General > General Technical Chat

Veritasium "How Electricity Actually Works"

<< < (119/185) > >>

Naej:
You could say the potential energy is qV so that there'll be two wires at different potentials connected to the resistor-lamp. So energy will flow from the wire into the lamp.

electrodacus:

--- Quote from: IanB on May 16, 2022, 08:01:25 pm ---
If current flows through the lamp/resistor while the capacitance is charging, then the lamp/resistor is consuming power and radiating it as heat and light. (Power = I2R) Where did that power come from?


Yes, and during that initial charging period the lamp lights up and consumes power. How did the power reach the lamp?


But in the transient the load receives power and therefore converts energy to heat and light. Where did that power come from during the transient?

--- End quote ---

You have a source (say battery) and you are charging a capacitor from that.
How if the energy from the source transferred to the capacitor ? Through wires.
There is some energy loss on the wires that ends up as heat in the wire and then is radiated to outside world.
Now in series on one of this wires you add a lamp. All you did was making one of those wires a higher resistance as lamp is also a wire.
You are asking how the energy is delivered from source to lamp and answer is through the wires.
All energy used by the lamp flowed through wires and when capacitor is fully charged so same voltage as the source lamp can no longer glow as no energy flows through it (basically a wire).

I guess the confusion is mostly related to what a capacitor is and the fact that energy flowing in capacitor is stored/conserved not used to do any work.

IanB:
Yes, OK. So let's put a capacitor on both sides, and put the lamp between the two capacitors. Now there is an air gap completely separating the lamp from the battery. But still the lamp lights up?

hamster_nz:

--- Quote from: IanB on May 16, 2022, 08:20:29 pm ---Yes, OK. So let's put a capacitor on both sides, and put the lamp between the two capacitors. Now there is an air gap completely separating the lamp from the battery. But still the lamp lights up?

--- End quote ---

We are going back to where we were in reply #266

electrodacus will say that the lamp (or in this case LEDs) are lit, but power doesn't flow through the capacitors, only in and out at the same time  :-//

electrodacus:

--- Quote from: IanB on May 16, 2022, 08:20:29 pm ---Yes, OK. So let's put a capacitor on both sides, and put the lamp between the two capacitors. Now there is an air gap completely separating the lamp from the battery. But still the lamp lights up?

--- End quote ---

There is no difference between one capacitor on one side of the lamp or capacitors on both sides.
An electron moved on one plate of a capacitor will push of an electron from the other plate and back in to battery with a single capacitor but if there are two capacitors in series then current will flow on the wire connecting the two capacitors as that one electron pushed off from the plate of first capacitor will on up on the plate of the other capacitor traveling through wire (is not the same electron that it will show up on the plate of the second capacitor but one displaced by that) then that electron on the second capacitor will push an electron in to the battery.
Again this current flow same as with just a capacitor will flow through wires only until capacitor/capacitors are charged. That energy stored in capacitor can be used latter to do work.
Like if you remove the battery and then connect a wire instead of the battery all energy that flowed into capacitors will now flow back out.

Navigation

[0] Message Index

[#] Next page

[*] Previous page

There was an error while thanking
Thanking...
Go to full version
Powered by SMFPacks Advanced Attachments Uploader Mod