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Veritasium "How Electricity Actually Works"

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bsfeechannel:

--- Quote from: electrodacus on May 09, 2022, 02:58:47 am ---I do not see how any engineer will be able to contest the above.

--- End quote ---

Then see. The energy resides in the dielectric between the plates of any of the charged capacitors and will happily migrate through space to the uncharged ones. The energy that goes into the wires and plates will be immediately dissipated.

Wires transfer charge. Space transfers energy.

hamster_nz:

--- Quote from: electrodacus on May 09, 2022, 03:53:23 am ---
--- Quote from: hamster_nz on May 09, 2022, 03:01:15 am ---
"Energy through a leaky capacitor"? What does "leaky" mean in the context of energy?

How long after removing from the circuit do you want me to leave the middle capacitor alone before I measure it? If it is 'leaky' and the energy is there because of such leaks, then it should self-discharge in a short period of time.

--- End quote ---

Leakage is an undesired effect of real capacitors and has nothing to do with anything related to the main question.
Energy flows in capacitors not through capacitors except for that pesky small current of few uA typical.
When you have a few mA or A of current in a loop with a capacitor in series then you have current going in to the capacitor not through it.
I sort of get frustrated and annoyed of explaining what in/out and trough means.

Se the problem above post #316 as that is proof energy only travels through wires not through the space around the wire or capacitors.

--- End quote ---

I too get frustrated when we talk about "energy flowing through" and then you reply talking about current. If you said something like "proof that charges only travels through wires not through the space around the wire or capacitors" I would agree with you (as long as it is at low voltages).

It is easily proven that we are able to move energy - actual Joules of measurable energy, that can do real work - from a battery into the middle capacitor of three in series. That energy will persists in the capacitor after it is removed from the circuit, and can be moved to the other side of the room. The rest of the original circuit could be disassembled, or even destroyed, and yet that energy still persists in what was the middle capacitor.

It should be undeniable proof that energy has been transferred from the battery into that middle capacitor, even though it has no other connection to the source except through the two capacitors. The only way for energy to get into that capacitor is either through the other two capacitors, or through the air around it. Both of which are things you say can't happen.

And the more 'ideal' the capacitors are the better this can be shown - an ideal capacitor with zero leakage could be left for months, and that energy would still be in there. Even with my less-than-ideal AliExpress components that energy sits there for minutes, if not hours.

It isn't as if my caps suddenly go all leaky when connected to a DC source - I have put a uA meter inline (once the caps are charged), and have measured exactly how little leakage is - under a uA at 10V.

hamster_nz:

--- Quote from: bsfeechannel on May 09, 2022, 04:31:52 am ---Wires transfer charge. Space transfers energy.

--- End quote ---

Oh.. that's good. I like that!

electrodacus:

--- Quote from: T3sl4co1l on May 09, 2022, 04:22:48 am ---If current "goes into" a capacitor and not "through" it then would you agree Vr = 0 for an initially discharged capacitor and voltage step Vs?


--- Code: ---     C
 +---||---+--o Vr
 |+       |
 Vs       R
 |-       |
 +--------+
_|_

--- End code ---

Tim

--- End quote ---

Electrons entering on one side of the capacitor to increase the density of free electrons means that electrons will need to leave the other plate.
But this energy is not used to do any work but stored and can be retrieved later to do work.
Say Vs is 3V and capacitor 1F the value of the resistor is irrelevant.
The capacitor will have a voltage potential of zero at the start so all 3V will drop across the resistor (will say resistor value is large enough that voltage source and capacitor DC resistance is low enough to to be of importance).
The charge current will start high then drop as the capacitor voltage increases until it gets to zero when voltage across the capacitor equal Vs (3V) then no current will flow so no energy will travel through the circuit (we ignore the capacitor small leakage current).
So now no energy can be transferred from Vs to resistor as current is basically zero.
Charging was inefficient but energy is stored in capacitor and can be used.
At all time no current (except that annoying leakage) has flown through the capacitor.
The way you charge a capacitor is by moving electrons from one plate to another but that is not done through dielectric but externally through wires.
There are now 4.5Ws of energy stored in the capacitor even if the power supply needed to deliver 9Ws as the other 4.5W was lost as heat on the resistor.
If you had a constant current power supply then no current resistor will have been needed for current limiting and power supply will have delivered 4.5Ws and all of that 4.5Ws will have been stored thus no work done at all / no heat loss.
Is like charging the capacitor with a linear regulator vs a DC-DC charger.

If you agree with the above (current flows in to capacitor as it is being charged and current flow stops when capacitor is fully charged) then you agree that energy is not flowing through the capacitor but in to capacitor.

hamster_nz:

--- Quote from: electrodacus on May 09, 2022, 05:02:17 am ---
--- Quote from: T3sl4co1l on May 09, 2022, 04:22:48 am ---If current "goes into" a capacitor and not "through" it then would you agree Vr = 0 for an initially discharged capacitor and voltage step Vs?


--- Code: ---     C
 +---||---+--o Vr
 |+       |
 Vs       R
 |-       |
 +--------+
_|_

--- End code ---

Tim

--- End quote ---

Electrons entering on one side of the capacitor to increase the density of free electrons means that electrons will need to leave the other plate.
But this energy is not used to do any work but stored and can be retrieved later to do work.
Say Vs is 3V and capacitor 1F the value of the resistor is irrelevant.
The capacitor will have a voltage potential of zero at the start so all 3V will drop across the resistor (will say resistor value is large enough that voltage source and capacitor DC resistance is low enough to to be of importance).
The charge current will start high then drop as the capacitor voltage increases until it gets to zero when voltage across the capacitor equal Vs (3V) then no current will flow so no energy will travel through the circuit (we ignore the capacitor small leakage current).
So now no energy can be transferred from Vs to resistor as current is basically zero.
Charging was inefficient but energy is stored in capacitor and can be used.
At all time no current (except that annoying leakage) has flown through the capacitor.
The way you charge a capacitor is by moving electrons from one plate to another but that is not done through dielectric but externally through wires.
There are now 4.5Ws of energy stored in the capacitor even if the power supply needed to deliver 9Ws as the other 4.5W was lost as heat on the resistor.
If you had a constant current power supply then no current resistor will have been needed for current limiting and power supply will have delivered 4.5Ws and all of that 4.5Ws will have been stored thus no work done at all / no heat loss.
Is like charging the capacitor with a linear regulator vs a DC-DC charger.

If you agree with the above (current flows in to capacitor as it is being charged and current flow stops when capacitor is fully charged) then you agree that energy is not flowing through the capacitor but in to capacitor.

--- End quote ---

If you agree with the above (current flows into the left hand side of the capacitor, and that current is flowing out of the right hand side of the capacitor, as it is being charged and both current flows stop when capacitor is fully charged) then you agree that during that time energy was flowing through the capacitor.

FTFY.

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