General > General Technical Chat
Veritasium "How Electricity Actually Works"
IanB:
--- Quote from: EEVblog on May 02, 2022, 12:31:57 pm ---I posted this comment on Derek's video, slighly expanded:
Riddle me this, a new thought experiment: Let's go extreme and say you have a 100mm diameter copper conductor at pure steady state DC delivering a small amount of power to a pure resitive load, say 1W, but go as low as you want. No transients, no skin effect, no nothing, just pure steady state DC into a resistive load.
Is there NO energy WITHIN this comicly large wire? None? It's all on the OUTSIDE of the wire in the fields at DC? Really? REALLY?
The classicial field theory math might work at DC, but I just can't get over the feeling that it doesn't pass the sniff test at DC. I don't get The Vibe I get with AC and transients. Quantum Electrodynamics and probability theory in the electron fields within the wire better passes the sniff test at DC.
Can someone please convince me that there is no energy flow within this 100mm diameter wire at all, and that all the energy flows outside the wire at DC.
--- End quote ---
I was thinking about this in terms of a water analogy.
Let's suppose we have a large lake, with a weir on the far side, and the lake is full to the top of the weir. Now let's suppose we turn on a hose, and start adding a small stream of water on the opposite side of the lake. By and by, water will start flowing over the weir at the same rate it is being added from the hose.
Clearly water is flowing across the lake from the hose to the weir. However, if we try to measure any gradient in the surface level of the lake, we will likely fail--it will be quite level from one side to the other to the precision of our instruments. Similarly, if we try to measure any change in pressure throughout the lake we will also fail, it will once again be the same everywhere to the precision of our instruments.
Yet for the water to flow across the lake, there must be a driving force. There must be a pressure/level gradient, no matter how infinitesimal, for the water to flow.
Perhaps the analogy here with the surface electric charge would be the gravitational field. The level of the lake surface must be slightly higher on the hose side than on the weir side, and this change in level in the presence of a gravitational potential field leads to a driving force that moves the water across the lake. The slightly higher level will simultaneously lead to a slightly higher hydrodynamic pressure below the surface, which in this analogy would correspond to a small electric field inside the wire.
electrodacus:
--- Quote from: aetherist on May 07, 2022, 07:13:19 am ---I am a retired civil engineer.
That a half of the energy is lost during transfer is interesting, i don’t have a view re that, but it smells fishy.
I have kept out of capacitor to capacitor stuff here, except that i did say that almost everything said/written here has looked wrong to me. Remember i suggested u needed a 2nd switch.
I have agreed with u that the Poynting Field carries zero energy – u must be confusing me with someone else.
The gas cylinder with a membrane is a standard analogy for capacitors – something to do with waving away the need for displacement current or something.
Electrons can't cross from one plate to the other, they would have to jump through the air or whatever.
[G] If we shorted across the gap from one plate to the other then electrons would flow through the short, & would deliver energy (whatever that means). But this is never done.
[C] If we shorted around the circuit from one plate to the other then electrons would flow through the wire, & would deliver energy (whatever that means). This is always done.
In both cases no energy is transferred via the air.
But the amount of energy transfer will usually not be the same in [G] & [C].
--- End quote ---
There is no strange smell :) This is a particular case where both capacitors are identical and you charge one form another. You can imagine that a real resistor has a DC series resistance and for transient/AC impedance.
Since this are equal (identical capacitors) you will build a divider when you parallel them so if you measure symmetrical you will always measure 1/2 of the charged capacitor voltage from the time you parallel them to when they are in steady state so no more current flow.
If capacitors are not identical then more or less than half of the energy will be lost as heat the exact half is for the identical capacitor case.
Both G and C will convert the same amount of energy to heat.
TimFox:
"That a half of the energy is lost during transfer is interesting, i don’t have a view re that, but it smells fishy."
You can do a very simple experiment about energy loss when connecting two capacitors.
Obtain (and measure) two 10 uF polypropylene capacitors. Make sure that each is discharged before proceeding.
Wire any available switch between the two capacitors. It's your choice: DPST or SPST and a wire.
Connect a 10 megohm voltmeter across one of them (time constant = 100 seconds), called C1.
Then, connect a reasonable and safe (say, 9 V battery) DC supply to that capacitor C1 and wait until the voltage stabilizes at voltage V1.
Disconnect the DC supply, note the voltage, close the switch, and note the voltage V2 before it decays due to the 10 megohm resistance.
The only reason to measure the capacitors before the test is to improve the accuracy.
Before closing the switch, the total energy in the system is (1/2) x C1 x V12.
After closing the switch, the total energy is (1/2) x (C1 + C2) x V22.
I don't need to repeat this test, since I already know the answer.
electrodacus:
I promised an explanation of what happens in the waveform captured by Derek.
Below with red is the voltage across the 1KOhm resistor with green is the supply voltage just after that R4 but it is an ideal voltage source so is basically the 20V supplied.
You can see that from simulation "LTspice" you get the same small power to resistor that is byproduct of charging the transmission line capacitance (I only simulated half the circuit as it is symmetrical).
Then at around 32ms current electron wave will travel half the line so at this time all line capacity is charged tho the part that charged first has already discharged not fully but significantly. Then as the electron travels on the other side of the transmission line the capacitance is being discharged starting with that far end continuing to maintain that limited amount of power for the lamp/resistor until about 65ns when the electron wave that left the source has finally arrived and rump up the lamp at full power (in fact slightly over as the remaining energy in capacitors is now basically fully discharged).
The inductors are also charged but at steady state they still contain significant amount of energy that will be delivered when switch is turned off.
Below is the power graph (area under the graph represents energy).
Here you can see that significantly more power is provided by the source than what it ends up as heat on the lamp/resistor in the first 65ns and that difference is stored energy. The wires are much lower resistance so while there is some thermal energy wasted there it is insignificant compared to the 1K resistor.
aetherist:
--- Quote from: electrodacus on May 07, 2022, 09:33:00 pm ---
--- Quote from: aetherist on May 07, 2022, 07:13:19 am ---I am a retired civil engineer.
That a half of the energy is lost during transfer is interesting, i don’t have a view re that, but it smells fishy.
I have kept out of capacitor to capacitor stuff here, except that i did say that almost everything said/written here has looked wrong to me. Remember i suggested u needed a 2nd switch.
I have agreed with u that the Poynting Field carries zero energy – u must be confusing me with someone else.
The gas cylinder with a membrane is a standard analogy for capacitors – something to do with waving away the need for displacement current or something.
Electrons can't cross from one plate to the other, they would have to jump through the air or whatever.
[G] If we shorted across the gap from one plate to the other then electrons would flow through the short, & would deliver energy (whatever that means). But this is never done.
[C] If we shorted around the circuit from one plate to the other then electrons would flow through the wire, & would deliver energy (whatever that means). This is always done.
In both cases no energy is transferred via the air.
But the amount of energy transfer will usually not be the same in [G] & [C].
--- End quote ---
There is no strange smell :) This is a particular case where both capacitors are identical and you charge one form another. You can imagine that a real resistor has a DC series resistance and for transient/AC impedance.
Since this are equal (identical capacitors) you will build a divider when you parallel them so if you measure symmetrical you will always measure 1/2 of the charged capacitor voltage from the time you parallel them to when they are in steady state so no more current flow.
If capacitors are not identical then more or less than half of the energy will be lost as heat the exact half is for the identical capacitor case.
Both G and C will convert the same amount of energy to heat.
--- End quote ---
Heat loss in G & C will depend on initial potentials. In the simplest case the heat loss would be equal, but in most cases it would not be equal (but this is a side issue)(not important).
Navigation
[0] Message Index
[#] Next page
[*] Previous page
Go to full version