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

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Alex Eisenhut:

--- Quote from: imo on July 02, 2022, 06:17:29 am --- kilowatts of energy

--- End quote ---

Yeah no.

electrodacus:

--- Quote from: imo on July 02, 2022, 06:17:29 am ---
Yep, that comes from the "simplistic" model used in primary education - the model of a "water pipe", energy transferred in the tubing through the water pressure, where the electrons are depicted like the water molecules.. The fact the small tiny lightweight electrons moving in a copper wire with a drift speed of an ant cannot transfer kilowatts of energy is hard to swallow then  :D

--- End quote ---

That analogy that you learned is way more correct than you think.
Electrons are small and have little mass but there are many. A single gram of copper metal contains about 9.5 * 10^21 copper atoms and so the same number of free electrons.

Do you understand that a capacitor is an energy storage device?
A discharged capacitor is just two parallel metal plates (say copper) with equal charge (same number of free electrons on both plates)
A charged capacitor will have more free electrons than copper atoms while the opposite plate has less free electrons than there are atoms.
The excess electrons will want to move to the plate with deficit of electrons but they can only do so if you connect the two plates with a wire/resistor (electrical conductor).
Since a copper conductor has a resistance to current flow that stored energy will end up as heat (radiated by the wire to the environment) and you call this electromagnetic radiation somewhere in the THz region so super high frequency.
The higher the resistance the longer it will take for the parallel plate to equalize and become neutral so fully discharge the energy stored in the capacitor.   

hamster_nz:

--- Quote from: electrodacus on July 02, 2022, 06:53:11 am ---
--- Quote from: imo on July 02, 2022, 06:17:29 am ---
Yep, that comes from the "simplistic" model used in primary education - the model of a "water pipe", energy transferred in the tubing through the water pressure, where the electrons are depicted like the water molecules.. The fact the small tiny lightweight electrons moving in a copper wire with a drift speed of an ant cannot transfer kilowatts of energy is hard to swallow then  :D

--- End quote ---

That analogy that you learned is way more correct than you think.
Electrons are small and have little mass but there are many. A single gram of copper metal contains about 9.5 * 10^21 copper atoms and so the same number of free electrons.

Do you understand that a capacitor is an energy storage device?
A discharged capacitor is just two parallel metal plates (say copper) with equal charge (same number of free electrons on both plates)
A charged capacitor will have more free electrons than copper atoms while the opposite plate has less free electrons than there are atoms.
The excess electrons will want to move to the plate with deficit of electrons but they can only do so if you connect the two plates with a wire/resistor (electrical conductor).
Since a copper conductor has a resistance to current flow that stored energy will end up as heat (radiated by the wire to the environment) and you call this electromagnetic radiation somewhere in the THz region so super high frequency.
The higher the resistance the longer it will take for the parallel plate to equalize and become neutral so fully discharge the energy stored in the capacitor.   

--- End quote ---

How much would that many electrons weigh? micrograms?

Naej:

--- Quote from: hamster_nz on July 01, 2022, 10:38:15 pm ---
--- Quote from: electrodacus on July 01, 2022, 10:13:05 pm ---
--- Quote from: hamster_nz on July 01, 2022, 10:04:38 pm ---
Fully agree. But I still cant decide if one unit of energy or two is flowing in the top wire. With the center wire in place it is 1, with it removed it is 2.

But as you say, removing that wire makes no change, so it must be only one or the other..

--- End quote ---

What flows through the wire are a stream electrons and that will be the definition for electrical current.
The current through that top wire has no reason to change as it will be the same with that center wire in placed or removed.
If we consider that internal DC resistance of the battery is zero (not realistic) then open circuit voltage of the battery will be the same as voltage under load.
Since voltage you will measure between the top and bottom wire will be the voltage of those two batteries in series in my example 20V the current through that 20Ohm of total resistance will be 1A.
At any one moment there are 20W of power dissipated as infrared radiation to the environment and in therms of energy you will need to specify a period like say for 1 second will be 20Ws = 5.55mWh but in 1ms it will be 1000x less so 20mWs = 5.55uWh   

--- End quote ---

That may all be true, but is 1x or 2x the energy being converted to infrared  by a resistor flowing in the top wire in each circuit?

Everybody agrees that both circuits have the same currents and voltages and powers and energy flows.

I would have though it is an easy qn. If energy flows only in wires, how much energy is flowing in the upper wire for both circuits? Because the currents and voltages are consistent, so should the energy...

Isn't it as simple as remover the top wire and see how much the power in the resistors changes?

--- End quote ---
As I said, 1x.

Your last question has an "interesting" logic. How much force is on a small part of a pressurized bottle? Isn't it as simple as removing the small part?
It works well with buildings too: how much force on a brick? Just remove it. And on a keystone? Etc.

PlainName:

--- Quote from: hamster_nz on July 02, 2022, 02:47:46 am ---
--- Quote from: electrodacus on July 02, 2022, 02:30:22 am ---
--- Quote from: hamster_nz on July 01, 2022, 11:43:19 pm ---
So you firmly believe that a different quantity of energy is flowing in the top wire, even though the currents and voltages are the same through each element?

And removing a wire that carries no energy is resposible for this difference?

--- End quote ---

What are you talking about ?
By top wire I hope you mean the wire between the positive of top battery and top resistor ?

If you remove that in first circuit then obviously can not be any current trough that wire as it is missing and so also no current from the top battery and no current through the top resistor.
But since there is a wire in the middle the lower resistor and battery will see a current 1A if battery is 10V on a 10Ohm resistor so 10W total as only half the circuit is closed circuit.

In second diagram since the middle wire is not present removing the top wire means no current at all and so no energy is being used.

--- End quote ---

The question is really simple. How much energy is flowing through the highlighted wire on the left and right diagrams. Sorry if I am not stating it clearly enough.

I say 1W (left) and 2W (right).

--- End quote ---

Soz for the lengthy quote but I think this is a good post to fall back to.

Perhaps the interested parties could just fill in appropriate values for voltage, resistance, current, etc., instead of rambling off about everyone else not understanding stuff. I've had a go.

Isn't the circuit on the left essentially two separate circuits, each burning 1W? The circuit on the right is a single circuit burning 2W overall. But the currents and voltages across the resistors are the same in each.

Suppose on the left you have just one circuit - let's delete the bottom battery, wires, resistor. So it is now a single circuit burning 1W. That top wire is passing 1A. On the right it is still passing 1A but the circuit is burning 2W, so how is that extra 1W getting around if the top wire is passing no more current (with no difference in voltage drop)?

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