General > General Technical Chat
Veritasium "How Electricity Actually Works"
electrodacus:
--- Quote from: Alex Eisenhut on May 27, 2022, 03:35:01 am ---So how does energy transfer from side of a vacuum tube diode to the other?
--- End quote ---
I'm not so old to have ever used a vacuum tube diode but a quick google revealed how that worked.
This is a direct quote that should provide you the answer:
"When the cathode of a vacuum tube is heated, electrons are emitted. If the anode has an electric potential higher than the cathode, the emitted electrons are attracted by it and an electric current start to flow "
So electrons do travel through vacuum in that particular case.
There are no electrons traveling from one wire to the other 1m apart wire in Derek's circuit thus energy travels through wire.
But the vacuum tube diode is a good example of energy traveling through vacuum so outside of the wire. It just that the vacuum tube diode has nothing to do with Derek's circuit where all energy delivered to lamp travels through wire.
SandyCox:
--- Quote from: electrodacus on May 26, 2022, 05:42:02 pm ---
--- Quote from: SandyCox on May 26, 2022, 05:32:37 pm ---What is your answer to my original question? How long will it take?
--- End quote ---
If as you say the shield is not connected to any point to the circuit then is not actually a shield so you still get energy through that lamp/resistor well before 65ns.
In simplest form what you say is
-| | |- a capacitor with a plate in the middle so like two capacitors in series. You can not call the middle plate a shield unless is connected to one of the sides.
--- End quote ---
A transmission line and a capacitor is not the same thing. Maybe you should think again.
What happens inside a perfectly conducting sealed enclosure stays inside the enclosure, and what happens outside stays outside. Magnetic and electric fields cannot exit or enter the enclosure.
So by shielding the electric and magnetic fields, you are forcing the "electricity" to go all the way round the loop.
electrodacus:
--- Quote from: SandyCox on May 27, 2022, 08:18:41 am ---
A transmission line and a capacitor is not the same thing. Maybe you should think again.
What happens inside a perfectly conducting sealed enclosure stays inside the enclosure, and what happens outside stays outside. Magnetic and electric fields cannot exit or enter the enclosure.
So by shielding the electric and magnetic fields, you are forcing the "electricity" to go all the way round the loop.
--- End quote ---
They are basically the same thing. Both the capacitor and the transmission line have resistance, inductance and capacitance just in different proportions.
If the shield as you say is not connected to any part of your circuit it will not do anything for this particular case.
The plate between the two capacitors plates unconnected to anything is useless.
-| | |-
While the simpler smaller shield I proposed will eliminate the energy being delivered to the lamp/resistor in the first 65ns as now the line capacitance is between that shield so current to charge that capacitor will flow through shield and not resistor.
SandyCox:
--- Quote from: electrodacus on May 27, 2022, 04:19:24 pm ---
--- Quote from: SandyCox on May 27, 2022, 08:18:41 am ---If the shield as you say is not connected to any part of your circuit it will not do anything for this particular case.
The plate between the two capacitors plates unconnected to anything is useless.
--- End quote ---
Nonsense!
Do you understand how shielding works? Have you heard of a Faraday cage?
--- End quote ---
electrodacus:
--- Quote from: SandyCox on May 27, 2022, 05:44:19 pm ---Nonsense!
Do you understand how shielding works? Have you heard of a Faraday cage?
--- End quote ---
What you have proposed is a special type of cage, is a rectangular donut :)
But leaving all this aside. Is there a way you can intrerup the energy flow from source to lamp/resistor without making a cut in the wire ?
Because if you can not that is another proof that energy travels through wire.
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