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Veritasium "How Electricity Actually Works"
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SiliconWizard:

--- Quote from: rfeecs on May 01, 2022, 08:06:38 pm ---
--- Quote from: SiliconWizard on May 01, 2022, 06:28:07 pm ---While the new video is a good effort and gets into a lot more details than the initial one, there still are points that are definitely not addressed.
Some of them come from - as was mentioned in the other thread early - a chicken-and-egg problem.

--- End quote ---

Something like do moving charges cause fields or do fields cause charges to move?
--- End quote ---

That's one of them, yes.


--- Quote from: rfeecs on May 01, 2022, 08:06:38 pm ---Perhaps he addressed that in saying electrons don't push each other.
He comes down on the side of fields cause charges to move.

--- End quote ---

So, if he said so, that must be settled? ;D

To extend the matter a bit, what about gravity?
electrodacus:

--- Quote from: rfeecs on May 01, 2022, 08:32:07 pm ---You are using a lumped model that is not adequate.

This circuit is three dimensional.  There is a one meter gap between the switch and the load.  You haven't explained how energy gets across that gap.

You can look at the simulation in the video.  When the switch is closed a spherical wave propagates out in all directions.  It should be very clear that some of that energy is going out away from the bulb.  Some of the energy is lost.  Your model doesn't account for the energy radiated away.

Consequently your equation is wrong.

--- End quote ---
There is no energy radiated away. This is a copy paste of a replay I just made to someone else but is applies to your question also.

Not sure how much you understand a battery so is best to replace that with a charged capacitor as it is simpler to understand than a battery.

----------------[RESISTOR]--------------------
-------------------{-CAP+}--s/ ------------------

The open loop above is just a charged capacitor "CAP" not connected to anything if the switch is open (ignoring the super small switch capacitance).
As soon as you close the switch "s/"  you are paralleling the charged "CAP" with the two series capacitors formed by the lines on each side and those caps are in series with the resistor but that is not very relevant (it is just like having a wire there).

So what you have when the switch is closed is a closed loop made up of 3 capacitors in series. You can consider those two discharged capacitors in series as a single capacitor and then simplification will be a charged capacitor in parallel with a discharged capacitor.
 _________I I__________
I                                         I
I                                         I
I                                         I
I_________I I__________I

There is no current flowing trough the capacitor dielectric and yes an electric field will be formed there as the capacitor charges but that is due to the electrons moving from the charged capacitor. There will not be a field at the discharged capacitor before the electrons from the charged capacitor get there.
rfeecs:

--- Quote from: electrodacus on May 01, 2022, 09:04:00 pm ---There is no energy radiated away.

--- End quote ---

I disagree.   |O

I wonder how you think radios work.  By energy storage I suppose.

electrodacus:

--- Quote from: rfeecs on May 01, 2022, 09:14:33 pm ---I disagree.   |O

I wonder how you think radios work.  By energy storage I suppose.

--- End quote ---

Have you properly read my replay ?
A constant electric field will not be detectable by a radio.
A radio works by charging and discharging energy. But the current discussion is much simpler than that.

The circuit simplifies to charging an empty capacitor with a charged capacitor.  If you disagree with the fact that this all that is two parallel capacitors (simplified) then let me know where do you think the omission is.
rfeecs:

--- Quote from: electrodacus on May 01, 2022, 09:31:54 pm ---
--- Quote from: rfeecs on May 01, 2022, 09:14:33 pm ---I disagree.   |O

I wonder how you think radios work.  By energy storage I suppose.

--- End quote ---

Have you properly read my replay ?
A constant electric field will not be detectable by a radio.
A radio works by charging and discharging energy. But the current discussion is much simpler than that.

The circuit simplifies to charging an empty capacitor with a charged capacitor.  If you disagree with the fact that this all that is two parallel capacitors (simplified) then let me know where do you think the omission is.

--- End quote ---

I have already done that.  I have said there is radiation.  You say there isn't.  Enough said.

The simulation shows what happens.

If I wanted to model it, I would start where the battery/switch is connected to two wires.  These wires form an antenna, or an odd looking transmission line.  They can be thought of as a skinny bi-cone antenna.  The infinite bi-cone looks like a transmission line and has constant impedance.  It has spherical symmetry and the wave propagates out spherically.  In this case, the wires are not conical but straight, but the propagation is approximately spherical and the impedance will change along the line but levels off to a slow increase in impedance.

When the switch closes, there is a transient voltage change, as in a Heaviside step function.  This transient is what starts the energy propagating out in all directions, roughly spherically.

After a period of time, the wave front hits the top pair of wires.  Now we have another antenna / transmission line.  The electric field across the load will cause a current and voltage wave to propagate along this line in a similar manner as the source antenna.  Clearly the signal is much smaller because the field has spread out spherically. 

The two antennae are clearly coupled and form another set of transmission lines, the twin line that has been often mentioned.  So to properly model this I would consider these to be coupled lines.  The odd mode impedance of the twin line is well known.  The even mode impedance will be formed by the bi-cone type lines.

Yes, this model is ridiculously more complicated that your capacitor model.  But it can model the fact that the lower pair of wires initially have a higher current than the upper pair of wires and give the correct current for both wires.

Your model cannot do this.
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