Veritasium "How Electricity Actually Works"

[electrodacus]

Like the fact that a 20nF capacitor has a very small capacity and so even a relatively short wire will have an inductance large enough that you form a resonant circuit.

Any indicance and capacitance will (at least in an ideal situation) form a resonant circuit. 

If you use a few mF then even with fairly long wires the inductance is super negligible and it is not a problem.

Only to the extent that the parasitic L and R of a real world 1 mF capacitor are likely to be larger.  But an ideal 1 mF capacitor plus a 10 nH inductor will resonate at about 50 kHz. A more realistic 100 nH inductance will lower the frequency to 15 kHz.

The DC-DC converter is fairly easy to do but likely not for last few people that replayed to me.

Everybody here agrees that the DC DC converter will do exactly as you have described.  Nobody is trying because there isn't any disagreement on this topic so no point arguing.  Our contention is that simply shorting a capacitor out will do something different and Tim did experiments to demonstrate that the circuit either oscillates or decays to V/2.

All are good points but you likely did not read all the earlier replays.
People did not understood that doing a test with superconductor materials is not trivial unless they work at some large university lab and even then it will be expensive. And did not got the idea that all the elements in the circuit will need to have zero resistance in order for all of the energy to remain in the two capacitors at the end of the test.

You are the first to mention that the DC-DC converter will do what I described. The others before where either not convinced as they where confusing charge with energy and thinking that everything is already conserved just with two normal capacitors.
And some others suggested that adding instructors and active components will somehow add energy from exterior.

The circuit will always decay to V/2 in a real setup (not ideal without resistance) as it was demonstrated in a video posted earlier.
But he concluded that energy was conserved as he confused charge with energy.
Exactly half of the energy was lost as heat in the circuit resistance and that will always be the case as long as resistance is different from zero no matter the resistance value.

The main point is that a transmission line is a capacitor or the other way around a capacitor is a transmission line and what Derek observed in his experiment as current trough that 1.1Kohm load resistor is just due to transmission line capacitance.
There will be a current trough the 1.1kOhm resistor because is in series with the supply the switch and two smaller value capacitors formed by the transmission line.
So his claim that energy travels outside the wires is false in all instances not just DC but even at initial transient when switch is closed or in AC
The electric field will only be present after charges are present.

Also the fact that half of the energy is lost as heat (heat in the conductors) should be clue enough that energy was transferred trough wires.
Electrons are those that transfer the energy from source to load and not as Derek claims electric field.

I'm also a bit rude and I apologies for that. I feel frustrated trying to explain this and I still have PTSD from a few months ago when I tried unsuccessfully to explain how the faster than wind direct down wind vehicle works and there the exact same type of mistake was made and energy storage was not considered.
The short answer to that problem is that energy is stored in pressure differential (air is a compressible fluid) while vehicle was well below wind speed and that stored energy is what allowed the vehicle to temporarily exceed wind speed. All experiments end before that stored energy is used up and so they concluded that vehicle will continue to accelerate basically forever.
Seeing people with phd in physics not understanding energy conservation was hard.

[T3sl4co1l]

All are good points but you likely did not read all the earlier replays.
People did not understood that doing a test with superconductor materials is not trivial unless they work at some large university lab and even then it will be expensive. And did not got the idea that all the elements in the circuit will need to have zero resistance in order for all of the energy to remain in the two capacitors at the end of the test.

I have just one more question.  You've seen the waveform; I ask you: after one half-cycle, where is the energy?  How much of it?  After one full cycle, where is the energy?  How much of it?  And this repeats on, ad infinitum; it is, as you say, an oscillator.

Where does the energy remain at the end of the test?

Tim

[electrodacus]

All are good points but you likely did not read all the earlier replays.
People did not understood that doing a test with superconductor materials is not trivial unless they work at some large university lab and even then it will be expensive. And did not got the idea that all the elements in the circuit will need to have zero resistance in order for all of the energy to remain in the two capacitors at the end of the test.

I have just one more question.  You've seen the waveform; I ask you: after one half-cycle, where is the energy?  How much of it?  After one full cycle, where is the energy?  How much of it?  And this repeats on, ad infinitum; it is, as you say, an oscillator.

Where does the energy remain at the end of the test?

Tim

Your was a normal circuit with resistance so it will not oscillate forever as energy will all be dissipated as heat after just a few oscillation cycles.
You seemed to have an almost matched resonant circuit meaning the capacitor and inductor energy storage capacity was almost identical thus the large fluctuation.
If you had order of magnitude larger capacitance than inductance then you will not even notice any oscillation unless you had super high vertical resolution oscilloscope.
If you had the means to build a superconductor experiment then you will have build the capacitors as in one of my earlier examples as below
____________________________
____________/   ______________

this sort of physical setup will have extremely low inductance relative to capacitance (very small distance between plates and a good dielectric material between)  then you will not see any oscillation unless you had super high vertical resolution equipment as oscillation amplitude will be some infinitesimal small value around the 0.707 Vi so maybe 4'th or 5'digit type fluctuation that will remain forever as there is no resistance for the energy to be dissipated as in non superconductor circuits.

An inductor opposes the change in current flow so the opposite of a capacitor that opposes the change in voltage.
The capacitor creates a electron imbalance between the plates that in turn result in an electric field inside the capacitor.
With the inductor an magnetic field is generated in the space surrounding the inductor and that field is where energy is stored (stored not dissipated).
So at constant current you will have a constant magnetic field (same as you have on a permanent magnet) but if the current reduces the magnetic field will reduce but the energy stored in that magnetic field is converted back to current flow.

The best analogy I can came up with will be a flexible stretchy hose where the elastic force represents the magnetic field.
If you flow water trough a long flexible hose (hose is already filed with water but not under any pressure) then when you open a valve connected say to a barrel filled with water (barrel the analogy for capacitor) the hose if super stretchy (high inductance) will oppose the flow of water as it will start to stretch before it allows water to flow on the other end and so at constant water flow (constant current flow) the hose is stretch is maintained at some level  but if you increase the flow the hose will stretch even more storing even more energy (it is not lost) and then that energy can be recovered as the flow drops as the hose will put pressure on the water trying to maintain the earlier flow.

Do not try to push this analogy further as there are differences beyond this. It is just an analogy for the inductor as an energy storage device.

[T3sl4co1l]

So you propose a superconducting resonator?  Yeah, those resonate too.

Examples that come to mind are the cavities used in particle accelerators, with Q factors up to 10^8 (it's not quite infinite because there's always some loss at AC), and superconducting qubits which, being small enough and cold enough that quantum mechanics is quite relevant, have ground states that are effectively resonators in perpetual motion (and for which, bulk measures like Q aren't so meaningful).

The bulk metal forms of these resonators might not be called high inductance, but the fact that they resonate at 100s or 1000s of MHz makes that irrelevant.

Low inductance is not no inductance!

The permeability of free space has units of per-length.  Anything that has nonzero physical size, must necessarily have inductance!  Even free space itself, or else waves wouldn't propagate (that, or some wierd causality shit that would be even more bonkers if true..).

It seems your gap in knowledge comes down to magnetic aspects:
- Length corresponds to inductance (notice I hinted earlier that the waveform and capacitance were sufficient to solve for the wire length -- evidently around 71m.  Hm, it's quite a bit less than that actually, I think; I was lazy and just coiled it up on a spool, magnifying the effect.)
- Energy is stored in the magnetic field, proportional to current flow.
- Energy conservation is true, AND charge conservation is true.  Both must be true jointly.  However, it happens to be a hell of a lot easier to lose energy to dissipation or radiation into the surroundings, than charge into the surroundings!
- We can assess the behavior of a series RLC circuit (which this is, necessarily: see points above) based on the ratio of Zo = sqrt(L/C) to R.  When Zo > R, some oscillation will be evident; when Zo = R, critically damped; Zo < R, overdamped (RC dominant).
- This is a continuum relationship and no distinction appears for R --> 0.
- As a special case, for R = 0, any combination of L and C will resonate; the damping factor is 0 regardless!

So I maintain that my waveform was obtained from a superconducting apparatus until proven otherwise.

I mean, how would you know?  Given the above information, can you solve for the resistance (if any) in my circuit?

And there's nothing wrong with the waveforms; half the time, the energy difference (the "missing" 0.5 Ei) is stored in the inductance as current flow.  The other half it's in one or the other capacitor, hence the voltages alternate between 0 and Vi.  Energy is always conserved!  And charge is always conserved too, which is why this process averages 0.5 Vi during the wave, and as the AC transient decays (when R > 0), the energy difference is dissipated as heat.  The fact that the capacitors end with 0.25 Ei each, 0.5 Ei total, is also no coincidence; perhaps less satisfying than having no dissipation, but the dissipation itself is a necessity (for such simple circuits; else, we must go to great lengths if we wish to avoid it -- such as DC-DC converters!) and so this is the result, no sqrt(2) to be found.


As for the sqrt(2), there is a separate chain of logic which should sound immediately.  Such special ratios are EXTRAORDINARILY rare from simple systems.  Impossible even, for suitable definition of "simple systems".  Such ratios are more likely to be found in, say, properties of signals -- take the peak to RMS ratio of a sine for example, or its integral which picks up a factor of pi -- but not from such simple, finite, geometric relationships like two capacitors rubbed together.  This is ultimately a deep truth about numbers themselves, you can't get an irrational (like sqrt(2)) from a rational (like 1/2) without going to some lengths first (sqrt(2) is an algebraic number).

Or, if we could easily construct such ratios -- it would certainly make transformer design easier.  We could easily and accurately match 50 to 75 ohms, for example: a 1.5:1 impedance ratio.  But we cannot: a 1.22474... turns ratio is needed.  We can only get arbitrarily close.  (The continued fraction representation of this ratio goes [1; 4, 2, 4, 2, ...]; large numbers in the continued fraction are desirable as they represent points of especially good (but still not perfect!) fit, but repeating sequences like this don't give any especially good stopping points.)

Tim

[electrodacus]

So you propose a superconducting resonator?  Yeah, those resonate too.

Examples that come to mind are the cavities used in particle accelerators, with Q factors up to 10^8 (it's not quite infinite because there's always some loss at AC), and superconducting qubits which, being small enough and cold enough that quantum mechanics is quite relevant, have ground states that are effectively resonators in perpetual motion (and for which, bulk measures like Q aren't so meaningful).

The bulk metal forms of these resonators might not be called high inductance, but the fact that they resonate at 100s or 1000s of MHz makes that irrelevant.

Low inductance is not no inductance!

The permeability of free space has units of per-length.  Anything that has nonzero physical size, must necessarily have inductance!  Even free space itself, or else waves wouldn't propagate (that, or some wierd causality shit that would be even more bonkers if true..).

It seems your gap in knowledge comes down to magnetic aspects:
- Length corresponds to inductance (notice I hinted earlier that the waveform and capacitance were sufficient to solve for the wire length -- evidently around 71m.  Hm, it's quite a bit less than that actually, I think; I was lazy and just coiled it up on a spool, magnifying the effect.)
- Energy is stored in the magnetic field, proportional to current flow.
- Energy conservation is true, AND charge conservation is true.  Both must be true jointly.  However, it happens to be a hell of a lot easier to lose energy to dissipation or radiation into the surroundings, than charge into the surroundings!
- We can assess the behavior of a series RLC circuit (which this is, necessarily: see points above) based on the ratio of Zo = sqrt(L/C) to R.  When Zo > R, some oscillation will be evident; when Zo = R, critically damped; Zo < R, overdamped (RC dominant).
- This is a continuum relationship and no distinction appears for R --> 0.
- As a special case, for R = 0, any combination of L and C will resonate; the damping factor is 0 regardless!

So I maintain that my waveform was obtained from a superconducting apparatus until proven otherwise.

I mean, how would you know?  Given the above information, can you solve for the resistance (if any) in my circuit?

And there's nothing wrong with the waveforms; half the time, the energy difference (the "missing" 0.5 Ei) is stored in the inductance as current flow.  The other half it's in one or the other capacitor, hence the voltages alternate between 0 and Vi.  Energy is always conserved!  And charge is always conserved too, which is why this process averages 0.5 Vi during the wave, and as the AC transient decays (when R > 0), the energy difference is dissipated as heat.  The fact that the capacitors end with 0.25 Ei each, 0.5 Ei total, is also no coincidence; perhaps less satisfying than having no dissipation, but the dissipation itself is a necessity (for such simple circuits; else, we must go to great lengths if we wish to avoid it -- such as DC-DC converters!) and so this is the result, no sqrt(2) to be found.


As for the sqrt(2), there is a separate chain of logic which should sound immediately.  Such special ratios are EXTRAORDINARILY rare from simple systems.  Impossible even, for suitable definition of "simple systems".  Such ratios are more likely to be found in, say, properties of signals -- take the peak to RMS ratio of a sine for example, or its integral which picks up a factor of pi -- but not from such simple, finite, geometric relationships like two capacitors rubbed together.  This is ultimately a deep truth about numbers themselves, you can't get an irrational (like sqrt(2)) from a rational (like 1/2) without going to some lengths first (sqrt(2) is an algebraic number).

Or, if we could easily construct such ratios -- it would certainly make transformer design easier.  We could easily and accurately match 50 to 75 ohms, for example: a 1.5:1 impedance ratio.  But we cannot: a 1.22474... turns ratio is needed.  We can only get arbitrarily close.  (The continued fraction representation of this ratio goes [1; 4, 2, 4, 2, ...]; large numbers in the continued fraction are desirable as they represent points of especially good (but still not perfect!) fit, but repeating sequences like this don't give any especially good stopping points.)

Tim

Maybe I was not clear enough.
If you have orders of magnitude larger capacity than inductance then you will not even be able to measure the signal (and I'm referring to amplitude not frequency).

So a sine wave with 0.001 V_i amplitude and a 0.707 V_i DC offset. An 8 bit scope will not be able to measure that and even with a 12bit scope the signal will be in the noise so you will read a DC voltage of abut 0.707 V_i

The setup that you build was something like 0.9 V_i waveform amplitude so an almost perfectly tuned LC oscillator due to using almost equal L and C in therms of energy storage.




Also all this is irrelevant as energy is conserved in LC vs the RLC where half of the energy ends up as heat when energy is transferred from one identical capacitor to another.  This is a fact both if you do the calculations using appropriate equations and if you measure the temperature rise of the conductors.
Now that I think about you can not get a perfect capacitor as there is no perfect dielectric (as far as I know) so there will be losses there and energy from this super conductor made capacitor will slowly dissipate over time as heat due to leakage in dielectric.

[electrodacus]

I mean, how would you know?  Given the above information, can you solve for the resistance (if any) in my circuit?

Tim

Any resistance value will result in half of the energy lost as heat on the resistance.
A smaller resistance will mean wires will heat faster and a larger resistance value means wires will heat slower but the exact same total amount will be wasted as heat.
With 20nF capacitors energy transferred is small so that half energy ending as heat is to small for you to be able to measure.
But use larger capacitors and a large value resistor in series then you will be able to measure that lost energy.

You know that half the energy was wasted by the end of the experiment
1F cap with 3V charging a discharged 1F capacitor
Start energy 4.5Ws end energy 2.25Ws
Current was limited by resistance so current trough a resistor means wasted heat.

You can add a 1Ohm, 100Ohm and 10kOhm between the two capacitors to increase the resistance and you will see that no matter what resistance you use you end up with the same 0.5 V_i on the two capacitors.

[T3sl4co1l]

Actually, power peaks for R = Zo.  If it were true that smaller resistance heats faster -- what would superconductors do?  Just explode?

Tim

[electrodacus]

Actually, power peaks for R = Zo.  If it were true that smaller resistance heats faster -- what would superconductors do?  Just explode?

Tim

Of course the speed of electron wave is not infinite so there is a limit to how fast the energy will be transferred from one capacitor to the other.
Yes a superconductor will explode if the temperature increases and it gets out of the super-conduction zone while high current is passing trough it.
Superconductor has no resistance not a small resistance but absolute zero.
Any resistance value no matter how small will dissipate the same amount of energy as heat in this two capacitor example assuming proper ratio of inductance and capacitance with inductance much, much smaller than capacitance.
If you have this resonant circuit as it was in your example energy will flow in multiple directions multiple times significantly increasing the losses then you no longer end up at 0.5V_i.
That is why industrial users of electricity are charged for apparent power so that they will try and correct the power factor.



If we use this same example with two parallel capacitors but the wires are super long so switch is super far from the two capacitors.
How long to you think it will take for the first unit of energy to get to the discharged capacitor?

Will energy get to the discharged capacitor based on the short distance between the two capacitors or it will be based on the long distance between capacitors and switch ?
This should demonstrate if energy travels inside or outside the wires.
 

[hamster_nz]

If we use this same example with two parallel capacitors but the wires are super long so switch is super far from the two capacitors.
How long to you think it will take for the first unit of energy to get to the discharged capacitor?

Will energy get to the discharged capacitor based on the short distance between the two capacitors or it will be based on the long distance between capacitors and switch ?
This should demonstrate if energy travels inside or outside the wires.

We already know the answer to that from the video - some energy will travel a shorter distance, via the EM fields, due to the changes in that fields. The bulk of the energy will start once the a stable field has built up around the whole length of the wire.

It doesn't answer if energy is transported inside or outside of the wires, just that energy can travel outside of wires (through the EM field), and that energy can travel along wires.

For me the interesting observation is that only the tiniest bit of EM energy travels the shortest distance, and after that the energy is then being supplied by changes in the EM field that are happening further and further away on the wires.

[Sredni]

If so: HOW MUCH ENERGY does travel along the path?

If we assume that potential energy is located in the rock, and arbitrarily say sea level is zero energy, then zero energy travels along the surface path, and negative energy travels back along the underground path.
Between the two paths (passing through a plane perpendicular to the paths), there is a net energy flow from A to B.  The amount of energy is the energy converted to heat.  So it depends on which hole it goes down.

This is equivalent to say that the path the energy follows depends on where we arbitrarily choose to set the zero reference. We have seen it with the electrical system as well: for a given convention we can have all energy traveling along the top wire, for another half in the top and half in the bottom wire, and for another one yet all energy travels along the bottom wire. They can't be all true, can they?
Now, in this particular example you elected the return path because it allows you to remove the chance element, but what if I put a second cusp on the return path? Do I have to wait till the weight has fallen in the return hole to know how much energy has traveled?

[Sredni]

The relativistic answer (convention) is that a momentum density p of matter corresponds to an energy flux of pc².
If you want general relativity then the answer is, well, complicated. https://math.ucr.edu/home/baez/physics/Relativity/GR/energy_gr.html
Classical physics is simple and says you have a potential energy of -GMm/r, you're not moving 0 energy, you're just converting: potential->kinetic->heat.

In the quasistatic idealization I had in mind, the mass takes an infinite time to travel from point A to point B. So, velocity is basically zero along the path and only changes during 'generation' (the machine raises the weight) and 'dissipation' (the weight does work against the gravitational field and it all becomes heat).

In my view the energy is in the gravitational field of the system planet + weight.  It is being added to the system during 'generation' and it is extracted from the system during 'dissipation'. The system occupies all space, so does it even make sense to ask if the energy is traveling?

[electrodacus]

We already know the answer to that from the video - some energy will travel a shorter distance, via the EM fields, due to the changes in that fields. The bulk of the energy will start once the a stable field has built up around the whole length of the wire.

It doesn't answer if energy is transported inside or outside of the wires, just that energy can travel outside of wires (through the EM field), and that energy can travel along wires.

For me the interesting observation is that only the tiniest bit of EM energy travels the shortest distance, and after that the energy is then being supplied by changes in the EM field that are happening further and further away on the wires.

What video ? The one with the wrong info made by Derek ?

The energy is transported trough wires that is why there is no energy transfer from battery to load until electron wave has the time to travel from source to load along the wire and not trough air.
It is also important I think to talk about the electric and magnetic field separately in this context.

Before you close the switch all you have is a constant electric field so no magnetic field.

Only the exact moment when the two conductors of the switch get close enough so that an electron can jump to the other conductor that has a different distribution of electrons the energy starts to be transferred or do any work.
If you just move the switch but not close enough for any electrons to move you have some local small variation in the electric field but no magnetic field and no work done so no transfer of energy.


But if you think you understand better the magnetic field think about at permanent magnet that has fields extending quite some distance outside the magnet but they can not do any work or transfer any energy.
if you move a conductive loop in a magnetic field then the energy to the system is provided by the one that moves the loop and as electrons move they create an equal and opposite field. It is not the magnetic field that moves the electrons is the person that moves the loop trough a magnetic field and that opposite field is the result of electrons moving.

[hamster_nz]

We already know the answer to that from the video - some energy will travel a shorter distance, via the EM fields, due to the changes in that fields. The bulk of the energy will start once the a stable field has built up around the whole length of the wire.

It doesn't answer if energy is transported inside or outside of the wires, just that energy can travel outside of wires (through the EM field), and that energy can travel along wires.

For me the interesting observation is that only the tiniest bit of EM energy travels the shortest distance, and after that the energy is then being supplied by changes in the EM field that are happening further and further away on the wires.

What video ? The one with the wrong info made by Derek ?

The energy is transported trough wires that is why there is no energy transfer from battery to load until electron wave has the time to travel from source to load along the wire and not trough air.
It is also important I think to talk about the electric and magnetic field separately in this context.

Before you close the switch all you have is a constant electric field so no magnetic field.

Only the exact moment when the two conductors of the switch get close enough so that an electron can jump to the other conductor that has a different distribution of electrons the energy starts to be transferred or do any work.
If you just move the switch but not close enough for any electrons to move you have some local small variation in the electric field but no magnetic field and no work done so no transfer of energy.


But if you think you understand better the magnetic field think about at permanent magnet that has fields extending quite some distance outside the magnet but they can not do any work or transfer any energy.
if you move a conductive loop in a magnetic field then the energy to the system is provided by the one that moves the loop and as electrons move they create an equal and opposite field. It is not the magnetic field that moves the electrons is the person that moves the loop trough a magnetic field and that opposite field is the result of electrons moving.

There will be a potential difference across the switch (otherwise no current will flow when the switch is closed). And as you point out it is a small capacitor - different charges separated by distance. There is then a force across the gap - and with simple geometries it can be calculated - https://physicstasks.eu/1535/force-acting-on-capacitor-plates. Force x distance = something  or other... You can listen to this on electrostatic speakers if you want.

For a constant magnetic field what you say is true. Likewise a transformer with DC over the primary windings has 0V over the secondary winding.

But if the magnetic field is changing, then you also have an electric field. How else would transformers work - the wires are not moving, yet energy is transferred?

As captured in the Faraday-Maxwell equation:

The Maxwell–Faraday equation states that a time-varying magnetic field always accompanies a spatially varying (also possibly time-varying), non-conservative electric field, and vice versa.

Connecting the battery causes a time-varying magnetic and electric field, and this transfers (a limited amount of) power across the 1m gap between wires.

[electrodacus]

There will be a potential difference across the switch (otherwise no current will flow when the switch is closed). And as you point out it is a small capacitor - different charges separated by distance. There is then a force across the gap - and with simple geometries it can be calculated - https://physicstasks.eu/1535/force-acting-on-capacitor-plates. Force x distance = something  or other... You can listen to this on electrostatic speakers if you want.

For a constant magnetic field what you say is true. Likewise a transformer with DC over the primary windings has 0V over the secondary winding.

But if the magnetic field is changing, then you also have an electric field. How else would transformers work - the wires are not moving, yet energy is transferred?

As captured in the Faraday-Maxwell equation:

The Maxwell–Faraday equation states that a time-varying magnetic field always accompanies a spatially varying (also possibly time-varying), non-conservative electric field, and vice versa.

Connecting the battery causes a time-varying magnetic and electric field, and this transfers (a limited amount of) power across the 1m gap between wires.

The potential difference is due to imbalance of charges and the electric field is the effect of that.

You mentioned electrostatic speakers but those same as electrostatic microphones/condenser microphones require a DC bias voltage / phantom voltage in order to work unlike dynamic microphones / speakers that generate energy by moving the coil installed on the membrane in a constant magnetic field.
(piezoelectric effect is a different story and plays no role here).

So you can not produce energy by moving the plates of a capacitor like example condenser microphone. You push the energy stored in the capacitor in and out in that DC bias (usually you measure the voltage drop on a resistor).
So you charge or discharge the capacitor and that phantom voltage / DC bias is there to cover any losses and keep the capacitor charged.

But even if there was a piezoelectric effect you will have energy generated just when there is motion so say air was piezoelectric (not the case) and when you moved the switch you produced some energy then as soon as you close the switch (relevant moment) no more energy is generated.

That is why I mentioned a few times that nothing moves in this circuit.

The transformer will not work with DC not that energy is transferred trough the field it is still transferred trough wires.
If there are no electrons to move in primary then you do not have any electron movement in secondary.
If you apply DC to a transformer you are using that coil as a heater. The transient at connection is not DC.


Yes the transient is due to electron wave that starts to form electrons just moving in random direction canceling out to starting to have a defined direction.
You are charging the line witch is a capacitor thus you have a current flow due to creating an electric field by moving electrons on the capacitor plate that results in an electron vacating the opposite plate and that happens at light speed so depends on the gap between capacitor plates.

[hamster_nz]

The transformer will not work with DC not that energy is transferred trough the field it is still transferred trough wires.
If there are no electrons to move in primary then you do not have any electron movement in secondary.

You might have missed a word in there somewhere.
Are you saying if I have 1A DC in the primary I will have current flowing in the secondary?
Or just that with no current in the primary windings there will be no current in the secondary?

The transient at connection [of a DC source to a transformer] is not DC.

Agreed - during the transient there is a changing magnetic field, and that changing field will causes a voltage across the transformer's secondary windings during the transient.

Once the magnetic field in the primary becomes stable then no more energy will be transferred to the secondary windings (apart from heat). That is until the DC source is disconnected. Then more as second pulse of energy will be transferred into the secondary (even though the DC source is disconnected!)

A resistor connected across the transformer's secondary windings will get (slightly) warmer during each of these transients - energy has been transferred from DC source to the resistor without them being electrically connected. That energy has definitely flowed between the two insulated wires that make up the transformer's winding (yes, through the insulation!).

The original Veritasium experiment has very long wires and so has a very long transient, during which a small amount of energy is transferred into the load through the space between the wires. But where along the wires this energy transfer is happening continuously changes depending on where the transient has got to. Only a tiniest initial amount is will be transferred directly over the 1m gap.

[SiliconWizard]

Only a tiniest initial amount is will be transferred directly over the 1m gap.

Yes. Which is basically what most of us had been considering all along, and that was later shown by a couple experiments. So, well, this is one of those topics that are kinda running in circles.

Has the whole thing shown that many people, including engineers, had misconceptions about "electricity" and that the model they frequently use, while working well enough in a large number of situations, is flawed? Yes. Has it really fully explained "how electricity works"? There are still some quirky corners there. I'm afraid that quite a few people, after having followed all this, will now understand that they had misconceptions, but will embrace a new concept that might itself lead to more misconceptions. Whichever is a better misconception among all those might not be so obvious in the end.

[electrodacus]

You might have missed a word in there somewhere.
Are you saying if I have 1A DC in the primary I will have current flowing in the secondary?
Or just that with no current in the primary windings there will be no current in the secondary?

No there will be no current in the secondary but there will be a current in the primary so basically a heating element.

Say you have a 600VA 120Vac 5A input transformer the output is irrelevant it can be anything as it will not be connected to anything.
So with primary connected to 120Vac and nothing on the output you maybe have a little bit of losses in the magnetic core so a bit of heat from there and the small associated IR losses in the primary copper.
With AC you just push energy in to the transformer to store as magnetic field in the ferromagnetic core but then that gets returned with just a bit of loss as the magnetic core is not perfect. So all energy is transmitted trough wires then converted to a magnetic field then due to ferromagnetic materials not being perfect you have some small loss then energy is converted back form magnetic to electric (electron flow) and sent back to your source so you have two way losses trough copper wires and a bit of loss in the ferromagnetic core.

You will sure not want to apply 120Vdc to that primary as you will then see that energy travels trough wires and not outside of them.

The transient at connection [of a DC source to a transformer] is not DC.

Agreed - during the transient there is a changing magnetic field, and that changing field will causes a voltage across the transformer's secondary windings during the transient.

Once the magnetic field in the primary becomes stable then no more energy will be transferred to the secondary windings (apart from heat). That is until the DC source is disconnected. Then more as second pulse of energy will be transferred into the secondary (even though the DC source is disconnected!)

A resistor connected across the transformer's secondary windings will get (slightly) warmer during each of these transients - energy has been transferred from DC source to the resistor without them being electrically connected. That energy has definitely flowed between the two insulated wires that make up the transformer's winding (yes, through the insulation!).

The original Veritasium experiment has very long wires and so has a very long transient, during which a small amount of energy is transferred into the load through the space between the wires. But where along the wires this energy transfer is happening continuously changes depending on where the transient has got to. Only a tiniest initial amount is will be transferred directly over the 1m gap.

Sorry I answered first part without reading this second part that shows you understand how a transformer works not quite sure why you do not realize that energy is delivered trough wires to the load.

[electrodacus]

Yes. Which is basically what most of us had been considering all along, and that was later shown by a couple experiments. So, well, this is one of those topics that are kinda running in circles.

Has the whole thing shown that many people, including engineers, had misconceptions about "electricity" and that the model they frequently use, while working well enough in a large number of situations, is flawed? Yes. Has it really fully explained "how electricity works"? There are still some quirky corners there. I'm afraid that quite a few people, after having followed all this, will now understand that they had misconceptions, but will embrace a new concept that might itself lead to more misconceptions. Whichever is a better misconception among all those might not be so obvious in the end.

I will say current models that engineers use are very accurate. Of course a transmission line modeled as finite number of LCR components will provide an approximation but even if you use analog computers you still need to read the result and that may be even less accurate due to measurement (reading) precision.
I still think that understanding what energy is and how energy storage is everywhere will be helpful for a better understanding.
Many including Derek just do not understand what energy is and it gets confused with all sorts of other things.

I guess I have the advantage that I work with energy generation and storage as a hobby.

[bsfeechannel]

I find it funny and at the same time annoying that these people who are reluctant to properly study math and physics, and end up struggling especially with electromagnetism, like to talk in the name of engineers, as if their misconceptions were the general mindset of our class so as to justify their position.

The hydraulic analogy, the origin of the energy in the wire idea, was dismissed right from the start by Maxwell himself in the 19th century. Derek only made this incontestable knowledge available to the masses.

[electrodacus]

I find it funny and at the same time annoying that these people who are reluctant to properly study math and physics, and end up struggling especially with electromagnetism, like to talk in the name of engineers, as if their misconceptions were the general mindset of our class so as to justify their position.

The hydraulic analogy, the origin of the energy in the wire idea, was dismissed right from the start by Maxwell himself in the 19th century. Derek only made this incontestable knowledge available to the masses.

Point to me where I have posted a wrong equation. All equations that I used to make the demonstration that energy travels trough wires and not outside the wires as Derek claims are used by engineers and physicist.

There are many posts as I try to refine/distill my explanation to the simplest form possible.

The two parallel capacitors are the simplest example I can give that demonstrate without any doubt that energy transfer is done trough wires and not outside of it.

[electrodacus]

I will try to make a resume of my demonstration that energy travels trough wires and not outside.



1) We have a capacitor on the right that is 1F and charged at 3V and one capacitor on the right also 1F but with no potential across 0V
Before the switch is closed there is a constant electric field inside the charged capacitor but there is no transfer of energy from the charged capacitor to the discharged capacitor.
By shorting the discharged capacitor at the beginning of the experiment we can ensure that there is no energy stored in that capacitor on the right and if the switch has any capacity then that switch capacity is at the same V_i potential as the capacitor on the left.

This proves that the constant electric field inside the charged capacitor can not transfer the energy to the discharged capacitor as long as the switch is open and alone should be sufficient proof that for energy to flow it requires a close loop.

2) If capacitors are 1m apart and the switch is 10m from both capacitors the time it takes from the moment the switch is closed to the moment any amount of energy is transferred to the discharged capacitor is the time it takes the electron wave to travel 10m.

This again is a proof that energy transfer from the source (charged capacitor) to the discharged capacitor travels trough wires.

3) The initial energy in this isolated system is:
0.5 * 1F * 3V² = 4.5Ws  (I prefer Ws as opposed to Joules as a unit but they are the same thing).

With a real circuit where resistance is higher than zero the voltage after the switch was closed and circuit reached steady state will be:
0.5 * 1F * 1.5V² = 1.125Ws  and the same amount in the other capacitor so total 2.25W

So only half of the initial energy is now found in the system the other half was dissipated as heat inside the wires and this can be verified with a thermal camera.

This is again another proof that energy has traveled inside the wire else it will not make sense to have lost half of the initial energy as heat and also see the wires heat up by the exact amount of energy missing.


I may add more points but I think this 3 points should be enough conclusive evidence that energy flows trough wires and not outside.
I'm open to any criticism just mention the point out of this 3 you do not agree with and what I did wrong.

[EEVblog]

2) If capacitors are 1m apart and the switch is 10m from both capacitors the time it takes from the moment the switch is closed to the moment any amount of energy is transferred to the discharged capacitor is the time it takes the electron wave to travel 10m.
This again is a proof that energy transfer from the source (charged capacitor) to the discharged capacitor travels trough wires.

If you put the switch at the end of the line then yes it will take the delay of the length of the line to start charging, exactly the same as Derek's original question would if the switch was at the end of the line. No difference at all, so that doesn't prove in any way that the energy flows inside the wire.

[electrodacus]

2) If capacitors are 1m apart and the switch is 10m from both capacitors the time it takes from the moment the switch is closed to the moment any amount of energy is transferred to the discharged capacitor is the time it takes the electron wave to travel 10m.
This again is a proof that energy transfer from the source (charged capacitor) to the discharged capacitor travels trough wires.

If you put the switch at the end of the line then yes it will take the delay of the length of the line to start charging, exactly the same as Derek's original question would if the switch was at the end of the line. No difference at all, so that doesn't prove in any way that the energy flows inside the wire.

Yes it proves that since energy is not flowing from the switch to the load but from the source to the load. The position of the switch should not be relevant if the energy from the charged capacitor will not need to travel trough wire.
If the switch is closer to charged capacitor / source then the wire after the switch forms another capacitor with the wire of the load thus effectively the small current Derek's is seeing is the one needed to charge that extra capacitor (the transmission line).

With the switch far away you also have a capacitor made by the transmission wires (that is fully charged before starting the test) and what the switch is doing is shorting that capacitor but from the far end of the capacitor thus electron wave that move the energy trough wires needs time to travel that distance.
 

[snarkysparky]

Re Energy not transferred in the wires.

1   The light bulb will not light without the wires.  Not even with the tiniest part of wire snipped out.

2   Any material not directly touching the wires may surround the wires and not affect the light bulb function in any way.

Energy is flowing in the wires.

[TimFox]

Alternative wording:  current flows through the wires into the light bulb.
The current flowing through the light bulb heats the filament to the point where it emits energy in the form of light and heat radiation.