The Big Misconception About Electricity

Cal

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Well if you go with the calculation the light should turn on in 3.333 nanoseconds.
Due to cable capacitance, the light might turn on in 3.3 ns. But for how long. This is a transient condition. The light might only remain on for 1 us. Then it will remain off till and slowly get brighter. After 142 seconds the light will have 63% of max power.
 

SpongeboB Sinewave

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They make the assumption that "on" means any none voltage across the light bulb's terminal. It takes nanoseconds for the RF/EMF wave + capacitance to across the 1 meter air gap.

It'll take 1 second for DC steady state voltage to reach the bulb and turn in on in the way we understand it.

Nanoseconds is technically correct by the definition of when the light bulb turns on but it's just dumb and misleading. Again, this video is the opposite of educational.
Okay thank you! That makes sense. Now if this is a incandescent light bulb, you are not going to see it turn on in that three nanoseconds in reality, even if the wire is superconducting.

So what you say makes perfect sense but The one /c answer is completely misleading not just partly misleading and my mind.

The c constant could then be the velocity of light or capacitance. If as you say it has to do with the capacitance of the gap between the wires then it was a trick question in my mind
 

Posplayr

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Well I'm (kind of) corrected. I did a steady state dc analysis. He did a transient analysis. I wish he used real values that represent the 1 m apart capacitance. I'm not sure the initial transient would have turned the light on.
This is a classic of example of "Those that can Do (EEVblog) ; those that can't teach(Veritasium)!

Based on the EEVblog's video, the answer to the question posed(by Veritasium) basically has nothing to do with the video (i.e. Poynting vectors and field theory) that describes the supposed problem. It is all distraction.

The light comes on instantly due to the rise time of the instantaneous change in voltage across the local 1 meter spaced wire parasitic capacitance.
 
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FilterGuy

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Wow.... I never expected the video to spark such an interesting debate!!!

A couple of observations:

1) There are many places in physics where a model works perfectly well in common applications but breaks down at the extremes.
  • The Newtonian gravity model works perfectly well for almost all we do, but completely misses the mark of Einsteins equations.
  • E=IR and P=IV work perfectly well for what we typically do, but it may not describe what is going on at the electromagnetic fields and waves level..
2) There can be multiple models that accurately describe the same physical event.
  • Light can be accurately described as both photons and waves.
  • There are two very different descriptions of what makes an airplane fly. They both match all empirical tests yet, physicists can't decide which is 'correct'
 

Posplayr

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2) There can be multiple models that accurately describe the same physical event.
This is speaking from a model theoretical standpoint (a part of formal logic), if two models predicts the same thing they necessarily include the same inner model.
 

robby

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That video just causes a whole lot of confusion over nothing.
Now that someone mentioned that it is really 1m/C it all becomes very clear that he is talking about the Capacitance between both of the lines and the Transient impedance that is created at the moment the switch is turned on. Assuming those wires are 1 meter apart.
If someone is really thinking that in the real world the bulb will be shinning brightly in a few nanosecond or whatever it works out to be they are going to be disappointed. These kinds of Transients can be ignored for the most part when working with DC and would only become an issue in high frequency work. Misleading again because he is using a battery in the example.
 

Cal

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If someone is really thinking that in the real world the bulb will be shinning brightly in a few nanosecond or whatever it works out to be they are going to be disappointed. These kinds of Transients can be ignored for the most part when working with DC and would only become an issue in high frequency work. Misleading again because he is using a battery in the example.

That's been my opinion. The initial turn-on transient won't have enough power to turn on the light. The video is quite disingenuous because when he closes the light switch the light turns on "instantaneously" and remains on. This is absolutely wrong.

After viewing the video again I need to make a correction to my steady state calculation. I used a wire distance of 300,000 km. The actual round trip wire distance is 4 * 300,000 km. The inductance of this wire is 6.27 kH. The wire in conjunction with the resistive light acts as a low pass filter. The time constant of the low pass filter is L/R. If the light has 10 ohm resistance, the time constant is 6.27 kH / 10 ohm = 627 seconds, or about 10 minutes. Once the switch is closed, voltage at the light slowly increases. After 10 minutes the voltage is 63% of battery voltage. The light would be still dimmed.

The guy also mentioned the wire has no resistance. That's extremely important for this model. He's relying on the 2 * 300,000 km dual wire length to be a perfect capacitor. Even without wire resistance, the cable still has inductance.

His ground rules are:
1. no wire resistance
2. no wire inductance
3. being 1 m apart, the wire act as a giant (perfect) capacitor

The turn-on transient turns the light on.
 

Posplayr

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That's been my opinion. The initial turn-on transient won't have enough power to turn on the light. The video is quite disingenuous because when he closes the light switch the light turns on "instantaneously" and remains on. This is absolutely wrong.
You have a far higher chance of lighting a bulb using static electrical charge (2-4Kv human model for ESD) and although I have seen blue ESD sparks there is not enough energy to heat an incandescent bulb filament.

This is the reason for his assumption that one election is enough to light the bulb forever presumably. And then he threatens to actually go test the theory in the desert so he has room to unroll these giant roles of wire. The whole thing would appear to be purposefully constructed to generate clicks. In other words, the knowing played a game of sleight of hand to create a controversial statement about something that ultimately is a fabrication.
 
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solarsimon

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1637859557829.png
My brain is now officially melted. And I got the question wrong.
Please tell me that all those (mainly Indian) videos are now true and all I need to power my house is some long cables, lossy rubber belts, a flywheel and a well trained hamster called Elvis.
 

SpongeboB Sinewave

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They make the assumption that "on" means any none zero voltage across the light bulb's terminals. It takes nanoseconds for the RF/EMF wave + capacitance to cross the 1 meter air gap.

It'll take 1 second for DC steady state voltage to reach the bulb and turn in on in the way we understand it.

Nanoseconds is technically correct by the definition of when the light bulb turns on but it's just dumb and misleading. Again, this video is the opposite of educational.

Yes ! I would call this kind of a trick question actually. Of course we all understand that a capacitor will allow a short pulse of current through the light bulb ! Now, had he had the battery and switch on one end of the pair of wires and the light bulb on the other end of this wire pair, the results would have been different and the explanation would have been different. Poynting vectors would still have been used in the explanation(s).

At my old-ish age, trick-ish questions like this make me question whether I am actually sane or not ! :) 😋

Next, do the same problem using a variable voltage supply cranked up slowly. :)
 

Hedges

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I saw the answer/explanation coming after the comments some of you guys gave. I didn't realize it would be 1/c, but that coupling to the nearby wire would occur almost immediately, long before signal reached far end of wire loop. (I see electromagnetic fields)

That's been my opinion. The initial turn-on transient won't have enough power to turn on the light. The video is quite disingenuous because when he closes the light switch the light turns on "instantaneously" and remains on. This is absolutely wrong.


His ground rules are:
1. no wire resistance
2. no wire inductance
3. being 1 m apart, the wire act as a giant (perfect) capacitor

The turn-on transient turns the light on.


Did he actually give all those rules? Maxwell must be rolling over in his grave.

You can't have "no inductance"
You can't have EM propagation without magnetic field, and you can't have magnetic field from moving charges without inductance.

Rising voltage in one wire would pull voltage up in other wire,
and current flowing in one wire would induce current in other wire.
For parallel transmission lines (e.g. traces in a PCB or cable) these effects tend to cancel at far end, reducing crosstalk.
But at near end, they add together increasing cross talk.

Both capacitance and inductance should contribute to pushing current through the light bulb. but given wire dimension and spacing I don't expect significant power transfer. Impedance of free space is 377 ohms. Whatever load that causes wire to present to battery, most of the energy (e.g. 12V^2/377ohm = ~ 0.5W or whatever) is going to be radiated for someone in another galaxy to detect. I don't think much will be gathered by the skinny wire 1 meter away.

Hit resonance, then current in the two wires contribute to cancel fields propagating a great distance.
A single wire would be an antenna. Two wires nearby would be a microwave coupler. Near resonant frequency driving source feels reduced impedance as power collected from second wire is carried away.



Undersea cable, wrapped in steel - signal lost in the steel? I don't think so.
If signal is sent across a twisted pair, they are more tightly coupled, and little energy spreads out to the lossy steel.
Transmission line characteristics are RLCG. These terms include loss in conductor and loss in dielectric. Propagation delay comes from inductance and (frequency dependent) capacitance. So propagation is frequency dependent, and shape of square pulses is not preserved.

Solution back then was to engineer cables (with magnetic materials?) for frequency independent propagation velocity, preserving wave shape.

Solution today is divide up data packet into multiple frequency bands (like multiple radio stations), and send portions of data over each band, which is narrow enough to preserve signal integrity. DSL works this way.
 

ejfluhr

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Electroboom also has a terrific video on this topic, plus he is hilarious to watch. Including here to make this thread more complete!
 

Cal

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I saw the answer/explanation coming after the comments some of you guys gave. I didn't realize it would be 1/c, but that coupling to the nearby wire would occur almost immediately, long before signal reached far end of wire loop. (I see electromagnetic fields)




Did he actually give all those rules? Maxwell must be rolling over in his grave.

You can't have "no inductance"
You can't have EM propagation without magnetic field, and you can't have magnetic field from moving charges without inductance.
That his setup requires no inductance is actually my opinion. He needs zero impedance transmission line for the light to turn on. The video with the lamp turning on when switch is closed is not what's going to happen. And it won't stay on. Just a voltage transient occurs when the switch is closed. Probably not enough current to light the lamp.

His question should ask when the first current arrives at the lamp. Not when it lights.
 

Hedges

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When the switch closes, current will flow into the RLC or lossy impedance of the transmission line. LC and lossless transmission line if superconductor. Some current will be coupled into the wire 1m away with lamp. As the rising edge propagates down the wire it may continue to induce current into the lamp. I haven't worked out the numbers but I think it is very low. My SWAG is 1 mA; if I build a model or test circuit that can be confirmed. Only tough part is 12V 1ns rise time (a TDR of higher voltage.)

Test circuit wouldn't be full length (of course! copper is too expensive ...), maybe 10m of wire then terminated in a resistor to ground. I think open-circuit would also work fine. 12" per nanosecond, that would let me measure a 30 ns pulse in the wire before driven signal hits short/open at far end. Actually 60 ns, because has to propagate back to be seen.

If someone has a TDR with a Transmit pod and separate receive pod, they can test this for us with a 1V step, see what current it drives back into 50 ohm input. Tek 11801, for instance.


1641347414388.png

Otherwise, maybe I'll do it in frequency domain with my VNA an convert to time domain. But would rather keep it closer to his (erroneous?) thought experiment and leave the extra math out.
 

Posplayr

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Electroboom also has a terrific video on this topic, plus he is hilarious to watch. Including here to make this thread more complete!
What happens when you ask home work questions with "simplifying assumptions"
 

robby

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There must be a dozen videos about Vertasium's video and it is good that he got peoples minds stimulated, but at the same time it was vexing to EE's who knew that in the real world that even if we followed his rules this was a load of BS.

The Laughable part is that in the end Veritasium was wrong!
He put in so many perfect conditions that he confused himself and ended up making his idea Invalid.

His Perfect Condition #2 was what did him in: "The light bulb has to turn on immediately when current passes through it"

So in the real world the light bulb would always be on since current is always passing through it. What made his theoretical plan work was what also undid it.
 
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