trying to do is not the law of conservation of charge.

Just integrate the current through the capacitors with respect to time to calculate the charge in each capacitor.

Think of it this way: You are in a sealed room with an empty box on the floor and a box full of tennis balls high op on a shelf. You move half of the balls from the full box to the empty box. In the process some of the gravitational potential energy of the balls is converted to another form (probably heat). The point is that the number of tennis balls stays the same. So we have the law of conservation of tennis balls. The law of conservation of charge is the same concept. It is about the number of charged particles, not their electric potential energy!

electrodacus:

--- Quote from: T3sl4co1l on May 14, 2022, 07:24:17 am ---You can just admit you don't know how superconductors behave, like, it's okay to not know things sometimes??  I know it's awfully late in the thread to do so [sunk cost fallacy], but late is still better than never?

The whole reason superconductors were brought up is you claim a conservation of energy, when none exists in any possible conceivable experiment that can be constructed, using such materials and no active circuitry (e.g. DC-DC).

You've had many "outs" presented to you throughout this thread:
- Energy goes into AC mode
- "Oops, I meant RMS voltage"
- Admit the trivial circuit (two capacitances and ""ideal"" wires) cannot be constructed
- Admit the trivial circuit (two capacitances and real wires, superconducting or otherwise) behaves as expected

After the first few pages, the main curiosity in this thread, I think, revolves around the peculiar psychology causing this reluctance; the technical topics are all well settled, after all.

Tim

--- End quote ---

I never did experiments with superconductors.
- The properties of superconductors are zero resistance to current flow (not super low resistance but zero).
- Conservation of energy in an isolated system is always true no matter if superconductor materials are used or not.

What are you calling an active circuit ? If the circuit is powered by the only source of energy at the start of the experiment meaning the charged capacitor in this case it is completely irrelevant what components you are using as long as the system stay isolated.
The inductor and simple diode is all that is needed as I have shown in spice simulation. Neither the diode nor the inductor have any stored energy at the beginning of the experiment.
If I can demonstrate that significantly less energy is lost by just adding the diode and inductor and sowed that conservation of charge is no longer a thing then what makes you so shure that can not be shown with superconductors where no conductor in circuit will have any resistance.

I also hope you understand the concept of an isolated system.

electrodacus:

--- Quote from: SandyCox on May 14, 2022, 08:25:48 am --- :palm: What you are trying to do is not the law of conservation of charge.

Just integrate the current through the capacitors with respect to time to calculate the charge in each capacitor.

Think of it this way: You are in a sealed room with an empty box on the floor and a box full of tennis balls high op on a shelf. You move half of the balls from the full box to the empty box. In the process some of the gravitational potential energy of the balls is converted to another form (probably heat). The point is that the number of tennis balls stays the same. So we have the law of conservation of tennis balls. The law of conservation of charge is the same concept. It is about the number of charged particles, not their electric potential energy!

--- End quote ---

In a circuit with resistance you have charged capacitor in series with resistor in series with discharged capacitor.
Since all this elements are in series current will be the same through the loop.
So at the start you have 3V in charged capacitor say 3Ohm total circuit resistance an 0V on discharged capacitor.
Thus almost all energy is dissipated as heat on the resistor as you have 3V / 3Ohm = 1A * 3V = 3W dissipated as heat on the resistor.
A bit latter you have 2V on the charged capacitor and 1V on the discharged capacitor as current exiting the charged capacitor can only be equal with the one entering the discharged capacitor thus the special case of charge being conserved and half of the energy lost as heat.
Now you will have (2V - 1V ) / 3Ohms = 0.333A * 1V = 0.333W loss on the resistor as heat.

If you do the integration either as a calculation or as a measurement you will see that no energy is missing and half of initial energy is still in the two capacitors while the other half was lost as heat on the resistor.  There is no energy unaccounted for so no paradox.

This is as basic of electrical knowledge as it gets.  If you want to claim that there is energy lost some other way other that as heat on the circuit resistance then you are welcome to prove that either with math or with an experiment.

TimFox:
Again, in that simple ideal circuit with a finite-value resistor connecting the second (initially uncharged) capacitor to the first (charged) capacitor, and no additional lossy elements (such as capacitor ESR or leakage) the total energy lost by the circuit itself (into heat) is independent of the resistance value, and my earlier statement holds, that it is the same value in the limit as R-->0.  Of course, the time required to transfer the charge is directly proportional to that resistor value and the limit as R-->0 of transfer time (to a specific settling fractional value such as 1 ppm, or 14 time constants) is zero.  Here, "limit" is the normal mathematical meaning of the term.

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