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Veritasium "How Electricity Actually Works"
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electrodacus:

--- Quote from: SiliconWizard on May 03, 2022, 08:54:26 pm ---If you assume the bottom 'wire' between the two capacitors has zero impedance, then it's essentially just one capacitor. Isn't it?

--- End quote ---


One capacitor is charged at Vi  so if you consider that the switch is a capacitor (with it is in real world) the capacitance is so small and negligible that all voltage will be on the (switch/capacitor) and the charge transferred to the discharged capacitor is extremely negligible.
So you can see a real switch as an ideal switch in parallel with a very small value capacitor.
When you close the switch you are basically shorting the small capacity capacitor (switch capacity).

For all intents and purposes you can consider the empty capacitor on the right as containing no charge.
Also if you where to add another switch on the bottom you are just adding another small capacitor in series and the difference between one switch and two switches is likely not even measurable with normal lab equipment.
aetherist:

--- Quote from: electrodacus on May 03, 2022, 08:33:40 pm ---
--- Quote from: aetherist on May 03, 2022, 08:27:47 pm ---
Two capacitors cant exist like this. This circuit needs 2 switches, not 1 switch.
--- End quote ---
:) what?
Why will you need a second switch ? redundancy :)
--- End quote ---
If u had 2 open switches, with a charged capacitor on leftside & non-charged capacitor on righthandside, then if u closed one switch i think that the 2 capacitors would not end up with the same (say half each) charge. They would have i think a very mixed arrangement of 4 different charges.
T3sl4co1l:

--- Quote from: electrodacus on May 03, 2022, 06:53:37 pm ---What is exactly that you want to test ?
Is that moving energy from one charged capacitor to an identical discharged capacitor using a DC-DC converter will result in final voltage being significantly more than 1/2 and close to 0.707 ?

--- End quote ---

As you said earlier, and as I said earlier: without a DC-DC converter.

Can you do it?



--- Quote ---If energy is conserved

--- End quote ---

Is a very, very big 'if'.

So can you do it?  With just wires, superconducting or otherwise?  As you said to do it earlier?  I'm really curious to see what you propose.  I haven't seen a 'no', no indication that it be impossible.

Granted, it's been a challenge getting anything of substance out of you at all...  I'm beginning to think you don't actually know what you're talking about. ;)

Tim
electrodacus:

--- Quote from: aetherist on May 03, 2022, 09:16:41 pm ---
--- Quote from: electrodacus on May 03, 2022, 08:33:40 pm ---
--- Quote from: aetherist on May 03, 2022, 08:27:47 pm ---
Two capacitors cant exist like this. This circuit needs 2 switches, not 1 switch.
--- End quote ---
:) what?
Why will you need a second switch ? redundancy :)
--- End quote ---
If u had 2 open switches, with a charged capacitor on leftside & non-charged capacitor on righthandside, then if u closed one switch i think that the 2 capacitors would not end up with the same (say half each) charge. They would have i think a very mixed arrangement of 4 different charges.

--- End quote ---

If you consider the switch ideal (no capacitance) then if you close just one of two switches absolutely nothing will happen.
Real switches do have some capacitance depending on how they are constructed but that will be so small so many orders of magnitude that they can be ignored.
If you where to go to extreme and say that switch capacitance is the same as the two main capacitors in the test say 1F as in the example then when you charge the capacitor on the left to 3V you are also charging the two series capacitors the one from the switch and that one drawn in diagram as discharged thus you will see 1.5V on the right side capacitor.

To get rid of even the smallest switch capacitance that you can ignore anyway. You can just short the empty capacitor before closing the switch to get rid of the infinitesimal amount of charge it got there when you charged the one on the left.
Adding another switch will not help with anything as it will be just another small capacitor in series.

electrodacus:

--- Quote from: T3sl4co1l on May 03, 2022, 09:28:02 pm ---
As you said earlier, and as I said earlier: without a DC-DC converter.

Can you do it?

--- End quote ---

Do you have the equipment to construct two capacitors form super conductive materials to do such a test ?
Also the equations show clearly what the result will be.


--- Quote from: T3sl4co1l on May 03, 2022, 09:28:02 pm ---
--- Quote ---If energy is conserved

--- End quote ---

Is a very, very big 'if'.

So can you do it?  With just wires, superconducting or otherwise?  As you said to do it earlier?  I'm really curious to see what you propose.  I haven't seen a 'no', no indication that it be impossible.

Granted, it's been a challenge getting anything of substance out of you at all...  I'm beginning to think you don't actually know what you're talking about. ;)

Tim

--- End quote ---

Energy is always conserved it is the law. If you do not agree with that then it is you that will need to prove it.

In the case of identical capacitors with resistance half of the energy is in the capacitors after the switch is closed and half is dissipated as heat from the conductors (that includes wires and capacitor plates).
You are lacking fundamental understanding thus you do not understand my replays.
It is sad to see influential science communicators and university professors not understanding the subject that they should learn others.   
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