General > General Technical Chat
Veritasium "How Electricity Actually Works"
<< < (29/185) > >>
T3sl4co1l:
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
electrodacus:

--- 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.
hamster_nz:
Discussing the case where you have switches in both the top and bottom wires...


--- Quote from: electrodacus on May 03, 2022, 09:28:10 pm ---If you consider the switch ideal (no capacitance) then if you close just one of two switches absolutely nothing will happen.

--- End quote ---

Well, not absolutely nothing. If only one switch is closed, there will always be 10V measured over the other switch, so the act of closing one of the switches changes the voltages on the right hand side.

With both switches open the voltages on the right hand side are not at all clearly defined, as the two circuits are not connected.

(I'm waiting for some "the quantum wave function collapsed because you need to a meter to make the measurements" handwaving... from the person who adds DC-DC converters)

electrodacus:

--- Quote from: hamster_nz on May 03, 2022, 10:01:59 pm ---Discussing the case where you have switches in both the top and bottom wires...


--- Quote from: electrodacus on May 03, 2022, 09:28:10 pm ---If you consider the switch ideal (no capacitance) then if you close just one of two switches absolutely nothing will happen.

--- End quote ---

Well, not absolutely nothing. If only one switch is closed, there will always be 10V measured over the other switch, so the act of closing one of the switches changes the voltages on the right hand side.

With both switches open the voltages on the right hand side are not at all clearly defined, as the two circuits are not connected.

(I'm waiting for some "the quantum wave function collapsed because you need to a meter to make the measurements" handwaving... from the person who adds DC-DC converters)

--- End quote ---

We are talking about energy transfer.
I think I mentioned before that your multi-meter is basically 1Mohm resistor so it is like closing the switch and measuring the voltage drop on that 1MOhm resistor.
When you have two switches if you had a high resolution oscilloscope you will see a spike when you connect the oscilloscope probe as the probe impedance closes that circuit and there is a circuit because the other open switch (a real one has capacitance) thus you rearrange some of the charges.


You are concentrating on insignificant details that you do not even understand instead of the main subject of how the energy is transferred (trough wire my claim or outside of the wire Derek's claim that you seems to be defending without any basis).
Also your do not provide any feedback on where my math is wrong. Math that shows the amount of energy in a capacitor.
 
hamster_nz:

--- Quote from: electrodacus on May 03, 2022, 10:16:57 pm ---Also your do not provide any feedback on where my math is wrong. Math that shows the amount of energy in a capacitor.

--- End quote ---

I have done that many times.... but one more time...

Can we agree that the total charge on the top half of the system is:

  Qtotal = Cc1 Vc1 + Cc2 Vc2

So here is the initial state of the system:

  Qtotal =   1F * 3V + 1F * 0V = 3 Coulomb.

I hope both agree that no charge can cross either capacitor regardless of the position of the switch. So the total charges on the top and bottom halves must remain the same before and after the switch is closed. If not, please let me know what mechanism is taking electrons out of the top half and moving them to the bottom half, as I am sure we both agree that no positively charged atoms are moving around either.

The switch is closed, and allowing for the incorrect assumption that everything settles down to a stable state when using ideal components:

  Qtotal = Cc1 Vc1 + Cc2 Vc2
  Qtotal =   1F * 1.5V + 1F * 1.5 = 3 Coulomb.

This is the same as you would have if you consider the two 1F capacitors now a single 2F capacitor:

  Qtotal =   2F * 1.5V = 3 Coulomb.

We have the same amount of charge, but are left with only half the voltage over each capacitor. The incorrect assumption that the system becomes stable has let us down, as half the energy is missing, but ideal capacitors and wires have no resistive element to dissipate it.

A simpler example that shows how flawed your position is

What about a single charged ideal capacitor (no ESR), ideal wires (zero ohms) and a single ideal switch (zero ohms when closed, perfect isolation when open), that shorts the capacitor.  If, as you say, energy must be conserved, what happens if the capacitor has been charged to 3V and the switch is closed?

What happens with "close to ideal wires", that have zero ohms resistance, but a small amount of inductance?

The inductance is important, as it gives a solution where the total energy in the system is conserved.


Navigation
Message Index
Next page
Previous page
There was an error while thanking
Thanking...

Go to full version
Powered by SMFPacks Advanced Attachments Uploader Mod