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Veritasium "How Electricity Actually Works"

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snarkysparky:
""The given problem is a dynamics problem.  Therefore we're only concerned with AC behavior.""

Does the video make this clear.  I missed it.

He does a poor job of separating the transient from the steady state conditions

Sredni:

--- Quote from: EEVblog on May 02, 2022, 01:37:11 pm ---
--- Quote from: bsfeechannel on May 02, 2022, 01:33:23 pm ---Double-posting, Dave?

https://www.eevblog.com/forum/chat/veritasium-(yt)-the-big-misconception-about-electricity/msg4150432/#msg4150432

--- End quote ---

Yes, because there are now two threads, and many people are not reading the other thread any more.

--- End quote ---

So let me double post this from the other thread, because I have yet to find a satisfying answer:

Ok, let me bring up this argument I put forward a few dozen pages ago (I will simplify it even more):

I have a mass of 1 kg in position P at 0 meters over sea level.
I take this mass 1000 meters away to drop it from a cliff into a hole deep 10 meters.
The potential energy of the mass is converted into kinetic energy and then this is uses to generate heat.  Let's say I 'generated' 1 joule of energy.

Has this energy traveled along the 1000 meters path?

What if I changed my mind and headed in a different direction and after 1000 meters I dropped the mass into a hole twice as deep?
Would the 2 joule energy have traveled instead?

Has energy ever traveled along the path?
How much?
1 joule? 2 joule? 100 joule? m c^2 joule?

electrodacus:

--- Quote from: hamster_nz on May 02, 2022, 05:20:19 am ---
Q1) How did that 10V measured over the switch get there  :-//?  My suggestion: as no charge has moved across the capacitor, no current is moving so it can't be magnetic. So unless I want to add a new fundamental field it must be electric field across the capacitor. This has caused charges to move around to balance out the electric field, until there is zero voltage gradient over the capacitor, resulting in 10V over the opened switch.

--- End quote ---

You closed the switch just not the one you are concentrating at. When you connected the multimeter you basically closed a switch that is in series with 1MOhm resistor so electrons will flow from the charged capacitor to the discharged one trough your multimeter.



--- Quote from: hamster_nz on May 02, 2022, 05:20:19 am ---I've measured this all on the bench - there is 10V across the switch.

You close the switch a small spark is heard and the 10V of potential difference is now gone.


--- Code: ---    +------------+
    |            |
  ---- 5V      ---- 5V
  ----         ----
    |            |
GND +------------+
           0V

--- End code ---
You now have 5V across both caps - and half the energy goes out of the system as heat.

What part am I missing?

--- End quote ---

You are missing the ESR.   The electron wave can move at the speed of light and how many of them can flow per unit of time depends on resistance in this case mostly the ESR and some of your wire resistance.
The reason you do not have 7.07V at the end of the experiment so same energy is because half of the energy was dissipated as heat in the ESR + wires.
If you could measure the temperature accurate enough you will see that capacitors and wires heated up by the exact amount representing half the original energy.

rfeecs:

--- Quote from: EEVblog on May 02, 2022, 05:43:26 am ---Hontas Farmer is back still saying the Derek is both right and wrong according to QFT/QED

--- End quote ---

She says QED is in fact simpler than classical field theory?

What if you need to actually calculate something for a macro system like this?  Given the dimensions of the wires (diameter, spacing), the input waveform (step function), the load impedance, what is the current at the load and the source (vs time)?

Clearly you can calculate this with classical field theory.  You can calculate the impedances from the dimensions.  You can do an accurate simulation.

She says for this problem, using classical theory is even harder.  You have to do all sorts of computer calculations.  But for QED its just probability.  Easy.

The fact is with classical theory you can actually get a numerical answer that you can use.  Can you do that with QED?  Or does the complexity immediately get out of hand for even a much simpler system, like a single molecule?  Here's a video discussing the difficulty of that:



There are lots of models.  Each has limitations.  Use the right tool for the job.

T3sl4co1l:

--- Quote from: Sredni on May 02, 2022, 03:27:30 pm ---So let me double post this from the other thread, because I have yet to find a satisfying answer:

Ok, let me bring up this argument I put forward a few dozen pages ago (I will simplify it even more):

I have a mass of 1 kg in position P at 0 meters over sea level.
I take this mass 1000 meters away to drop it from a cliff into a hole deep 10 meters.
The potential energy of the mass is converted into kinetic energy and then this is uses to generate heat.  Let's say I 'generated' 1 joule of energy.

Has this energy traveled along the 1000 meters path?

What if I changed my mind and headed in a different direction and after 1000 meters I dropped the mass into a hole twice as deep?
Would the 2 joule energy have traveled instead?

Has energy ever traveled along the path?
How much?
1 joule? 2 joule? 100 joule? m c^2 joule?

--- End quote ---

Potential energy is just, what it is.  It's a property of the mass's altitude (in this case, or velocity too if you want to count that), not some internal state that comes along for the ride.  It's not like it's got a battery in it. :)  However, if you include yourself in the system, you're a battery (of sorts), that's transporting energy.

Or if you have a definition to apply it to, like: "transporting energy" sounds like integral path length times potential energy along that path, in which case, you'd be transporting potential energy with respect to that 10m drop any time you're above its floor, and negative potential while below.

Tim

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