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Veritasium "How Electricity Actually Works"

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PlainName:

--- Quote ---A capacitive dropper works because capacitor is always charged and discharged.
--- End quote ---

You are confusing charge with energy.


--- Quote ---No energy passed through capacitor
--- End quote ---

On the contrary, you mean no charge passed through the capacitor. But power went around the the whole circuit for the duration that the capacitor charged.

snarkysparky:
""So with DC you have just one pulse as the capacitor chargers and with AC you constantly charge and discharge the capacitor..""


Yes and in that one charging pulse there was a finite amount of energy available to drive a load on the other side of the capacitor.




electrodacus:

--- Quote from: dunkemhigh on May 21, 2022, 06:58:04 pm ---
--- Quote ---A capacitive dropper works because capacitor is always charged and discharged.
--- End quote ---

You are confusing charge with energy.


--- Quote ---No energy passed through capacitor
--- End quote ---

On the contrary, you mean no charge passed through the capacitor. But energy went around the the whole circuit for the duration that the capacitor charged.

--- End quote ---

Charging a capacitor means storing energy in to it.
No energy went around the capacitor but in to capacitor.

So when you are charging your phone battery is the energy going around you battery or in to your phone battery ?

The concept of energy storage seems to be completely missing from your understanding.

You have the concept that you can transfer electrical energy with just one wire so no closed loop.
 
You are probably thinking at a compressed air tank and how you take air from environment and push it in storing energy.
A more accurate way to think will be to have the tank split in two isolated chambers and you are moving air from one half of the cylinder to the other half to charge it and then to discharged it so use the energy you do the reverse let the air from the pressurized half go back in to the half with lower pressure.
This last example is much more accurately describing a rechargeable battery or capacitor.
For that air pump to work you need to connect both halves of the cylinder so you need to have a closed loop.
Now say you have a similar compressed air tank with two chambers that is identical but discharged so both sides contain air but at same pressure.
If you now connect this charged cylinder to the discharged one with two hoses the pressure delta will be lower and equal in both cylinders and now you have lost half of the energy as you only have a quarter of initial energy in each tank because this sort of transfer was 50% efficient.

electrodacus:

--- Quote from: snarkysparky on May 21, 2022, 07:10:43 pm ---""So with DC you have just one pulse as the capacitor chargers and with AC you constantly charge and discharge the capacitor..""

Yes and in that one charging pulse there was a finite amount of energy available to drive a load on the other side of the capacitor.

--- End quote ---

I'm trying very hard to understand what sort of thing you imagine happens.
Why was the amount of energy available finite ?
Why do you say the other side of the circuit ? It is irrelevant on what side of the circuit you load is.
You can connect the lamp on any side of the capacitor it will work exactly the same.

With that example that I provided you will have 9.5Ohms on the side you connect the lamp 9Ohms lamp + 0.5Ohms wire and on the other side you only have the 0.5Ohms wire.
On one side you have 72mJ * 0.95 = 68.4mJ and on the other side you have 3.6mJ so loss on wires as heat is now just 7.2mJ in total the rest is useful work assuming you wanted an incandescent lamp (you can replace the lamp with an electric motor or any other load you like).
There is a big difference between the 64.8mJ that ended on the lamp and the 72mJ that ended in capacitor.
The 64.8mJ already converted that energy in something that you wanted (light) while the 72mJ in the capacitor did nothing it is still available for you in original form as stored electrical energy.

PlainName:

--- 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!

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