General > General Technical Chat
"Veritasium" (YT) - "The Big Misconception About Electricity" ?
TimFox:
Skin-effect states that the AC resistance of a length of wire is always higher than the DC resistance, since there is less current density in the center of the wire at AC than at DC, where the current density is uniform.
The manufacturer produces wires with a thin cladding of copper over a steel core that have nominal 30% or 40% of the conductivity of pure copper at DC.
No electons need apply.
adx:
--- Quote from: aetherist on March 14, 2022, 09:15:01 pm ---
--- Quote from: adx on March 14, 2022, 01:27:12 am ---
--- Quote from: aetherist on March 13, 2022, 10:07:27 pm ---Problem 6. Electons are a surface dweller, hence u would think that doubling the dia of a wire would halve the resistance. If doubling the dia results in a ¼ resistance then electons are in trouble. For DC current.
--- End quote ---
Well spotted.
--- End quote ---
Looking ahead for good excuses. If new (electon) electricity was very sensitive to temperature then that could explain the further doubling of resistance, to make it 1:4 instead of 1:2 (if indeed the 1:4 exists).
--- End quote ---
No because you can vary the temperature and test for that either whole or as individual metals. Also heat output (and resistance) can be measured without much rise in temperature, either by heatsinking the wire, or not putting much power in; resistance measured at 1mA say on a µV reading meter is very close to resistance measured at 1A. It won't get hot enough to double in resistance, which is far too hot to touch, or even molten, for most (all?) metals.
But I was originally thinking that doesn't preclude some other physical effect, like the electons pairing with some internal electrons, so that their energy loss is proportional to the cross sectional area. But how does an electon lose energy? Not a proof, but the electron drift model is a vastly simpler (and more direct) way to explain resistance.
TimFox:
Note that the CopperweldTM resistance data that I linked above are tabulated at two different current levels for each wire part number.
aetherist:
--- Quote from: TimFox on March 14, 2022, 10:42:29 pm ---Skin-effect states that the AC resistance of a length of wire is always higher than the DC resistance, since there is less current density in the center of the wire at AC than at DC, where the current density is uniform.
The manufacturer produces wires with a thin cladding of copper over a steel core that have nominal 30% or 40% of the conductivity of pure copper at DC.
No electons need apply.
--- End quote ---
That a copper center gives 100% conductivity, but a steel center gives only 30% (for say t=r/10) is a worry for electons.
Electons suggest that a wire with a steel center should give say 99% (koz all of the electons live on the Cu).
I would love to see a test for a steel clad copper wire. Old (electron) electricity might say that the conductivity for DC should be say 95% (if area of steel is say 10% of total area)(koz most of the drifting electrons would live in the Cu), whereas new (electon) electricity might say 15% (koz all of the electons would live on the Fe)(& based on Fe having 6.00 times the resistance of Cu).
So, why 30% IACS & not 99% IACS? (for copper clad steel).
Still thinking.
aetherist:
--- Quote from: adx on March 14, 2022, 10:54:36 pm ---
--- Quote from: aetherist on March 14, 2022, 09:15:01 pm ---
--- Quote from: adx on March 14, 2022, 01:27:12 am ---
--- Quote from: aetherist on March 13, 2022, 10:07:27 pm ---Problem 6. Electons are a surface dweller, hence u would think that doubling the dia of a wire would halve the resistance. If doubling the dia results in a ¼ resistance then electons are in trouble. For DC current.
--- End quote ---
Well spotted.
--- End quote ---
Looking ahead for good excuses. If new (electon) electricity was very sensitive to temperature then that could explain the further doubling of resistance, to make it 1:4 instead of 1:2 (if indeed the 1:4 exists).
--- End quote ---
No because you can vary the temperature and test for that either whole or as individual metals. Also heat output (and resistance) can be measured without much rise in temperature, either by heatsinking the wire, or not putting much power in; resistance measured at 1mA say on a µV reading meter is very close to resistance measured at 1A. It won't get hot enough to double in resistance, which is far too hot to touch, or even molten, for most (all?) metals.
But I was originally thinking that doesn't preclude some other physical effect, like the electons pairing with some internal electrons, so that their energy loss is proportional to the cross sectional area. But how does an electon lose energy? Not a proof, but the electron drift model is a vastly simpler (and more direct) way to explain resistance.
--- End quote ---
Temp aint temp.
Temp for an electon is the temp of the Cu atoms closest to the electon, ie the skin of the wire.
Temp for an electron is the temp of the Cu atoms closest to the electron, ie the temp of the whole wire.
The hottest Cu atoms on the surface of the skin might be 10 times the temp of the Cu atoms inside the wire.
Electon energy loss keeps me awake at night.
I am starting to think that when an electon gives energy to heat a wire then it is not energy loss.
This might sound strange.
But, look at an ordinary photon, it radiates energy for ever, without losing energy. Or, put another way, the lost energy is immediately replenished by the aether.
And an electron radiates energy for ever.
And an orbiting electron radiates energy for ever.
Why shouldn’t electons heat a wire for ever.
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