General > General Technical Chat
Veritasium "How Electricity Actually Works"
Naej:
--- Quote from: bsfeechannel on May 13, 2022, 03:23:17 am ---
--- Quote from: Naej on May 13, 2022, 12:21:39 am ---Really, where then?
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Now you're trolling.
--- Quote ---You gave no example of an alternative breaking any of this ;D . Coincidence?
--- End quote ---
If you couldn't recognize it, this means you didn't go very far in your understanding of the problem.
--- End quote ---
Right so there are plenty of problems with S=JV but you are forbidden to discuss them.
--- Quote from: bsfeechannel on May 13, 2022, 03:23:17 am ---
--- Quote ---Sure so in the 20th century and 21st physicists never heard of antennae. ::)
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Who knows? What is important to understand is that the alternatives never prospered.
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Prosperity is knowledge. Or science I don't know. Or is it scientology?
And influencers like Derek are needed to keep physicists in the right direction. It's definitely how science works.
--- Quote from: hamster_nz on May 13, 2022, 02:20:20 am ---
--- Quote from: Naej on May 13, 2022, 12:43:42 am ---And it doesn't bother you at all that energy is supposedly flowing through the region A yet there is no circuit you can put in region A in the diagram that can extract energy from it?
(Hopefully the answer is no)
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No, with no voltage potential over region A, there is no way you can extract energy from the field in just that region.
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So what really matters is the potential, while Poynting's energy flow give no useful information.
--- Quote from: hamster_nz on May 13, 2022, 02:20:20 am ---
--- Quote from: Naej on May 13, 2022, 12:43:42 am ---Are you bothered???
Do you think that if every electron could communicate with far away electrons, humans could make machines hacking this property to communicate messages?
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It did bother me just a little - Every charge being constantly aware of every other charge in the universe does not have the feel of being fundamental to the universe.
An electric field can still be used to communicate messages. "Ripples in the fields" is preferable to "very small pushes and shoves over great distances".
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Every night, you can check that electrons in your eye are "aware" of what's going on with electrons in far away stars.
--- Quote from: hamster_nz on May 13, 2022, 02:20:20 am ---
--- Quote from: Naej on May 13, 2022, 12:43:42 am ---A wire is an inductor. With 2 you have an inductor and half a capacitor. And it works even for infinitesimal wires. 8)
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Fair call. But I still feel that a lumped model of a continuous thing is (very useful) approximation.
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Yes it's quite bizarre: the lumped model approximation has no reason to be a good approximation with light-speed processes, yet it is an extremely precise one with transmission lines.
I guess the main reason is they hardly behave as antennae, due to their counter-currents.
TimFox:
--- Quote from: electrodacus on May 13, 2022, 08:00:30 pm ---
--- Quote from: TimFox on May 13, 2022, 07:42:41 pm ---(PS: when I distinguished active and passive circuits, I mean that an active circuit has an external power source. So a diode in the dark is passive, but it has losses.)
What happens if you leave S2 on for a long time in your simulation?
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What do you think it will happen ?
The energy will flow back and forth between the two capacitors and inductor (RLC resonant circuit) until half of the initial energy will be dissipated as heat so you end up with half the voltage.
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Yes, with two equal capacitors and any passive two-terminal network between them, that is what I expect. Each capacitor has 1/2 the original voltage, and therefore 1/2 the original charge and 1/4 the original total energy in both capacitors, until the leakage resistance not included in the model eventually discharges both of them. I'm busy this weekend, but I will do a very simple experiment next week with polypropylene capacitors and a 10 megohm voltmeter.
If I use two 10 uF capacitors, and then one 10 uF (original charge) and a second 20 uF (originally discharged), I will have discharge time constants of 200 sec (both caps) and 300 sec (both caps), so I will lose 1% of the charge due to the parasitic voltmeter in 2 sec or 3 sec, respectively, after I throw the switch, first disconnecting C1 from the battery, and then connecting C1 to C2 (DPDT CO, L&N huge honking switch).
electrodacus:
--- Quote from: TimFox on May 13, 2022, 08:51:37 pm ---Yes, with two equal capacitors and any passive two-terminal network between them, that is what I expect. Each capacitor has 1/2 the original voltage, and therefore 1/2 the original charge and 1/4 the original total energy in both capacitors, until the leakage resistance not included in the model eventually discharges both of them. I'm busy this weekend, but I will do a very simple experiment next week with polypropylene capacitors and a 10 megohm voltmeter.
If I use two 10 uF capacitors, and then one 10 uF (original charge) and a second 20 uF (originally discharged), I will have discharge time constants of 200 sec (both caps) and 300 sec (both caps), so I will lose 1% of the charge due to the parasitic voltmeter in 2 sec or 3 sec, respectively, after I throw the switch, first disconnecting C1 from the battery, and then connecting C1 to C2 (DPDT CO, L&N huge honking switch).
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Not sure I understand what you want to test.
Will you parallel two 10uF capacitors so that you get a 20uF and use that 20uF capacitor as the discharged capacitor and then use a 10uF charged capacitor from witch you will charge the 20uF one ?
If that is the case then you will see that voltage will drop down to 33% so 1V if your charged capacitor is at 3V initially.
So a lot more energy will be lost as heat compared to the two identical capacitor test.
You start with 45uWs of energy 0.5 * 10uF * 32
And you end up with 15uWs of energy 0.5 * 30uF * 12
The other 30uWs ends up as heat in the conductors.
TimFox:
A simple demonstration at maybe 2% accuracy:
Connect voltmeter across C1, and discharge C2
Charge C1 = 10 uF to 9 V (battery), measure the voltage immediately after opening the battery-C1 switch (must be faster than a few seconds with the 100 sec time constant of 10 uF & 10 megohm).
Close C1 to C2 switch and measure voltage as soon as it settles, paying attention to the 200 sec discharge time constant.
Do again with C2 = 20 uF to demonstrate unequal capacitor case.
It should agree with your calculation, to show that total charge rather than total energy in the open system is conserved.
electrodacus:
--- Quote from: TimFox on May 13, 2022, 09:25:21 pm ---A simple demonstration at maybe 2% accuracy:
Connect voltmeter across C1, and discharge C2
Charge C1 = 10 uF to 9 V (battery), measure the voltage immediately after opening the battery-C1 switch (must be faster than a few seconds with the 100 sec time constant of 10 uF & 10 megohm).
Close C1 to C2 switch and measure voltage as soon as it settles, paying attention to the 200 sec discharge time constant.
Do again with C2 = 20 uF to demonstrate unequal capacitor case.
It should agree with your calculation, to show that total charge rather than total energy in the open system is conserved.
--- End quote ---
Total energy is always conserved in an isolated system.
So in this case 1)
charged 10uF charging a discharged 10uF half of the energy will remain in the two combined capacitors and half of the energy will be dissipated as heat.
In case 2)
charged 10uF charging a discharged 20uF one third of the energy will remain stored in the capacitors while two thirds will be dissipated as heat so again energy is conserved.
There is no rule about charge being conserved it just happen due to symmetry as there is resistance everywhere including the capacitor plates.
If you add a discharged inductor in the circuit you are not adding either charge or energy and you are just adding another energy storage device as an intermediary.
The point is that energy is always conserved in an isolated system and using this equation is easy to see where energy ended up (in the conductors) from witch should be easy to conclude what path the energy has traveled.
If energy transfer was not done through conductors then there will be no reason to have all the energy loss (exactly all) inside the conductors.
The symmetry is broken when you add an inductor even if you are actually not bringing any charge from outside and system is still isolated. Inductor is basically a fairly long wire.
Edit:
I'm thinking of ways you can break the symmetry but unless one of the capacitors has superconductor plates I do not see how it can be done other than by adding an inductor as intermediary energy storage device.
Parallel capacitors will always have this symmetry as they have resistance (they are like a transmission line).
Adding an inductor will just break this symmetry reducing the loss as heat in wire.
I will try to think of a mechanical analog.
It will not make sense to do the experiment as you probably agree with me on what the result of that test will be.
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