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Veritasium "How Electricity Actually Works"

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electrodacus:

--- Quote from: dunkemhigh on May 21, 2022, 07:59:47 pm ---
--- Quote ---Charging a capacitor means storing energy in to it.
No energy went around the capacitor but in to capacitor.
--- End quote ---

So what caused the resistor to heat up? If it had been an LED it would emit light. If it were a motor it would turn. You cannot charge that capacitor without those things happening and yet, according to you, there was no energy available to them. They would be perpetual motion machines running on nothing!

--- End quote ---

I do not get where did I say anything that looked like a perpetual motion machine ? It is a supper inefficient circuit only 50% efficient.
Supply needs to deliver 144mJ just to charge the capacitor with 72mJ so the other 50% 72mJ are lost on the wires or if you add a lamp or motor they do some work.

Just wires and capacitor 144mJ = 72mJ + 72mJ
Adding a 9Ohm incandescent lamp 144mJ = 64.8mJ + 7.2mJ + 72mJ

Where do you see a perpetual motion machine ?  144mJ are provided and some of that is lost as heat the rest is doing some work and half of it in this case remains stored in the capacitor.
I need to know where do you see the perpetuum mobile ?

snarkysparky:
Lets do a simple question.

the circuit here

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

If a DC voltage ( 20 V) is switched into X1 and X1 and left connected indefinitely will there be any current through R4 and R5 temporarily?

Simple yes /no question,

electrodacus:

--- Quote from: snarkysparky on May 21, 2022, 08:18:50 pm ---Lets do a simple question.

the circuit here

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

If a DC voltage ( 20 V) is switched into X1 and X1 and left connected indefinitely will there be any current through R4 and R5 temporarily?

Simple yes /no question,

--- End quote ---

Yes there will be a short current spike trough R4 and R5 as the C1 is charged but current will be very small maybe almost impossible to measure because there is a much larger capacitor C2 that will charge while being discharged across R3,R4 and R5
And there is that R2 that will continue to let some small current pass even after this initial spike.
If you remove R2 then it will be just that initial spike.

Why use that much more complex example when you have the super simple capacitor + wires + voltage source that we are discussing ?

If energy was to go through an ideal capacitor instead of in to it when charging then voltage across the capacitor will remain zero.  We know voltage across capacitor increases while it is being charged as all energy goes into the capacitor and remains there as stored energy.

snarkysparky:
Well I used it cause the simple example seems to be mixed up with a lot of phoohey.

So we know ...  and you agreed ..  that some energy made it through the circuit to R4 and R5 with the application of a dc source to the circuit input.

Do you dispute this ?

This energy seems to have "gotten by" the capacitor.  It is a series element in the circuit. 

So what do you say about this energy.   We know that if an AC source is connected the circuit functions as a power supply allowing energy to be extracted from X3 and X4.   A suddenly applied DC signal has some AC in it. 

If you still say that NO energy is transferred by the capacitor where did the energy for R4 and R5 come from?
Where does it come from if AC is applied to X3 and X4.




electrodacus:

--- Quote from: snarkysparky on May 21, 2022, 08:42:18 pm ---Well I used it cause the simple example seems to be mixed up with a lot of phoohey.

So we know ...  and you agreed ..  that some energy made it through the circuit to R4 and R5 with the application of a dc source to the circuit input.

Do you dispute this ?

This energy seems to have "gotten by" the capacitor.  It is a series element in the circuit. 

So what do you say about this energy.   We know that if an AC source is connected the circuit functions as a power supply allowing energy to be extracted from X3 and X4.   A suddenly applied DC signal has some AC in it. 

If you still say that NO energy is transferred by the capacitor where did the energy for R4 and R5 come from?
Where does it come from if AC is applied to X3 and X4.

--- End quote ---

Those resistors are part of the return for the capacitor that is being charged.
You can not charge that capacitor by just connecting one of the terminals from a battery or ideal voltage supply.
You can not have just electrons leaving the battery without same number returning and for that to happen you need a circuit.

How come that after capacitor is charged no more current is flowing ?  Why no energy can go "through" the capacitor after the first few ms ?
The answer should be obvious and that is energy went in to capacitor not trough.
All evidence points to this correct conclusion but for some reason you chose to ignore all evidence.
Once you have 72mJ stored in the 1000uF capacitor there is no more current flow and if you take the capacitor out of the circuit you can still measure 12V across and can calculate that 72mJ are contained as stored energy and you can also discharge that energy and measure that is all there all 72mJ. 

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