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Veritasium "How Electricity Actually Works"

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electrodacus:

--- Quote from: hamster_nz on April 30, 2022, 04:11:21 am ---Um, you can't do that "if there are no losses" bit.

https://en.wikipedia.org/wiki/Two_capacitor_paradox

--- End quote ---

I can do whatever I want :) and I sure can imagine a circuit with no losses.
There is no paradox so not quite sure why that wiki page even exists.
The energy in each capacitor will just be half of the energy stored in the charged capacitor at the beginning of the test that means the same energy when you add the energy in both capacitors.

"Thus the final energy Wf is equal to half of the initial energy Wi. Where did the other half of the initial energy go?"

The above quote from Wikipedia is just silly and written by someone that did not understood what he was writing.
The final energy is the same as the initial energy and can be tested even with real components (there will be a small deficit due to restive losses) but it will still be very close to the same energy.
Just get two identical capacitors with low self discharge charge one of them then parallel that with the discharged one and measure the voltage to get the final energy.

hamster_nz:

--- Quote from: electrodacus on April 30, 2022, 07:36:28 am ---
--- Quote from: hamster_nz on April 30, 2022, 04:11:21 am ---Um, you can't do that "if there are no losses" bit.

https://en.wikipedia.org/wiki/Two_capacitor_paradox

--- End quote ---

I can do whatever I want :) and I sure can imagine a circuit with no losses.
There is no paradox so not quite sure why that wiki page even exists.
The energy in each capacitor will just be half of the energy stored in the charged capacitor at the beginning of the test that means the same energy when you add the energy in both capacitors.

"Thus the final energy Wf is equal to half of the initial energy Wi. Where did the other half of the initial energy go?"

The above quote from Wikipedia is just silly and written by someone that did not understood what he was writing.
The final energy is the same as the initial energy and can be tested even with real components (there will be a small deficit due to restive losses) but it will still be very close to the same energy.
Just get two identical capacitors with low self discharge charge one of them then parallel that with the discharged one and measure the voltage to get the final energy.

--- End quote ---

(Assuming that I am not missing something, and the joke isn't on me)

I don't like playing with high-value charged capacitors, but the math is strongly on my side.

Capacitance (F) is Volts (V) per Coulomb of charge (Q). 

   C = V / Q

Energy stored in a capacitor is

    P = Q * V / 2

This can be proven with Calculus if you want.  NOTE Voltage is not Energy - a high voltage spark from static does not have more energy than an arc welder even though the voltage much, much higher!

So a 1F capacitor charged to 1V holds 0.5 Joule of Energy.

An experimenter charges a 1F capacitor to 1V, and then connects a second 1F capacitor to the first one in a loss-less way.

They now have two 1F capacitors, each with half a Coulomb each, and because V = C * Q they each will measure at 0.5V.

The energy in each capacitor is now 0.5 * 0.5 / 2 = 0.125 J, and for both capacitors that gives 0.25J in total.

Half of the energy has disappeared (well, in reality it will be lost into the environment somehow, maybe with a loud POP or a sizzle followed with swearing).

The only way out of this is if you can magic up additional 0.42 Coulomb of charge out of nowhere to make each 1F capacitor measure at 0.71 V  - because V = 0.71 and Q = 0.71 then P = V * Q = 0.5J.  But that can't happen because charge is a physical quantity and is conserved..

EDIT: sed s/power/energy/g

SandyCox:
Your reasoning is wrong from the start:
1. C = q/V not V/q
2. P is power. It is not the amount of energy stored in the capacitor. The stored energy is E = 1/2CV^2.

hamster_nz:

--- Quote from: SandyCox on April 30, 2022, 10:01:53 am ---Your reasoning is wrong from the start:
1. C = q/V not V/q
2. P is power. It is not the amount of energy stored in the capacitor. The stored energy is E = 1/2CV^2.
3. The final voltage across each of the capacitors will be 1/sqrt(2) V.

--- End quote ---

Yes, shouldn't post after dinner. A half pint of cider wipes me out.

(1) Yeah, my bad. Was standing on my head at the time of writing.

(2) Both are the same thing - see http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capeng.html if you want to double-check that

    U = 1/2*Q^2/C = 1/2 * Q * V = 1/2 * C * V^2

(3)  So each 1U capacitor will have V = 0.71, and as C = q/V and so q = 0.71? Where does the extra 0.21 Coulombs of charge come from for each capacitor?

hamster_nz:
Just went out to test.... definitely halves the voltage, not x 0.71



Not done under rigorous conditions - can't be bothered heating up the soldering iron. The caps are just bits and bobs from the junk draw with the same value and rating, so aren't perfectly matched, and so on.

EDIT: Updated video to have two meters.

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