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Electroboom: How Right IS Veritasium?! Don't Electrons Push Each Other??

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PlainName:

--- Quote from: dunkemhigh on July 02, 2022, 07:54:06 pm ---
--- Quote ---The circuit on the left is one circuit.
--- End quote ---

Yes, so another post has made it two distinct circuits. They are then joined by the top/bottom wires to give the left circuit.

--- End quote ---

To be absolutely clear, and bearing in mind people want actual real wires (albeit simulated), this is what I meant.

electrodacus:

--- Quote from: imo on July 02, 2022, 07:58:02 pm ---
Ok, based on the wiki and 1A current (https://en.wikipedia.org/wiki/Drift_velocity)

E = 1.5E-21 Joules/s [Watts] for v = 23um/sec

This is just illustrate the energy transfer from the battery to the lamp is not done by the kinetic energy of moving electrons..

--- End quote ---

I will do my best to simplify things for you but keep in mind that any analogy has limitations.

You have two closed volumes say they are a cube 1m * 1m * 1m so 1m3 in volume and one contains 9900 gas molecules (free electrons) and the other one 10100
So there is a pressure differential (same as potential differential).  They are the equivalent of a charged capacitor each closed volume represent one of the plates.
Now you take a pipe that is very long and thin say a square section pipe that is 0.1m * 0.1m * 100m and the pipe is also filed with the same gas but it contains 10000 gas molecules so the average pressure between the two you can call this neutrally charged.

You connect the two gas filled cubes with this 100m long pipe.
Now gas particles will start to enter the pipe on one end from the gas volume containing 10100 molecules as pressure is higher there than in the pipe and at the exact same time gas particles from the pipe will enter that volume with lower pressure and just 9900 gas molecules.

Gas molecule is accelerated by the pressure differential so the speed increases until it hits another molecule or the pipe walls.
The energy travels at the speed of sound for this particular gas but the average drift velocity will be way slower that the speed of sound in this gas.

Can you now see why average drift velocity can not be used to calculate the energy that was radiated as heat from this system?
 
 

iMo:

--- Quote from: electrodacus on July 02, 2022, 08:34:41 pm ---
--- Quote from: imo on July 02, 2022, 07:58:02 pm ---
Ok, based on the wiki and 1A current (https://en.wikipedia.org/wiki/Drift_velocity)

E = 1.5E-21 Joules/s [Watts] for v = 23um/sec

This is just illustrate the energy transfer from the battery to the lamp is not done by the kinetic energy of moving electrons..

--- End quote ---
..
Can you now see why average drift velocity can not be used to calculate the energy that was radiated as heat from this system?

--- End quote ---

Sure, that is exactly what I am trying to show you since my first replay to your post..  :D

hamster_nz:
I'm in full agreement with Dunkemhigh - here's why I think the energy flows in the two circuits are different, even though the voltages and currents are the same.

The circuit on the left can be made more clear by:

- Making the middle wire very think & chunky.

- then reduce the length of the wire, till it is as long as a standard wire.

To me it is very clear that energy from the top battery can only get to the top resistor, and energy from the bottom battery can only get to the bottom resistor. For further proof, this can be calculated by cutting the a wire and measuring the open-circuit voltage, and then measuring the current - 1V & 1A. This is very different to the circuit on the right, where if you cut a wire and measure the open-circuit voltage you get 2V & 1A.

So the top wire in the two circuits have different amounts of energy flowing through them, depending on a presence (or absence) of a wire carrying zero current and with zero voltage over it.

The point of this whole exercise being that the top wire, batteries and resistor all have the same voltages and current, but that top wire is enabling the transfer of twice as much energy than in the other. The presence of a wire carrying zero current makes a difference to the energy flow - in this case, removing it doubles the energy being transferred via the top wire, while voltages and currents remain the same.

Naej:

--- Quote from: imo on July 02, 2022, 06:48:48 pm ---
--- Quote from: electrodacus on July 02, 2022, 06:02:12 pm ---
--- Quote from: imo on July 02, 2022, 05:39:53 pm ---
And the kinetic energy of that amount of electrons would be something like 3*E-14 Joules..

--- End quote ---

That is wrong but please provide the details of how you came up with that number.
The electrical potential is 2V the current is 1A and over one second you get 2Ws that is the same with 2 Joules. So no matter how you make the calculation if you do things correctly you will get 2 Joules.

--- End quote ---

The kinetic energy of that N electrons per second (the amount N=6.25E18 you calculated) is

E = 1/2 * (N * me) * v^2, where me is the mass of an electron, and the v is the drift speed of the electrons in the copper, let say 1cm/s to be extremely optimistic..

E = 3E-16 Joules (each second) = 3E-16 Watts

--- End quote ---
No the speed of electrons is ~ 1600 km/s so you must multiply by 2e16 if you want to know the kinetic energy of electrons.

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