#562 – Electroboom!

[Jesse Gordon]

That's precisely your, Mehdi's, Dave's and everybody-else-that-insist-that-KVL-always-hold's mistake.

KVL says that the voltages measured along the path of a mesh add up to zero.

But Lewin showed at least one circuit where the voltages do not add up to zero.

Therefore KVL doesn't always hold. Period.

There's nothing you can do to remedy this. It is not about Lewin. It is the way nature works. Go to your lab, set up the same experiment, measure the voltages around the circuit and see for yourself. We are glad that those who repeated the experiment were precisely the guys who thought that Lewin had somehow cheated or blundered the experiment.

They're all dead inside now. All of them.

You cannot solve such circuit using KVL. It is impossible. You'll have to resort to the full monty and use Maxwell's equations at least to calculate the EMF produced by the varying magnetic field.

After that, you can of course devise an equivalent circuit where instead of a varying magnetic field producing the extra EMF upon the entire circuit, you have a battery, a generator, a transformer, or any other equivalent lumped (i.e. localized) component to produce the same EMF and get the exact same effect on the other components. In that case, you can solve the equivalent circuit using KVL because you theoretically removed the varying field from the circuit and stashed it away in the equivalent component.

But that is just a theoretical trick that has a lot of caveats.

One of them is that you will not find this extra EMF on the circuit. Not even if you fart your way through the Bohemian Rapsody.

So how's that possible? How can we have, so to speak, a spooky "component" that produces an EMF in a circuit, but is not present there? This seems to violate the principle of conservation of energy, doesn't it? Those are legit questions. But to answer them you need to abandon KVL, which these people are not prepared to accept. To reconcile their cognitive bias with the real phenomenon that contradicts it, they create all the irrational arguments I listed above and more.

I must say, I found the above post charming when I read it. I didn't get sucked into this thread  until late in the game when Thinkfat invited me.

When I read the part about all those sad people who went to their labs and experimented and were then dead inside, all of them, I was like huh? Then I realized that I must be one of them because I went to my lab to experiment too hahahaha! 

But my experments showed that KVL holds up just fine! Even at 40Khz in an air core transformer!

If this is being dead inside, I'm sure having a blast. Maybe being dead inside isn't so bad after all.

And when I see how the people who are alive inside behave, dodging questions, jumping all over the place, making the most absurd predictions -- I guess I don't feel too bad about being dead inside! 

KVL says that the voltages measured along the path of a mesh add up to zero.

But Lewin showed at least one circuit where the voltages do not add up to zero.

And Lewin did that by not using a MESH.  As you said, KVL requires a MESH - i.e., a loop with no sub loops inside it.

And Lewin used a loop that contained a subloop. KVL didn't fail Lewin, Lewin failed KVL. Lewin failed Physics if you ask me, but you didn't. 

If you want to use KVL on nested or multiple loops, you have to reduce each child loop to a lumped element, then reduce each next child loop to a lumped element, until ultimately you have only a single loop, or, as you said, a mesh.

Lewin's loop system had multiple active loops which all had voltage differences and he simply did not account for a bunch of them.

If he had considered each of his loops as individual nested meshes,  (i.e. an inner mesh and then his outer loop) then KVL would have held up just fine.

(HEHE and if he hadn't got one of his VOLT METERS ON BACKWARDS!) 

One of them is that you will not find this extra EMF on the circuit. Not even if you fart your way through the Bohemian Rapsody.

So how's that possible? How can we have, so to speak, a spooky "component" that produces an EMF in a circuit, but is not present there?

All you gotta do is measure the output voltage of the transformer, and there's your missing EMF.

If you're talking about trying to reach in through the core with your probe to measure the EMF, here's your answer: When you do that, you're creating ANOTHER SECONDARY with the same phase and voltage:




Anyway, I guess I'm glad I found out I was one of the "Dead inside" so I know to have a somber moment once in a while in honor of, oh wait, if I'm not dead inside, then who the tarzan is?

[bsfeechannel]

Too bad you didn't get an engineering degree, you might have been able to debunk KVLers if you had!   

The bogus claims of your nonsense “experiments” were debunked fair and square 9 days ago. Since then you have been trying to conduct “interviews” with forum members to see if you can somehow disqualify them. Let me tell you, it’s not working.

I know the right questions to ask, but you won't answer them.

Your questions be like “Wait a minute! Do you believe tomatoes are not animals?”

We can’t help you.

[jesuscf]

\$
\begin{array}{l}
 V_2  = R_2  \cdot I = 900\Omega  \cdot 1mA = 0.9V \\
 V_1  =  - R_1  \cdot I =  - 100\Omega  \cdot 1mA =  - 0.1V \\
 \end{array}
\$

As for V_AD: any two nodes that are half a circle apart have half of the total EMF.  In this case |V_AD|=0.5V.

I see you used Ohm's law to compute the voltage drop along the resistors. So Ohm's law works. Wonderful.
Can you apply Ohm's law to compute the voltage drop across the wire that goes from R1 to point A?
Because we know that from A to D we have a voltage of half a volt, while across R1 we have what 0.1 V (fix the signs as you please while I put popcorns in the microwave). So, what gives?

(you can use the resistivity of copper, or a very low resistance for the wires, for example 1 milliohm, if zero resistance is a problem)

This is, by large, the most telling evidence of your ignorance of what is going on in the circuit (or any circuit by that matter!) in the whole thread.    How do you expect me to take you seriously when you don't know the circuital difference between a resistor and a voltage source?  That was explained before in this thread several times so I'll not do it again.

[jesuscf]

Too bad you didn't get an engineering degree, you might have been able to debunk KVLers if you had!   

The bogus claims of your nonsense “experiments” were debunked fair and square 9 days ago. Since then you have been trying to conduct “interviews” with forum members to see if you can somehow disqualify them. Let me tell you, it’s not working.

I know the right questions to ask, but you won't answer them.

Your questions be like “Wait a minute! Do you believe tomatoes are not animals?”

We can’t help you.

Hey bsfeechannel when I was asking you if you have an electrical engineering degree of some sort, it was because I had (and I still have) the impression you didn't know what you were talking about.  Will I ask Jesse Gordon if he has an electrical engineering degree? No!  Jesse Gordon knows what he is talking about!  I don't care where he learned the stuff; he just knows.

[bsfeechannel]

Jesse Gordon knows what he is talking about!

We also know what he’s talking about.

[bsfeechannel]

How do you expect me to take you seriously when you don't know the circuital difference between a resistor and a voltage source?

The problem is that wires don’t make that distinction. So, you’re not gonna be fool enough to make it.

[jesuscf]

Jesse Gordon knows what he is talking about!

We also know what he’s talking about.

No you don't.  You don't even know the basics of electric circuits.

[jesuscf]

How do you expect me to take you seriously when you don't know the circuital difference between a resistor and a voltage source?

The problem is that wires don’t make that distinction. So, you’re not gonna be fool enough to make it.

Before this thread you knew nothing.  Sadly, there is no progress.  That is the problem of arguing with somebody whose fundamentals are not there.

[bsfeechannel]

You don't even know the basics of electric circuits.

Says every single debunked KVLer.

[bsfeechannel]

That is the problem of arguing with somebody whose fundamentals are not there.

I’m not versed in pseudo science. Sorry.

[Kalvin]

I guess this problem could be modeled and solved using a 3D EM-simulator/solver. Then we could say: Case closed.

Any 3D EM-simulator gurus here?

[thinkfat]

This has already been done. The video by "Silicon Soup" has been linked in this thread at least twice. It is a very good video as it also shows the electric field component, without which you can not understand the problem at all.

[Sredni]

Again, you're quoting me out of context.

So many words that have nothing to do with the actual discussion. It's as if you were trying to create a smoke screen to hide the inconsistencies in your approach. The problem is that in almost every paragraph you write there is either a wrong concept, or a misconception, or a red flag betraying your limited knowledge. And the same is true for the two very long posts you wrote yesterday. I will try to collect all (well, most) the misconceptions and the false statement in separate posts by category, so that instead of an endless tit for tat, other users can benefit from a more organic discussion.

Let's start with a simple red flag.

Don't be afraid of the signs!

Regarding the following part of my picture, you wrote (emphasis mine):


https://i.postimg.cc/Fs4SRYQL/screenshot-21.png

 

I see you noticed that you made the same mistake Lewin did and you got one of the volt meters backwards, but at least you caught it in time to put a text note that says I can flip the volt sign.
So yes, I'd like to flip the sign on the left-hand meter which reads 1V. You have to keep the meters all pointing clockwise or counter clockwise -- but all the same way for a given test -- around the loop. That's just the way KVL is. If you mix up your volt meter signs even with pure batteries and resistors, KVL will appear to fail then too. 

Dude, seriously? You cannot handle the relativity of polarities? Apart from the fact that we are using voltmeters in an AC circuits, so the phase information is lost on us... what problem can you have in a circuit with batteries if one voltmeter is flipped? If it measure -0.5V with the probes in that position, it will measure +0.5V with the probes inverted.

You know I can choose whatever sign I want for the current in a circuit, then solve for its value and if it comes down negative it just means that it is flowing in the opposite direction to the one I supposed at first?
I hope you know at least some of the basic, such as how a current divider works, or what it means to load or shunt a generator, because otherwise this will be a very long discussion.

You dragged the discussion we were having here down about ten notches.

[Sredni]

The ambiguously 'unambiguos' voltages in the two branches

Jesse, referencing this figure



You wrote (still showing you do not understand what the term 'lumped' really means, but I will get to that in a later, dedicated, post):

You clearly show different possible paths within the element, the 5v and 0v paths inside that element are unambiguous,  unvarying, and if that is the path in that element, then it can be considered a lumped element in a KVL loop and KVL will hold. 

It appears you are referencing the bottom figure, where both 5V and 0V paths are present. Good, Let's forget about the 'ambiguous' orange paths and let's focus on the unambiguous green (5V) and purple (0V) paths. Well, the path in the left branch of the circuit is a purple path that has the unambiguous voltage of 0V; the path in the right branch of the circuit is a green path that has the unambiguos voltage of 5V.

What exactly is the excuse you invoke to deny that two branches of the same circuit, with the same starting and ending points will have different values for the unambiguos voltage associated to the path that goes through them?

[Sredni]

\$
\begin{array}{l}
 V_2  = R_2  \cdot I = 900\Omega  \cdot 1mA = 0.9V \\
 V_1  =  - R_1  \cdot I =  - 100\Omega  \cdot 1mA =  - 0.1V \\
 \end{array}
\$

As for V_AD: any two nodes that are half a circle apart have half of the total EMF.  In this case |V_AD|=0.5V.

I see you used Ohm's law to compute the voltage drop along the resistors. So Ohm's law works. Wonderful.
Can you apply Ohm's law to compute the voltage drop across the wire that goes from R1 to point A?
Because we know that from A to D we have a voltage of half a volt, while across R1 we have what 0.1 V (fix the signs as you please while I put popcorns in the microwave). So, what gives?

(you can use the resistivity of copper, or a very low resistance for the wires, for example 1 milliohm, if zero resistance is a problem)

This is, by large, the most telling evidence of your ignorance of what is going on in the circuit (or any circuit by that matter!) in the whole thread.    How do you expect me to take you seriously when you don't know the circuital difference between a resistor and a voltage source?  That was explained before in this thread several times so I'll not do it again.

Good: "the soldier that flees is good for another battle".

Can you shed some light on when exactly the wire becomes a 'voltage source'? Do you believe, like Jesse Gordon, that the wires acquire this magical voltage when they cross the magic portal of the core, or are they always voltage sources? For examples, is a quarter turn wire (open, just a quarter turn) a voltage source when it is inside the hole of the toroidal core? Does it ceases to be a voltage source when it is in front of the toroidal core but does not pass through the hole in the donut?

Teach me.

[jesuscf]

\$
\begin{array}{l}
 V_2  = R_2  \cdot I = 900\Omega  \cdot 1mA = 0.9V \\
 V_1  =  - R_1  \cdot I =  - 100\Omega  \cdot 1mA =  - 0.1V \\
 \end{array}
\$

As for V_AD: any two nodes that are half a circle apart have half of the total EMF.  In this case |V_AD|=0.5V.

I see you used Ohm's law to compute the voltage drop along the resistors. So Ohm's law works. Wonderful.
Can you apply Ohm's law to compute the voltage drop across the wire that goes from R1 to point A?
Because we know that from A to D we have a voltage of half a volt, while across R1 we have what 0.1 V (fix the signs as you please while I put popcorns in the microwave). So, what gives?

(you can use the resistivity of copper, or a very low resistance for the wires, for example 1 milliohm, if zero resistance is a problem)

This is, by large, the most telling evidence of your ignorance of what is going on in the circuit (or any circuit by that matter!) in the whole thread.    How do you expect me to take you seriously when you don't know the circuital difference between a resistor and a voltage source?  That was explained before in this thread several times so I'll not do it again.

Good: "the soldier that flees is good for another battle".

Can you shed some light on when exactly the wire becomes a 'voltage source'? Do you believe, like Jesse Gordon, that the wires acquire this magical voltage when they cross the magic portal of the core, or are they always voltage sources? For examples, is a quarter turn wire (open, just a quarter turn) a voltage source when it is inside the hole of the toroidal core? Does it ceases to be a voltage source when it is in front of the toroidal core but does not pass through the hole in the donut?

Teach me.

At this point I am pretty sure you have no formal education in electrical engineering.  But here is the answer anyhow: every piece of wire and every resistor in the loop has an inductance.  That should be enough to understand what is going on.

[Kalvin]

This has already been done. The video by "Silicon Soup" has been linked in this thread at least twice. It is a very good video as it also shows the electric field component, without which you can not understand the problem at all.

Thanks! Checked the video once, and looks really good. I need to dedicate more time watching it again doing the math.

[Jesse Gordon]

How do you expect me to take you seriously when you don't know the circuital difference between a resistor and a voltage source?

The problem is that wires don’t make that distinction. So, you’re not gonna be fool enough to make it.

Before this thread you knew nothing.  Sadly, there is no progress.  That is the problem of arguing with somebody whose fundamentals are not there.

Look on the bright side! He did learn that his paradigm is so wrong that he dare not answer questions lest his lack of understanding be revealed! 

[Jesse Gordon]

That is the problem of arguing with somebody whose fundamentals are not there.

I’m not versed in pseudo science. Sorry.

Like, for example, the pseudo science that which says KVL measures as holding with a loop formed by resistors and the output windings of closed-magnetic-circuit-core transformers, like this:?

https://i.postimg.cc/15gbsCmz/20211119-232948.jpg

Do you really think it's pseudo science for me to say that if I measure the voltage difference across each element like KVL states and sum up the voltages that it will sum to zero?

Because I tried it. It did sum to zero.

You can't just call observable reality pseudo science because you don't understand it.

IN REALITY, as MEASURED with a volt meter, sum(V1, V2, V3)=0 which means KVL holds in this scenario.

[Jesse Gordon]

You don't even know the basics of electric circuits.

Says every single debunked KVLer.

"Says every single lewinite."

There. Do you see how incredibly feeble we both sound saying that? I'd have to be dead inside to say that. 

Instead, I'm going to say this:

I personally used a transformer and a volt meter and I measured around a loop of elements like this:


I found, by observation, that KVL HOLDS in this scenario.

Tell me, can you at least agree that KVL at least APPEARS to hold in the case of closed-magnetic-circuit-core transformer secondary windings as elements in a loop, as shown above?

Dude, all the voltage differences across each element sum to zero! That's literally the definition of KVL!

How can you not admit that for closed-magnetic-circuit-core transformer secondary windings, KVL holds with that winding as an element in a loop with other elements?

[Jesse Gordon]

I guess this problem could be modeled and solved using a 3D EM-simulator/solver. Then we could say: Case closed.

Any 3D EM-simulator gurus here?

Simulators are a wonderful thing, but we should not neglect real-world observed data either.

Here I test with an open air core transformer: https://youtu.be/nAsZFP8Cfxk

Here I test with a closed-magnetic-circuit-core transformer (EI Core) https://youtu.be/iDWv8QJrzUo

And this one is more humor than anything, but it shows what Lewin's mistake would have been like had he been using batteries instead of transformer secondary windings: https://youtu.be/V_Gs8tSqBRY

And here's a longer demonstration showing how you can run your probe leads if you do want to measure voltage differences in a changing magnetic field without having your reading influenced by said changing magnetic field: https://youtu.be/_mzxE_p5dN8


Of course all the accurate simulations give the same results as the real world observations.

I'm really not sure what the argument is even about.

The biggest point seems to be to defend Lewin's honor, and has nothing to do with the technical facts.

Those defending Lewin seem to have little understanding of actual electromagnetics, and generally refuse to even accept that KVL does hold in a loop of elements one of which is a closed-magnetic-circuit-core (toroidal) transformer secondary winding, as shown here:


I built a working model and measured around the loop, getting the voltage differences across each element per KVL and they summed to zero.

But the people defending Lewin won't even admit that "Yes, in that case, KVL appears to hold" because they feel they are somehow betraying their master.

What's really odd, is that Lewin wasn't even using a closed-magnetic-circuit transformer, he was using an open air core transformer which complicates things for the uninformed.

So knowing that, I really have no idea why they won't accept that KVL can at the very least appear to hold with an element which is the output of such a transformer as I show in my diagram above.

[Jesse Gordon]

Again, you're quoting me out of context.

So many words that have nothing to do with the actual discussion. It's as if you were trying to create a smoke screen to hide the inconsistencies in your approach. The problem is that in almost every paragraph you write there is either a wrong concept, or a misconception, or a red flag betraying your limited knowledge. And the same is true for the two very long posts you wrote yesterday. I will try to collect all (well, most) the misconceptions and the false statement in separate posts by category, so that instead of an endless tit for tat, other users can benefit from a more organic discussion.

Let's start with a simple red flag.

Don't be afraid of the signs!

Regarding the following part of my picture, you wrote (emphasis mine):


https://i.postimg.cc/B6WVgpBm/screenshot-21.png

 

I see you noticed that you made the same mistake Lewin did and you got one of the volt meters backwards, but at least you caught it in time to put a text note that says I can flip the volt sign.
So yes, I'd like to flip the sign on the left-hand meter which reads 1V. You have to keep the meters all pointing clockwise or counter clockwise -- but all the same way for a given test -- around the loop. That's just the way KVL is. If you mix up your volt meter signs even with pure batteries and resistors, KVL will appear to fail then too. 

Dude, seriously? You cannot handle the relativity of polarities? Apart from the fact that we are using voltmeters in an AC circuits, so the phase information is lost on us... what problem can you have in a circuit with batteries if one voltmeter is flipped? If it measure -0.5V with the probes in that position, it will measure +0.5V with the probes inverted.

Are you dyslexic? or using a screen reader?

I complained about you putting the sign wrong on the "LEFT-HAND meter which reads 1V" and you zoom in on the RIGHT HAND meter which reads 1V?

Dude, SERIOUSLY? Look at your original picture, to which I was replying -- except this is the whole thing:


https://i.postimg.cc/R04QGyHs/KVL-works-if-I-leave-out-the-magnetic-region.jpg

Take note of this:

<-----LEFT      RIGHT------>

I asked about the LEFT-Hand 1V meter, and you cropped it off and showed the RIGHT 1V meter. You're quoting me out of context even if you have to cut away the part of the diagram to which I'm replying!

You know I can choose whatever sign I want for the current in a circuit, then solve for its value and if it comes down negative it just means that it is flowing in the opposite direction to the one I supposed at first?
I hope you know at least some of the basic, such as how a current divider works, or what it means to load or shunt a generator, because otherwise this will be a very long discussion.

You dragged the discussion we were having here down about ten notches.

Of course I understand signs.

But it just so happens that when applying KVL, you have to have all the volt meters pointing the same clock direction around the loop, and you have the LEFT-HAND 1V (<---) meter pointing the wrong way, which would cause KVL to appear to fail even if it was strictly resistors and batteries.

Lewin ALSO made this same error, having one of HIS volt meters backwards when he purported to measure test KVL. That's why one of his scopes read a positive spike and the other read a negative spike. If he'd had them both clocked the same direction, they'd have both read positive (or both negative), and the voltage across the resistors would have summed to the negative of his induced EMF, and KVL would have held.

So of course I know how signs work. And of course I know that if you hook a volt meter up backwards you can just negate whatever it reads. But when we're modeling, if you hook up the volt meter backwards but then don't tell the model that it's backwards, then you failed KVL, KVL didn't fail you.


You drew 6 volt meters in your loop: One had no polarity indicated, two were positive-leading-clockwise and three of them were negative-leading-clockwise except two.

Remember how they hammer into students to pay attention to the sign? It's got to be second nature by the time you graduate. But you draw a KVL loop and you're over the place.

So sure, you can put your volt meters whatever way you want then flip the signs as needed - but no engineer would draw what you drew.

There's just no way you're an engineer of any sort of the meaning. Unless.... Unless.. Do you drive a locomotive? That might suite you, since you don't have to think about whether you drive on the right or the left. 

You complain about "So many words that have nothing to do with the actual discussion."

Fine, let's talk about the actual discussion, if you will:

In the diagram below, assuming that all of the volt meters are polarity-clocked the same way, and assuming you ignore the 0.5v meters, will KVL hold?
Will the algebraic sum of all the voltage differences of each element around the loop be zero?


https://i.postimg.cc/R04QGyHs/KVL-works-if-I-leave-out-the-magnetic-region.jpg

[Jesse Gordon]

The ambiguously 'unambiguos' voltages in the two branches

Jesse, referencing this figure



You wrote (still showing you do not understand what the term 'lumped' really means, but I will get to that in a later, dedicated, post):

You clearly show different possible paths within the element, the 5v and 0v paths inside that element are unambiguous,  unvarying, and if that is the path in that element, then it can be considered a lumped element in a KVL loop and KVL will hold. 

It appears you are referencing the bottom figure, where both 5V and 0V paths are present. Good, Let's forget about the 'ambiguous' orange paths and let's focus on the unambiguous green (5V) and purple (0V) paths. Well, the path in the left branch of the circuit is a purple path that has the unambiguous voltage of 0V; the path in the right branch of the circuit is a green path that has the unambiguos voltage of 5V.

What exactly is the excuse you invoke to deny that two branches of the same circuit, with the same starting and ending points will have different values for the unambiguos voltage associated to the path that goes through them?

What do you mean "two branches of the same circuit?" One's the secondary winding on a transformer and the other's not. You no longer have a single loop -- you have multiple loops each with their own voltage sources. Like this:


https://i.postimg.cc/LX6wZVFJ/20211121-235058.jpg

Like I keep saying, MODEL REALITY. Don't slip in extra elements that you're not modeling for. Don't call it a MESH when you have multiple nested or networked loops.

[Sredni]

This has already been done. The video by "Silicon Soup" has been linked in this thread at least twice. It is a very good video as it also shows the electric field component, without which you can not understand the problem at all.

Thanks! Checked the video once, and looks really good. I need to dedicate more time watching it again doing the math.

Here's a map to navigate these troubled waters:


https://i.postimg.cc/QdRGMwMM/Map-to-the-Ring-Quest.png

It shows the fields inside the ring, alone.

[Sredni]

Don't be afraid of the signs!

You have to keep the meters all pointing clockwise or counter clockwise -- but all the same way for a given test -- around the loop. That's just the way KVL is. If you mix up your volt meter signs even with pure batteries and resistors, KVL will appear to fail then too.
...
I complained about you putting the sign wrong on the "LEFT-HAND meter which reads 1V" and you zoom in on the RIGHT HAND meter which reads 1V?

Well, on the right there is a loop made of voltmeters to which you can apply KVL and one of the voltmeter has a flipped polarity (according to you), I though you referred to that one.
But thank you for confirming that the left voltmeter would read 1V and therefore there is no voltage in the wires.

But it just so happens that when applying KVL, you have to have all the volt meters pointing the same clock direction around the loop,

This is probably true in schools for children with special needs. I can assure you that among adults, we take whatever polarity comes with the circuit and adapt to that without batting an eye.
You are indeed afraid of the signs.

Lewin ALSO made this same error, having one of HIS volt meters backwards when he purported to measure test KVL. That's why one of his scopes read a positive spike and the other read a negative spike.

That's not an error. It's a no brainer to automatically flip the signs in your head when the loops are these simple.
I told ya, this was a red flag.

Fine, let's talk about the actual discussion, if you will:
In the diagram below, assuming that all of the volt meters are polarity-clocked the same way, and assuming you ignore the 0.5v meters, will KVL hold?

Do you really need to ask? Did you not read what I wrote in the top right rectangle?
"KVL works here. Works for every path in the shaded region". Here path means 'closed path' or 'loop' to which apply KVL.
Why do you need to ask, I wonder...

Have you seen the rectangle in the other picture, though? 
Never mind, I will repost it with due highlighting.