Veritasium "How Electricity Actually Works"

[electrodacus]

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

Where is the confusion for you?

Do you understand the difference between energy that was stored and energy that did work ?
The process of storing energy in capacitor was in this case 50% efficient meaning that the 50% that was lost as heat on the wire is gone and the 50% stored in the capacitor is just that stored energy it did not do any work but it can be used to do work.
If instead of capacitor you will have had an electric motor that use 72mJ to move a weight from point A to point B then that will have been energy that has done work and it will have been gone no more 72mJ for you to use as you want.

If someone can not understand the difference between stored energy and energy that did work then I can not call him or her anything other than confused.

[bsfeechannel]

If someone can not understand the difference between stored energy and energy that did work then I can not call him or her anything other than confused.

You seem to lack basic high-school knowledge of physics and electricity. That's why you don't understand what these people are talking about. Then you project your confusion on them, saying they are wrong.

Maybe it's time to revisit the basic concepts and try to understand what they're saying.

[electrodacus]

You seem to lack basic high-school knowledge of physics and electricity. That's why you don't understand what these people are talking about. Then you project your confusion on them, saying they are wrong.

Maybe it's time to revisit the basic concepts and try to understand what they're saying.

Or maybe high-school physics and even university level is badly done.
School in general seems to be tailored to those that can memorize a lot of facts with no effort put into understanding them.

Here is an example to illustrate the difference between energy used to do work and stored energy.
So say you are the source of energy and you are pushing an electric vehicle on a flat road.
Vehicle is in neutral but there is of course friction loss so you put in 200Wh worth of energy and push the vehicle for 1km.
Now this 200Wh of energy did work.
Then at the same time I also push an identical vehicle and also expend 200Wh on the same flat road but I moved the vehicle just 100m = 0.1km because the regenerative brakes on the vehicle where engaged and 180Wh ended as energy stored in the battery.

When you look at this two systems the input energy was exactly the same 200Wh but in first case all energy input in the system did work while in the second case only 10% was used to do work the other 90% was stored as electricity and that stored energy can be used to do work but at this point in time it is stored energy and anything can be done with that not necessarily move the vehicle the other 900m but maybe just use the energy to listen to music on the car radio.

In this example there was an energy conversion from mechanical to electrical before storing unlike the capacitor problem where no conversion was done.
The 144Ws of energy delivered by the source was used to charge the capacitor with an fairly bad efficiency of just 50% so you end up with 72Ws stored in the capacitor but those 72Ws in the capacitor are still there in the original form of electrical energy ready to be used to do work.
Capacitor is an energy storage device that will not let current pass through thus no energy is passing through (the displacement current is just a mathematical concept not a physical current).
Something like displacement current or displacement energy is a fictional math concept for calculation purposes as maybe energy storage was harder to explain.

[hamster_nz]

You seem to lack basic high-school knowledge of physics and electricity. That's why you don't understand what these people are talking about. Then you project your confusion on them, saying they are wrong.

Maybe it's time to revisit the basic concepts and try to understand what they're saying.

Or maybe high-school physics and even university level is badly done.
School in general seems to be tailored to those that can memorize a lot of facts with no effort put into understanding them.

Here is an example to illustrate the difference between energy used to do work and stored energy.
So say you are the source of energy and you are pushing an electric vehicle on a flat road.
Vehicle is in neutral but there is of course friction loss so you put in 200Wh worth of energy and push the vehicle for 1km.
Now this 200Wh of energy did work.
Then at the same time I also push an identical vehicle and also expend 200Wh on the same flat road but I moved the vehicle just 100m = 0.1km because the regenerative brakes on the vehicle where engaged and 180Wh ended as energy stored in the battery.

When you look at this two systems the input energy was exactly the same 200Wh but in first case all energy input in the system did work while in the second case only 10% was used to do work the other 90% was stored as electricity and that stored energy can be used to do work but at this point in time it is stored energy and anything can be done with that not necessarily move the vehicle the other 900m but maybe just use the energy to listen to music on the car radio.

In this example there was an energy conversion from mechanical to electrical before storing unlike the capacitor problem where no conversion was done.
The 144Ws of energy delivered by the source was used to charge the capacitor with an fairly bad efficiency of just 50% so you end up with 72Ws stored in the capacitor but those 72Ws in the capacitor are still there in the original form of electrical energy ready to be used to do work.
Capacitor is an energy storage device that will not let current pass through thus no energy is passing through (the displacement current is just a mathematical concept not a physical current).
Something like displacement current or displacement energy is a fictional math concept for calculation purposes as maybe energy storage was harder to explain.

You do realize that you will be pushing 10x harder in the second case? 0.1 times the distance means 10x the force...

[electrodacus]

You do realize that you will be pushing 10x harder in the second case? 0.1 times the distance means 10x the force...

I already mentioned that same amount of energy will be used 200Wh so the fact that you need to push harder is obvious.

In both cases the same amount of energy was used just that in first case all that energy was used to do work while in second case small amount was used to do work the rest was stored.
The stored energy can not be called work.

[hamster_nz]

The person pushing the car has done the same amount of work. Just in the first case more got 'lost' to the environment.

[electrodacus]

The person pushing the car has done the same amount of work. Just in the first case more got 'lost' to the environment.

We are not discussing the source.
Yes 144mWs where provided by the source in capacitor charging example that is the equivalent to the person providing 200Wh in both cases.

The result was 72mWs of heat loss in the wires transporting the energy to capacitor and 72mWs of stored energy in the capacitor.

The result for vehicle examples was in first case 200Wh of work for moving the vehicle 1km
For second example there was 20Wh of work to move the vehicle 100m and 180Wh of energy stored in the EV battery.
Those 72mWs stored in capacitor and those 180Ws stored in battery are are not work. The energy stored there can be used to do work at any time but at that moment it is just stored electrical energy which is not work.

[hamster_nz]

Are you really arguing that work is not required to store energy in capacitor (or in this case a cars battery)?

Are we in Humpty Dumpty land again, where words only mean what you say the do?

[electrodacus]

Are you really arguing that work is not required to store energy in capacitor (or in this case a cars battery)?

Are we in Humpty Dumpty land again, where words only mean what you say the do?

If transfer could be done at 100% efficiency then yes no work will be done to transfer energy from source to capacitor.
In the particular example efficiency was just 50% so half of the energy ended as heat during transportation.
You can use a DC-DC converter and transport the energy with about 90% efficiency so that only 10% is lost for moving the charge from a source to a capacitor (energy storage device).

What part exactly is not clear.   

[PlainName]

What part exactly is not clear

How the resistor manages to consume energy without there being any transfer from one side to the other.

This diversion to how capacitors charge and at what cost, etc, is simply your smokescreen that you hope will obscure your inability to deal with the real issue.

[hamster_nz]

Are you really arguing that work is not required to store energy in capacitor (or in this case a cars battery)?

Are we in Humpty Dumpty land again, where words only mean what you say the do?

If transfer could be done at 100% efficiency then yes no work will be done to transfer energy from source to capacitor.
In the particular example efficiency was just 50% so half of the energy ended as heat during transportation.
You can use a DC-DC converter and transport the energy with about 90% efficiency so that only 10% is lost for moving the charge from a source to a capacitor (energy storage device).

What part exactly is not clear.

I'm not quite clear how you reconcile that with the actual formal definition of work for an electric field:

https://en.wikipedia.org/wiki/Work_(electric_field)

But then again, since you reject the existence of electric fields, that would hardly be surprising.

[electrodacus]

What part exactly is not clear

How the resistor manages to consume energy without there being any transfer from one side to the other.

This diversion to how capacitors charge and at what cost, etc, is simply your smokescreen that you hope will obscure your inability to deal with the real issue.

You are charging an energy storage device.
There are electrons flowing through wires that have resistance in to the capacitor plates. This electrons remain on the plate do not jump the dielectric and this electrons can flow back delivering energy when you want to use that stored energy.
Will thinking at a rechargeable battery be easier ?
When you charge your phone battery say a LiCoO2 type that is charged from 3V to 4.2V trough a linear regulator can even be a resistor form a 5V supply.
Will you not have a part energy dissipated as heat on the wire resistance and series linear regulator and one part stored energy in the battery.
The energy that went in to battery did no work it is stored energy and can be used latter to supply your phone.
Capacitor same as the rechargeable battery are energy storage devices so energy will not flow through them but in them.
When Maxwell was alive he had no idea such things as electrons exist.  He was also a mathematician not a physicist. He made an important contribution at that time showing electricity and magnetism are related.


Is the word store not different to you from the word work?
If you stored some quantity of energy then it is stored it did not do any work but it is ready to do work when you want to use it.
In the process of moving that energy from a source to the storage device some energy will be lost and in this case just happened to be 50% lost as heat but it could have been 20% lost as heat and 30% used to move an object so do work and then the other 50% stored.
It could also have been 10% lost as heat and 90% stored if you will have used a DC-DC constant current regulator.

So stored just means moving energy available in one place to another place and that amount of energy that was stored has done no work and in this case energy is in the exact same form as the original not converted.     

[electrodacus]

I'm not quite clear how you reconcile that with the actual formal definition of work for an electric field:

https://en.wikipedia.org/wiki/Work_(electric_field)

But then again, since you reject the existence of electric fields, that would hardly be surprising.

That definition you linked to has little to do with what we are discussing.

In this case a particle moves from source to capacitor (energy storage device) through a wire that has a resistance.
When I say the electron moves from the source to capacitor that will be incorrect as it is not the same electron that exits the source with the particle that enters the capacitor plate.
It is like a domino effect that happens at the speed of light but in a wire that has resistance this creates loss of energy so the electron that exits the source has more energy than the electron entering the capacitor with the difference resulting in heat.
So from source 144mWs worth of energy will leave with half that ending as heat due to wire resistance and the other half ends stored in the capacitor.

I think the main problem is to understand what stored means and it is not the same thing with work (quite the opposite). Stored energy has the ability to do work but it has not done so as long as it remains stored.

[PlainName]

What part exactly is not clear

How the resistor manages to consume energy without there being any transfer from one side to the other.

This diversion to how capacitors charge and at what cost, etc, is simply your smokescreen that you hope will obscure your inability to deal with the real issue.

You are charging an energy storage device.

Actually, I am heating  up a resistor. Where did that energy come from, and how did it get there? The only energy source available is the power supply upwind of the capacitor. So, from what I can gather from your 'explanation' it goes something like this:

PSU: "Hey Cap+, gonna shove you some lovely joules."

Cap+: "Eh? Oh, hang on a mo..."
Cap+: Oy, Cap-, you wanna dump your old electrons and get some new ones?"

Cap-: "Sounds fab to me."

Cap+: "OK, let's have them then"

PSU: "Here you go. Enjoy!"

Cap-: "Hey, resitor, 50gazillion elect(r)ons coming your way!"

Resistor: "WTF? Now?!?! Shit, have to burn them up."

I don't think energy storage has meaning in this context. The cap is series with the resister and you talk as if it's in parallel and providing the juice. It isn't - the PSU is. If anything, the cap is acting as a blockage.

Bringing in a battery complicates things, just as inserting a diode would. Let's keep it simple with the minimal parts necessary to show the problem.

[electrodacus]

Actually, I am heating  up a resistor. Where did that energy come from, and how did it get there? The only energy source available is the power supply upwind of the capacitor. So, from what I can gather from your 'explanation' it goes something like this:

PSU: "Hey Cap+, gonna shove you some lovely joules."

Cap+: "Eh? Oh, hang on a mo..."
Cap+: Oy, Cap-, you wanna dump your old electrons and get some new ones?"

Cap-: "Sounds fab to me."

Cap+: "OK, let's have them then"

PSU: "Here you go. Enjoy!"

Cap-: "Hey, resitor, 50gazillion elect(r)ons coming your way!"

Resistor: "WTF? Now?!?! Shit, have to burn them up."

I don't think energy storage has meaning in this context. The cap is series with the resister and you talk as if it's in parallel and providing the juice. It isn't - the PSU is. If anything, the cap is acting as a blockage.

Bringing in a battery complicates things, just as inserting a diode would. Let's keep it simple with the minimal parts necessary to show the problem.

What do you think the major difference is between a capacitor and a rechargeable battery ?
Thy are both energy storage devices.

Yes 144mJ where provided by the source out of this 50% ended up stored in capacitor and the other 50% ended up heating the wires.

How come you did not used Res+ and Res- just Cap+ and Cap- ?
To have a current flow you need a close circuit. Connecting just the positive wire of a supply to a resistor or a capacitor will have no effect.

Maybe one of the people that read this posts can do a better job than me explaining the difference between energy that did work and stored energy.

Ideal case with no resistance anywhere in the circuit 72mJ will flow from the ideal source in to capacitor and system form an energy balance point of view has no charge.
Financial equivalent will me moving $72 from one bank account (yours) to another of your back accounts with no transaction fee.  In the end you still have $72 is just in another bank account but other than that there is no difference.
But if transferring $72 costs you another $72 in fees then you pay $144 from your first bank account and you end up with $72 in another bank account thus you lost $72 just to move $72 from one account to another.
You still have $72 but you used to have $144.   You are acting like all $144 were lost.
 

[snarkysparky]

"half that ending as heat due to wire resistance"

It took power to heat the wire.  That power came from the flow of electrons into and out of the capacitor wires.

We consider the heat in the wires to be useful work.   

[snarkysparky]

Are we still really arguing that power can be transferred through a capacitor???


If anybody says NO then would they please explain how this circuit gets power out the right side when connected to other power source through only a capacitor.

https://en.wikipedia.org/wiki/Capacitive_power_supply

[electrodacus]

"half that ending as heat due to wire resistance"

It took power to heat the wire.  That power came from the flow of electrons into and out of the capacitor wires.

We consider the heat in the wires to be useful work.

Yes electrons flowed in one side of the capacitor and other unrelated electrons flowed out of the other side but this means energy was stored in the capacitor.
When you will discharge the capacitor say by shorting the terminals of that capacitor electrons that flowed in will flow out and in to the side where electrons flowed out resulting in 72mJ of heat (the amount that was stored in the capacitor).

[snarkysparky]

"Yes electrons flowed in one side of the capacitor and other unrelated electrons flowed out of the other side but this means energy was stored in the capacitor."

And that energy that heated the wires flowed because of charge moving into and out of the capacitor.

Do you agree that a capacitor can not only store energy but can also pass through energy?
If not how does the capacitor power supply above work?

[electrodacus]

And that energy that heated the wires flowed because of charge moving into and out of the capacitor.

Do you agree that a capacitor can not only store energy but can also pass through energy?
If not how does the capacitor power supply above work?

Yes wires had resistance to current flow so part of total energy 144mJ ended as heat. 72mJ to be exact ended as heat and the other 72mJ ended as stored energy in the capacitor.
No energy passed through capacitor as there is nothing left 144mJ = 72mJ lost as heat + 72mJ stored in the capacitor.
A capacitive dropper works because capacitor is always charged and discharged.
For example let say you want to power an incandescent lamp in the example I provided before.
You just add a lamp in series with the ideal DC voltage supply and the capacitor and say the lamp resistance is 9 Ohms.

Then you will have 144mJ from the DC supply form that 72mJ will end up stored in the capacitor 
The other 72mJ will be lost on the wire's and lamp total 10Ohms = (9Ohm + 0.5Ohm + 0.5Ohm)
So 90% will be delivered to the incandescent lamp 64.8mJ (mostly infrared photons and some visible light photons) and the other 10% will end as heat loss on the wires 7.2mJ
But that is all done in a few ms then you can wait for an hour and there will be no more energy flowing just this 144mJ from which half was stored and half was used.
But if you reverse the polarity of the power supply or you even just remove the supply and close the circuit the capacitor will deliver the stored 72mJ to the lamp and wires then you can repeat this.
So with DC you have just one pulse as the capacitor chargers and with AC you constantly charge and discharge the capacitor.

[PlainName]

A capacitive dropper works because capacitor is always charged and discharged.

You are confusing charge with energy.

No energy passed through capacitor

On the contrary, you mean no charge passed through the capacitor. But power went around the the whole circuit for the duration that the capacitor charged.

[snarkysparky]

""So with DC you have just one pulse as the capacitor chargers and with AC you constantly charge and discharge the capacitor..""


Yes and in that one charging pulse there was a finite amount of energy available to drive a load on the other side of the capacitor.

[electrodacus]

A capacitive dropper works because capacitor is always charged and discharged.

You are confusing charge with energy.

No energy passed through capacitor

On the contrary, you mean no charge passed through the capacitor. But energy went around the the whole circuit for the duration that the capacitor charged.

Charging a capacitor means storing energy in to it.
No energy went around the capacitor but in to capacitor.

So when you are charging your phone battery is the energy going around you battery or in to your phone battery ?

The concept of energy storage seems to be completely missing from your understanding.

You have the concept that you can transfer electrical energy with just one wire so no closed loop.
 
You are probably thinking at a compressed air tank and how you take air from environment and push it in storing energy.
A more accurate way to think will be to have the tank split in two isolated chambers and you are moving air from one half of the cylinder to the other half to charge it and then to discharged it so use the energy you do the reverse let the air from the pressurized half go back in to the half with lower pressure.
This last example is much more accurately describing a rechargeable battery or capacitor.
For that air pump to work you need to connect both halves of the cylinder so you need to have a closed loop.
Now say you have a similar compressed air tank with two chambers that is identical but discharged so both sides contain air but at same pressure.
If you now connect this charged cylinder to the discharged one with two hoses the pressure delta will be lower and equal in both cylinders and now you have lost half of the energy as you only have a quarter of initial energy in each tank because this sort of transfer was 50% efficient.

[electrodacus]

""So with DC you have just one pulse as the capacitor chargers and with AC you constantly charge and discharge the capacitor..""

Yes and in that one charging pulse there was a finite amount of energy available to drive a load on the other side of the capacitor.

I'm trying very hard to understand what sort of thing you imagine happens.
Why was the amount of energy available finite ?
Why do you say the other side of the circuit ? It is irrelevant on what side of the circuit you load is.
You can connect the lamp on any side of the capacitor it will work exactly the same.

With that example that I provided you will have 9.5Ohms on the side you connect the lamp 9Ohms lamp + 0.5Ohms wire and on the other side you only have the 0.5Ohms wire.
On one side you have 72mJ * 0.95 = 68.4mJ and on the other side you have 3.6mJ so loss on wires as heat is now just 7.2mJ in total the rest is useful work assuming you wanted an incandescent lamp (you can replace the lamp with an electric motor or any other load you like).
There is a big difference between the 64.8mJ that ended on the lamp and the 72mJ that ended in capacitor.
The 64.8mJ already converted that energy in something that you wanted (light) while the 72mJ in the capacitor did nothing it is still available for you in original form as stored electrical energy.

[PlainName]

Charging a capacitor means storing energy in to it.
No energy went around the capacitor but in to capacitor.

So what caused the resistor to heat up? If it had been an LED it would emit light. If it were a motor it would turn. You cannot charge that capacitor without those things happening and yet, according to you, there was no energy available to them. They would be perpetual motion machines running on nothing!