Explaining Veritasium Electricity Video: Energy doesn't flow in wires (with Eric Bogatin)
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- čas přidán 19. 06. 2024
- How long does it take to turn on a light if the switch, battery and light are on the Earth, but the cable connecting them goes to the Moon and back? Super interesting discussion about Veritasium electricity misconception video. Thank you Eric
Links:
- Eric Bogatin: / eric-bogatin-368860
- Signal Integrity Academy (use FEDSI - 3 months free): www.bethesignal.com/bogatin/3...
- FEDEVEL Academy (use VERIT100 to save $100USD on Online and Download): courses.fedevel.com/
Chapters:
00:00 What this video is going to be about
00:58 What is the circuit we are going to talk about
02:59 Explaining transmission line - the basics needed to understand our problem and solution
10:55 LED placed at the end of transmission line - explaining what will be happening
33:23 Explaining what will be happening if the cable goes to the moon and back
48:18 The answer
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- Robert
Instead of dismissing/waving hands like Dave from EEVblog did, you guys made the whole thing meaningful. And it's actually more helpful than not for even experienced electrical engineers.
Dave from EEVblog is stuck with his own opinion which he claims to be fact.
@@donmab The difference for me is that I can actually understand what Dave is saying. Maybe it's because I'm also Australian, but the point still stands.
@@brugj03 isn't that pretty much what one's opinion is supposed to be? Weird thing to say.
@@BOT_0x76DE45AB Honestly, I think that by introducing complexities like resistance and loss, EEVBlog may be more practical, but he is using it in a way that it just clouds the actual working of the system and saying that everyone already knows about this is just not true. That's not the point of the original video which is trying to explain first principles of EM. I do agree that Derek may have done a poor job by starting with non-ideal components and just claiming them to be ideal, whereas you would just choose a PEC in a lecture, but I guess this makes it more relatable to people.
@@BOT_0x76DE45AB I vant to knov vay you call yourself a "Ruski robot" ven you are Australian?
The video Dave did on the Varitasium was not too detailed he pretty much agreed with most of what was said by Varitasium in relationship to principles of transmission lines and propagation. Actually it was probably one of two of his videos I was able to sit through right to the end, normally I have to turn him off because he tends to just go overboard and overcomplicate a generally simple subject.
By far the best explanation of this chapter in electromagnetism in just one hour! Highly recommended!
I finally understand this now after 45 years. It's never too late! The topic he explains is the course that washes out a lot of would-be EEs.
As always, Eric is a pleasure to watch and listen to. I love how this video brings together a lot of different aspects of physics and engineering. Great lecture!
I watched a lot of videos, and finally understood what is happening due to this great man, Eric. Amazing !
I've seen a few reaction videos on the same subject already but yours is the best 😀
It's great that you explained the difference between the two cases, that was extremely important.
Thank you both 😊
Finally the correct explanation! It infuriated me to no end that soooo many electrical engineering youtubers' explanations were so misleading to the point of being wrong! This person is probably the very first one to correctly explain the precise physics of signal propagation! Thank you! Now my soul can finally rest knowing that the world (or at least your viewers ;-) ) have received the correct, physically-accurate explanation.
He did it as an EE engineer. Taking the coaxial 50ohms cable as an example was a perfect idea, even though it's also quite confusing. (the original question is very high impedance pair, 1m apart)
Physics engineers do explain it with their own formulas. That doesn't make them wrong.
Eric Bogatin is my PCB Mentor
Holy crap this guys taskbar is panic inducing 😵💫🤪
This is a nice explanation of the problem.
It remined me that in my first job after college many years ago, I was testing a circuit that was built from TTL and I wanted to look at the output of a 7474 D flipflop. So I fastened a wire twisted pair to the output and ground and ran it several feet to an oscilloscope to look at it. to my surprise, the Flipflop would not stay set when it got appropriate inputs. This is how I remember it, but it was a long time ago.
After further investigation, I determined that the Flip Flop was actually setting then resetting in a very short time. It did set properly when I put the scope probe directly on the output with the wire detached, and with wire attached you could see that it was momentarily set.
Then I realized that the scope probe was a very high impedance and the line was reflecting the initial edge from the Flipflop setting back to the chip pin. AND unfortunately the D output was internally connected to the opposite side input of the output latch (not buffered), and the reflection reset the flip flop.
So that was a lesson for me. Care is needed in higher speed measurements.
It is not correct to treat the Veritasium arrangement as a pair of single impedance transmission lines. You aren't driving the two ends of your "transmission" lines with the voltage source. For the first few nanoseconds, the lower pair looks like a dipole antenna. The bulb line, also looks like a dipole antenna, receiving the signal. A dipole arrangement does not have a single Zo - the step response is actually pretty challenging to figure out. My point is that you cannot assume the current in the transmitting side is equal to the receiving side - there will be coupling, but I contend a fairly small current - not enough to realistically light the bulb. The beef with the original video was not that some signal arrives 1m/c seconds after switch closure, but that there was implied full power transfer. It won't be.
@bobwhite137 They've got it right. It's not an antenna, because when we place two antennas that close together, we create a highly-coupled transformer, not antennas anymore. (Same thing would happen with loop antennas: if we put them right on top of each other, we now have a transformer, not a xmit/rcv antenna pair.) We also get a capacitor where the plates are formed by two close-spaced wires.
So which is it, a transformer, or a capacitor?
It's both, and Heaviside discovered, when the two concepts are combined, the reactance of the transformer and the capacitor cancel out to form a virtual resistance. If we could use a DVM to measure Veritasium's million-meter ideal wires, we'll see a 700-ohm resistance, rather than 50-ohms of standard coax cable.
To act like dipole antennas, the wires need to be separated by about 1/4th the wire length, so the coupling isn't anywhere near 100%.
Yes, this appeals to my intuition too. The arrangement is like antennas or even a transformer. But probably the antennas make the most sense. This was pointed out in another video discussing the original video. All of it is instructive though!
@@wbeaty: "One fourthTH"? Oops!
I think that current starts after 1/c (in seconds) and it is slowly building up until the RC circuit practically is charged after 5 times RC. Since we have a switch, we have to study the step performance (transient) of the circuit and treat it as an RC circuit (until it reaches the steady state which is 5 times RC practically). If we assume R=0 then RC is zero then the circuit charges in 0 seconds. Inconsistencies arise when we assume R=0 and C some value. Using the notion of the displacement current, the transmission line, the transformer, the antenna, the cable impedance is a good tool to understand the circuit. But i consider it as an RC circuit since we have a switch and therefore i am analyzing the step performance of the circuit. The problems arise when we assume a practically infinite and distributed capacitance with no resistance...
@@wbeaty 'a highly-coupled transformer' Not even that highly coupled a meter apart, you'd consider any low voltage transformer a failure if you move the coils one meter apart..
The electrons are literally IN the wire. They don't actually touch, so the energy is in colliding the field shells. The field extends from the electron, and is both inside and outside the wire. That you can tap the energy of the moving electrons both from inside the wire, and somewhat outside the wire, doesn't mean the energy source isn't still the energy in the moving electrons, and is still basically stuck to the wire. Make the wire take a 90 degree turn and the outside field takes that same turn, you can't make it just go straight off the wire because it's not really flowing outside the wire.
It takes special pains like many coils and a core to efficiently couple the magnetics of the flowing electrons of one wire into another, and efficiently transfer power from one wire to another. It takes other special pains to use the flow in an antenna and peel the energy off the wire into EM radio waves and similar.
If it were really just that easy and all the energy was really outside the wire, transformers and antennas wouldn't be their own specialized fields and take pretty careful planning and adjustment to make sure they even work in an OK manner..
Route a wire all over a room, bending it in different paths, and try to get the waves to do anything that isn't strictly confined to following the path of the wire. Won't happen. Just because you can tap the energy and affect it from the outside doesn't mean that energy isn't coming from the parts that are stuck in the wire, it just means they have a field around them.
Great talk! Thanks to both of you for taking the time to do this.
Excellent video, thank you so much for posting it. I have seen others with Dr Bogatin and after watching Veritasium's I hoped that in the responses that were likely to be posted we could see his discussion on the subject. Wish granted, what a treat!
To top it up 3 months of free access to his lectures, for "our patience". Dr Bogatin, I get your passion and it takes no patience at all on my part to watch your talks and lectures so I cannot thank you enough for this.
What this explains is how so many of us are conditioned to view things in a defined way. The first circuit makes us feel safe and knowledgeable as we are so used to seeing it this way but when confronted by the second circuit we suddenly develop anxiety and stress because we are conditioned not to really understand circuits but memorize them.
The memorization over understanding is part of the game. Understanding that electricity is like sound and can be amplified to any value without loading the source or adding more energy would make investors in power plants freq. out. You don’t need a new power plant, just an unbalanced transformer.
Another awesome video that is not just providing some useful information but is helping develop an intuitive understanding of signal propagation. Great work, Robert!
Fantastic job explaining this Eric. This is exactly how I learned these concepts back in college.
Fantastic video, I really enjoyed this. Thank you for uploading.
This video has changed my level of thinking towards electronics..
Thank you so much Prof. Eric Bogatin and Robert for these wonderul talks. Looking forward to more.
Thank you for giving us a great learning opportunity from someone who literally wrote the book!
Brilliant! Thanks for your work Robert !
Thank you both Eric and Robert!
This is amazing ! Just watched Veritasium's video few days back and wanted to know this very perspective. Thanks Robert !!
Thank you very much Girish
"sometimes a lie tells more about the truth than the truth itself"..........."strategic simplification"........love it!
Great video! Now understand much better why placement on PCB are important. It was nebulous to me at first, but it's much clearer now! Thank you for you time guys ! Love how internet brings us together. It help a lot !
I saw the Veritasium video and was waiting for folks like Eric to peel the onion back more.
He didn't break out a smith chart. I always liked those in school.
Great video! I Was waiting for it
I love his way of explaining things. There is even a risk that I might start understanding transmission lines even though I have been avoiding them thus far.
Thank you very much Eric and Robert! Loved it
One of your best videos yet !
The over-simplifications in Veritasium's video did exactly the opposite of the expected results: More confusion.
What a pleasure to watch this. Thank you Eric.
Oh wow thank you Eric for your offer I am glad I watched to the end :)
This was a really good explanation. I definitely have a deeper understanding about current now.
This is the best explanation of this subject I have listened to so far.
I'm so lucky that I came across this video, much love and appreciation.
Thanks Robert and Eric! Great video!
Great video! You guys really did this topic justice
_Robert, I noticed that both you and Eric were sitting in outer space, albeit on opposite sides of the planet._ 🚀
_So I was waiting for you guys to string up two light-seconds of superconducting wiring for the "bench test."_ 🤣
Lol 😂
Wow, I think this is the fourth evaluation of the original video I've watched, and this is probably the best. Thanks. One thing: I thought of the long parallel transmission lines as a sort of transformer or even antennas but I guess that's just transmission line theory in a special case....
This was a really great example and very well explained by Eric, and I only know the very basic electrics.
Excellent explanation, Thank you very much.
Thanks for making this! I was really struggling to understand what was going on
Sounds like Eric was really getting into the details there, would love to hear him discuss more on how Maxwell's equations have been simplified to the circuit elements we use today.
There's an article in IEEE Spectrum in 2014 called The Long Road to Maxwell's Equations. The article tracks the progress of understanding electromagnetics from Maxwell's predecessors to Oliver Heaviside's transformation of the equations to their modern form.
It's frequency that transmits he didn't mention that. So I give Eric thumb 👎
@@fyfoh Missed this, I'll check it out, thanks!
As soon as I saw Veritasium Electricity video it reminded me about Eric Bogatin. Thank you Robert for the video!
Haha, finally a brilliant explanation from a real physicist. Thank you guys!
Excellent. A piece of test equipment called Time Domain Reflectometer uses this. You mentioned this as I write. If you know the impedence of the transmission line and put a resistor of the same value at the end the TDR will give you the distance of all the impedence changes down the line. Near shorts and opens and the distance of the same down the line.
Great Video, expect more!
Thanks for the video!
Wow, great video. Using an example makes studying Electromagnetism and Transmission lines much more interesting.
Excellent video about the topic.
Thanks, best explanation of this idea.
Breathtakingly clear explanation of how Electricity works there by Eric Bogatin (I'm at 7:57 right now). I won't forget this explanation. Kudos to Robert Feranec as our host! thanks for posting! :D p.s. and of course to Dr. Derek Muller for having started the conversation in the first place! :D
I was building my first PCB circuit for wired data transmission over RS-485 line (no electrical engineering education), and the terminology was very confusing: termination resistors, impedance matching, etc. While I did all the things "by the book" and followed various instructions, this explanation opened my eyes of what is really going on and what is actually important.
I learn from you since I took your pcb courses. Great thanks.
Unfortunately, any photonic coupling between two wires 1 meter apart is going to be negligible. Photons are generated from the surface of the wire and then spread out before they hit the second wire. If the radius of the wire is 1 millimeter, then the photon cloud will have been attenuated 1000-squared = 1,000,000 times when it hits the second wire. So there's no getting around it. You're really going to have to wait a full second before the bulb lights.
But the assumption from the origional videowas any current turns the light on. Technically correct but misleading.
Edit i do wish he had actually gone over the "capacitance"of 2 cables 1m apart to show the true current across it mathematically. Not sure the standard equations apply though so might have to derive for custom circumstances.
The two transmission lines extending left and right have a characteristic impedance of around 800 ohms. When the switch is closed those transmission lines look exactly like an 800 ohm resistor each.
Therefore, the circuit at t-0 is equal to a battery and two resistors going to a bulb.
This means that the bulb will light dimly after the time delay because of the 1 m separation. This is easy to see if you replace the two transmission lines with resistors that are 1 m long.
The light will then flicker as the reflections bounce back and forth and eventually equalize to the simple DC circuit of the bulb and battery.
@@Observ45er this cannot be true. If it was your computer would recieve noticable current every time you plugged it in. The power supply circuits recieve power and then nearly instantly induce or via capacitance would destroy all the ic circuits as they recieve the induced 12V power from the rails. The induced current or capacitance here is negligible and if it wasnt computers as work in life would fail immediately.
Exactly. I can not see how he claims there will be a 20mA return. For an example, if I placed several wires in parallel near the second "receiving" wire, then would all of them have 20mA? Surely that would violate some conservation principle. Neither wire would know the circuit specifics in the time the pulse would propagate over, so what makes this specific wire special?
@@MrThedumbbunny it is however correct, that is just how it goes. It is what it is because it is a distributed system with a uniform structure and hence it obeys those laws.
Really nice discussion - loved the musing on Maxwell's equations!
Amazing video, I learned a lot 😮
The best explanation I have ever seen of signals and transmission theory. @Dr Bogatin: Very well done. And thank you to Robert Feranec for inviting him!
Unfortunately, the problem presented to Dr. Bogatin, is NOT the problem presented by Veritasium. Though I totally understand why it was presented in the manner it was, nevertheless, the Veritasium problem has a Physical Demonstration to it that people who view the video get to see. That demonstration makes certain things "tangible" to the viewers and also modifies how we perceive it and how it should be perceived.
Let's review:
1. The demo shows a (car) battery, an LED bulb, insulated conductors. These conductors are then given the hypothetical "no internal resistance" and they are 300,000km long, and separated by 1 meter. When the switch is thrown in the demo, the LED turns Solid ON. It simply remains On.
There is literally NO mention of Blinking on. Clearly the definition of "ON" here is suspect if one were to consider "blinking" on as "ON".
The question offered by Veritasium was after the switch closes, when does the LED turn on? (0.5sec, 1s, 2, 1/c (sic), and None of the above).
After listening to Dr. Bogatin, (who clearly knows the material), one would believe that current flow across the bulb would be enough to power the LED. However, let's review some of the facts that everyone seems to leave out.
#1 These are insulated conductors not capacitor plates sitting on dielectrics. The conductors are also not Bare Wires, they are insulated. Please get that fact straight.
The impedance of insulated conductors 1 meter apart will be markedly different than the simulation because the simulations do not include the fact that there is insulation on the wires.
#2 The LOAD is a 12 volt LED Bulb. These "bulbs" are internally made of many LEDs in series. Each LED in series has a forward voltage that MUST BE exceeded in order for the LED to turn on. Nowhere in any simulation are you taking into account the forward voltage requirement of the circuit.
#3 THERE IS NOT AC SIGNAL. Dr. Bogatin refers to the signal frequently because he is equating the circuit with typical signal. BUT, there is NOT AN AC Signal in this example. There is only DC, and the instantaneous turning on of the Switch. There will be a very brief pulse. That is it.
#4. If you take all these elements: Insulated Wires, DC, Not AC, 12 volts, 12 Volt LED, Forward Voltages of the Diodes, No signal. It results in at best... the microscopic blink of an LED, nothing more, for at best a moment in time. I am sorry, but I simply cannot equate this to "The Light Turns ON".
Please show a simulation that corrects the above noted issues, and let's see how it plays out.
This was my understanding after dismissing veritasiums video as an interesting application of Maxwell's equations in an "accessibile" manner. Maybe a bit too on the "clickbaity" side as it completely failed to mention anything other than stuff you learn about pretty early on if you take such a course.
My intuitive answer was greater than 1 second (estimate was between 1-15s) in which you would see the light fully turn on and stay on assuming it was possible in the first place. I then did consider the 1/c answer and yeah it would "turn on" but it felt a little disingenuous as it would also turn off in short order. My conclusion was that half the question was missing and this was equivalent to asking a vague question that can be interpreted in many ways for the sake of generating impressions.
I even simulated it using a capacitor on a breadboard and exactly what you would expect happens. I used 10mm wires which ends up at a capacitance of ~2 milli ferad. The LED was on for a decently long while before turning off as it takes a while to charge such a large capacitor (relatively speaking). However, I still wouldn't call that "turned on" if it turns off without you turning the circuit off. But I do suppose it's long enough to prove the point. Made me extremely curious as to a scaled up version of the real life demo.
Again, I feel as though part of the overall question is missing which is why it's genius. We're engaging with it regardless of whether we want to or not. If you want the question to be what is the length of time before the lightbulb emits a photon (assuming it could given the voltage) then I would have with zero hesitation said "almost instantly" likely completely ignoring the 1m thing (which does matter here) as that's what I was taught over and over. But that doesn't generate a massive amount of impressions given different people's experiences and backgrounds now does it? Veritasium knows what he's doing. (◠‿・)
@@LiveType I am not sure. I guess it depends on whether he values his integrity. Because, the reason why I watched his videos was to get to the deeper truth of things. If this is how the future of his videos will play out, he lost only 1 viewer. It may only register as a blip on the magic light bulb hehehe but for me, I value the integrity of the process. It really bothers me to see this level of misdirection by someone I really thought knew better.
Thank you very kindly
Amazing video
Thank you so much! I am more confused than ever which is good :-D
Hello Robert,
great idea to cooperate with Dr. Bogatin in this topic!
I have one question regarding this issue:
For me both cases described are pretty clear, but what would happened if two wires instead of layout from top picture (transmission line like PCB) and bottom ( from veritasium) would be placed as one big loop (signal lanes very far apart from each other and LED far from voltage source)? The coupling between signal and return would be extremely low. How this signal would propagate then?
I would bet that principle is the same but due to very high impedance between lines there would be extremely small displacement current and due to very long risetime of this signal we would need more time for voltage to become stable.
What is your/Dr. Bogatin's opinion?
Thank you for this video :)
That was really interesting, when Mr Eric spoke about Maxwell equations and quantum mechanics. I wouldn't even mind if he continued the talk and then later came back to the explanaition of the circuit below.
I think this is the best explanation about this paradox.
Excellent.
Excellent explanation using the capacitor. Changing electric and magnetic fields occur at the speed of EM radiation (ie light).
So, everyone has a clock speed in our brains of the speed of light.
Amazing!!!
I have a fast ohm meter; GHz scope and ns pulses from a photodiode illuminated by a laser. Me yelling “half current before reaching full current!” at the end! Very much enjoyed!
Big thanks to Eric! This strategic simplifacition made me understand the thing. Even there was left out distance between transmissionline conductors and impedance of the led/light. (And that Veritas' transmissionline wasn't ideal).
Now I can sleep again, so to say :)
Thank you!!!
While agree with John that this video is better than anything else I must say that I am rather frustrated by the original video, the EEVBlog video and also this video, and the overwhelming number of comments. Why is everyone ignoring the simple facts given in the original video: the lines are 1m apart, they do NOT form a 2m thick coax cable, the copper inside the lines must be in the order of 1mm thick, the lamp is big and will consume in the order of 1W, the energy source is a "normal" battery 6 or 12 V, not 1000000V or so.
I have by now seen so many "physic professors" that apparently due to their mission to correctly explain the physics behind a transmission line completely oversee or deliberately ignore the problem statement.
A decent (and therefore attentive) physics professor should also understand the following very simple facts. a) Coulomp's law according to which the electric field strength decays with 1/r^2, implying that also the time derivative of field strength decays with 1/r^2. b) The time derivative of a constant is zero.
Now be my guest to calculate how much energy can be transmitted from some short, e. g. 30cm long section of the wire near the switch to a corresponding wire section near the lamp by a single pulse of e. g. 1ns. Even if one doesn't have that minimum amount intuition to understand that in the order of 1W will certainly not be transmitted (just think what consequences that would have on every day life), it is anyway clear that after 1n the light would be off until the field wave has travelled through the cable.
I am also astonished why none of the "professors of physic" goes to their lab, take e. g. 40 m cabele (instead of 4*300000km) , pick a pulse generator to generate a (off->on) pulse, a lamb or resistor and a scope and do the experiment. This takes 10min. The result shows that the answers given in all videos is wrong!
I better stop here...
I truly loooooove this this video!!! I do like varitasium but this explained it sooo much better and complete. I could follow the explaination with ease. It really explained what would happen if this and that!
I do have a question, in the video at 35:59 Eric mentioned that the switch would be the source of the signal.
I answer my own question below:
If you would have your switch on the moon and your led and battery next to each other and you turn on the switch the wave front wouldn't start at the battery but it would start at the switch cause that is where you open the "gate" let's say. So the source of the signal si still the battery BUT the source of the voltage wave is the switch!
Can I also say that Eric, you are so incredibly cool! Your knowledge and your way of explaining such complex systems in a very engaging and just lovely way! It get's me fired up! And I would LOOOVE to be in one of your classes one day... Maybe, it's a little far away but if I ever get the chance, I would take it in a hartbeat!!! Mind you I would never be able to finish a entire class but I would LOVE to be one in or two lessons!
A;sp for Robert! The things you can learn from this channel! I started watching you to learn to use altium and it's MANY MANY features, and now I am learning a little bit about the quantum electric behaviour on transmition lines?!?!? Like what? This is just insanity! just incredibly! Love every single video that I have watched!!
Bogatin: a great teacher!
I loved this video!! unfortunately the last part lacked some detail... The circuit created in the original video, as Eric perfectly described has 2 transmission lines and the reflections can make the situation quite complicated. Depending on the Zo of the Tx lines (in the video two simple cables separated 1m by air) can be in the order of 800 ohms (a 1mm diameter cable and 1m separation using the standard equation for a bifilar tx line), if the propagation speed is considered to be C (time delay to reach the end is 1 sec), and just to take a number consider that the light bulb is 100 ohms, the multiple reflections (and transmissions) from one Tx line to the other will create a time constant that can last tens of seconds (with steps every 2 seconds) to get to the steady state value. This is quite simple to simulate in Pspice. The instantaneous current through the light bulb is V/(R+2*Zo) (neglecting the V internal resistance), which naturally will be extremely small because is displacement current between two wires spaced 1m appart.
I've still not seen anyone explain this correctly, as Federico Di Vruno has mentioned, and now even the Master of SI himself has ignored the details that so many engineers are bothered about on this problem!
@@HelloWorlds__JTS see here for a bit more detail…
And how energy is transferred if wires are closely shielded ?
@@naszadynastia a wire with a shield around it is nothing more than another transmission line. Displacement current will still occur, depending on how the shield is connected to the other parts of the circuit it can have some effect… conduction current (ie stationary regime)is not affected at all by shielding.
Exactly! That was my first answer, the LED will start instantly but very dim. Also Dave on eevblog missed that out.
In the section "Explaining transmission line..." there's an over-complication. The reason why you have instant current is because this is where you inject the power and make a measurement. The transmission line is irrelevant. What you measure takes another path through your measuring lines. (I didn't have patience to watch further than 10:33 so it may come up later, but there is no excuse for confusing the issue at the beginning)
wonderful
The best one!
Wow this cleared everything up and I learned a lot... fantastic, great and informative discussion, Eric is a great teacher, thanks Robert!
Now i do understand what's happen, thank you very much, veritasium said the answer but it didn't explain as it should.
cracks are forming in the brick wall of my lack of education in this area.
Thank You Both
YEEEEEEEESSSSSSSSSSS!!!! i watched veritasiums video and many responses to that video and this is by FAR the best explanation. I have been lucky enough to sit in seminars with Rick Hartley and Dan Beeker who stress this idea that "energy doesn flow in the wires but the space around them". thanks for posting this video.
It's depending on the time (stable vs dynamic state) that we examine the signal I guess?
Hey Robert, thanks for the video! This is exactly what I thought given an earlier video you did with Eric Bogatin on transmission lines.
I still have a couple of questions though. I found the formula for characteristic impedance of a transmission line online as (120/sqrt(er))*acosh(s/d). Here er is the relative dielectric constant, s is wire separation and d is wire diameter.
If I plug in a wire diameter of 2mm, wire separation of 1m and dielectric constant of 1, the transmission line impedance is around 830 ohms. If I want an impedance of 8300 ohms, these same wires would have to be around 10^27m apart. Is this correct or is this formula incomplete?
Furthermore, if we take 2 transmission lines of 830 ohms and a load of 12.1k (an led + 12k series resistance with a 230V voltage source) the current through the load before reflection can reach 88% (12100/(12100+2*830)) of the final current, but how would it get there? How long would it actually take before the current through the load reaches, say 80% of that 88% figure? Surely the current will start rising at t=1/c seconds, but won't reach 88% until some time after that. Maybe you could do a physics based simulation of some sort?
In terms of this explanation, the "Veritasium" question then becomes "What is the input impedance of two conductors spaced 1m apart (even allowing Derek his 0 ohms of resistance), and what would be the current flow when a single digit voltage is applied to that transmission line, and would those femtoAmps be enough to light the bulb in 3ns as he claimed?"
Thanks for this version of the explanation.
Exactly, his example question has actually nothing to do with the point he was trying to make (the poynting vector). I guess he tried to come with a practical example.. and thats the best he could manage, because for DC the whole poynting vector just doesn't matter, so he couldn't find an example where the standard (simplified) model of "power travels in the wire" breaks down.
It get interesting in high AC with the skin effect, so why for example silver coating cables can make sense, while it doesn't for DC etc. but thats again slightly different.
@@georgelionon9050 I dunno... I was taught, decades ago, the transmission lines could be considered an almost infinite number of stacked "low pass" RCL ccts... Even giving Derek his 0 ohm conductor, the miniscule Farads of "capacitance" between the conductors wouldn't cause many electrons to move ("displacement current"?) to establish a field that transfers energy... My gut says the load wouldn't have a noticeable reaction until the fields that travel the length of the conductor meet at his bulb.
Consider electrons streaming out of the -ve pole of the battery (DC)... As their "virtual" members charge up one plate of the virtual bridging capacitors, their electric field will repel other virtual electrons across the gap of the virtual capacitor... Those will flee, "seeing" one path to the resistance of the bulb's filament, OR an alternate path in the opposite direction at the exact same voltage... Meanwhile, the travelling magnetic field following the leading edge of the pulse is coaxing electrons to flow in the other direction. Would the two effects cancel ? I dunno...
@@rustycherkas8229 Honestly I don't have a good calculation at hand how long the LED would light up and how strongly. One meter is quite far apart. However note the way Derek sneaked in the conditions for his questions, he said "light up" not "starts lighting" and that he says in the beginning his ideal bulb will light for "any energy flow". So under this conditions the answer is 1m/c. Also the answer is d/c for any such bulb in the universe, no wires required. Thats why I say, his example was bad for the effect/argument he wanted to make.
Thank you for the interesting video!
I wonder what would happen if you increased the distance between the transmission lines. For example if the switch was on the north pole and the LED on the south pole.
Wouldn't the capacitive coupling be negligible?
Hello Robert, Does that mean,even if the wire is open in the moon, we'll see the LED illuminates a few seonds (and then turns off)? (If the LED is located close to the switch).
And if we carry the open end-point to a further place, would we measure 50 ohms for a longer time period?
It is really amusing to watch all these iconic people fail explaining this thought experiment. First of all, it is a thought experiment with no solid rules being set, at all. We can assume a bunch of things because Derek, in all his ignorance, failed to strictly define the rules for his thought experiment. This gives all those debunking videos a chance to focus on something that was not the focus of the thought experiment. Hell, I can't even guess what Derek tried to prove: current propagation under ideal conditions focusing on sub-atomic phenomena, or realistic in-depth analysis of such a circuit... He just stumbled upon this topic, found out it is a trick question and tried to create the "wow" effect on his audience. Wow, indeed.
This is a great explanation of what to expect from a transmission line model, and it all sounds sensible based on everything being a 50 ohm transmission line, with a 50 ohm source power supply with the example of a short, open, or led at the end. But the case in the original veritassium video didn't really say that the loops to the moon are transmission lines with any particular characteristic impedance. If there is no measurable interaction between the wire going to the moon and back (Think capacitor with infinite space between plates), do we then have to wait for the 3 second round trip for current to start to flow in the light?
At 34:20 Dr B says there is displacement current in the transmission line on the right, with current flowing from the top (just right of the bulb) to the bottom (-ve terminal of the battery). However, before the switch is thrown, the *entire* circuit (other than the small section betn the batt +ve and the switch) is at 0V (the batt -ve vge). Why would there be disp current from the right of the bulb to batt -ve?
Nice lecture, thank you! Could you continue the discussion on how energy flows if an AC source is used instead of DC
It's the same, but electromagnetic fields change vectors of direction 50Hz.
In electromagnetic theory you have LOCAL ENERGY CONSERVATION the relationship between energy density and WORK(fields on charges and charges on fields) and FLUX.
Poynting vector is needed for calculation of flux.
Poynting vector points in direction it does so that surface integral for flux gives correct answer.
Poynting vector is part of local energy conservation and that's the point to remember.
Local energy conservation applies everywhere of course, current or no current.
The impedance measured at the input end is Zo only if the far end is terminated with a load of Zo Ohm. If you remove the load, you won't have a 50 Ohm to take your 20mA at the switch.
48:35 What would happen current-wise at the LED if you moved the resistor in the 2nd example to where the LED is in the first example? Normally the order of these components doesn't matter in practice, but would they in this example?
Great video and really helped me understand what is going on. One question though, how does the position of the battery and the switch affect the outcome? If they are far apart does the 20mA current propagate from the switch instantaneously?
The position of the battery doesn't matter. Only the switch to bulb distance.
So when there is a break at the Moon, we will see the light near the switch to be glowing for 3 seconds at about 1/2 of its full intensity and then it turns off, as it gets the circuit break reflection? And in case there is not the break but the shortcut, the light will double its initial intensity?
Great video, and I love Bogatin, but I think he’s not quite right at the end. The “ideal-ish” LED will blink briefly as the displacement current activates it from 1m away, and then it will turn back off until the circuit begins to settle ~1s later. After the displacement wavefront passes the LED, what transient condition exists across the LED terminals? There’s no longer any current flow until after ~1s when we start to see the LED turn on permanently as the reflections settle out.
I'm replying just so that I'll be notified if this gets answered. The wave is propagating down the line, but it isn't clear to me that there are new changes at the origin until a reflection gets there. I told myself it was current being conducted back from new displacement at the wavefront, but ... I'm not confident of that, nor am I confident that a Doppler-like effect wouldn't at least decrease it. Of course, I also don't quite get why it comes up in "10 steps", instead of after 1 step, or at least in a zeno's paradox of 1/2 the way each step forever.
If we're specifying ideal transmission lines, then there will be no reflections.
@@jimjjewett I think the displacement current will be enough to sustain the LED, and there's many ways to explain it, but this is my layman understanding using lumped elements (please correct me if I'm wrong):
As displacement current is produced through each section of wire, charges are depleted in that section, which causes current to flow from the adjacent section. In the top circuit, only one wavefront is produced, which means there's displacement current in only one section of wire at any given time, but conduction current is supplied by the source, which produces a continuous current between the source and the wavefront.
In the bottom circuit, there are two wavefronts producing displacement current in the same direction, starting at the same spot. So immediately the displaced charge from one wavefront is picked up by the other wavefront, such that there's always a continuous current between the two wavefronts.
EDIT: Might be completely off-base. This explanation seems to imply that you can throw the switch, remove the source, and the LED will stay lit from the displacement current due to the wavefronts alone. Something's definitely missing here lol
Displacement current is related to transmission line. It flows between the capacitors. Led is far from source. So displacement current is flows between transmission line first. After some time it conventional current reaches to led then led will glow.
When we have a look at Eric's graph, it says, 50OHM all the time until it's back ... so, that would mean 20mA flowing all the time when the edge is travelling the way there and back. If we are talking about the bottom circuit (led, battery, switch, resistor placed in the middle), the current would be flowing through the LED when the edges would be travelling. Or?
In the final explanation where the lamp and switch are in the centre and he says that the lamp turns on immediately, I suggest that it actually turns on slowly (in relative terms) because until the wave has propagated far enough down the line to where the capacitance is high enough to couple sufficient current across the lamp can't light. Comments welcome.
32:09 Thank you for explaining the approximation.
Eric quickly mentions that there are two transmission lines in the second diagram, however, if so, that statement should also apply for the first diagram. In any case, for the second diagram he says that two signals emanate from the switch when it is closed. However, the voltage across the ideal cell remains at 2 volts at all times, and the voltage along the conductor from the cell's negative terminal around the circuit and past the LED remains unchanged, so there is not any associated displacement field to the right of that cell, so no displacement current would flow. Does the pulse also travel through the cell raising the voltage on the 'negative" terminal of the cell? With regards to the voltage divider concept, the voltage on the load side of the source impedance (across the switch) is initially two volts, then as soon as it closes the voltage on the load side of the source impedance drops to one volt with respect to the cell's negative terminal, so what happened to the electric field (represented by the extra volt) that was across the switch contacts? The associated field must have dissipated its energy some place. Was it radiated as an E-M wave? Can anyone clarify? It seems that Eric's comments assume an "instantaneous steady state condition" at various points in time as the pulse travels down the transmission line, but remains unclear what happens after the switch is closed, but before the pulse starts its journey down the conductor. Does the pulse travel in both directions?
Yes, the pulse does travel in both directions, and it lowers the voltage on the negative side of the battery
You seem to be assuming the 'return' current in the oppoaing wire always matches the top wire current.How does that change with distance between the wires? How does that change if i placed a third wire on top of the first one? It cannot just induce equal current in every case.
In case the cables are perfectly shielded like in faraday cage, what is the time for lighting the bulb?