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Veritasium "How Electricity Actually Works"

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electrodacus:

--- Quote from: hamster_nz on May 09, 2022, 02:24:39 am ---
--- Quote from: electrodacus on May 09, 2022, 01:21:21 am ---The entire discussion is about energy traveling through wires or outside of wires. Same can be summarized as energy traveling through a capacitor (not the leakage through dielectric) that is basically what Derek will say vs energy flows in and out of the capacitor as capacitor is an energy storage device and that is what I and everyone that has correct understanding of reality is saying.

--- End quote ---
Is that statement compatible with an LED + resistor between two physically small capacitors emitting light for a few seconds - a physical demonstration that energy can flow through the dielectric of the capacitors, even though electric charge can't.

If a glowing LED doesn't suffice, how about you put a third identical capacitor in the middle (so three capacitors in series). apply 12V. Then remove the middle capacitor, walk it to the other side of the room and measure the voltage on it with a DMM. It will read 4V, and will have  1/2 C V^2 Joules of energy in it. That energy is far higher than a microamp or so of leakage current count account for.

Where did it this energy come from?  How does that energy get into the middle capacitor if energy can only flow in wires and not through capacitors?

Maybe your definition of "energy traveling through [a capacitor]" doesn't match mine?

Acutally, what is your definition of "energy traveling through" something?

--- End quote ---

That was energy through a leaky capacitor (as all electrolytic capacitors are). But if you are referring to initial high current through LED as you seems you do that is due to capacitors charging but no current passes through capacitor. Current flows in or out of the capacitor.

I may actually post a problem with 3 different case and provide the solution also to see if you agree with the results.
Yes the capacitor in the middle will charge as all 3 capacitors in series are viewed as a single capacitor.
But the fact that you measure 4V means that all energy that flowed in that circuit with 3 identical capacitors in series supplied by 12V power supply flowed into the capacitors and not through them.

Maybe is a matter of understanding the definition of flowing in or flowing through.
Like again an analogy with limited scope just so we batter define flowing through vs flowing in/out of.
Water flows in a bucket and what will mean for water to flow through a bucket will be a bucket with a hole on the bottom so water can flow through.
Those electrolytic capacitors have a leakage current same as a bucket with a very small leak (the bucket is still useful to transfer liquid from some part to another but no long storage).
The capacitor made by the long transmission wires are more like a small coffee cup so low capacity but very well build with no measurable leakage.

The small current in the first 65ns or so Derek observes in his test is due to current flowing in to the transmission line capacitance (as I showed in spice simulation) and is not due to leakage with for that transmission line with 1m of air is not measurable (way to small).

Also back to those two identical parallel capacitors. If you add an incandescent lamp instead of the switch to transfer the energy from one capacitor to the other you will get the exact same half voltage and half the energy remaining in the capacitors while you get some visible photons (maybe depends of how large the capacitors are vs the energy need for filament to glow) but the fact remains that you can add anything in between the capacitors to do some work and the energy left in the capacitors at the end of the experiment is the same as directly paralleling the capacitors to transfer the energy.

electrodacus:

--- Quote from: bsfeechannel on May 09, 2022, 02:33:36 am ---I just did it. From a contradiction you can prove whatever, remember? So the flaw of your entire reasoning lies in the fact that you don't understand the most fundamental tenets of electricity. Go back to the books and if you have any question, we're here to help.

--- End quote ---


OK here is a problem and let me know if it the results are correct as I will provide the answer to the problem also.

A) Two identical 1F capacitors one fully discharged so 0V and one charged at 3V that will be paralleled.

Initial energy 0.5 * 1F * 32 = 4.5Ws
End energy 1.125Ws in each capacitors so just half of the total initial energy the rest ended as heat.
You get the same result if you insert at resistor or an incandescent lamp instead of the switch.

B) Three identical 1F capacitors two of them charged at 3V and one discharged 0V that will all be paralleled

Initial energy 9Ws
End energy 6Ws so all capacitors will be at 2V (0.5 * 3F * 22) = 6Ws
There is just 2Ws in the capacitor that started empty so that came from the two charged ones 1Ws from each.
They started with 4.5W each and ended up with 2Ws each so the 1Ws that each provided to discharged capacitor resulted in 1.5W of loss so worse transfer energy than first case.
Again you can add any resistor or incandescent bulb and result will be the same.

C) Three identical 1F capacitors just one of them charged at 3V the other two empty.

Initial energy 4.5Ws
End energy 1.5Ws (just 1V across each capacitor). So 3Ws ended up as heat double compared to what is left stored.



If you agree that the calculations above are correct and you also agree that all that wasted heat is in the wires/capacitor plates and lamp or resistor plus radiated as IR especially with the lamp as filament has low thermal mass and it is in vacuum then you agree that all energy traveled through wires/conductors.


I do not see how any engineer will be able to contest the above. You can measure and see that is what will happen including the wasted energy ending up as heat.

hamster_nz:

--- Quote from: electrodacus on May 09, 2022, 02:41:56 am ---That was energy through a leaky capacitor (as all electrolytic capacitors are). But if you are referring to initial high current through LED as you seems you do that is due to capacitors charging but no current passes through capacitor. Current flows in or out of the capacitor.

--- End quote ---
"Energy through a leaky capacitor"? What does "leaky" mean in the context of energy?

How long after removing from the circuit do you want me to leave the middle capacitor alone before I measure it? If it is 'leaky' and the energy is there because of such leaks, then it should self-discharge in a short period of time.



--- Quote ---But the fact that you measure 4V means that all energy that flowed in that circuit with 3 identical capacitors in series supplied by 12V power supply flowed into the capacitors and not through them.

--- End quote ---
So how did the energy get into the middle capacitor? If it didn't go through the other two capacitors then it must have gone around them? Those are really the only two options. Pick one (or both even). Energy not being able to pass through capacitors is demonstratable false.



--- Quote ---Maybe is a matter of understanding the definition of flowing in or flowing through.
Like again an analogy with limited scope just so we batter define flowing through vs flowing in/out of.
Water flows in a bucket and what will mean for water to flow through a bucket will be a bucket with a hole on the bottom so water can flow through.
Those electrolytic capacitors have a leakage current same as a bucket with a very small leak (the bucket is still useful to transfer liquid from some part to another but no long storage).
The capacitor made by the long transmission wires are more like a small coffee cup so low capacity but very well build with no measurable leakage.

--- End quote ---
In this case I think you are confusing the model as being reality, not being representative of reality.

If I made this test out of three transmission lines (say 100m rolls of coax, would my results be different?

If I went out and sourced the lowest possible leakage capacitors in the known universe, would my results be different?


--- Quote ---Also back to those two identical parallel capacitors.

--- End quote ---

I'ld rather not - exploring how you explain that energy gets into the middle capacitor is far more interesting and enlightening.

I just left a charged capacitor on the bench for 10 minutes before I put the meter over it. Still has energy in it, so leakage doesn't seem to be an major issue.

electrodacus:

--- Quote from: hamster_nz on May 09, 2022, 03:01:15 am ---
"Energy through a leaky capacitor"? What does "leaky" mean in the context of energy?

How long after removing from the circuit do you want me to leave the middle capacitor alone before I measure it? If it is 'leaky' and the energy is there because of such leaks, then it should self-discharge in a short period of time.

--- End quote ---

Leakage is an undesired effect of real capacitors and has nothing to do with anything related to the main question.
Energy flows in capacitors not through capacitors except for that pesky small current of few uA typical.
When you have a few mA or A of current in a loop with a capacitor in series then you have current going in to the capacitor not through it.
I sort of get frustrated and annoyed of explaining what in/out and trough means.

Se the problem above post #316 as that is proof energy only travels through wires not through the space around the wire or capacitors.

T3sl4co1l:
If current "goes into" a capacitor and not "through" it then would you agree Vr = 0 for an initially discharged capacitor and voltage step Vs?


--- Code: ---     C
 +---||---+--o Vr
 |+       |
 Vs       R
 |-       |
 +--------+
_|_

--- End code ---

Tim

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