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Veritasium "How Electricity Actually Works"
electrodacus:
--- Quote from: T3sl4co1l on April 30, 2022, 06:21:52 pm ---Then you will agree the final voltages will be sqrt(2)/2, not 0.5, or 1 (initial)?
Please provide waveforms showing this.
Tim
--- End quote ---
Are you confusing voltage with energy ?
For an ideal pair of identical capacitors (no ESR) one charged and one discharged, when paralleled the energy will remain the same thus voltage will drop to sqrt(2)/2
For a real pair of identical capacitors (ESR different from zero), when you parallel them the total energy will be 0.5 as the other half was lost as heat in the ESR the voltage will also be half in this particular example but is not to be confused with energy that also just happened to be half due to the symmetry of the example.
hamster_nz:
--- Quote from: electrodacus on April 30, 2022, 04:22:51 pm ---They will have half of energy each if you do not have ESR and capacitors have the same capacity.
--- End quote ---
No you won't. You need to take a long bath, relax and reflect on this. In particular how the resistance (in Ohms) between the two caps makes no difference in the outcome, only the time it takes to get there.
--- Quote from: electrodacus ---If capacity is different then energy stored in each will be proportional (still no ESR).
Only in the special case where you have identical capacitors with identical ESR that you end up with half the initial energy as the other half was lost as heat due to ESR.
--- End quote ---
With idea components (2 ideal capacitors and two ideal wires), there is only one place the excess energy can go - into the EM field, causing oscillation between all the energy being in one cap, and then being in the other, until the radiated EM waves removes half the energy from the system
pepsi:
Personally, I think the inital question Veritasium posed was misleading and that is what caused all the controversy. He shouldn't have asked how long before the light bulb lit up because for a practical engineer that is when you have adequate load current across the bulb.
electrodacus:
--- Quote from: hamster_nz on April 30, 2022, 07:54:29 pm ---No you won't. You need to take a long bath, relax and reflect on this. In particular how the resistance (in Ohms) between the two caps makes no difference in the outcome, only the time it takes to get there.
--- End quote ---
Take a piece of paper and draw a diagram showing two ideal capacitors with a series resistance. Then make the calculations and see what happens.
If you have 1Ws stored in the one capacitor then you parallel that with another identical capacitor that has no energy stored the energy will be split in half so each capacitor will have 0.5Ws total in both will be 1Ws as there is no loss.
Real capacitors have ESR thus half of the energy will be lost as heat thus in that case at the end each capacitor will have just 0.25Ws total 0.5Ws for both and the other 0.5Ws will have ended as heat.
--- Quote from: hamster_nz on April 30, 2022, 07:54:29 pm ---With idea components (2 ideal capacitors and two ideal wires), there is only one place the excess energy can go - into the EM field, causing oscillation between all the energy being in one cap, and then being in the other, until the radiated EM waves removes half the energy from the system
--- End quote ---
What excess energy ? In ideal case where you parallel two capacitors that have no ESR there will be no lost energy. The energy will just be redistributed in the two capacitors.
hamster_nz:
I suggest exactly the same, but you focus on charge (the actual conserved quantity), not energy.
Draw +s and -s on the plates it that helps.
Half the charge in each cap gives half the voltage, giving each capacitor a quarter of the energy of the original capacitor's charge.
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