Veritasium "How Electricity Actually Works"

[electrodacus]

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 ?

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]

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:

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]

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).

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...

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?

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.

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]

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.

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]

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.

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.

[PlainName]

And the word trough means something completely different than in / out.

Suppose you have a pipe with a diaphragm separating the left from right sides, both filled with water. A piston on the left side is pushed in, and as a consequence a piston on the right is pushed out.

Water is pushed in, but no water transfers from left water to right water. However there is energy transfer through from one piston to the other.

[electrodacus]

And the word trough means something completely different than in / out.

Suppose you have a pipe with a diaphragm separating the left from right sides, both filled with water. A piston on the left side is pushed in, and as a consequence a piston on the right is pushed out.

Water is pushed in, but no water transfers from left water to right water. However there is energy transfer through from one piston to the other.

That is a bad example.
I already provided a better example but as any analogy will still have limitations.
A cylinder separated with a diaphragm but filled with air or any other gas.
If pressure is equal on both side it will be equivalent to a discharged capacitor so there is no stored energy.
You can install a pump that takes air molecules from one side and moves them in to the other side.
Say you start with 1000 molecules on each side so 2000 molecules in total.  Then you use the pump so you need to put in some mechanical energy to push air from one side in to the other side charging this energy storage device with energy.
Say you move 500 air molecules from one side to the other so now you have 500 on one side and 1500 on the other.
Now say you want to charge an identical discharged cylinder from this charged one.
You will connect two hoses (hoses also have air in them just at normal pressure) so when you connect to the discharged cylinder
pressure will want to equalize.
You start with an excess of 500 air molecules one one side and a deficit of 500 molecules on the other side so 250 molecules will be transferred for the pressure to equalize.
You end up with 1250 / 750 in both cylinders. So was there any work done other than the heat loss due to friction of moving the molecules from one cylinder to another ?

But as mentioned you can not push any analogy to far.

[IanB]

Energy stored in capacitor is not work.

Work has to be done to store energy in a capacitor. So the energy stored in a capacitor results from the electricity doing work.

There is either heat or work. If the electricity is not generating heat then it is doing work. If it is not doing work then it is generating heat.

So in my example capacitor stored 72mJ so half of what the source provided the other half was converted in to heat.

Half of what the source provided was used to do work on the capacitor, the other half was converted to heat.

[electrodacus]

Work has to be done to store energy in a capacitor. So the energy stored in a capacitor results from the electricity doing work.

There is either heat or work. If the electricity is not generating heat then it is doing work. If it is not doing work then it is generating heat.


Half of what the source provided was used to do work on the capacitor, the other half was converted to heat.

You are confusing work with stored energy.
In fact you completely ignore energy storage as if it was not still there.
Out of the 144mJ of energy from the supply half was converted to heat the other half was stored in the same form that it was delivered so electrical energy.
That other half is stored in the capacitor and you can do whatever you want with it as it was not been used to do any work.

Not sure why is so hard to get that energy is in the same form electrical energy and it is available it was not used.
The capacitor is now a source containing 72mJ.


I think you accused me of violating the energy conservation but it looks like that is you.
I say 144mJ delivered by source = 72mJ as heat in the wires + 72mJ stored as electrical energy.
You say 144mJ = 72mJ as heat in the wires + 72mJ work + 72mJ stored

[IanB]

I think you accused me of violating the energy conservation but it looks like that is you.
I say 144mJ delivered by source = 72mJ as heat in the wires + 72mJ stored as electrical energy.
You say 144mJ = 72mJ as heat in the wires + 72mJ work + 72mJ stored

You didn't read what I said. You must learn to read.

Energy = Heat + Work

144 mJ (Energy) = 72 mJ (Heat) + 72 mJ (Work)

It's not complicated, is it?

[Alex Eisenhut]

I think you accused me of violating the energy conservation but it looks like that is you.
I say 144mJ delivered by source = 72mJ as heat in the wires + 72mJ stored as electrical energy.
You say 144mJ = 72mJ as heat in the wires + 72mJ work + 72mJ stored

You didn't read what I said. You must learn to read.

Energy = Heat + Work

144 mJ (Energy) = 72 mJ (Heat) + 72 mJ (Work)

It's not complicated, is it?

You're incredibly patient. I don't quite know what dacus's problem is, but he's long past the point of entertaining for me.

It's like telling him that it takes time to charge a capacitor and he replies that time is not energy. 

[electrodacus]

You didn't read what I said. You must learn to read.

Energy = Heat + Work

144 mJ (Energy) = 72 mJ (Heat) + 72 mJ (Work)

It's not complicated, is it?

So there is no stored energy ? A 1000uF capacitor with 12V across contains no stored energy ?
What did that work you are talking about did ?

[hamster_nz]

You didn't read what I said. You must learn to read.

Energy = Heat + Work

144 mJ (Energy) = 72 mJ (Heat) + 72 mJ (Work)

It's not complicated, is it?

So there is no stored energy ? A 1000uF capacitor with 12V across contains no stored energy ?
What did that work you are talking about did ?

The battery. Those electrons didn't just get into the capacitors by themselves now did they?

[electrodacus]

The battery. Those electrons didn't just get into the capacitors by themselves now did they?

Losing half the energy as heat is not enough ? You can say they went downhill.

[IanB]

You're incredibly patient. I don't quite know what dacus's problem is, but he's long past the point of entertaining for me.

It's like telling him that it takes time to charge a capacitor and he replies that time is not energy. 

It's fascinating seeing how many different ways he can come up with to argue that black is white, but yes, it does get boring after a while.

[hamster_nz]

The battery. Those electrons didn't just get into the capacitors by themselves now did they?

Losing half the energy as heat is not enough ? You can say they went downhill.

Oh, on second read your question it makes no sense.... I missed the extra "did" on the end.

[electrodacus]

Oh, on second read your question it makes no sense.... I missed the extra "did" on the end.

What question are you referring to?
Storing 72mJ in the capacitor resulted in 72mJ lost as heat so you can say charge efficiency is 50%
This charge efficiency can be improved to at least 90% if if you add as a minimum an inductor and diode as demonstrated in the past.
Also that 50% loss can be converted to something other than heat like maybe visible light by adding a lamp in series (just one example of many posible).

[electrodacus]

It's fascinating seeing how many different ways he can come up with to argue that black is white, but yes, it does get boring after a while.

You are making the black white.
Energy stored is just that and not work.
Claiming a charged capacitor has no stored energy is just absurd.

[IanB]

Claiming a charged capacitor has no stored energy is just absurd.

Of course. That is why nobody has ever claimed that.

Energy stored is just that and not work.

On the contrary, work is energy stored, heat is energy lost.

Here's a reference:

https://openstax.org/books/university-physics-volume-2/pages/8-3-energy-stored-in-a-capacitor

To move an infinitesimal charge dq from the negative plate to the positive plate [of a capacitor] (from a lower to a higher potential), the amount of work dW that must be done on dq is dW=Vdq=(q/C)dq.

This work becomes the energy stored in the electrical field of the capacitor. In order to charge the capacitor to a charge Q, the total work required is...

[Alex Eisenhut]

It's fascinating seeing how many different ways he can come up with to argue that black is white, but yes, it does get boring after a while.

You are making the black white.
Energy stored is just that and not work.
Claiming a charged capacitor has no stored energy is just absurd.

I don't think anyone claimed that.

Does a capacitor store time?

[electrodacus]

Of course. That is why nobody has ever claimed that.

On the contrary, work is energy stored, heat is energy lost.

You are very confused.

Doing work means using energy not storing energy.
Once you have done the work that energy to do the work is used so no longer available.

As an example a vehicle using 1kWh to travel 5km is work done.
That 1kWh is no longer available it was used to cover friction thus lost as heat.

You either admit energy is stored in the capacitor or you can claim that some work was done. You can not claim both are true at the same time as that is not how this universe works.
Getting those 72mJ in the capacitor was already very inefficient at just 50% with another 72mJ lost as heat.

[hamster_nz]

Of course. That is why nobody has ever claimed that.

On the contrary, work is energy stored, heat is energy lost.

You are very confused.

Doing work means using energy not storing energy.
Once you have done the work that energy to do the work is used so no longer available.

As an example a vehicle using 1kWh to travel 5km is work done.
That 1kWh is no longer available it was used to cover friction thus lost as heat.

You either admit energy is stored in the capacitor or you can claim that some work was done. You can not claim both are true at the same time as that is not how this universe works.
Getting those 72mJ in the capacitor was already very inefficient at just 50% with another 72mJ lost as heat.

Wikipedia disagrees... see "Work (physics)"

[SandyCox]

Of course. That is why nobody has ever claimed that.

On the contrary, work is energy stored, heat is energy lost.

You are very confused.

You appear to be surrounded by confused people. What's the common factor? It's you!

[Alex Eisenhut]

Doing work means using energy not storing energy.

That is correct. For example, moving charge from a battery (energy storage, yes?) to a capacitor. Now the energy is in the capacitor, that took work.

Sort of like slogging through your obtuse ramblings.

Does a capacitor store time?

[electrodacus]

That is correct. For example, moving charge from a battery (energy storage, yes?) to a capacitor. Now the energy is in the capacitor, that took work.

Sort of like slogging through your obtuse ramblings.

Does a capacitor store time?

The work that was done to move that charge from the voltage source to capacitor resulted in 72mJ of heat on the wires.
So efficiency of moving electrical energy from the ideal voltage source to capacitor was 50%
 Thus you end up with 72mJ of heat due to work you performed to move the charge and 72mJ of electrical energy stored in to capacitor.
The 72mJ stored in the capacitor did not do any work but they are there to do whatever work you want.