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Veritasium "How Electricity Actually Works"
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electrodacus:

--- Quote from: hamster_nz on May 14, 2022, 09:39:53 pm ---
--- Quote from: electrodacus on May 14, 2022, 09:17:08 pm ---But if the circuit is not isolated meaning you have some other loop in range of this and that has resistance then energy will be lost to that eventually all the initial energy will be lost in that magnetically coupled loop.

--- End quote ---
Are you saying if there was a suitable loop in range (say with a resistor in it), then all the energy would be lost from this circuit? That all the energy would be transferred through the air to this loop, and it would end up heating the resistor?

--- End quote ---

Yes but it may take hundreds of years if the distance of that loop is relatively far.  And that will not be considered an isolated system. The discussion of isolated system can be just theoretical in real tests it will be close enough to an isolated system in order to draw some conclusions depending on what the test wants to show.
It is very similar with a superconductor ring where you induce a current at that remains there indefinitely  https://www.open.edu/openlearn/science-maths-technology/engineering-technology/superconductivity/content-section-2.2
Of course you can not have an ideal capacitor as you can not have a dielectric without losses so the LC even if made of superconductor materials it will eventually dissipate the energy as heat in the dielectric material.
hamster_nz:

--- Quote from: electrodacus on May 14, 2022, 09:59:58 pm ---
--- Quote from: hamster_nz on May 14, 2022, 09:39:53 pm ---
--- Quote from: electrodacus on May 14, 2022, 09:17:08 pm ---But if the circuit is not isolated meaning you have some other loop in range of this and that has resistance then energy will be lost to that eventually all the initial energy will be lost in that magnetically coupled loop.

--- End quote ---
Are you saying if there was a suitable loop in range (say with a resistor in it), then all the energy would be lost from this circuit? That all the energy would be transferred through the air to this loop, and it would end up heating the resistor?

--- End quote ---

Yes but it may take hundreds of years if the distance of that loop is relatively far.  And that will not be considered an isolated system. The discussion of isolated system can be just theoretical in real tests it will be close enough to an isolated system in order to draw some conclusions depending on what the test wants to show.
It is very similar with a superconductor ring where you induce a current at that remains there indefinitely  https://www.open.edu/openlearn/science-maths-technology/engineering-technology/superconductivity/content-section-2.2
Of course you can not have an ideal capacitor as you can not have a dielectric without losses so the LC even if made of superconductor materials it will eventually dissipate the energy as heat in the dielectric material.

--- End quote ---

You seem to be saying:

"A portion of the energy is outside of the wires, and can be transferred to completely galvanically isolated circuits. The amount of energy transferred between the two depend on the physical geometry of their physical arrangement - so size, distance, orientation and so on."

Do you get dielectric losses if you use a vacuum as the dielectric in capacitors? What heats up if there is nothing there?
PlainName:

--- Quote --- like Derek did in the Direct downwind faster than wind video where what he basically presented there was an overunity device so getting more power out than in.
--- End quote ---

You were, and are, wrong about that. You wanna pop back over to that thread and answer the questions you conveniently forgot about?
electrodacus:

--- Quote from: hamster_nz on May 14, 2022, 11:15:56 pm ---You seem to be saying:

"A portion of the energy is outside of the wires, and can be transferred to completely galvanically isolated circuits. The amount of energy transferred between the two depend on the physical geometry of their physical arrangement - so size, distance, orientation and so on."

Do you get dielectric losses if you use a vacuum as the dielectric in capacitors? What heats up if there is nothing there?

--- End quote ---

You can only transfer energy while the magnetic field strength is variable so it will not work at DC.
I try to stay on the subject and magnetic field is not involved in what happens in those first few ns when switch is closed.
The line capacity is what induces that current through the lamp/resistor.
Vacuum will also have losses as there is no such thing as real vacuum (But I'm not a physicist so I will not pretend to know all the subatomic particles that may or not pop up or out of existence at random).

Main subject is electrical energy and if it flows through wires or not. Since we are ignoring any super small leakage current through air or even vacuum the main electric current will travel through wires and you need current on top of electrical potential in order to have power with integrated over time will be energy.   
electrodacus:

--- Quote from: dunkemhigh on May 14, 2022, 11:18:34 pm ---You were, and are, wrong about that. You wanna pop back over to that thread and answer the questions you conveniently forgot about?

--- End quote ---

I will go there and check but you will need to also go there and point to me the question you are referring to as I'm sure there are a lot of them.
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