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Veritasium "How Electricity Actually Works"
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hamster_nz:

--- Quote from: electrodacus on May 20, 2022, 10:05:11 pm ---
--- Quote from: hamster_nz on May 20, 2022, 09:50:24 pm ---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?

--- End quote ---

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.

--- End quote ---

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:

--- Quote from: dunkemhigh on May 21, 2022, 08:13:02 am ---
--- Quote ---What part exactly is not clear
--- End quote ---

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.

--- End quote ---

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:

--- Quote from: hamster_nz on May 21, 2022, 09:05:09 am ---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.

--- End quote ---

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:

--- Quote from: electrodacus on May 21, 2022, 03:29:05 pm ---
--- Quote from: dunkemhigh on May 21, 2022, 08:13:02 am ---
--- Quote ---What part exactly is not clear
--- End quote ---

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.

--- End quote ---

You are charging an energy storage device.

--- End quote ---

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:

--- Quote from: dunkemhigh on May 21, 2022, 05:07:13 pm ---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.

--- End quote ---

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.
 
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