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Veritasium "How Electricity Actually Works"
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PlainName:

--- Quote ---You are ignoring the capacitors formed by the transmission line wires one on each side of the lamp.
--- End quote ---

While in theory there would be some pumping via line capacitance, are you really saying that the amount shown in the experiment is solely due to the capacitance of two wires 1m apart? Seriously?
Naej:

--- Quote from: bsfeechannel on May 15, 2022, 04:15:22 pm ---
--- Quote from: Naej on May 13, 2022, 08:26:38 pm ---Right so there are plenty of problems with S=JV but you are forbidden to discuss them.

--- End quote ---
Of course there's a conspiracy of mainstream scientists whose aim is to silence any dissenters.

--- End quote ---
A good thing Derek is there to stop it. Youtubers shall prevail over mainstream scientists!
electrodacus:

--- Quote from: dunkemhigh on May 15, 2022, 06:56:14 pm ---
--- Quote ---You are ignoring the capacitors formed by the transmission line wires one on each side of the lamp.
--- End quote ---

While in theory there would be some pumping via line capacitance, are you really saying that the amount shown in the experiment is solely due to the capacitance of two wires 1m apart? Seriously?

--- End quote ---

Yes that is exactly what I'm saying and the simulation for that showed the exact same result. I did rough estimate of the line capacitance based on two pipes 2.5cm diameter 1 meter apart and so values I used in the simulation should be fairly equivalent to real test Derek made and there result where also very similar.
You need to remember that we are talking about 1Kohm resistor as the lamp/load and about 65ns
PlainName:
Wouldn't such a small capacitance be quickly discharged, so the scope should show a downward trend right after the leading edge? The time constant would be really really small, after all.

Edit: let's say the wire is 5mm across (actually a tube) and 5m long. The capacitance would be 0.21pf and the time constant 2.10e-12 aka 2.1femtoseconds.
electrodacus:

--- Quote from: dunkemhigh on May 15, 2022, 08:16:50 pm ---Wouldn't such a small capacitance be quickly discharged, so the scope should show a downward trend right after the leading edge? The time constant would be really really small, after all.

Edit: let's say the wire is 5mm across (actually a tube) and 5m long. The capacitance would be 0.21pf and the time constant 2.10e-12 aka 2.1femtoseconds.

--- End quote ---

It is a distributed capacitance and there is inductance involved so charging of each capacitor (seens a series of lumped components) will be delayed.
You can see the downward trend when the switch is open after was just 30ns closed but the wave and associated storage both as inductance and capacitance will travel all around the transmission line until it gets to lamp after about 65ns or so.
So the transmission line is not just a single capacitance and inductance plus resistance it is a continuous set of this.
I think I used 100 RLC elements for the simulation of a 10m transmission line so I divided the total line capacity in 100 small pcs then use that as a repeating circuit.
You where probably considering the capacitance of a single wire while that is fairly small in Dereks experiment you have the capacitance between two parallel wires that are 1m apart and I assumed 25mm diameter pipes (looked like 1" copper pipe).
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