General > General Technical Chat
Veritasium "How Electricity Actually Works"
TimFox:
--- Quote from: electrodacus on May 25, 2022, 11:07:04 pm ---
--- Quote from: TimFox on May 25, 2022, 07:10:44 pm ---
As I remember it, the aluminized Mylar sheets were roughly 1 by 2 m, and the double-sided metallization was extremely thin (< 1 um). Power is a strange concept in electrostatic speakers (capacitive load), but we were planning to drive them with quite high voltages directly from the plates of 6DQ5 tubes. In our initial calculations, we had forgotten that the material available (for free) was double-sided. We were trying to take advantage of the tooling for stringing the wires on a glass-epoxy frame.
In your example case B, when the charge flows from the left to the right face of the central solid conductor, it flows through a huge cross-sectional area (width times height) compared with the Mylar case where it flows through an area that is wide but very thin.
--- End quote ---
That was a large sheet (not something I had in mind) and the double metalization will not have been helpful quite the opposite as mylar as outside layer will have helped reduce the risk of a discharge when sheet got to close to the grid. You also needs some sort of transformer to add audio signal.
Yes in my case b) the heat loss due to conduction is fairly small.
--- End quote ---
Huge sheets: these were large high-energy physics contraptions. Full-size electrostatics (Quad and KLH) were common then, 6DQ5s were still in production, and lots of military-surplus high-voltage transformers were available.
A transformer is not needed if you can couple directly to the plates of the push-pull driver, using the Vbb and another bias voltage.
We were younger then, and full of large ambitions.
I was contrasting this situation to your case B.
electrodacus:
--- Quote from: TimFox on May 26, 2022, 02:44:42 am ---Huge sheets: these were large high-energy physics contraptions. Full-size electrostatics (Quad and KLH) were common then, 6DQ5s were still in production, and lots of military-surplus high-voltage transformers were available.
A transformer is not needed if you can couple directly to the plates of the push-pull driver, using the Vbb and another bias voltage.
We were younger then, and full of large ambitions.
I was contrasting this situation to your case B.
--- End quote ---
Yes it can work with some high voltage push-pull mosfet or IGBT and optically isolated gate driver but is more complex than just a transformer with center tap.
I did Electrical Engineering so high voltage / high energy stuff and we had laboratories for lightning simulation tho I was more passionate about low voltage automation electronics.
As a hobby I never did projects exceeding 60Vdc open circuit the safe low voltage standard limit.
Case b) and c) where done to show that the internal plates that are fully isolated from supply have no net charge. You need to go to something like d) and then disconnect the two surfaces in order to isolate the sides with different charges.
For example the setup in Derek's test was more like
f) -| |_[resistor]_| |-
So capacitors on each side and resistor/lamp in between (that is the transient part).
So while current flows through resistor it is electrons from one plate moving to the other plate as the capacitors charge from the source.
If source is then removed and you short the capacitors all stored energy will flow back through resistor to have those plates neutral again.
So amount of energy through the resistor is directly proportional with the amount of charge being stored in the capacitors and then when they are discharged the same amount will flow back.
So electrons are the ones that travel through wire and transport the energy.
If we shield one quarter of Derek's transmission line and connect that shied on the opposite side of the resistor this capacitive energy storage will be done around the resistor so no more energy before those 65ns or so but then despite this shielding energy arrives through wires at the resistor/lamp.
And it can be DC or AC it will still get to that lamp resistor through wires despite the shield that removes the capacitive effect of the line.
PlainName:
--- Quote ---I put in the effort to read what each one of you has to say.
--- End quote ---
That may be, but what you write in response to a post is usually irrelevant. It's just waffle as a diversion. For instance, just above this snarkysparky asked about three capacitors in series and clearly stated that it was the middle one that was interesting. Your response is 16 lines rabbiting on about parallel capacitors.
Either you didn't understand their short and to the point post, so just latched onto 'multiple capacitors' and made a guess at what it was about, or you deliberately obfuscated by diverting to a parallel setup (which is completely irrelevant).
Then, just to rub it in you then asked three questions designed to steer the discussion away from the series setup, and you will act miffed when those aren't answered.
electrodacus:
--- Quote from: dunkemhigh on May 26, 2022, 06:18:42 am ---
--- Quote ---I put in the effort to read what each one of you has to say.
--- End quote ---
That may be, but what you write in response to a post is usually irrelevant. It's just waffle as a diversion. For instance, just above this snarkysparky asked about three capacitors in series and clearly stated that it was the middle one that was interesting. Your response is 16 lines rabbiting on about parallel capacitors.
Either you didn't understand their short and to the point post, so just latched onto 'multiple capacitors' and made a guess at what it was about, or you deliberately obfuscated by diverting to a parallel setup (which is completely irrelevant).
Then, just to rub it in you then asked three questions designed to steer the discussion away from the series setup, and you will act miffed when those aren't answered.
--- End quote ---
What is the difference between this two ?
1) -| | | |-
2) -| |_| |_| |-
If you can not disconnect those links at 2) it is exactly the same as 1)
At 2) the electrons flow through that wire link to another plate but the net charge is zero as no electrons from battery get there.
As soon as you disconnect the links you have 3 separate and isolated capacitors and if all equal each contains 1/3 of the energy that battery delivered.
But you needed to take the action to cut the connections after the capacitor was charged.
No energy ever flowed from battery to the middle capacitor.
If you get all the middle plates out either on 1) or on 2) (without disconnecting the connections at 2) then the remaining two outer plates will contain the initial amount of energy. We will assume that the dielectric thickness remains about the same so plates have negligible thickness.
And yes I post long replays to explain how things actually work.
You still did not answered the main question for the last few posts. How will energy travel outside wire if you shield the wire from electric fields?
And also avoided to answer if proving that to you will be sufficient evidence or will you find some other invented theory to deny that electrical energy travels through wires.
PlainName:
--- Quote ---And also avoided to answer if proving that to you will be sufficient evidence or will you find some other invented theory to deny that electrical energy travels through wires.
--- End quote ---
If you can prove it one way or the other I would be happy to accept the definitive explanation.
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