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

[Alex Eisenhut]

kilowatts of energy

Yeah no.

[electrodacus]

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 

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]

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 

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.   

How much would that many electrons weigh? micrograms?

[Naej]

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..

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   

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?

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]

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?

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.

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).

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)?

[PlainName]

For a wire we can calculate the power, but we cant calculate the amps, koz we cant calculate the resistance.
We don’t know what electric resistance iz or iznt. Hence there is no possible calculation from first principles.

For that matter, we (rather, you) don't know how electons stick to conductors and not non-conductors, so there is no possible calculation from first principles for electon theory either.

[m k]

Better duplicate the middle wire and split the circuit so that GND is bottom of the higher part.
Lower part remains below GND, its voltage is negative compared to GND.

[electrodacus]

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)?

I see that is not just hamster_nz that is confused. You seems to have the same confusion.

The two circuits due to symmetry are not different in any way. From an electrical point of view they are identical as that wire in the middle has no electrical role (no current meaning no electrons flow through that wire).

But I guess same as hamster_nz you agree with the above fact is just that you do not understand what power and energy is.
Both power and energy are calculated and not measured.
What you measure is just voltage (pressure will be the analog for fluids) even the current is not measured but calculated as the voltage drop on a length of wire with known resistance.

Also just to be clear there is no current flow through that piece of wire connecting the GND symbol.

One electron has a charge of about 1.6 * 10^-19C
So 1A for one second 1As a number of 1/1.6*10^-19 = 6.25 * 10¹⁸ electrons flow each second trough that wire.
That is a huge number of electrons that flow through the wire each second.

So that number of electrons leave the negative of the bottom battery every second and travel all the way around the wires/resistors (wires and resistors are the same thing) and enters the positive of the top battery.
At the same time same number of electrons leave the negative of the top battery and enter the positive of the bottom battery.

The above is valid for both diagrams thus there is no difference between the two.

 

[PlainName]

If you take two identical circuits (battery, resistor, 2 wires) which are not connected together, they each burn 1W. Now, if you connect the top wire of one to the bottom wire of the other, you have the circuit on the left. Now, you may say that the left circuit is identical to the right circuit, but in that case just pretend (to save having to draw it) that the left circuit is the two identical but separate circuits I just described.

Don't we now have 2 x 1W circuits with 1A going through each of the top wires in one case, and in the other 1 x 2W circuit with 1A going through the sole top wire?

[electrodacus]

If you take two identical circuits (battery, resistor, 2 wires) which are not connected together, they each burn 1W. Now, if you connect the top wire of one to the bottom wire of the other, you have the circuit on the left. Now, you may say that the left circuit is identical to the right circuit, but in that case just pretend (to save having to draw it) that the left circuit is the two identical but separate circuits I just described.

Don't we now have 2 x 1W circuits with 1A going through each of the top wires in one case, and in the other 1 x 2W circuit with 1A going through the sole top wire?

You are confused by the fact that wires and resistors are the same thing. Also the "wires" in your diagram are superconductors meaning they have zero resistance to current flow.

It will be useful if you make the drawing at scale and just use wires with say 1Ohm per some random unit of length that chose say 10cm or maybe 100pixels
Then all the wires you use will have resistance and you will probably understand where your thinking is wrong.
You should realize that the two schematics are the same.
Or continue to use those superconductor wires but add a resistor for every wire you have.

So in the schematic on the left you need to add 4 more resistors and on the one of the right 1 more resistor.
 

[iMo]

..
One electron has a charge of about 1.6 * 10^-19C
So 1A for one second 1As a number of 1/1.6*10^-19 = 6.25 * 10¹⁸ electrons flow each second trough that wire.
That is a huge number of electrons that flow through the wire each second.

So that number of electrons leave the negative of the bottom battery every second and travel all the way around the wires/resistors (wires and resistors are the same thing) and enters the positive of the top battery.
At the same time same number of electrons leave the negative of the top battery and enter the positive of the bottom battery.

The above is valid for both diagrams thus there is no difference between the two.

And the kinetic energy of that amount of electrons would be something like 3*E-14 Joules..

[electrodacus]

And the kinetic energy of that amount of electrons would be something like 3*E-14 Joules..

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.

[gnuarm]

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)?

Current is not power.  Do there's no reason to think there is a contradiction in the same current resulting in two different power levels in the two circuits.  The power is doubled because the voltage is doubled.  Or you could just as well say the power is doubled because the resistance is doubled.

What if there were no batteries or resistors, just zero ohm wires with a 1A current?  Same power?  I think not.

[iMo]

And the kinetic energy of that amount of electrons would be something like 3*E-14 Joules..

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.

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

[PlainName]

It will be useful if you make the drawing at scale and just use wires with say 1Ohm per some random unit of length that chose say 10cm or maybe 100pixels

I'll make it easier: each wire is 1mR. I don't think that affects the displayed values very much unless you're wanting to use some distraction.

[eugene]

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.

The circuit on the left is one circuit. The fact that there is zero voltage between the two ends of the center wire (between the two junction dots) means there is zero current flowing in that wire. It has no effect on any calculations of current, power, energy, or any other relevant property.

That's true only because of the symmetry. If you make one battery droop a little, or separate the circuit into two, or change anything in any way, then you have something different.

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)?

Do you understand that you're talking about two different circuits? Each can be analyzed using Ohm's Law. No QM required. Analyze the two circuits using Ohm's Law and you will discover the currents flowing in each part. They're different circuits, so there may or may not happen to be the same magnitude of current flowing through different parts, but the analysis of either does not depend on the other in any way whatsoever. They're separate circuits.

[eugene]

And the kinetic energy of that amount of electrons would be something like 3*E-14 Joules..

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.

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

This is not true. The drift velocity is just the average velocity of all the electrons, which happen to be going in all different directions at different speeds. Most of that averages out to the drift velocity.

The average kinetic energy is much greater than 1/2 mv² where v is the drift velocity. It depends on, amongst other things, temperature. The transfer of kinetic energy of electrons is responsible for most of the heat flow in materials that are good electrical conductors. This is covered by a large field of study in physics called statistical mechanics. A tutorial is outside the scope of this forum.

[electrodacus]

The kinetic energy of that N electrons per second (the amount N 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).

That is not how it is done. You guessing stuff.
d is the distance.

KE = F * d = E * Q * d = V/d * Q * d = V * Q = 2V * 1As = 2Ws = 2 Joules

The electron drift velocity is probably not what you think it is.

There is a very small average time between collisions of electrons with ions and during that time the electron is accelerated by the electric potential.

Edit: The eugene reply is better articulated than mine so read that also. 

[PlainName]

The circuit on the left is one circuit.

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

[iMo]

And the kinetic energy of that amount of electrons would be something like 3*E-14 Joules..

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.

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

This is not true. The drift velocity is just the average velocity of all the electrons, which happen to be going in all different directions at different speeds. Most of that averages out to the drift velocity.

The average kinetic energy is much greater than 1/2 mv² where v is the drift velocity. It depends on, amongst other things, temperature. The transfer of kinetic energy of electrons is responsible for most of the heat flow in materials that are good electrical conductors. This is covered by a large field of study in physics called statistical mechanics. A tutorial is outside the scope of this forum.

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..

[PlainName]

The circuit on the left is one circuit.

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

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

[electrodacus]

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..

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]

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..

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

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

[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]

And the kinetic energy of that amount of electrons would be something like 3*E-14 Joules..

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.

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

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.