General > General Technical Chat
Veritasium "How Electricity Actually Works"
bsfeechannel:
--- Quote from: SandyCox on May 12, 2022, 08:07:01 am ---Don't let the pretty pictures distract you.
--- End quote ---
Like this one by Poynting himself, for instance, showing the energy going from a battery to a resistive wire loop through the empty space.
Don't look at that. It'll shatter your faith in the wires.
hamster_nz:
--- Quote from: electrodacus on May 13, 2022, 02:29:58 am ---
--- Quote from: hamster_nz on May 13, 2022, 02:23:00 am ---You are a hard person to goad something out of, but eventually it works. :)
--- End quote ---
Is that your replay ? How about you understand now that if you transfer energy more efficiently from one capacitor to the other that is identical you get close to 0.707 the voltage of the charged capacitor in both.
It should not be me that is doing the work.
--- End quote ---
You keep on bringing up DC-DC converters
Reply 118:
--- Quote from: electrodacus on May 03, 2022, 04:36:08 am ---Just test with a DC-DC converter and you will see very close to ideal is possible. If the DC-DC converter was 100% efficient (so ideal) then you get 0.707 Vi but you do not need a DC-DC converter if there is no resistance to get the same result.
--- End quote ---
Everybody (even I) agree that it is possible:
Replay 119:
--- Quote from: hamster_nz on May 03, 2022, 04:52:27 am ---(And of course if you have an ultra efficiency DC-DC convertor to make the transfer then you might get close to 71%, but we don't. We have two ideal caps, some ideal wire and an ideal switch).
--- End quote ---
But you keep on tuning it back to "oh, if I use a DC/DC converter it proves something". So I called your bluff.
So finally we have the proof that everybody else missed. That if you transfer energy to and from a magnetic field, using an inductor, (thus taking the energy "out of the wire") that energy only flows in the wires. :-//
electrodacus:
--- Quote from: hamster_nz on May 13, 2022, 04:37:33 am ---
You keep on bringing up DC-DC converters
...
So finally we have the proof that everybody else missed. That if you transfer energy to and from a magnetic field, using an inductor, (thus taking the energy "out of the wire") that energy only flows in the wires. :-//
--- End quote ---
At first you and others mentioned that you can not have 0.707 * Vi because that will violate the conservation of charge witch likely you confused with energy.
Then after pointing out the formula for energy stored in a capacitor it was still argued that 0.707 * Vi is not possible in an circuit made with superconductors so zero resistance.
Then I mentioned something that is way easier to test (a DC-DC converter) and claim remained the same that voltage at the end of the experiment can not be higher than 0.5 Vi.
The energy seen through the bulb in the first 65ns in Derek's experiment is related to line capacity so no energy travels through air.
Any transmission line has both inductance and capacitance (both are energy storage devices) so the transmission line model that any electrical engineer uses is a realistic and accurate way to predict what happens.
I simplified the two capacitors spice model even more and below is the schematic and the graph
In the schematic you have a 3V supply to charge one of the 1000uF capacitors then the supply is disconnected by the switch S1 after 20ms and then starting at 40ms the S2 switch is closed for just 6ms then open.
The charged capacitor C1 will discharged from 3V down to around 2V while the C2 that started fully discharged is charged using that energy at 2V so you start with C1 at 3V and you end up with both identical capacitors charged at 2V using only energy available in C1
The only other two components are the 47mH inductor and a diode. The inductor is an intermediary energy storage helping reduce the losses due to circuit resistance. There is no outside energy coming in to the circuit and some is still lost as heat but much less than just paralleling the two capacitors directly.
The red graph is the control for the S2 switch showing the period switch was closed. The blue is the voltage across C1 and green voltage across C2.
SandyCox:
Here's another one:
Let's connect and ideal DC voltage source V_d to a fully discharged capacitor C at t=0s:
1. How much energy is delivered by the source?
2. How much energy is stored in the capacitor at t=1s?
3. How much charge is delivered by the source?
4. How much charge ends up in the capacitor?
You can try the same with a current source and an inductor.
SandyCox:
--- Quote from: electrodacus on May 13, 2022, 05:04:17 am ---
--- Quote from: hamster_nz on May 13, 2022, 04:37:33 am ---
You keep on bringing up DC-DC converters
...
So finally we have the proof that everybody else missed. That if you transfer energy to and from a magnetic field, using an inductor, (thus taking the energy "out of the wire") that energy only flows in the wires. :-//
--- End quote ---
At first you and others mentioned that you can not have 0.707 * Vi because that will violate the conservation of charge witch likely you confused with energy.
Then after pointing out the formula for energy stored in a capacitor it was still argued that 0.707 * Vi is not possible in an circuit made with superconductors so zero resistance.
Then I mentioned something that is way easier to test (a DC-DC converter) and claim remained the same that voltage at the end of the experiment can not be higher than 0.5 Vi.
The energy seen through the bulb in the first 65ns in Derek's experiment is related to line capacity so no energy travels through air.
Any transmission line has both inductance and capacitance (both are energy storage devices) so the transmission line model that any electrical engineer uses is a realistic and accurate way to predict what happens.
I simplified the two capacitors spice model even more and below is the schematic and the graph
In the schematic you have a 3V supply to charge one of the 1000uF capacitors then the supply is disconnected by the switch S1 after 20ms and then starting at 40ms the S2 switch is closed for just 6ms then open.
The charged capacitor C1 will discharged from 3V down to around 2V while the C2 that started fully discharged is charged using that energy at 2V so you start with C1 at 3V and you end up with both identical capacitors charged at 2V using only energy available in C1
The only other two components are the 47mH inductor and a diode. The inductor is an intermediary energy storage helping reduce the losses due to circuit resistance. There is no outside energy coming in to the circuit and some is still lost as heat but much less than just paralleling the two capacitors directly.
The red graph is the control for the S2 switch showing the period switch was closed. The blue is the voltage across C1 and green voltage across C2.
(Attachment Link)
(Attachment Link)
--- End quote ---
You cannot add inductors and diodes or DC-to-DC converters. The problem is about connecting two capacitors in parallel without any other components.
How do you think energy is transferred through a capacitor? Let's take a parallel plate capacitor with vacuum between the plates.
What about the airgap of a transformer?
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