General > General Technical Chat
Veritasium "How Electricity Actually Works"
snarkysparky:
"half that ending as heat due to wire resistance"
It took power to heat the wire. That power came from the flow of electrons into and out of the capacitor wires.
We consider the heat in the wires to be useful work.
snarkysparky:
Are we still really arguing that power can be transferred through a capacitor???
If anybody says NO then would they please explain how this circuit gets power out the right side when connected to other power source through only a capacitor.
https://en.wikipedia.org/wiki/Capacitive_power_supply
electrodacus:
--- Quote from: snarkysparky on May 21, 2022, 05:37:03 pm ---"half that ending as heat due to wire resistance"
It took power to heat the wire. That power came from the flow of electrons into and out of the capacitor wires.
We consider the heat in the wires to be useful work.
--- End quote ---
Yes electrons flowed in one side of the capacitor and other unrelated electrons flowed out of the other side but this means energy was stored in the capacitor.
When you will discharge the capacitor say by shorting the terminals of that capacitor electrons that flowed in will flow out and in to the side where electrons flowed out resulting in 72mJ of heat (the amount that was stored in the capacitor).
snarkysparky:
"Yes electrons flowed in one side of the capacitor and other unrelated electrons flowed out of the other side but this means energy was stored in the capacitor."
And that energy that heated the wires flowed because of charge moving into and out of the capacitor.
Do you agree that a capacitor can not only store energy but can also pass through energy?
If not how does the capacitor power supply above work?
electrodacus:
--- Quote from: snarkysparky on May 21, 2022, 06:06:05 pm ---And that energy that heated the wires flowed because of charge moving into and out of the capacitor.
Do you agree that a capacitor can not only store energy but can also pass through energy?
If not how does the capacitor power supply above work?
--- End quote ---
Yes wires had resistance to current flow so part of total energy 144mJ ended as heat. 72mJ to be exact ended as heat and the other 72mJ ended as stored energy in the capacitor.
No energy passed through capacitor as there is nothing left 144mJ = 72mJ lost as heat + 72mJ stored in the capacitor.
A capacitive dropper works because capacitor is always charged and discharged.
For example let say you want to power an incandescent lamp in the example I provided before.
You just add a lamp in series with the ideal DC voltage supply and the capacitor and say the lamp resistance is 9 Ohms.
Then you will have 144mJ from the DC supply form that 72mJ will end up stored in the capacitor
The other 72mJ will be lost on the wire's and lamp total 10Ohms = (9Ohm + 0.5Ohm + 0.5Ohm)
So 90% will be delivered to the incandescent lamp 64.8mJ (mostly infrared photons and some visible light photons) and the other 10% will end as heat loss on the wires 7.2mJ
But that is all done in a few ms then you can wait for an hour and there will be no more energy flowing just this 144mJ from which half was stored and half was used.
But if you reverse the polarity of the power supply or you even just remove the supply and close the circuit the capacitor will deliver the stored 72mJ to the lamp and wires then you can repeat this.
So with DC you have just one pulse as the capacitor chargers and with AC you constantly charge and discharge the capacitor.
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