Veritasium "How Electricity Actually Works"

[electrodacus]

Demonstrations such as placing an ammeter on both legs of a capacitor as it charges, showing the same current flowing both terminals, at the same time are complete fabrications...


Unless it has to be the same electron flowing out as flowed in - in which case long wires are out due to the slow drift velocity. On a 1m x 2mm diameter copper wire,  carrying 1A DC , it will take on average about 12 hours for an electron to travel down that wire.

It is absolutely not the same electron not even the same electron after some delay that you find living the other plate of the capacitor.
On wires is not the electron that flows into the wire that gets on the other end at the speed of light but the electron that enters the wire will create a wave that pushes out an electron on the other end after about the time light needs to travel that distance.
Electron drift speed is a different subject.

So when you charge a capacitor from a battery none of the electrons entering one plate will be found on the other side.
The electrons leaving from the other plate where already on that plate and they leave in order for that plate to become positively charged.

[electrodacus]

Something goes through to use up energy across that resistor (or lamp, or anything else that consumes power). Where does it come from? It is energy of some form, isn't it?

No energy will not go through but in to capacitor.
Those electrons that travel through the resistor (basically a wire) where never in the battery or the wire connected on the other side of the capacitor.

The analogy I can make is a compressed air cylinder that is split in two by an rubbery elastic membrane.
Discharged cylinder means the membrane is relaxed in the middle then when you push more air molecules in one of the sides same amount of air molecules will leave from the other side but they will not be the same as that elastic membrane (dielectric) will prevent any air molecules passing from one half of the cylinder to the other.
Electrons can not pass through dielectric so no current no energy flows through.
As I mentioned before a switch is also a capacitor where close circuit it means you shorted the two plates. As long as switch is open so just a capacitor no energy will flow through it.

[PlainName]

Something goes through to use up energy across that resistor (or lamp, or anything else that consumes power). Where does it come from? It is energy of some form, isn't it?

No energy will not go through but in to capacitor.

Well, hang on a minute. There is 4mA going through that resistor, enough to light up an LED. Energy must be involved there, no? Where did that come from?

The analogy I can make is a compressed air cylinder that is split in two by an rubbery elastic membrane.

In that case the air on one side pushes against the membrane and that pushes against whatever is on the other side. You've transferred energy from the pushing side to the pushed side, haven't you?

[electrodacus]

Well, hang on a minute. There is 4mA going through that resistor, enough to light up an LED. Energy must be involved there, no? Where did that come from?

In that case the air on one side pushes against the membrane and that pushes against whatever is on the other side. You've transferred energy from the pushing side to the pushed side, haven't you?

Yes those 4mA are due to electrons leaving that plate of the capacitor (the discharged capacitor has equal amount of free electrons on each plate) so when you charge a capacitor you push electrons on to one plate while the same amount of electrons leave the other plate same as in that compressed air tube analogy air molecules enters one side and others are leaving from the other side.

So a discharged compressed air cylinder with two partitions will have air in it say 500 molecules one one side and 500 molecules on the other side.
Say you charged this discharged cylinder from another one that is charged but 10x larger that has 8000 in one side and 2000 on the other side.
Now with two hoses you connect the large one to the small one to charge it
You will get almost 800 on one side and more than 200 on the other side (to lazy to do the exact calculation) then you are left with about 7700 on the large one and about 2300 on the other side. You get the idea the pressure (voltage) will equalize.
After pressure has equalized there is no flow so no energy transfer.  You can add some turbine (lamp) in between the two and it will work during transfer but stop when pressure is equalized.
As any analogy you can not go to far and still be correct but for the limited purpose of explaining why energy flows in or out of capacitors but not through should be good enough.   

[hamster_nz]

Something goes through to use up energy across that resistor (or lamp, or anything else that consumes power). Where does it come from? It is energy of some form, isn't it?

No energy will not go through but in to capacitor.

Well, hang on a minute. There is 4mA going through that resistor, enough to light up an LED. Energy must be involved there, no? Where did that come from?

The analogy I can make is a compressed air cylinder that is split in two by an rubbery elastic membrane.

In that case the air on one side pushes against the membrane and that pushes against whatever is on the other side. You've transferred energy from the pushing side to the pushed side, haven't you?

Careful, you may encourage them to go full Humpty Dumpty mode on you:

'When I use a word,' Humpty Dumpty said, in rather a scornful tone, 'it means just what I choose it to mean — neither more nor less.'

[PlainName]

Well, hang on a minute. There is 4mA going through that resistor, enough to light up an LED. Energy must be involved there, no? Where did that come from?

In that case the air on one side pushes against the membrane and that pushes against whatever is on the other side. You've transferred energy from the pushing side to the pushed side, haven't you?

Yes those 4mA are due to electrons leaving that plate of the capacitor (the discharged capacitor has equal amount of free electrons on each plate) so when you charge a capacitor you push electrons on to one plate while the same amount of electrons leave the other plate same as in that compressed air tube analogy air molecules enters one side and others are leaving from the other side.

OK, I guess we are on the same page, then. The energy going into the capacitor on the plus side is pushing energy out of the capacitor on the negative side, right? In a simplistic manner of speaking.

So, back to our capacitor with 1m of free air in the middle... there is a net energy transfer from one side to the other. The PSU is pushing some energy into the cap which is consequently pushing the same amount of energy out to the resistor which is consuming it (aka converting to heat/light). The PSU has lost it, the resistor has used it, yes?

[SandyCox]

Please look at Equation 18. Zo is real. It is a resistance and has no imaginary part. Note that the two transmission lines and the resistor form a resistive divider. Let's assume that the line is perfectly matched, i.e. R = 2Zo. Before the first reflection arrives back, 3/4 of the energy delivered by the battery is sorted in the two transmission lines. 1/4 is dissipated in the resistor. No energy is transferred from the transmission line to the resistor.

[electrodacus]

Not sure why I did not get notifications but likely google AI trying to protect me from wasting more time
In any case you all seems to completely ignore energy storage is like that does not even exist for you.
It all gets down to you thinking energy pases trough a capacitor that will mean current passing through a capacitor instead on in to a capacitor as that energy is stored energy doing no work.
The wires connecting the capacitor to the supply will have a loss and you can add resistor or lamps in series with those and have even more loss but all the energy that will flow is to charge the capacitor not going through capacitor.
That is why with DC current will only flow for a short time as much as it is needed to charge the capacitor. If there is any current after that it will be super small and is the leakage current for that capacitor that is the equivalent for a high value resistor in parallel with the capacitor and has nothing to do with the results measured by Derek as that leakage current is so small it will not register on a 8 or even 12bit oscilloscope.

The definition of an electrical current is "a stream of charged particles like electrons or ions moving through an electrical conductor".
There are no electrons let alone ions traveling through that 1m of air between the lines.
If there is no electrical current then there is no electrical power and thus no energy traveling outside the wire.

[PlainName]

It all gets down to you thinking energy pases trough a capacitor...

So you're agreed that energy does indeed pass through a capacitor? That 1m gap without any wires is actually allowing energy transfer through it?

[electrodacus]

It all gets down to you thinking energy pases trough a capacitor...

So you're agreed that energy does indeed pass through a capacitor? That 1m gap without any wires is actually allowing energy transfer through it?

No electrical energy is transferred through that 1m gap.
There are no electrons flying from one wire to the other wire that is 1m apart.

[PlainName]

It all gets down to you thinking energy pases trough a capacitor...

So you're agreed that energy does indeed pass through a capacitor? That 1m gap without any wires is actually allowing energy transfer through it?

No electrical energy is transferred through that 1m gap.
There are no electrons flying from one wire to the other wire that is 1m apart.

Then what is the resistor burning up?

[electrodacus]

Then what is the resistor burning up?

The resistor is in series with the capacitor so is basically a high resistance wire connecting the capacitor to the source.
There are capacitors on both sides in Derek's experiment but it is no different is just electrons moving from plate of one capacitor to plate on the other capacitor so still a high resistance wire connecting the two capacitors in series.
 
Do you agree that no electrons travels through the 1m space from one wire to the other?
If you do then you must understand that no electrical energy travels through that 1m gap.

[IanB]

If you do then you must understand that no electrical energy travels through that 1m gap.

If no electrical energy travels through the gap, what kind of energy does travel through the gap? What name will you give it?

[hamster_nz]

Then what is the resistor burning up?

The resistor is in series with the capacitor so is basically a high resistance wire connecting the capacitor to the source.
There are capacitors on both sides in Derek's experiment but it is no different is just electrons moving from plate of one capacitor to plate on the other capacitor so still a high resistance wire connecting the two capacitors in series.
 
Do you agree that no electrons travels through the 1m space from one wire to the other?
If you do then you must understand that no electrical energy travels through that 1m gap.

I agree that no electrons travel through the 1m space from one wire to the other, but it does not follow that no electrical energy travels through that 1m gap.

I am sure you agree that no electrons will pass through the two capacitors either side of the LED+Resistor. However, electrical energy does pass through the capacitors, and it does light the LED.  That energy is coming from the DC supply!



And as pointed out before, I can replace the LED+R with a third capacitor in series, and charge the middle capacitor from the DC supply, remove it form the circuit and measure it. It does have charge (and therefore energy), even though no electrons have moved through any of the capacitors.

[electrodacus]

If you do then you must understand that no electrical energy travels through that 1m gap.

If no electrical energy travels through the gap, what kind of energy does travel through the gap? What name will you give it?

If you ignore a few infrared photons then there is no energy that travels through that gap.
All energy travels through wires.

Here is a good video about capacitors. He should have mentioned that the initial positive charge plate that he starts with is charged relative to ground

[electrodacus]

I agree that no electrons travel through the 1m space from one wire to the other, but it does not follow that no electrical energy travels through that 1m gap.

I am sure you agree that no electrons will pass through the two capacitors either side of the LED+Resistor. However, electrical energy does pass through the capacitors, and it does light the LED.  That energy is coming from the DC supply!


And as pointed out before, I can replace the LED+R with a third capacitor in series, and charge the middle capacitor from the DC supply, remove it form the circuit and measure it. It does have charge (and therfore energy), even though no electrons have moved through any of the capacitors.

No enenrgy pases trough capacitors.
What you see as work done on the LED is the result of electrons moving from the plate of first capacitor to a plate on the second capacitor as the two capacitors in series are charged from the supply.
If you disconnect the battery and then short the wires that were before connected to battery you are discharging the capacitors so electrons will flow back between the two plates making them neutral again.

[IanB]

No enenrgy pases trough capacitors.
What you see as work done on the LED

Since it takes energy to do work, and the energy did not pass through the capacitors, how did it get from the battery to the LED?

[electrodacus]

Since it takes energy to do work, and the energy did not pass through the capacitors, how did it get from the battery to the LED?

The LED is in series with the capacitor/capacitors so that is wasted energy during charging process.
Is the same as waste energy in the wires as they carry the energy from the source to the capacitors (energy storage device) you just add an LED and thus extra loss to the process of charging the capacitor.

Electrical energy travels through wires and travels through LED/lamp/resistor but does not travels through capacitor but in capacitor.

[PlainName]

There are capacitors on both sides in Derek's experiment

Let's not complicate matters unnecessarily.

There is a power source, cap, resistor, return. That's it. We've decided the cap is the two wires 1m apart, and we've gone with your values for it. We know the resistor is converting energy to heat.

The only issue is that there is the 1m gap without wires in that circuit. What, exactly, is passing from the power supply side of the capacitor to the other side that lets the resistor heat up? It must be energy, mustn't it, since that's what everything ultimately reduces to.

So, if energy is not being transferred across that gap, what IS going across? That resistor isn't using nothing to heat up, you know.

[electrodacus]

There are capacitors on both sides in Derek's experiment

Let's not complicate matters unnecessarily.

There is a power source, cap, resistor, return. That's it. We've decided the cap is the two wires 1m apart, and we've gone with your values for it. We know the resistor is converting energy to heat.

The only issue is that there is the 1m gap without wires in that circuit. What, exactly, is passing from the power supply side of the capacitor to the other side that lets the resistor heat up? It must be energy, mustn't it, since that's what everything ultimately reduces to.

So, if energy is not being transferred across that gap, what IS going across? That resistor isn't using nothing to heat up, you know.

Electrons from the source flow through wires in to one plate of the capacitor. Same number of free electrons leave the plate on the other side of the capacitor and return to the source.
All energy flows through wires in to the plates that are also wires and no energy flows through the capacitor dielectric in this case 1m of air.
The resistor is nothing more than a wire with higher resistance so energy flows through it same as it flows through wires.
if source is disconnected this charge imbalance remains there and if you short this wires then electrons from one plate can travel to the other plate that has a deficit of electrons in order for both plates to become neutral so you are discharging the capacitor and current will flow in reverse through wires.

Electrons (charged particles) are the ones that carry the energy and they do so through wires.
While air even 1m of it can become conductive the potential (voltage) will need to be way higher than 20V so no energy was transferred outside the wires in Derek's experiment. 

[IanB]

So let's stop talking about energy and start talking about power.

The battery is a source of power. The LED/resistor is a consumer of power. The LED lights up, so power was transferred from the battery to the LED. The capacitor is between the battery and the LED. What kind of power went through the capacitor on its journey from the battery to the LED?

[electrodacus]

So let's stop talking about energy and start talking about power.

The battery is a source of power. The LED/resistor is a consumer of power. The LED lights up, so power was transferred from the battery to the LED. The capacitor is between the battery and the LED. What kind of power went through the capacitor on its journey from the battery to the LED?

Why it is hard to understand the concept of energy storage.
There is nothing going "through" capacitor but in to capacitor.

You need wires to transfer energy from the source to the capacitor and if you also add an LED to the wires it makes no much difference other than reduce the speed at with the energy is transferred to the capacitor.

Is sort of like you want to believe that electrical energy can pass through in order to deny that energy storage exist.
There was a good reason I replaced the energy source with a charged capacitor as I was thinking one way or the other you can understand that capacitor is an energy storage device.

Electrical power is product of electrical current and electrical potential. I think is clear that you can not have electric current flow through air at least not with 20V and 1m of distance so there can not be any power.

You only have an electric field inside the capacitor once you have an imbalance of charge and not the other way around.
All this are well known facts but thanks in part to Derek facts are no longer relevant if you can make absurd claims like "energy doesn't flow in wires" and back them with no evidence. Show a waveform that you do not understand and say here is the proof.

His channel will more appropriately be named mendacium even if he's intent is veritas.   

[hamster_nz]

Show a waveform that you do not understand and say here is the proof.

Walking away from the bench with the middle capacitor of three in series, charged with stored energy is all the proof you should need.

If energy was only able to flow in wires, and capacitors only stored energy and prevented it from passing through, then that would be impossible.

[electrodacus]

Walking away from the bench with the middle capacitor of three in series, charged with stored energy is all the proof you should need.

If energy was only able to flow in wires, and capacitors only stored energy and prevented it from passing through, then that would be impossible.

Proof for what ?
I explained that electrons from the plate of once capacitor will be pushed in the other capacitor plate through wire when you have multiple capacitors in series.

[PlainName]

I explained that electrons from the plate of once capacitor will be pushed in the other capacitor plate

Across 1m of air? What is it that the plus side can cause the negative side to rise to the same voltage and put out some current?