Veritasium "How Electricity Actually Works"

[SandyCox]

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.

What happens when the electric field isn't conservative? Does current always involve electrons?

[electrodacus]

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?

No electrons and thus no energy will travel that 1m gap.
When you have two capacitors in series connected to a supply each capacitor will have a plate connected to supply one to the positive and the other to the negative terminal of the supply.
Those plates will act as if it was a single capacitor then the other two plates that are connected with a wire between them will always have the same number of electrons when discharged and when charged is just that when charged electrons from one plate travels through the wire to the other plate as capacitors are charged.
If you have a lamp there in series there will just be extra energy wasted as capacitors charge since is like wire has higher resistance to electron flow.

Seen from the outside as a black box with just two wires the two capacitors in series will look like a single capacitor with lower capacity.
There is no difference between one capacitor or two or 3 capacitors in series.
From outside you will see that the amount of energy going in to black box will be equal with the amount going out plus the energy lost as heat on the DC ESR and the DC ESR includes the wire connecting the two capacitors in series inside the black box. If you add a lamp in that black box that will just further contribute to increase in the DC ESR and you will not be able to know form the outside that there is a lamp in the box or if there are two capacitors in series or just one.
You will not be able to know that from outside of the black box because no matter what measurements you make there will be no difference from a single capacitor or multiple in series and or lamps or resistors added to that in series.

[electrodacus]

What happens when the electric field isn't conservative? Does current always involve electrons?

In an isolated system the field will be conservative. Electric current will always require some sort of charged particle either electrons or ions. In a wire the electrons will be the ones that move. In a battery the ions will move from one plate to another.

So to transport electrical energy from one plate to another you need electron flow usually in a wire but with very high voltages the electrons can also travel through air but not the case here with 20V and 1m of air.
So electric and magnetic field are conservative in Derek's experiment.

[PlainName]

No electrons and thus no energy will travel that 1m gap.
When you have two capacitors in series connected to a supply each capacitor will

Hey, please try not to introduce dead cats. Just stick with the ultra-simple circuit we were discussing: PSU->cap->res->return.

So, if no energy is passed from one side to the other, what is the resistor burning?

Hang on... O. M. G.... Maybe you are converting to aetherwind?

Joking aside, explain how the resistor can show a voltage across it and sink some current. Where is that coming from?

[snarkysparky]

""Electric current will always require some sort of charged particle either electrons or ions""

Yes but the charged particles need not "transfer" to transmit energy.

[electrodacus]

Hey, please try not to introduce dead cats. Just stick with the ultra-simple circuit we were discussing: PSU->cap->res->return.

So, if no energy is passed from one side to the other, what is the resistor burning?

Hang on... O. M. G.... Maybe you are converting to aetherwind?

Joking aside, explain how the resistor can show a voltage across it and sink some current. Where is that coming from?

No energy passes from one side of the capacitor to the other. Energy is being stored in the capacitor.
Current will from from the source in to capacitor through wires and obviously through the resistor that is also a wire.
Current will decrease as the capacitor is charged until no current flows as capacitor is full.

Say capacitor when discharged start with 1000 free electrons on one plate and 1000 electrons on the other plate.
When battery is connected to capacitor if you connect just the positive nothing will happen same if you connect just the negative.
When battery is connected to capacitor on both sides electrons will flow from battery in to the wire and when that happen there will be an electron wave that will push out electrons on the other side of the wire in to the capacitor plate while at the same time electrons from the other plate will get out of the capacitor plate in to the wire and that will push electrons in to battery.
At the end when capacitor potential equal battery potential current flow stops.
At this point you may have 1100 electrons on one plate and 900 electrons on the opposite plate.
The 100 extra electrons are due to battery pushing those through the wire and the deficit of electrons on the other plate when trough wire in to battery.
No electrons have traveled through the capacitor dielectric.
Now the capacitor contains energy that can be used to do work. So you will say energy flowed in the capacitor where it is stored not through capacitor.

[electrodacus]

""Electric current will always require some sort of charged particle either electrons or ions""

Yes but the charged particles need not "transfer" to transmit energy.

What do you mean by "transfer"
Charged particles as electrons repel each other thus if you have a region with higher electron density it will want to migrate to a region with lower density.
Since there is not enough energy for them to travel through air even for a few mm distance between switch contacts the contacts need to be close enough so that 20V in this case has enough energy to push an electron from the negatively charged region in to the region with deficit of electrons.

So what you likely call "transfer" has to do with repelling forces that negative charged particles feel relative to one another. That is not high enough at 20V to have electrons fly from the wire trough 1m of air to the other wire and for that you will need way higher voltage about 3400000V

[IanB]

No energy passes from one side of the capacitor to the other. Energy is being stored in the capacitor.

Let's stop bringing up the energy stored in the capacitor. Everyone knows energy is stored in capacitors. Everyone agrees energy is stored in capacitors. Nobody has ever denied that energy is stored in capacitors. Stop acting like this a big mystery.

If we look at an energy balance on the system, we have:

1. The energy leaving the source (battery) is E_S
2. The energy stored in the capacitor is E_C
3. The energy consumed (dissipated) by the load resistor/LED is E_L

Since you tell us that energy is always conserved, we know that  E_S = E_C + E_L

Clearly some energy from E_S ended up as E_L, which means some energy was transferred from the battery to the load, and some energy was stored in the capacitor.

[snarkysparky]

Lets take the issue to the instantaneous case.

So the usual battery ,  switch,  capacitor,  resistor    all in series.

Lets say batter V  = 12V

Capacitor is 1000 uf.

Resistor is 10 ohms.

Initially capacitor is discharged.

At the instant the switch is closed we have.

I = 120 ma
Resistor is dissipating power of  12*12/ 10   = 14.4 watts.

capacitor energy = ce =  0.5C * Vc^2

diff that by time

Dce/Dt  =  Vc 

Vc  is zero at the beginning.    Capacitor is storing No energy.

In this instantaneous point would you say that the 14.4 watts the resistor is dissipating is traveling "through" the capacitor??
If not how is it getting there?

 

[electrodacus]

Lets take the issue to the instantaneous case.

So the usual battery ,  switch,  capacitor,  resistor    all in series.

Lets say batter V  = 12V

Capacitor is 1000 uf.

Resistor is 10 ohms.

Initially capacitor is discharged.

At the instant the switch is closed we have.

I = 120 ma
Resistor is dissipating power of  12*12/ 10   = 14.4 watts.

capacitor energy = ce =  0.5C * Vc^2

diff that by time

Dce/Dt  =  Vc 

Vc  is zero at the beginning.    Capacitor is storing No energy.

In this instantaneous point would you say that the 14.4 watts the resistor is dissipating is traveling "through" the capacitor??
If not how is it getting there?

As soon as an electron passes through the resistor an electron will also end up in the capacitor so voltage across the capacitor is no longer zero (it will be super close to zero but not zero).
There are capacitors everywhere not just that huge by comparison 1000uF capacitor. The wires and even resistor will for a capacitor with the return wire.
So if 1 million electrons (absolute random number but can be calculated exactly) end up stored in the capacitor then that number has passed through wires and the resistor witch is also a wire.

[electrodacus]

Let's stop bringing up the energy stored in the capacitor. Everyone knows energy is stored in capacitors. Everyone agrees energy is stored in capacitors. Nobody has ever denied that energy is stored in capacitors. Stop acting like this a big mystery.

If we look at an energy balance on the system, we have:

1. The energy leaving the source (battery) is E_S
2. The energy stored in the capacitor is E_C
3. The energy consumed (dissipated) by the load resistor/LED is E_L

Since you tell us that energy is always conserved, we know that  E_S = E_C + E_L

Clearly some energy from E_S ended up as E_L, which means some energy was transferred from the battery to the load, and some energy was stored in the capacitor.

You claim that everyone understand that energy is stored in the capacitor but if that was to be true there will have been no discussion about energy passing through capacitor. Including you based on this post you just made.

And is not me that is saying energy is conserved that is a fact that has never been disproven.

Yes energy was transferred only through wires and some of that energy ended up as lost energy in the wires and load resistor that is also a wire just with higher resistance.

[PlainName]

As soon as an electron passes through the resistor an electron will also end up in the capacitor

What causes that (and all the other) electrons to pass through the resistor just when voltage is applied to the capacitor and not at other times?

[electrodacus]

As soon as an electron passes through the resistor an electron will also end up in the capacitor

What causes that (and all the other) electrons to pass through the resistor just when voltage is applied to the capacitor and not at other times?

Like I mentioned many times before. Electrons are charged particles with negative charge so they repel each other.
You can say they are under pressure if you compare this with gas molecules.

Say a discharged capacitor that is basically an energy storage device has 500 free electrons on both plates that means 0V so discharged capacitor and then say capacitor is charged so you have 750 electrons on one side and 250 on the other side. What will happen if capacitor terminal are shorted ?
The wire shorting also has free electrons so the electrons on the plate with 750 electrons when in contact with the wire (closed loop) will want to go to the lowest energy state so electrons will be pushed out of that plate in to the wire and the other end of the wire will release an electron in the capacitor plate that has a deficit thus 749 and 251 will be the total after the first electron did the work.
In the end plates will get back to 500 and 500 electrons so no more stored energy and all that stored energy will end up as heat in the wire (wire temperature will increase with the exact amount that energy stored contained).

[PlainName]

OK, forget it. It's a lost cause and I am out.

[snarkysparky]

Energy is conserved.  At the instant the switch is closed the battery is providing 14.4 watts of power.

That 14.4 watts need to be accounted for everywhere.  Assuming zero resistance wires.   So  for the capacitor we have Joules / second  = watts and same for the resistor.

The capacitor is gaining zero energy at the instant the switch is closed.   

The capacitor IS passing energy through it.   Suppose the capacitor was infinite in capacitance.  No energy would ever be stored in it but load current could flow into one plate an out of the other plate with no transfer across the dielectric.

Maybe we are saying the same thing.

What is happening is that the for the plates of the capacitor to "charge" relative to each other current must flow into and out of the plates but NOT across the dielectric barrier.

And instantaneously the energy balance at the start is  that no energy is being stored in the capacitor.  And somehow energy shows up at the resistor.

[electrodacus]

OK, you're on.

Here is the waveform and the circuit that created it. You may have seen it before.

Now, the green line is the voltage applied to the circuit. Clearly, it is the cause of red line, which is the voltage across the resistor. (Or, alternatively, the voltage over the resistor - shown by the red line - is caused by the voltage applied to the circuit.)

There's the proof. Now, perhaps you can say just how that happens? How does the one cause the other?

You forgot to draw the source that is also connected to GND. Maybe that is what confuses you.
electrons from the source will flow in to resistor and in to the capacitor plate while on the other side that other capacitor plate will send the exact same number of electrons from that plate to the source.
Resistor is nothing else than a wire helping to complete the circuit. You need a closed loop to charge that capacitor and maybe because that is not seen in your schematic you are confused about how the energy is stored in to capacitor.

You also understand that red waveform also represents the current through the resistor and as you can see it will fast trend to zero meaning once the capacitor plate has enough electrons on one side with similar deficit on the other the capacitor is fully charged and at same voltage as the source thus no more energy transfer from the source to capacitor.

[electrodacus]

Energy is conserved.  At the instant the switch is closed the battery is providing 14.4 watts of power.

That 14.4 watts need to be accounted for everywhere.  Assuming zero resistance wires.   So  for the capacitor we have Joules / second  = watts and same for the resistor.

The capacitor is gaining zero energy at the instant the switch is closed.   

The capacitor IS passing energy through it.   Suppose the capacitor was infinite in capacitance.  No energy would ever be stored in it but load current could flow into one plate an out of the other plate with no transfer across the dielectric.

Maybe we are saying the same thing.

What is happening is that the for the plates of the capacitor to "charge" relative to each other current must flow into and out of the plates but NOT across the dielectric barrier.

And instantaneously the energy balance at the start is  that no energy is being stored in the capacitor.  And somehow energy shows up at the resistor.

Exact moment that switch is closed the electron from the tip of the switch will jump to the other side on the wire so there will be no power through the resistor and no energy in the capacitor.
You will also need to consider the energy stored in wire inductance as wire has some inductance that opposes electron flow converting that in to magnetic field around the conductor.
At about the speed of light there will be an electron passing through resistor and arriving in the capacitor at witch point you can no longer say there is no energy delivered to capacitor.
Yes the initial part is super inefficient as say capacitor is at 0.001V and battery 12V that means 11.999V drops across the resistor and that means terrible charging efficiency with a small fraction of a percent efficiency and it will take the capacitor to get to 6V for the efficiency to get to 50% still horrible but not as crazy.
Even if you do not have the capacitor there just the resistor with just one terminal connect to battery the other terminal just  not connected to anything and then on the other side of the battery a switch and a piece of wire again that wire not connected to anything due to the capacitance between the resistor and that not connected wire after the switch you will have some capacitance maybe a few pF but still enough so that when you close the switch energy will flow through the resistor for a very short period of time. 

[electrodacus]

OK, forget it. It's a lost cause and I am out.

It is a lost cause for you as you are the one that has no understanding of the subject.

[hamster_nz]

OK, forget it. It's a lost cause and I am out.

It is a lost cause for you as you are the one that has no understanding of the subject.

As a rough rule, would you say that capacitors allow voltage & current transients to pass through, and but do not pass steady voltages and currents?  (deciding what is a transient and what is a steady voltage can be computed based on the Rs Cs & Ls in the circuit)

So at DC they have infinite resistance, and with signals of high enough frequency they offer very, very little resistance?

[TimFox]

The elementary features I was taught in the first week of electronics 101 were that you can't change the voltage across a capacitor instantaneously, and you can't change the current through an inductor instantaneously.
Analysis of most simple AC and pulse circuits can start from this, and then introduce the finite resistances and time constants.

[PlainName]

OK, forget it. It's a lost cause and I am out.

It is a lost cause for you as you are the one that has no understanding of the subject.

Yes, my lost cause is getting you to a) understand the question and b) answer that without going off on a tangent to confuse things.

Tell you what, just say which of these facts you think are not actually facts:

1. The supply provides energy to the circuit.

2. The resistor consumes energy in the circuit.

Now, everything and anything between those to is completely irrelevant. It doesn't matter if there are electrons, electons, pixies, BFO Tesla(tm) batteries, whatever. None of that matter a jot to those two facts.

3. There's a 1m air gap in that black box between the supply and resistor.

OK? Which of 1-3 do you think is not a fact?

[electrodacus]

As a rough rule, would you say that capacitors allow voltage & current transients to pass through, and but do not pass steady voltages and currents?  (deciding what is a transient and what is a steady voltage can be computed based on the Rs Cs & Ls in the circuit)

So at DC they have infinite resistance, and with signals of high enough frequency they offer very, very little resistance?

No it will be wrong to say that electrical current passes through a capacitor even as transient. What you think of as transient current passing through is actually current going in not passing trough.
With say AC you have current going in as capacitor is charged and going out as capacitor is discharged.
At no point there is any current going trough the capacitor thus no energy goes through capacitor.

[electrodacus]

The elementary features I was taught in the first week of electronics 101 were that you can't change the voltage across a capacitor instantaneously, and you can't change the current through an inductor instantaneously.
Analysis of most simple AC and pulse circuits can start from this, and then introduce the finite resistances and time constants.

Yes that will be basic introduction to have some roule to memorize. But the reason voltage cannot change instantaneously across a capacitor is because energy will be going in the capacitor and stored.
Same thing with inductor where energy is also stored.
All energy that went in except for the amount that was lost in the process as heat will come out of this energy storage devices.
So there is no electrical enenergy that pases trough a capacitor.
Electrical energy is the integral of electrical power over time.  To have power you need to have a current different from zero and you can not have current through a dielectric like plastic or air.   

[electrodacus]

Yes, my lost cause is getting you to a) understand the question and b) answer that without going off on a tangent to confuse things.

Tell you what, just say which of these facts you think are not actually facts:

1. The supply provides energy to the circuit.

2. The resistor consumes energy in the circuit.

Now, everything and anything between those to is completely irrelevant. It doesn't matter if there are electrons, electons, pixies, BFO Tesla(tm) batteries, whatever. None of that matter a jot to those two facts.

3. There's a 1m air gap in that black box between the supply and resistor.

OK? Which of 1-3 do you think is not a fact?

What you call me going on a tangent is essential in understanding what happens. Energy going trough a capacitor will involve electrons traveling through an electrical insulator like air and that is not the case.

1. Yes.
2. Yes.
3. OK

This part is wrong:
"Now, everything and anything between those to is completely irrelevant. It doesn't matter if there are electrons, electons, pixies, BFO Tesla(tm) batteries, whatever. None of that matter a jot to those two facts.
"

It does matter if there are electrons traveling through that 1m air gap as if they where then you and Derek will be right in saying that energy travels outside the wire but for that you need as I mentioned almost three and a half million volt so that air becomes a conductor.

[PlainName]

This part is wrong:
"Now, everything and anything between those to is completely irrelevant. It doesn't matter if there are electrons, electons, pixies, BFO Tesla(tm) batteries, whatever. None of that matter a jot to those two facts.
"

It does matter if there are electrons traveling through that 1m air gap as if they where then you and Derek will be right in saying that energy travels outside the wire but for that you need as I mentioned almost three and a half million volt so that air becomes a conductor.

No, it doesn't matter. We are not interested in the how at this point, only whether it does or not. If there are electrons magically jumping the gap that's fine. It's also fine if they don't. All we need to agree on is that energy is provided on one side, consumed on the other, and there is significant clear air between.

Are you happy agreed with that?