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

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eugene:

--- Quote from: dunkemhigh on July 02, 2022, 10:22:44 am ---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.

--- End quote ---

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.


--- Quote ---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)?

--- End quote ---

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:

--- 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 ---

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:

--- Quote from: imo on July 02, 2022, 06:48:48 pm ---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).

--- End quote ---

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:

--- 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.

iMo:

--- Quote from: eugene on July 02, 2022, 07:44:30 pm ---
--- 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 ---

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.

--- End quote ---

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..

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