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| Electroboom: How Right IS Veritasium?! Don't Electrons Push Each Other?? |
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| iMo:
--- Quote from: Naej on July 02, 2022, 09:19:55 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 --- 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. --- End quote --- That 1570km/sec is the Fermi speed of electrons in the "electron gas" (random vectors inside the conductor) without any field imposed (no current). The average is zero. With a current the electron gas starts to drift with the speed "v". Anyhow, to make the long story short - the energy in the circuits is not transferred via the conductors, but outside the conductors - along the conductors - via the Fields (the Electromagnetic Fields), where the strongest field is in a close vicinity of the conductor, and the intensity of the field decreases with distance off the conductor (but it spreads over entire Universe theoretically). The fields propagate with the speed of light (in vacuum). The Poynting vector shows the direction where the energy flows. The energy of the field is transferred to the "heat" when the field starts to enter the conductor with a higher "resistance" (it means the density of free electrons in such a "resistive" conductor is lower compared to a good conductor) and while the field enters the resistive conductor (and the Poynting vector bends towards the conductor) it comes to an energy transfer from the field to phonons (phonons are the "heat particles") also known as Joule's losses/heating (there is a change of the speed of the field propagation when it enters the conductor - speed slows down significantly - and transfers the energy into the phonons). Therefore the lamp starts to lit, or a resistor starts to heat up, while the good conductors wired to them stay almost cold.. |
| eugene:
--- Quote from: hamster_nz on July 02, 2022, 09:15:52 pm ---[...] here's why I think the energy flows in the two circuits are different, even though the voltages and currents are the same. --- End quote --- Should I conclude from that statement that you believe there's a way for electrical energy to flow other than via current? |
| electrodacus:
--- Quote from: imo on July 02, 2022, 09:10:20 pm --- Sure, that is exactly what I am trying to show you since my first replay to your post.. :D --- End quote --- Have you read and understood the analogy of free electrons with gas molecules? In that example the speed of sound is the limit not the speed of light. Also only 100 gas molecules were exchanged between the two volumes (capacitor plates) but none of the 100 molecules that left the higher pressure volume ever got on the other side. The drift velocity in that example is not the speed of sound or anywhere close to that and using that will give you a wrong result. The drift velocity is the mean speed of all molecules in that pipe (wire equivalent) so not 100 molecules but 10000 in that example moved at that average speed. Each molecule traveled at much higher speed but due to collisions and direction change drifted very little. So kinetic energy is transferred from one molecule to another while they collide. So you can not just multiply the very slow average drift speed velocity with the mass of just 100 molecules as all 10000 molecules in the pipe moved at that average drift speed. Same with the electrons. |
| iMo:
--- Quote from: electrodacus on July 02, 2022, 10:27:14 pm --- --- Quote from: imo on July 02, 2022, 09:10:20 pm --- Sure, that is exactly what I am trying to show you since my first replay to your post.. :D --- End quote --- Have you read and understood the analogy of free electrons with gas molecules? .. --- End quote --- No.. --- Quote from: imo on July 02, 2022, 09:48:50 pm --- --- Quote from: Naej on July 02, 2022, 09:19:55 pm ---.. 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. --- End quote --- That 1570km/sec is the Fermi speed of electrons in the "electron gas" (random vectors inside the conductor) without any field imposed (no current). The average is zero. With a current the electron gas starts to drift with the speed "v". Anyhow, to make the long story short - the energy in the circuits is not transferred via the conductors, but outside the conductors - along the conductors - via the Fields (the Electromagnetic Fields), where the strongest field is in a close vicinity of the conductor, and the intensity of the field decreases with distance off the conductor (but it spreads over entire Universe theoretically). The fields propagate with the speed of light (in vacuum). The Poynting vector shows the direction where the energy flows. The energy of the field is transferred to the "heat" when the field starts to enter the conductor with a higher "resistance" (it means the density of free electrons in such a "resistive" conductor is lower compared to a good conductor) and while the field enters the resistive conductor (and the Poynting vector bends towards the conductor) it comes to an energy transfer from the field to phonons (phonons are the "heat particles") also known as Joule's losses/heating (there is a change of the speed of the field propagation when it enters the conductor - speed slows down significantly - and transfers the energy into the phonons). Therefore the lamp starts to lit, or a resistor starts to heat up, while the good conductors wired to them stay almost cold.. --- End quote --- |
| gnuarm:
--- Quote from: eugene on July 02, 2022, 07:38:34 pm --- --- 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. --- End quote --- You can also look at it as the circuit has two currents flowing in the center wire, equal and opposite. |
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