| General > General Technical Chat |
| Veritasium "How Electricity Actually Works" |
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| electrodacus:
--- 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. |
| PlainName:
--- 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 --- I wish you luck and great patience :popcorn: |
| hamster_nz:
--- Quote from: electrodacus on May 26, 2022, 03:44:34 pm ---So you need to replace that right side with coaxial cable or just use at least a half pipe (half cylinder or U shaped) to shield the wire form the electric field generated by the lower wire. --- End quote --- um, what "electric field generated by the lower wire"? You were arguing that the energy is all in the wires. So why the need to shield from what you consider a non-existent electric field? |
| electrodacus:
--- Quote from: hamster_nz on May 26, 2022, 11:58:03 pm --- --- Quote from: electrodacus on May 26, 2022, 03:44:34 pm ---So you need to replace that right side with coaxial cable or just use at least a half pipe (half cylinder or U shaped) to shield the wire form the electric field generated by the lower wire. --- End quote --- um, what "electric field generated by the lower wire"? You were arguing that the energy is all in the wires. So why the need to shield from what you consider a non-existent electric field? --- End quote --- The lower line and top line form a capacitor that will be charged. There is one capacitor made by the lines on each side. If you add that shield then the shield will be the other plate of that capacitor. If you connect that to the other side of the resistor the current to charge the line capacitance will no longer flow through the lamp/resistor. Since the only so called "proof" Derek had was that energy arriving earlier than 65ns. Adding this shield will get rid of that energy and still energy will get to resistor and travel through wires. Derek's claim "energy doesn't travel through wires" and I can say just make a cut in the wire and see if any energy gets to lamp/resistor. |
| Alex Eisenhut:
So how does energy transfer from side of a vacuum tube diode to the other? |
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