General > General Technical Chat
Veritasium "How Electricity Actually Works"
electrodacus:
--- Quote from: snarkysparky on May 19, 2022, 08:17:07 pm ---I tell ya what I think it means.
The resistance of the wires got to dissipate as much energy as is stored in the capacitor.
And the current that made this possible went into one terminal of the capacitor and out the other terminal. I call this "through" what do you call it ?
--- End quote ---
Thanks for doing the calculation and providing the numbers.
What it means is that energy from the source was split in two with one part ending as heat in wire/resistor and the other as stored energy in capacitor thus no energy passed through capacitor.
You can prove that that amount of energy 72mJ is stored in capacitor either by calculation or by testing.
If any energy will have been going "through" capacitor it will have done some work so end up as heat instead it was all stored so going in to capacitor.
An example where energy was going through something will be if you replaced the capacitor with a fuse that say melted with 72mJ but then all 144mJ provided by the source will have ended as heat nothing stored with 72mJ passing through wires and another 72mJ passing trough fuse before it melted all ending up as heat so 144mJ of heat.
If energy was to pass through capacitor then you could not store anything. It will mean your capacitor is defective and plates are shorted.
But if plates where shorted you will have dissipated way more energy as heat in one second on the wires 144J instead of 2000x less at 72mJ
aetherist:
--- Quote from: HuronKing on May 19, 2022, 10:08:34 am ---Sorry be posting 3 comments in a row but enough nonsense has been perpetrated here about conduction current being the only way that energy can move from A to B in space. Here is a very simple experiment measuring displacement current as equal to the conduction current but there are NO electrons moving across any gaps whatsoever:
--- End quote ---
Interesting.
1. I see that the induced wave on the left of the capacitor was stronger than on the right of the capacitor.
2. And i suppose that if he removed the right half of the capacitor including its wire connection then he would have gotten an even stronger wave on the left part of the circuit (stronger than with the right side in place).
3. Here (1) & (2) support my new (elekton) elekticity.
4. The electricity on the rhs of the capacitor is an (induced) electron electricity.
5. On the lhs its an elekton elekticity.
6. Tween the plates there is no electricity, no current, there is a radio signal, ie em radiation, which induces an electron electric current on the rhs. [/quote]
--- Quote from: dunkemhigh on May 19, 2022, 11:52:04 am ---
--- Quote --- And i suppose that if he removed the right half of the capacitor including its wire connection then he would have gotten an even stronger wave on the left part of the circuit (stronger than with the right side in place).
--- End quote ---
Easy to suppose, but unless you can show that then "Here (1) & (2) support my new (elekton) elekticity" is just building on quicksand.
Hell, we can all do that. I suppose that if he removed the right side the left side would have doubled. No, tripled! Now just let me make up a theory to account for that...
--- End quote ---
Not so fast. We already have some data. The lhs was stronger than the rhs.
How does old electron electricity explain that?
How was there more change (in current) on the lhs?
--- Quote from: dunkemhigh on May 19, 2022, 12:43:08 pm ---Yes, I saw that. What I'm disputing is that you can extrapolate and say you would see anything with the right side missing. Maybe you would, but that wasn't shown and we can't tell from the experiment.
--- End quote ---
Yes. And i said that my new (elekton) elekticity could explain. But that old (electron) electricity couldn’t explain. But praps old (electron) electricity can explain. Lemmeseenow.
Old (electron) electricity would have to say that there is electric energy lost in the air gap, whilst the electric energy was going from left to right.
Which raises a question – is the AC energy going from left to right? I reckon that it is.
If the AC energy is alternating & coming from the left & then from the right etc then i can't say that energy is lost in the air gap – koz if alternating then the signal would have the same strength on lhs & rhs.
[I doubt that electrodacus is correct that the energy loss across the air gap is due to a problem with scope grounding.]
Anyhow, old (electron) electricity might be correct if it says that energy is flowing from left to right & if it says that much energy is lost in the air gap (or including the plates too).
The em radiation around a wire has intrinsic energy (which can be called Poynting Field energy if u like) in the non-transient electricity case, & the em radiation (Poynting Field if u like) transmits energy from the wire in the transient case.
All of this energy has come from or is coming from the source (eg battery). Via wires (via elektons & electrons).
Old (electron) electricity (i reckon) demands an energy loss in the air gap.
My new (elekton) elekticity duznt need an energy loss, it just needs an inefficient energy transmission across the air gap. This inefficient process involves elektons on one plate (lhs) inducing an electron voltage on the other plate (rhs).
In old (electron) electricity the process would involve electrons (mainly surface electrons i reckon)(not so much internal electrons, which old (electron) electricity reckons) on both plates.
A large energy loss in the air gap (including the plates) would demand a rise in temperature, & or some kind of new radiation.
Not forgetting that there is a small energy loss in every inch of every wire all the time anyhow, including gaps.
What everyone around here (cept me) probly duznt know, is that the em radiation around a wire duznt accord with conservation of energy. The field is due to the radiation from elektons & electrons. The radiation from elektons & electrons is perpetual. Standard science has no explanation for this. It duznt have an explanation koz it duznt need an explanation. It duznt need an explanation koz it duznt recognise the question. Aether theory has the answer.
The above duznt apply to transient electricity. Transient electricity involves the em radiation, but it also involves conservation of energy.
Alex Eisenhut:
I'm afraid to use a light switch now. Will the light turn on? Will the universe split in two?
IanB:
--- Quote from: electrodacus on May 19, 2022, 10:43:18 pm ---If any energy will have been going "through" capacitor it will have done some work so end up as heat instead it was all stored so going in to capacitor.
--- End quote ---
You have a problem with your terminology.
In thermodynamics, the First Law can be roughly paraphrased to say: Energy = Heat + Work
So the energy dissipated in the resistor was Heat, and the energy stored in the capacitor was Work. It takes work to store energy, because that's where the stored energy comes from, it is the accumulation of work done.
If the energy stored in the capacitor were not Work, then it would be Heat, and the capacitor would get hot. The capacitor does not get hot, so the electricity is not generating Heat, it is doing Work.
electrodacus:
--- Quote from: IanB on May 19, 2022, 11:40:45 pm ---
You have a problem with your terminology.
In thermodynamics, the First Law can be roughly paraphrased to say: Energy = Heat + Work
So the energy dissipated in the resistor was Heat, and the energy stored in the capacitor was Work. It takes work to store energy, because that's where the stored energy comes from, it is the accumulation of work done.
If the energy stored in the capacitor were not Work, then it would be Heat, and the capacitor would get hot. The capacitor does not get hot, so the electricity is not generating Heat, it is doing Work.
--- End quote ---
Energy stored in capacitor is not work.
It can do work as any other stored energy if you want to use it but it is not work.
So in my example capacitor stored 72mJ so half of what the source provided the other half was converted in to heat.
You can then remove the source and just add a wire where the source was since that way energy stored in capacitor can be converted to heat in that resistor so another 72mJ and now all the energy that source provided was used.
And the word trough means something completely different than in / out.
If I say water went in to bucket or through the bucket is not the same thing.
Where through the bucket will mean bucket is damaged has a hole so water goes through instead of in.
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