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| Electroboom: How Right IS Veritasium?! Don't Electrons Push Each Other?? |
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| iMo:
--- Quote from: electrodacus on July 02, 2022, 04:37:01 pm ---.. 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. --- End quote --- And the kinetic energy of that amount of electrons would be something like 3*E-14 Joules.. |
| electrodacus:
--- 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. |
| gnuarm:
--- Quote from: dunkemhigh on July 02, 2022, 10:22:44 am ---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)? --- End quote --- 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:
--- 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 |
| PlainName:
--- Quote ---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 --- End quote --- 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. |
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