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| Veritasium "How Electricity Actually Works" |
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| T3sl4co1l:
--- Quote from: electrodacus on May 03, 2022, 09:50:37 pm --- --- Quote from: T3sl4co1l on May 03, 2022, 09:45:59 pm ---I do. I performed the test. I observed an oscillating waveform, with the average voltage being 0.5 Vi, and the peak sine amplitude being 0.5 Vi. Put another way, the capacitors slosh alternately between 0 and 1 Vi. At no point are they simultaneously 0.71 Vi; and the current in the wire connecting them (inductor, really), when one is at 0.71 Vi, is nonzero, so opening the switch at that instant would result in energy loss. Can you explain my findings? Tim --- End quote --- Let me guess. You used some simulation tool that was not setup properly or the simulation tool was wrongly designed. --- End quote --- I mean, you're welcome to scrutinize whatever models I use. But my oscilloscope speaks for itself: From the left, the voltage steps down corresponding to the switch turning on. Subsequently, the capacitor voltages (Ch1, Ch3) oscillate as predicted. I'm not seeing any trend towards 0.71 Vi here. Tim |
| rfeecs:
--- Quote from: hamster_nz on May 04, 2022, 12:18:26 am ---The unasked question so far is: Those pipes are hollow. What if you put a thick wire inside those pipes, electrically connected only at both ends of the pipe. You balance the cross-sectional area such that the wire and pipe have the same mm^2. After a constant current power supply is connected and the system reaches a steady state. How much current would flows through the inner wire, and how much through the outer pipe? --- End quote --- As with everything it seems, there is a video for that: He says the currents starts out on the outside of the conductor, then moves inside with time. A constant current source would be high impedance. Not sure that's the best choice for a high speed signal. So who's going to test it? |
| electrodacus:
--- Quote from: T3sl4co1l on May 04, 2022, 01:07:52 am ---I mean, you're welcome to scrutinize whatever models I use. But my oscilloscope speaks for itself: From the left, the voltage steps down corresponding to the switch turning on. Subsequently, the capacitor voltages (Ch1, Ch3) oscillate as predicted. I'm not seeing any trend towards 0.71 Vi here. Tim --- End quote --- What do you have on channel 1 and 3 and also what are you triggering on channel 4? Also what is the exact circuit ? |
| T3sl4co1l:
Channels 1 and 3 are the voltages at the two capacitors in the circuit: The scope is grounded to one side of the switch, hence the voltage step at turn-on. This also indicates the initial voltage. Ch4 is a sync signal triggering the switch. Digital averaging and bandwidth limiting (100MHz) have also been used to improve signal quality. The cursor at 4.2µs shows the half wave time. Tim |
| electrodacus:
--- Quote from: T3sl4co1l on May 04, 2022, 01:23:38 am ---Channels 1 and 3 are the voltages at the two capacitors in the circuit: The scope is grounded to one side of the switch, hence the voltage step at turn-on. This also indicates the initial voltage. Ch4 is a sync signal triggering the switch. Digital averaging and bandwidth limiting (100MHz) have also been used to improve signal quality. The cursor at 4.2µs shows the half wave time. Tim --- End quote --- A photo of the setup will be super useful. I do not get how your setup is made and there are to many unknowns. What capacitors are you using value and type. How are the capacitors connected (wires if so how long). How isolated is the system ? Like any power supply connected (even if just one of the wires from the power supply is connected). What sort of switch do you use ? Is that maybe a mosfet (what model) and how is that controlled ? Is is a photo diode optocoupler ? It may be supper important and likely the reason for osculations if your circuit is not properly isolated. |
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