Veritasium "How Electricity Actually Works"

[T3sl4co1l]

As you asked someone else a few messages back, "I need to ask what is your qualification." Seems that literally everyone is wrong except for you, so what is your qualification for having the definitive knowledge about energy?

Fair turnabout; ah, but qualifications only matter if anyone was taking him seriously. 

Tim

[electrodacus]

I guess those two are professors at MIT and same as the other professor at University of California Alex have a poor understanding of energy storage.

Tim

I will need to correct this as it is unfair since I have no read the book written by them. In case of Alex I have seen his inability to understand energy by being unable to explain how that vehicle works and win the bet with Derek.

As you asked someone else a few messages back, "I need to ask what is your qualification." Seems that literally everyone is wrong except for you, so what is your qualification for having the definitive knowledge about energy?

I'm an electrical engineer but work with energy storage and energy generation (is both my job but more importantly my hobby).
I looked probably at all forms of energy generation and energy storage so maybe that gives me a better understanding of energy.

It seems in this particular problem people confuse charge with energy. I do not think I seen the energy equation in Derek's video yet the main claim he makes is that "energy doesn't travel through wires" with is completely wrong.
He made the same mistake with the "faster than wind direct downwind vehicle with is powered by the wind only".  That is more than a perpetuum mobile is an overunity device claim (obviously not true).
Alex the profesor knew that since such a machine will violate the conservation of energy but was unable to figure out that energy storage was involved so that vehicle energy storage device was charged while below wind speed using wind energy and that stored energy is what allowed the vehicle to temporarily exceed wind speed (not indefinitely as claimed).
To make the claim fit they used wrong equations (equation Derek or the guy that "invented" the machine came up just to fit their silly explanation).
So while Alex did not made any false claim on camera he as a physics professor should have understood how the vehicle works and debunk the claim instead of offer it even more credibility.

[aetherist]

Circuit theory isn't sufficient to solve the two-capacitor problem. We need to take the electrodynamic behavior of the system into account.

So let's do the thought experiment of connecting a charged and a discharged capacitor in parallel at t=0s. We assume that all the conductors are perfect and that the two capacitors are of the parallel plate type with vacuum between the plates. The two capacitors can be modelled as two lossless transmission lines (see for instance House and Melcher example 14.2.1). After the two capacitors are connected in parallel, an electromagnetic wave will bounce back and forth between the two capacitors. The amount of energy stored in each capacitor will fluctuate, but the total amount of stored energy in the two capacitors will remain the same.

If we go to the next level and take electromagnetic radiation into account, then al the energy will be radiated into space over time.

U suggest a gedanken where a charged capacitor & a discharged capacitor are suddenly connected in parallel.
One way to connect the 2 capacitors in parallel is to have 2 switches, which is what i asked for earlier.

Yes, a capacitor can be modelled as a parallel pair of transmission lines.

I don’t agree that an em wave will bounce back & forth tween the 2 capacitors. Electricity will bounce back & forth. And if electricity is an em wave then yes an em wave will bounce back & forth.
But i suspect that your em wave is Veritasium's Poynting Vector or Poynting Field. The Poynting Vector or Field are not electricity. I don’t know what they are. Praps some kind of description of what exists. But certainly not a description of what makes things happen.

No, circuit theory is not sufficient to solve the 2 capacitor problem, & no, the electrodynamic behaviour of the system is not sufficient. Both fail. Both are wrong. It comes down to what is electricity. 
If we use a bad theory for electricity then we are unlikely to solve an electricity problem. If we do solve an electricity problem then that would be due to good luck (the scientific term is i think     equivalence).  But, we know that circuit theory plus electrodynamic behaviour plus luck fail every time.

If lossless, then all of the energy will not be radiated into space over time.
There will be no heat losses, there will be no radiation losses.

Listen. Photons do not looz energy over time (ignoring things which rob energy from a photon). Photons are eternal (ignoring things). Electricity consists of photons (not electrons)(not em waves)(not silly Poynting stuff), hence in your gedanken the electricity will be eternal.

Hey, just noticed, this is my 400th posting/reply/comment on this forum. 
I have cast 400 pearls.

[TimFox]

"Photons do not looz energy over time (ignoring things which rob energy from a photon)."
Tautological, yet mis-spelled.
Photons are emitted and absorbed all the time.  In Compton scattering of x-ray photons, the photon loses a fraction of its energy.

[aetherist]

"Photons do not looz energy over time (ignoring things which rob energy from a photon)."
Tautological, yet mis-spelled.
Photons are emitted and absorbed all the time.  In Compton scattering of x-ray photons, the photon loses a fraction of its energy.

A photon left well alone duznt change in any way.
I will spell proper when u & all talk proper (ie i will use archaic spelling when u & all use archaic pronunciation).

Photons are eternal (almost). Once created they exist forever (until they are killed)(there are at least 2 ways, which i wont go into today).

Yes, photons can be emitted, & absorbed, & praps cut up (Compton), & stretched, & bent (& in the end annihilated).
But until they are annihilated they are eternal.

So too electrons. But, electrons are photons. Hence in a way old (electron) electricity is photonic.
My new (electon) electricity is of course purely photonic (my electons propagating at the speed of light).

[TimFox]

When radioactivity was first discovered, around 1900, three types were identified:  alpha, beta, and gamma.
They were initially named in order of their penetrating power through matter.
Since their first discovery, it was found and demonstrated experimentally that beta particles were negatively-charged electrons, and gamma rays were uncharged photons.
In 1900, Becquerel determined that the charge/mass ratio of beta particles equaled that of electrons in cathode rays.
Around 1914, it was demonstrated that gammas are electromagnetic radiation, i.e., photons.
How, then, can electrons be photons and vice-versa?  Electrons have charge and mass, photons have neither charge nor mass.

[aetherist]

When radioactivity was first discovered, around 1900, three types were identified:  alpha, beta, and gamma.
They were initially named in order of their penetrating power through matter.
Since their first discovery, it was found and demonstrated experimentally that beta particles were negatively-charged electrons, and gamma rays were uncharged photons.
In 1900, Becquerel determined that the charge/mass ratio of beta particles equalled that of electrons in cathode rays.
Around 1914, it was demonstrated that gammas are electromagnetic radiation, i.e., photons.
How, then, can electrons be photons and vice-versa?  Electrons have charge and mass, photons have neither charge nor mass.

Good questions.
Gammas are photons. Radio waves are em radiation. Radio waves are not photons. Gammas are not em radiation. EM radiation is emitted by photons, it is a part of every photon.

An electron is a photon that has formed a loop by biting its own tail. At which time the em radiation splits, some going out (negative charge), some going in (which is then annihilated)(Williamson explains)(he says that the loop has a twist, hence all of the negative em radiation always goes outwards)(or most of it).
Jeans called an electron "bottled light".
Catt i think called an electron "a rolled up photon".
A positron has the positive charge going out, negative charge going in (where it is annihilated).

Electrons do not orbit a nucleus. Photons orbit a nucleus. In that sense we have 2 kinds of electron. One kind orbits nothing (& has the form of a loop). The other kind orbits a nucleus (& need not be a complete loop)(ie it need not bite its own tail).

Mass is a bit of a mystery. Firstly photons do have mass. The question arises -- how come a photon gains lots of mass when it becomes an electron. Williamson mentions a possible way.
Mass is the ability to annihilate aether. Something that has more mass annihilates more aether.
However, i have my own theory. A photon propagates along a line, hence the inflow of aether giving us what we call mass flows in perpendicular to the line, ie the inflow streamlines converge in 2 dimensions, while the guilty photon leaves the scene at the speed of light.
Meanwhile back at the ranch, an electron is a photon that is constrained in 3 dimension, the inflow streamlines converge in 3 dimensions, to a (stationary) point rather than to a line (which is fleeing at c).
I think that the convergence of streamlines, & the stationary versus fleeing stuff, might explain the difference in gravitational mass tween a photon & an electron.

My new (electon) electricity, ie my electons, consist of photons that hug the surface of a wire, rather than hugging a nucleus. Electons are photons that are constrained in 2 dimensions. Free photons (light) are constrained in 1 dimension. Electrons are photons that are constrained in 3 dimension (usually called confined photons).

Everything that we feel & see is made of photons. Photons are the fundamental (quasi) particle.
When photons form loops they give us every other kind of elementary particle (electrons quarks etc).

There are 3 forms of electricity. My electons (on a wire). Free surface electrons (on a wire).  Possibly drifting electrons (inside a wire).

The charge on a capacitor consists of my electons on the negative plate, & induced free-surface-electrons on the other plate. However, my new (electon) electricity is a work in progress. I need to tick all of the boxes.

My electons immediately & simply explain why the discharge of a capacitor takes twice the time predicted by the flawed standard circuit theory.  In the meantime everyone around here is talking rubbish about capacitors.

[TimFox]

So photons form a loop, and gain mass to become electrons.
Whence comes the charge?
I'm glad to see you admit the possibility of electrons drifting:  TI and the other semiconductor manufacturers can now continue their processes.
Electrons accelerating emit electromagnetic radiation:  see antenna theory and synchrotron radiation (both of which work).
Photons interacting with electrons can increase the energy of the electrons:  see atomic structure theory (not the archaic "orbits" you keep harping on) and spectroscopy (both of which work).

[Naej]

It certainly deserves to be separated  because the physical behavior is quite different:

Separated.

Since the 19th century, everybody thought that everything electric and magnetic, from DC to cosmic rays, through radio waves, heat, light, ultraviolet, X-rays and whatnot, is the manifestation of the same freaking physical phenomenon.

Now you're saying that they are different. I wonder why the Nobel Committee has not noticed you yet.

Oh they are both implied by Maxwell's equations.
But because their physical behavior is quite different, they are called differently. Much like DC current and gamma rays are called differently. Duh.
You don't call wires, capacitors, coils and antennae "Maxwell stuff" do you?

Yes I can. Energy flows in wires and reappear in a lightbulb/engine/LED.

Yes energy is transferred from the battery to the short, through the wire.

If you say so, it must be true.

Well you say energy flows through vacuum just to mock engineers, why can't I say it flows through wires?

[bsfeechannel]

The circuit theory is perfectly capable in solving the two parallel capacitor problem.

There is no such thing as conservation of charge. The conservation of energy is a law.

Kirchhoff's circuital laws (KCL/KVL) are an extension of the principle of conservation of charge. So your assertions above are contradictory.

I looked probably at all forms of energy generation and energy storage so maybe that gives me a better understanding of energy.

You lack fundamental understanding though.

[electrodacus]

You lack fundamental understanding though.

I proved that energy travels through wires in multiple ways sumarized in this post  https://www.eevblog.com/forum/chat/veritasium-how-electricity-actually-works/msg4156462/#msg4156462
Let me know what part of that is wrong.
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.

[aetherist]

So photons form a loop, and gain mass to become electrons.
Whence comes the charge?
I'm glad to see you admit the possibility of electrons drifting:  TI and the other semiconductor manufacturers can now continue their processes.
Electrons accelerating emit electromagnetic radiation:  see antenna theory and synchrotron radiation (both of which work).
Photons interacting with electrons can increase the energy of the electrons:  see atomic structure theory (not the archaic "orbits" you keep harping on) and spectroscopy (both of which work).

I like the ideas (re charge) coming from Williamson, & also from Conrad Ranzan.  Here are some links to Williamson.

Is the electron a photon with toroidal topology.              Williamson & van der Mark.    1997.
http://home.claranet.nl/users/benschop/electron.pdf

On the nature of the photon and the electron.            J.G.Williamson     2015?
https://www.researchgate.net/publication/281749668_On_the_nature_of_the_photon_and_the_electron

A new theory of light and matter.                J.G.Williamson.            (Dated: July 18, 2014)
https://www.researchgate.net/publication/323873485_A_new_theory_of_light_and_matter

I think i am ok with that electrons accelerating stuff. And i am ok with that photons interacting with electrons stuff.
Re electricity, one can in many cases just change electron to electon.
In antennas it is the electons sloshing up&back, not electrons. And the electons emit em radiation.
In synchrotrons it is i suppose electrons going round & round, emitting em radiation.
I am ok with that.

[hamster_nz]

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.

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, with out switching off the 12V, 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?

[bsfeechannel]

Oh they are both implied by Maxwell's equations.
But because their physical behavior is quite different, they are called differently. Much like DC current and gamma rays are called differently. Duh.

Precisely. Theories like the Newton's law of gravitation or Maxwell's equations are breakthroughs because they show that things that on the surface appear to be different are in fact just aspects of the same crap.

So, to the untrained eye, radiation, capacitive coupling and "transmission of energy through wires" look like different and uncorrelated things.

But those in the know know that they're the same thing.

You don't call wires, capacitors, coils and antennae "Maxwell stuff" do you?

Why not? Will they get offended?

Well you say energy flows through vacuum just to mock engineers, why can't I say it flows through wires?

You don't get it. I AM an engineer. If you want to contradict the experimental data, knock yourself out.

[bsfeechannel]

I proved that energy travels through wires in multiple ways sumarized in this post  https://www.eevblog.com/forum/chat/veritasium-how-electricity-actually-works/msg4156462/#msg4156462
Let me know what part of that is wrong.

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.

[electrodacus]

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.

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?

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]

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.

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]

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.

"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.

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.

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.

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.

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?

Also back to those two identical parallel capacitors.

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]

"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.

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: [Select]

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

Tim

[bsfeechannel]

I do not see how any engineer will be able to contest the above.

Then see. The energy resides in the dielectric between the plates of any of the charged capacitors and will happily migrate through space to the uncharged ones. The energy that goes into the wires and plates will be immediately dissipated.

Wires transfer charge. Space transfers energy.

[hamster_nz]

"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.

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.

I too get frustrated when we talk about "energy flowing through" and then you reply talking about current. If you said something like "proof that charges only travels through wires not through the space around the wire or capacitors" I would agree with you (as long as it is at low voltages).

It is easily proven that we are able to move energy - actual Joules of measurable energy, that can do real work - from a battery into the middle capacitor of three in series. That energy will persists in the capacitor after it is removed from the circuit, and can be moved to the other side of the room. The rest of the original circuit could be disassembled, or even destroyed, and yet that energy still persists in what was the middle capacitor.

It should be undeniable proof that energy has been transferred from the battery into that middle capacitor, even though it has no other connection to the source except through the two capacitors. The only way for energy to get into that capacitor is either through the other two capacitors, or through the air around it. Both of which are things you say can't happen.

And the more 'ideal' the capacitors are the better this can be shown - an ideal capacitor with zero leakage could be left for months, and that energy would still be in there. Even with my less-than-ideal AliExpress components that energy sits there for minutes, if not hours.

It isn't as if my caps suddenly go all leaky when connected to a DC source - I have put a uA meter inline (once the caps are charged), and have measured exactly how little leakage is - under a uA at 10V.

[hamster_nz]

Wires transfer charge. Space transfers energy.

Oh.. that's good. I like that!

[electrodacus]

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: [Select]

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

Tim

Electrons entering on one side of the capacitor to increase the density of free electrons means that electrons will need to leave the other plate.
But this energy is not used to do any work but stored and can be retrieved later to do work.
Say Vs is 3V and capacitor 1F the value of the resistor is irrelevant.
The capacitor will have a voltage potential of zero at the start so all 3V will drop across the resistor (will say resistor value is large enough that voltage source and capacitor DC resistance is low enough to to be of importance).
The charge current will start high then drop as the capacitor voltage increases until it gets to zero when voltage across the capacitor equal Vs (3V) then no current will flow so no energy will travel through the circuit (we ignore the capacitor small leakage current).
So now no energy can be transferred from Vs to resistor as current is basically zero.
Charging was inefficient but energy is stored in capacitor and can be used.
At all time no current (except that annoying leakage) has flown through the capacitor.
The way you charge a capacitor is by moving electrons from one plate to another but that is not done through dielectric but externally through wires.
There are now 4.5Ws of energy stored in the capacitor even if the power supply needed to deliver 9Ws as the other 4.5W was lost as heat on the resistor.
If you had a constant current power supply then no current resistor will have been needed for current limiting and power supply will have delivered 4.5Ws and all of that 4.5Ws will have been stored thus no work done at all / no heat loss.
Is like charging the capacitor with a linear regulator vs a DC-DC charger.

If you agree with the above (current flows in to capacitor as it is being charged and current flow stops when capacitor is fully charged) then you agree that energy is not flowing through the capacitor but in to capacitor.

[hamster_nz]

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: [Select]

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

Tim

Electrons entering on one side of the capacitor to increase the density of free electrons means that electrons will need to leave the other plate.
But this energy is not used to do any work but stored and can be retrieved later to do work.
Say Vs is 3V and capacitor 1F the value of the resistor is irrelevant.
The capacitor will have a voltage potential of zero at the start so all 3V will drop across the resistor (will say resistor value is large enough that voltage source and capacitor DC resistance is low enough to to be of importance).
The charge current will start high then drop as the capacitor voltage increases until it gets to zero when voltage across the capacitor equal Vs (3V) then no current will flow so no energy will travel through the circuit (we ignore the capacitor small leakage current).
So now no energy can be transferred from Vs to resistor as current is basically zero.
Charging was inefficient but energy is stored in capacitor and can be used.
At all time no current (except that annoying leakage) has flown through the capacitor.
The way you charge a capacitor is by moving electrons from one plate to another but that is not done through dielectric but externally through wires.
There are now 4.5Ws of energy stored in the capacitor even if the power supply needed to deliver 9Ws as the other 4.5W was lost as heat on the resistor.
If you had a constant current power supply then no current resistor will have been needed for current limiting and power supply will have delivered 4.5Ws and all of that 4.5Ws will have been stored thus no work done at all / no heat loss.
Is like charging the capacitor with a linear regulator vs a DC-DC charger.

If you agree with the above (current flows in to capacitor as it is being charged and current flow stops when capacitor is fully charged) then you agree that energy is not flowing through the capacitor but in to capacitor.

If you agree with the above (current flows into the left hand side of the capacitor, and that current is flowing out of the right hand side of the capacitor, as it is being charged and both current flows stop when capacitor is fully charged) then you agree that during that time energy was flowing through the capacitor.

FTFY.