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Veritasium "How Electricity Actually Works"
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electrodacus:

--- Quote from: hamster_nz on May 03, 2022, 03:09:30 am ---
Really? This? Again? Charge is the conserved quantity. In a capacitor the voltage is proportional to the charge (due to the very definition of Capacitance) so in this case voltage is conserved too. Electrical energy does not have to be conserved in an electric circuit - it is often converted to some other form of non-electrical energy (heat. light, motion, radio waves and sometimes smoke).

You will never, never, never, never, never, never, never, ever get a stable state with 0.707 * Vi.

If you don't believe me take it up with this random internet guy from Princeton:

https://physics.princeton.edu/~mcdonald/examples/twocaps.pdf

"If there were no damping (dissipative) mechanism, the circuit would then oscillate forever"

"[if there is a damping (dissipative) mechanism] eventually a static charge distribution results, with charge Qi/2 and voltage Vi/2, on each capacitor."

--- End quote ---

Seems like understanding what energy is is a big problem in general. This leads to a lot of confusion.

Here I will provide you with a full example for the identical parallel capacitor problem.


Vi = 3V
C = 1F

So energy contained in the charged capacitor = 0.5 * 1F * 3V2 = 4.5Ws (4.5J if you prefer).
In ideal case so no resistance in the circuit half of this energy will be transferred to the empty capacitor that is otherwise identical.
Now you can say you have a 2F equivalent capacitor that contains 4.5Ws (energy can not go anywhere else as there is no resistance in this ideal circuit).
or if you open the switch after current flow has stopped you have two 1F capacitors each containing 2.25Ws
Voltage on the now parallel capacitors in ideal case will be 3V * 0.707 = 2.121V

You can test that by calculating the energy stored in now the 2F equivalent capacitor
0.5 * 2F * 2.1212 = 4.5Ws

So energy is conserved because it is an ideal circuit with no losses.

Now for the real circuit will need to simplify things to keep them simple as a real capacitor will be exactly the same as a transmission line so capacitance inductance and resistance.  We can ignore the inductance as that also is an energy storage device so there will be no losses in this circuit because of that or not significant anyway.
Say the capacitors have 1Ohm equivalent resistance and this includes the terminals and wires to keep things super simple.
I do not have a drawing but you can just imagine 1Ohm resistor in series with each capacitor.

When you close the switch the voltage at the switch will already be half of the charged capacitor voltage so 1.5V in this example that is thanks to the resistor voltage divider.
Should be simple to understand why half of the energy will be dissipated on this resistance in the circuit no matter what the resistor value is.
It is half only because the two capacitors are identical else it will be more or less than half the energy that will be lost during transfer.
With half the energy remaining in the capacitors after current has stopped flowing you have just 1.5V

0.5 * 2F * 1.5V2 = 2.25Ws
The other 2.25Ws where dissipated on the resistors as energy was flowing trough the conductors and they have a resistance.

hamster_nz:

--- Quote from: electrodacus on May 03, 2022, 03:50:46 am ---You can test that by calculating the energy stored in now the 2F equivalent capacitor
0.5 * 2F * 2.1212 = 4.5Ws

--- End quote ---
Yes, I do understand why in all cases half the energy is lost. The 'ideal' case is not realizable.

0.707 Vi is never a solution, even with ideal components. You can take a different path through the math of the system and end up with a different answer, which is 0.50 Vi. With ideal components the system is either in a constant state of change (oscillating) or cannot be solved to a consistent answer.

Edit: Sorry about the random edits
electrodacus:

--- Quote from: hamster_nz on May 03, 2022, 04:00:40 am ---Is the "ideal" system in a steady state (i.e. without oscillation?)

If so can you please tell me how much charge is in each of those ideal 1F capacitors, (or the ideal 2F combined one, if you like that better)? I am pretty sure the formula is Q = CV.

As we have no source of charges in this system (aside from the initial charged cap) if this is more than the initial charge, where these additional charges have appeared from? No charges are able to cross either capacitor, and we can't just magic up +s and -s out of nowhere.

And yes, I do understand why in all cases half the energy is lost. The 'ideal' case is not realizable, and your solution of 0.707 Vi is inconsistent, because the Lumped Element Model is just an approximation of reality.

--- End quote ---

Yes there is no oscillation.

You can get super close to ideal if you use an efficient DC-DC converter to transfer the energy from the charged capacitor to the discharged capacitor.
I think you did the experiment with the parallel capacitors (unless I remember wrong).
Get maybe two super capacitors as they contain more energy and it will be easier to find a DC-DC converter to work with those.
There may be some energy harvesting ultra low power DC-DC converters. The parallel capacitors is like using a linear regulator.

Please understand that energy conservation can not be broken and you do not care about the charge but about the energy. You have the impression that charge is linear but you need more energy to push the second electron in compared to the first one.
electrodacus:

--- Quote from: hamster_nz on May 03, 2022, 04:00:40 am ---
--- Quote from: electrodacus on May 03, 2022, 03:50:46 am ---You can test that by calculating the energy stored in now the 2F equivalent capacitor
0.5 * 2F * 2.1212 = 4.5Ws

--- End quote ---
Yes, I do understand why in all cases half the energy is lost. The 'ideal' case is not realizable.

0.707 Vi is never a solution, even with ideal components. You can take a different path through the math of the system and end up with a different answer, which is 0.50 Vi. With ideal components the system is either in a constant state of change (oscillating) or cannot be solved to a consistent answer.

Edit: Sorry about the random edits

--- End quote ---

You used the wrong math and likely also why you think the system will oscillate.

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.
hamster_nz:

--- Quote ---Please understand that energy conservation can not be broken.

--- End quote ---
Total energy is conserved, but Electrical Energy isn't. If you drive a lightbulb, you convert electrical energy into heat and light. If you drive a speaker you get noise and heat. If you have a solar cell you can convert light into Electrical Energy. If you have a turbine you can convert gravitational potential energy of water into Volts and Amps.


--- Quote from: electrodacus on May 03, 2022, 04:19:34 am ---You have the impression that charge is linear but you need more energy to push the second electron in compared to the first one.

--- End quote ---

That is exactly the point. The half of the charge that comes out of the charged capacitor has 3/4th of the energy in the capacitor.

What you are saying is we take half the energy (which is the 'last' 29% of the charge that took it from 0.707 Vi to Vi ) and put it in the other capacitor and they are now balanced! It isn't - because there is 71% of the charge in one capacitor and 29% of the charge in the second, causing a potential differences and potential differences cause more charge to flow. Once the switch is pushed the only stable state is with 50% of the charge in each capacitor, and by very nature of capacitance 0.5 Vi.

(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).
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