Author Topic: Does Kirchhoff's Law Hold? Disagreeing with a Master  (Read 67168 times)

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Offline Zucca

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #250 on: November 25, 2018, 01:12:14 am »
But I think it's worth watching if you are interested and have half an hour to kill.

This was excellent. Thanks rfeecs, I feel like now I understand everything and the earth is one more time not flat.
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Offline ogden

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #251 on: November 25, 2018, 01:27:02 am »
To say there is "EMF in the wires" makes no sense. According to Faraday's law, in this case the EMF is the time rate of change of the magnetic flux through a surface.  The surface defines the EMF.  It is not located at specific points in the path that defines the surface.

Wait... So you say that there is no EMF induced in the straight wire which is located in the changing magnetic field and only complete/closed loop results in EMF?
 

Offline rfeecs

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #252 on: November 25, 2018, 02:03:15 am »
To say there is "EMF in the wires" makes no sense. According to Faraday's law, in this case the EMF is the time rate of change of the magnetic flux through a surface.  The surface defines the EMF.  It is not located at specific points in the path that defines the surface.

Wait... So you say that there is no EMF induced in the straight wire which is located in the changing magnetic field and only complete/closed loop results in EMF?
I'm saying this is Faraday's law:




The right hand side of the equation is the EMF.  It is defined by a surface bounded by a closed contour.  The closed contour doesn't have to be a wire.  It can be any closed contour.

For the EMF to do work, you need a circuit.  So that forms a closed contour.  A straight wire by itself is not a closed contour.  You can't say it results in an EMF all by itself, or that it contains an EMF.

So what can you say about a straight wire that is located in a changing magnetic field?  There will be a defined charge distribution in the wire.  There may even be current in the wire as the charge moves back and forth.  There will be an electric field around the wire, induced by the magnetic field.  The current in the wire may induce it's own magnetic field.  You can say lots of things, but not that it contains an EMF.

You can connect a meter across the ends of the wire.  Now you may measure a voltage and call that an EMF.  Because you created a closed contour with the wires of the meter.
« Last Edit: November 25, 2018, 02:16:01 am by rfeecs »
 

Offline sectokia

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #253 on: November 25, 2018, 12:36:13 pm »
To say there is "EMF in the wires" makes no sense. According to Faraday's law, in this case the EMF is the time rate of change of the magnetic flux through a surface.  The surface defines the EMF.  It is not located at specific points in the path that defines the surface.

Wait... So you say that there is no EMF induced in the straight wire which is located in the changing magnetic field and only complete/closed loop results in EMF?

That's exactly right.

And this is something Mehdi gets wrong when he measures the emf across the loop with the gap in it. By closing the gap with the meter probes only now does the emf exist.

Those is not just a theoretical thing. It is verifiable. If there was an emf there would be a static build up at each end of the wire that you could measure the e field of. But there isn't. It only becomes measurable when you close the loop. If you close it with a resistor there is a static build up on either side of it that you can measure.


 

Online GeorgeOfTheJungle

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #254 on: November 25, 2018, 01:53:38 pm »
Those is not just a theoretical thing. It is verifiable. If there was an emf there would be a static build up at each end of the wire that you could measure the e field of. But there isn't. It only becomes measurable when you close the loop. If you close it with a resistor there is a static build up on either side of it that you can measure.

I think the charge moves in the wire even if it's not a closed loop, if you could measure it you'd see a + q in one end and a minus q in the other end.
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Online Berni

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #255 on: November 25, 2018, 02:48:49 pm »
Those is not just a theoretical thing. It is verifiable. If there was an emf there would be a static build up at each end of the wire that you could measure the e field of. But there isn't. It only becomes measurable when you close the loop. If you close it with a resistor there is a static build up on either side of it that you can measure.

I think the charge moves in the wire even if it's not a closed loop, if you could measure it you'd see a + q in one end and a minus q in the other end.

Yes the charges move in a open loop of wire too. Just that they quickly bunch up at the end of the wire, this causes a electric field inside the wire that starts pulling the changes back into a equalized state. At some point the force put on the changes by the magnetic field balances out with the force from the electric field and the charges stop moving. This means the electric field is equal to the magnetic EMF. So there is indeed a electric field inside the wire (Happens even on a superconducting wire), but you can't directly use Faradays law to calculate it as it requires a loop. You have to calculate it using more fundamental math of applying forces to electrons (Faradays law is mostly an application of that fundamental math in a more useful form). It all depends on how you think about voltage. Yes there are more bunched up electrons on one end of the wire and if you generated a few kV of EMF you would even have the electrons fly off into the air and ionise it, but there is also an opposing magnetic EMF present. So if you add up all forms of EMF (electric and magnetic) you indeed get 0V. Electrons feel both types of EMF so the formal definition of voltage across the ends of the wire is 0V. However if you connect the two ends using a wire that travels in such a way that it generates no magnetic EMF you will get current flow proportional to the electric field of those bunched up electrons. This wire would close the loop in such a way that Faradays law would calculate a EMF voltage equal to the voltage of the electric field the bunched up electrons created.

So yes the open piece of wire does push electrons much like a open circuit battery would, but due to the definition of voltage its still 0V because the magnetic EMF is included.

The way definitions are set up we get confusing things when dealing with open loops of wire, but that's kinda okay since a open loop of wire can't do anything useful, it needs to be connected to something on the ends to do something meaningful and at that point the loop got closed anyway.
 

Offline Sredni

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #256 on: November 25, 2018, 03:19:36 pm »
I think the charge moves in the wire even if it's not a closed loop, if you could measure it you'd see a + q in one end and a minus q in the other end.

And which end will have a plus and which will have a minus?
Imagine your AB segment in a uniformly distributed time-varying B field. It is all the same, spatially.
Which end will get the plus, and which end will get the minus?

Can't decide?
Well, let's create a square loop with AB as its side. Use the right hand rule to find out how the current will flow with that flux varying configuration. Now you can tell me which extreme of AB is plus and which is minus, correct?
Except...

That it all depends which 'side of the loop' AB is on.

Think of two square clocks with a common AB side. Is the seconds hand going up or down? Does it depends on which clock the hand belongs to?
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Online GeorgeOfTheJungle

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #257 on: November 25, 2018, 03:34:19 pm »
With a strong enough (variable) magnetic field and/or enough loops, you should (in theory) be able to light a LED (for example, to say something) even with the ends of the loop open, simply because there's charge being pushed/moving along into the wire and that's what a current is.
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Online Berni

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #258 on: November 25, 2018, 03:34:33 pm »
For those that are disappointed to not see Kirrhhoffs laws be broken here is a way to break them.



This is the schematic of the experiment for it. We have a AC signal source that powers a circuit of two resistors. Circuit analysis tells us this is essentially a 50% restive divider. So we expect the blue voltmeter to always read half of what the yellow voltmeter is showing.





Here is the test setup. Voltmeters are again represented by an oscilloscope. The RF signal generator (set to +10dBm) is out of the shot, but you can see the coax cable coming in from it. The red coiled up wire is a 1m crocodile clip lead to serve as the 1m monopole antenna.






Here is the closeup of the test circuit. It is built using SMD resistors to provide better behavior at RF frequencies while trace lengths are kept to a minimum. Active scope probes ( 0.6pF loading) are used to probe the circuit without drastically affecting it. Additionally a 6dB attenuator is used before the signal enters the board to prevent standing wave issues in the long coax cable to the RF synthesizer.



Antenna disconected


This is what we get once we turn it on.

Acording to cirucit analysis we should be getting
504mV / 2 = 252mV

We measure 259mV so an error of 2.8% .This is well within reason given resistor, probe and scope tolerances.




Antenna connected


So now we connect the antenna to the midpoint between the resistors and let it hang over the edge of the table so that it is far away from objects it could potentially capacitively couple to. Other end is not connected to anything and is floating in mid air.





This results in the flowing image on the scope. The overall loading caused the input signal to sag a bit so we need to recalculate it:
470mV / 2 = 235mV

We measured it to be 150mV this is an error of 57%. This is certainly not inside a reasonable margin of error! Clearly the currents in the two resistor can't be identical if the voltages across them are so different. Well the solution is easy, the missing current is simply flowing into the antenna.

Okay, but where is that current returning? The antenna has only one connection. All currents flow in loops, so where does this loop return the current back?

The frequency of 76MHz was not just randomly picked. The calculated quarter wavelength for a 1m long monopole is 71MHz. But mine ended up measuring to be 76MHz due to the way its bent, its insulation, evnivorment... etc. This means that 1m piece of wire is very good at radiating energy out as radio waves for that particular frequency. Its sucking energy out of our circuit to be able to do so.





But what if this is just capacitive loading from the wire dragging down the signal? Well capacitors always create a phase shift. If you look at the previous screenshot you can see the two signals are still perfectly in phase with each other. So to prove this is not the case we leave the wire connected to change the frequency to 152MHz. This is double the frequency before and as a result turns out quarter wavelength of wire into a half wavelength of wire. This makes it very difficult for the wire to radiate the energy because all of it is simply bouncing back into it.

So we do the math again for the expected result:
472mV / 2 = 236mV

And we measured it to be 233mV so an error of 1.3%. This is once again well within our margin of error. So the circuit is again acting like circuit analysis tells us. This proves that capacitive loading of the wire was not at fault here. But because it no longer was able to radiate out energy as radio waves means it was no longer stealing energy out of the circuit so it was allowed to operate as usual.


Conclusion:
Kirchhoffs laws indeed stop working when your circuit starts to emit radio waves. This is very difficult to model and is very frequency dependent. This is not caused by parasitic inductance and capacitance so the circuit model can't be fixed by simply adding those in.

This is the reason why a lot of literature states that Kirchhoffs laws only work in low frequency AC circuits. The low frequency here is considered to be one witch has a wavelength significantly shorter than the circuits physical size. This prevents parts of the circuit becoming antennas and radiating away energy. This is a known limitation of his law.

So are you happy now all of you that want to see Kirchhoffs being wrong?
« Last Edit: November 25, 2018, 03:37:20 pm by Berni »
 
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Online Berni

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #259 on: November 25, 2018, 03:49:16 pm »
I think the charge moves in the wire even if it's not a closed loop, if you could measure it you'd see a + q in one end and a minus q in the other end.

And which end will have a plus and which will have a minus?
Imagine your AB segment in a uniformly distributed time-varying B field. It is all the same, spatially.
Which end will get the plus, and which end will get the minus?

Can't decide?
Well, let's create a square loop with AB as its side. Use the right hand rule to find out how the current will flow with that flux varying configuration. Now you can tell me which extreme of AB is plus and which is minus, correct?
Except...

That it all depends which 'side of the loop' AB is on.

Think of two square clocks with a common AB side. Is the seconds hand going up or down? Does it depends on which clock the hand belongs to?

Sure you can decide. You follow the usual rules of direction with magnetic fields.



This example shows the wire moving trough the magnetic field, but you can just as well keep the wire stationary and move the magnetic field instead.

So i see no issue with determining what end is positive, its the one that has current is flowing towards it.
 

Offline Sredni

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #260 on: November 25, 2018, 04:27:27 pm »
I think the charge moves in the wire even if it's not a closed loop, if you could measure it you'd see a + q in one end and a minus q in the other end.

And which end will have a plus and which will have a minus?
Imagine your AB segment in a uniformly distributed time-varying B field. It is all the same, spatially.
Which end will get the plus, and which end will get the minus?

Can't decide?
Well, let's create a square loop with AB as its side. Use the right hand rule to find out how the current will flow with that flux varying configuration. Now you can tell me which extreme of AB is plus and which is minus, correct?
Except...

That it all depends which 'side of the loop' AB is on.

Think of two square clocks with a common AB side. Is the seconds hand going up or down? Does it depends on which clock the hand belongs to?

Sure you can decide. You follow the usual rules of direction with magnetic fields.


And you are the one who does not shift goalposts, eh?

There is no motional emf in Lewin's circuit.
In your example, the velocity of the moving bar is breaking the symmetry.
Please answer the question:

How do you decide which extreme of the bar has the plus and which has the minus when your bar is STATIONARY with respect to a SPATIALLY UNIFORM but TIME-VARYING B field?
Are you capable of answering THIS question without changing the problem?

Quote
This example shows the wire moving trough the magnetic field, but you can just as well keep the wire stationary and move the magnetic field instead.

Not the same problem. Please answer the question above.

Example, if I ask
"What it 9 divided by 4"
you should not answer
"9 divide by 3 is 3".

How do you decide which side is plus and which is minus in the case of a bar, stationary with a spatially uniform, time-varying magnetic field?

Same question applies to GeorgeoftheJungle, of course. He too did not answer.

EDIT: As for the antenna example, of course that breaks KVL as well, but we are trying to keep things simple here. So our circuit is in the domain of quasi-static electrodynamics, where the d/dt of the field is so small that the concatenation of B and E fields dies off in a very short distance. We are in fact disregarding the displacement current in Ampere-Maxwell's equation. So, no radiation.

EDIT: removed emoticon.
« Last Edit: November 25, 2018, 04:38:11 pm by Sredni »
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Online Berni

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #261 on: November 25, 2018, 05:16:16 pm »

And you are the one who does not shift goalposts, eh?

There is no motional emf in Lewin's circuit.
In your example, the velocity of the moving bar is breaking the symmetry.
Please answer the question:

How do you decide which extreme of the bar has the plus and which has the minus when your bar is STATIONARY with respect to a SPATIALLY UNIFORM but TIME-VARYING B field?
Are you capable of answering THIS question without changing the problem?

Sorry, my mistake there. I forgot about the uniform varying field in the original question.

In this case you can simply apply the left hand rule to determine the positive end. Align your thumb with the field and the fingers show the direction of current.

The only case where this is problematic is when a wire is straight since we can't determine if a straight line is bending towards the clockwise or counter clockwise direction. But such a straight wire would not generate any EMF so it doesn't have a direction.

If you prefer to solve this with Faradays law you can also just connect the ends of the wire with a straight line in this case. That straight line will not generate any EMF but will close the loop so that you can take an integral of the field going trough its surface area, so whatever the result of Faradays law is the EMF of that wire segment.(This only works in a uniform field tho as otherwise you can get EMF on a straight wire)

EDIT: Made a mistake here on how open loops work, see this thread for explanation: https://www.eevblog.com/forum/chat/does-kirchhoffs-law-hold-disagreeing-with-a-master/msg1993139/#msg1993139

EDIT: As for the antenna example, of course that breaks KVL as well, but we are trying to keep things simple here. So our circuit is in the domain of quasi-static electrodynamics, where the d/dt of the field is so small that the concatenation of B and E fields dies off in a very short distance. We are in fact disregarding the displacement current in Ampere-Maxwell's equation. So, no radiation.

I do agree it is a more complex example and is more difficult to reproduce, but it does show a case of something that you can't properly model using lumped circuit meshes. Because you can't produce a lumped circuit that also means you can't use Kirchhoffs laws.

Dr. Lewins example can be easily modeled accurately using lumped circuit meshes. Once you have a lumped circuit you can apply Kirchhoffs laws and they work perfectly fine.

So i suppose RF engineers can indeed say "Kirchhoffs law is for the birds", most other engineers, not so much.
« Last Edit: November 26, 2018, 07:48:12 pm by Berni »
 

Online GeorgeOfTheJungle

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #262 on: November 25, 2018, 05:25:04 pm »
How do you decide which side is plus and which is minus in the case of a bar, stationary with a spatially uniform, time-varying magnetic field?
Same question applies to GeorgeoftheJungle, of course. He too did not answer.

A varying magnetic field pushes q in one direction, that's how you know where q is going to move to. Berni has even drawn it for you. What am I missing?
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Online Berni

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #263 on: November 25, 2018, 05:46:03 pm »
How do you decide which side is plus and which is minus in the case of a bar, stationary with a spatially uniform, time-varying magnetic field?
Same question applies to GeorgeoftheJungle, of course. He too did not answer.

A varying magnetic field pushes q in one direction, that's how you know where q is going to move to. Berni has even drawn it for you. What am I missing?

Yes but that picture is not very helpful for a varying uniform field. When you have a varying non uniform field you get induction in that straight piece of wire because the field lines appear to be moving in relation to the wire.

When you have a varying uniform field its only the magnitude of the field that changes, the actual field lines stay in the same place. This field will try to push electrons in a circle around the field lines. This can be imagined as every field line trying to get electrons to circle around it at the same time. On a straight wire this makes the electron get pulled in both directions simultaneously since some field lines are on one side and some on the other side of the wire. Once you put a bend in the wire this makes the electrons easier to move in the direction the wire is bending. So as a result the fields on the outside of the bend are mostly pushing the electrons into the side of the wire while the fields on the inside of the curve are pushing electrons more along the direction of the wire. This makes the fields on the inside of the bend win out and start moving electrons along the wire according to the left hand rule.

Sorry if this explanation is not very scientific but i think it makes the concept easier to grasp.

EDIT: Made a mistake here on how open loops work, see this thread for explanation: https://www.eevblog.com/forum/chat/does-kirchhoffs-law-hold-disagreeing-with-a-master/msg1993139/#msg1993139
« Last Edit: November 26, 2018, 07:47:57 pm by Berni »
 
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Online GeorgeOfTheJungle

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #264 on: November 25, 2018, 06:37:26 pm »
How do you decide which side is plus and which is minus in the case of a bar, stationary with a spatially uniform, time-varying magnetic field?
Same question applies to GeorgeoftheJungle, of course. He too did not answer.

A varying magnetic field pushes q in one direction, that's how you know where q is going to move to. Berni has even drawn it for you. What am I missing?

Yes but that picture is not very helpful for a varying uniform field. When you have a varying non uniform field you get induction in that straight piece of wire because the field lines appear to be moving in relation to the wire.

When you have a varying uniform field its only the magnitude of the field that changes, the actual field lines stay in the same place. This field will try to push electrons in a circle around the field lines. This can be imagined as every field line trying to get electrons to circle around it at the same time. On a straight wire this makes the electron get pulled in both directions simultaneously since some field lines are on one side and some on the other side of the wire. Once you put a bend in the wire this makes the electrons easier to move in the direction the wire is bending. So as a result the fields on the outside of the bend are mostly pushing the electrons into the side of the wire while the fields on the inside of the curve are pushing electrons more along the direction of the wire. This makes the fields on the inside of the bend win out and start moving electrons along the wire according to the left hand rule.

Sorry if this explanation is not very scientific but i think it makes the concept easier to grasp.

Oh, ok, (I think) I get it. Thanks!

When I said "the charge moves in the wire even if it's not a closed loop, if you could measure it you'd see a + q in one end and a minus q in the other end" I was thinking in ElectroBooms' setup @ 9m13s: youtu.be/0TTEFF0D8SA?t=9m13s
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Online Berni

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #265 on: November 26, 2018, 06:50:32 am »
You do realize you are making up rules on the fly, do you?
So, let me get this straight (pun intended).
Let's consider a square loop, with AB side perfectly straight. According to your model, there is no 'partial emf' on that side, right? And if you bend it a bit, the sign of the 'partial emf' will change according to the curvature of that side?
Or, let's just consider an AB segment alone: according to your rule: when it is straight there is no charge build up, but if it is bent in one way the charges are + on A and - on B, while if it is bent the other way the charges are - on A and + on B, correct?

Man, am I glad I do not live in the same universe as you. Looks pretty much more complicated than the universe I am in.

Its not the curvature itself that causes voltage, but the overall trend the wire is taking. A square is still bending around to form a loop that creates the usual direction. An example that seams to go both ways is a S shaped line. If you connect the ends of it you get a clockwise and counterclockwise loop so the surface area integrates to zero if both are symmetrical. If you did connect the ends you would also get zero current due to the area being zero. Electrons in the positive side of the S curve can't magically know about the ones in the negative side and not move because of it. The path the wire takes causes fields on the inside of the curves overall trend to overlap causing one side of the wire to start having more effect than the other.

I will admit i don't fully understand the underlying magic that determines why things move in the specific directions inside fields but the overall effects this causes seam to point towards this.

I would love to live in a simpler universe but magnetic and electric effects are linked trough the effects of Einsteins relativity and that stuff does all sorts of weird things.

KVL breaks in the case of radiation, correct. But it also breaks in the case of induction.

It worked just fine for me in the case of Dr. Lewins experiment so do we perhaps need a different experiment to show it not working with induction?



So for the case of open loops still producing charge separation i propose the flowing experiment:


We break a loop by putting a capacitor in series with it. This capacitor can be formed by two metal plates just like the symbol so there is a real air gap between the ends of the wire. No electrons can jump the gap between the plates. The capacitor requires a change in voltage to push electrons onto the plate and off the other plate in order to create the illusion of current flowing trough it.

So what would happen in this circuit when the uniform magnetic field is suddenly turned on?




But one could argue that that's just a small gap so we could just ignore it. So lets take it a step further and introduce a 2nd capacitor on the other side to completely cut the loop in half.

Can the 2 completely separated segments of wire generate a voltage on the capacitor to get the current flowing?


By the way this example is also possible to build as a real life experiment. There is even a way to measure the current in this loop without actually connecting any test equipment to the loop. Circuit analysis tells us that this is a LC circuit, if it indeed is one it will hold on to the pulse of energy and use it to oscillate back and forth, generating an AC magnetic field around the loop even after we remove the original field. We can measure this field as proof of current in the loop.
 

Offline Sredni

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #266 on: November 26, 2018, 03:47:19 pm »
Its not the curvature itself that causes voltage, but the overall trend the wire is taking. A square is still bending around to form a loop that creates the usual direction.

It looks like you are trying to say you can define emf only if you can identify an area to associate to it.
Well, that's progress.  That's what Faraday has tried to tell you since the nineteenth century.

Quote
I will admit i don't fully understand the underlying magic that determines why things move in the specific directions inside fields but the overall effects this causes seam to point towards this.

Yes, it points towards an area, encircled by a closed (possibly fictional, in the sense it does not have to be all inside real conductors) loop.
As the formula stating Faraday's law has always said.

Quote
I would love to live in a simpler universe but magnetic and electric effects are linked trough the effects of Einsteins relativity and that stuff does all sorts of weird things.

Correct, but not as weird as you think.
I suggest to brush up your physics on some good book, like Purcell (Electricity and Magnetism, second volume of the Berkeley physics series), and then look up the practical applications of the basic concepts in books like Ramo, Whinnery, VanDuzer (Fields and Waves in Communication Electronics).

As for the further goalpost shifting at the end of your post, please... Leave caps out - we are trying to keep things simple here. If we are having trouble understanding each other with a simple circuit like that, what do you think would happen if you introduce another paradox generating element, like the two caps back to back?

And no, KVL did not work with Lewin's circuit. You had to introduce that magical emf term to make your numbers check. That's Faraday at work. In fact, you cannot locate that voltage anywhere with a voltmeter, can you? I am talking about that circuit, do not try to modify it. Let down those scissors, I tell you!!!

EDIT: Repetitia juvant.
What happens if we pull the resistors out of the loop and make sure we cannot interfere with the flux that is generating the emf? That we have a series of two resistors and a black box with two terminals. Now you can call that the secondary of a hidden transformer. Now you can located the voltage it 'generates' with a voltmeter. It's right there, at its two terminals! Now you can delude yourself KVL works, and call it, instead of Faraday's Law, "extended KVL" or "modified KVL" or "modern KVL". Lumped circuit theory works, all voltages we can measure are uniquely defined. Now the quarrel "KVL vs Faraday" is just a language barrier.
But when the resistors are inside the loop, say goodbye to lumped circuit theory and uniquely defined voltages. You have to take paths into account.
« Last Edit: November 26, 2018, 03:56:46 pm by Sredni »
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Online Berni

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #267 on: November 26, 2018, 07:44:12 pm »

It looks like you are trying to say you can define emf only if you can identify an area to associate to it.
Well, that's progress.  That's what Faraday has tried to tell you since the nineteenth century.

Alright i have to apologies because i did get one detail about voltage in an open loop wrong. Straight wires require a special case to produce no EMF in the field (I will go back and add a note about this in my previous posts)

I finally found an article that explains how voltages in stationary open loops work when exposed to a varying uniform field:
Induced voltage in an open wire by K. Morawetz, M. Gilbert, A. Trupp
www.eevblog.com/forum/chat/does-kirchhoffs-law-hold-disagreeing-with-a-master/?action=dlattach;attach=581270




It is indeed solved by closing the loop using a wire flowing a path that generates no EMF. However such a wire does not directly connect the ends of the open loop wire segment. What is required instead is two straight wires that travel to the geometric origin of the uniform magnetic field. Such straight wires that pass trough or touch the geometric origin point of the field generate no EMF. This was causing problems for me because i did not realize that even ideal uniform fields have a center point. However in the case of a close loop coil the center point becomes irrelevant because it affects the entire loop and averages out by the time you get all the way around. So technically the beloved closed loop is just a special case of a open loop that closes in on itself(Or the open loop being a special case of a closed loop, whatever way around you want to think about it).

So there you do get voltage induced in a open loop of wire (Proven using Maxwell even).



Correct, but not as weird as you think.
I suggest to brush up your physics on some good book, like Purcell (Electricity and Magnetism, second volume of the Berkeley physics series), and then look up the practical applications of the basic concepts in books like Ramo, Whinnery, VanDuzer (Fields and Waves in Communication Electronics).

I do have at least some understanding on all of those areas, just that some i don't know well in to the detail and especially not down into the deep math behind it. Its just not something i deal with on a regular basis. Electronics engineering has so many abstractions in place that there is no need to delve this deep into the fundamental math under it all. Hence why most electronics engineers know about Maxwell and what he did, but they never used his equations on the job. Its just easier and faster to use the derived "easy bake" equations for calculating everything you would need, but if you dig down and dissect a lot of those equations you tend to find some Maxwells equation somewhere in there.

Contrary to popular belief engineers are mostly not math geniuses, they are just really good at looking up the right equation and quickly punching it into a calculator. Its simply the fastest way to get work done on a deadline.


As for the further goalpost shifting at the end of your post, please... Leave caps out - we are trying to keep things simple here. If we are having trouble understanding each other with a simple circuit like that, what do you think would happen if you introduce another paradox generating element, like the two caps back to back?

The capacitor in an inductive loop is just my proposed experiment to show that an open loop can generate a voltage. Circuit analysis is well understood for RLC circuits so we can easily use it to predict the behavior, then actually do the experiment and see if the results match. Capacitors don't create any more of a paradox than inductors.  Can you propose a simpler experiment that shows or disproves the presence of voltage in open loops of wire (aka fractional turns)? If the experiment can be done with equipment and materials found in a reasonable electronics lab i will recreate it.

The purpose of this experiment was to show that fractional turns can indeed take part in a circuit and pick up EMF just like complete loops can.

And no, KVL did not work with Lewin's circuit. You had to introduce that magical emf term to make your numbers check. That's Faraday at work. In fact, you cannot locate that voltage anywhere with a voltmeter, can you? I am talking about that circuit, do not try to modify it. Let down those scissors, I tell you!!!
Its not introducing a magical emf out of nowhere.

The cirucit mesh model simply needs to know about the properties of a wire. You do agree that a coil of wire with 100 turns placed across two points in a circuit is modeled using an inductor symbol right? Well these inductors are not closed loops as they have two terminals that connect to other components of a circuit (just like a straight wire).

A straight piece of wire is basically the same thing except with much less inductance since magnetic flux is not being reused multiple times on the same wire by coiling. This website provides a helpful calculator for this: https://www.eeweb.com/tools/wire-inductance . Among other things it also provides a calculator for loop inductance that comes useful later(mutual inductance)

If you place two such 100 turn coils in close proximity you can get some of the same magnetic field passing trough both coils. This turns them into coupled inductors where they not only have self inductance, but also something called a mutual inductance (This is essentially a transformer). The value of self and mutual inductance for each is all that is needed to describe the magnetic properties of them. Any number of coils can be added to this magnetically coupled inductor, not just two. The coupled inductor model is the "mathematical adapter" that brings Maxwells equations into a form that fits into circuit analysis theory. Once it fits inside the circuit analysis abstraction all other circuit analysis tools can be applied(Kirchhoff being only one of them). Its sort of like a software API, but with math rather than code. By putting an inductor into the mesh we simply create an instance of the inductor model that deals with magnetic effects for us.

So since a straight piece of wire is simply a inductor with less turns than a coil of wire we can model a piece of wire in the exact same way. Tho due to it having essentially zero turns means the self inductance is pretty low(but NOT zero) while the mutual inductance is likely significantly larger as soon as it forms a larger loop with other components of the circuit that it connects to. This mutual inductance is where the loop inductance is if you connect multiple segments of wire together into a complete loop.

We are still using Maxwells equations and Faradays law deep inside the equation that calculated the inductance value in Henrys. So why is modeling a length of wire as an inductor incorrect? Is using Maxwells equations as part of another equation forbidden?


EDIT: Repetitia juvant.
What happens if we pull the resistors out of the loop and make sure we cannot interfere with the flux that is generating the emf? That we have a series of two resistors and a black box with two terminals. Now you can call that the secondary of a hidden transformer. Now you can located the voltage it 'generates' with a voltmeter. It's right there, at its two terminals! Now you can delude yourself KVL works, and call it, instead of Faraday's Law, "extended KVL" or "modified KVL" or "modern KVL". Lumped circuit theory works, all voltages we can measure are uniquely defined. Now the quarrel "KVL vs Faraday" is just a language barrier.
But when the resistors are inside the loop, say goodbye to lumped circuit theory and uniquely defined voltages. You have to take paths into account.

Yes that is correct, see its not that hard to think in terms of circuit mesh models. You can indeed fix things by adding a black box transformer into the circuit. However Dr. Lewins experiment is about the voltage on points A and B. Once we lump all of the loop inductance into one black box we loose points A and B.

But wait! We can fix that. Instead of lumping all of it into one black box we can just lump each wire segment into its own black box. This way we get 4 such black boxes that are located between the resistors terminals and the points of interest.



This way points A and B are maintained while each black box is now one of the 4 secondary coils to the solenoid coil creating the original field. They all know about each other trough mutual inductance. If we suddenly also want a point C that's halfway between A and the left resistor we simply cut the black box into two blackboxes with half the inductance value. Point C now pops out as the midpoint between the two new blackboxes. This is why i was trying to explain above that open wire segments can interact with magnetic fields just as much as wire loops can, the black boxes are models of a open wire segment.

As i said KVL is not just a different form of Faradays law. Its is like comparing an apple to a car, they are two completely different things. Faradays law calculates the relationship between voltage and magnetic flux change in loops. Kirchhoffs voltage law just calculates voltage relationships in abstract electrical circuit meshes. It has nothing to do with magnetic or electric fields. Its just a law that is part of circuit analysis methods, those will then call upon Faradays law whenever circuits have to deal with inductance. Circuit analysis wraps Faradays law into the form of an inductor. The process of mesh analysis eventually marries together the equation for an inductor model (That contains Faradays law deep inside it) with Kirchhoffs voltage and current laws to produce a mathematical model of the circuit. Both KVL and Faradays law exists together in that resulting circuit equation, you can't see it in the form that Faradays law is written but if you expand the equations backwards far enough you could eventually get it to pop out Faradays law in textbook form. KVL can't deal with anything other than voltages so it relies on other laws to do it instead. It can't even understand resistance, it needs help from Ohms law to tell it the voltage on a resistor. Because of that we don't say KVL is for the birds because Ohms law handles it better.

So the only thing that Faradays law and KVL have in common is that during circuit analysis KVL makes use of Faradays law to be able to understand magnetic fields. They are great friends even if circuit analysis sometimes demands it to do unusual things such as fractional loop segments(That still work fine, see above). Is it forbidden to plug the results from one law into another law?

Circuit models are not meant to describe the underlying physics, they are about as real as the imaginary part of complex numbers is real. But circuit models use just enough physics to accurately describe the behavior of the circuit on a macroscopic level.
« Last Edit: November 26, 2018, 07:56:33 pm by Berni »
 
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Offline radioactive

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #268 on: November 27, 2018, 11:17:29 pm »
I've modified one of the openEMS examples to try and simulate this experiment numerically via the openEMS FDTD field solver.  In the attached image of the geometry, you can see the physical model of the excitation coil with feedline (copper) and resistor,  the wire loop (copper), two resistors (100/900 ohms)  / voltage/current probe elements, and two infinite resistance voltage probes measuring the voltage on the wire segments on either side of the resistors. The infinite resistance flags openEMS not to add any material properties for these probes (so they are unaffected by the fields and cause no loading).  The mesh is something close to 4 million cells which is probably a bit overkill, but wanted to make sure everything got integrated.  The simulation is configured to dump the volume of the magnetic field (H-field) and results in about 12GB of vtr files that you can view as an animation in paraview.  The attached matlab file can be used to create the geometry file and run the openEMS simulation.

[edit] removed sim files.  I'm an idiot.  I'l post a working sim when I have it finished.
« Last Edit: December 05, 2018, 12:07:56 am by radioactive »
 
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Offline bsfeechannel

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #269 on: November 28, 2018, 12:41:01 am »
I will admit i don't fully understand the underlying magic that determines why things move in the specific directions inside fields but the overall effects this causes seam to point towards this.

I would love to live in a simpler universe but magnetic and electric effects are linked trough the effects of Einsteins relativity and that stuff does all sorts of weird things.

Our brains are not hardwired to admit space being "warped" by any kind of changing field, be it magnetic, electric, gravitational or whatever.

Fortunately our brains are equipped with reason, so we can painstakingly break free from our basic intuitions and build a reasoning that will lead us to new levels of understanding. It takes time and effort, but it's worth the journey.

You might know that Maxwell is a theory that accounts for our four little happy dimensions of the spacetime in which we are immersed. The road to Maxwell starts with calculus, which is nothing more than the mathematical (i.e. formal) study of change. You know, we live in an ever changing universe. So someone had to invent a theory to give us tools to deal with that.

I happen to have a happy little video about calculus.

Calculus For Young Players - BSFEEChannel #2



That won't make anyone an expert but is going to give viewers an idea of what it is and how to immediately apply it to simple electronic circuits. However, if the viewer got interested in learning more, a good introduction to calculus is the series of 31 lectures given by Professor Richard Delaware at UMKC, which inspired my video above.



Then you'll come to vector calculus, also known as vector analysis. This is the theory that will give you tools to deal with change over whatever lines, surfaces, and volumes, because things are not all in the same place at the same time. There is no absolute simultaneity in the universe (blame Einstein).  A good start is the book Electromagnetic Waves and Radiating Systems by Edward C. Jordan and Keith G. Balmain. The first chapter is dedicated to introduce you to vector analysis. Interestingly enough, the first section discusses the limitations of circuit theory (i.e. Kirchhoff) without neglecting its importance, while demonstrates why the more complicated field theory (a.k.a Maxwell) is worth the effort.

Now that you upgraded your brain to think fourth-dimentionally, you can tackle Maxwell with ease. And that's what that book does when you reach chapter 4.

The cool thing about electromagnetism is that, since the electromagnetic force is 10³⁶ times stronger than gravity, mind boggling things that defy common sense happen right on top of your bench at human scale compared to gravitational phenomena that are only relevant at astronomical scale. This video shows just an example.

 

Offline bsfeechannel

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #270 on: November 28, 2018, 12:56:32 am »
I do have at least some understanding on all of those areas, just that some i don't know well in to the detail and especially not down into the deep math behind it. Its just not something i deal with on a regular basis. Electronics engineering has so many abstractions in place that there is no need to delve this deep into the fundamental math under it all. Hence why most electronics engineers know about Maxwell and what he did, but they never used his equations on the job. Its just easier and faster to use the derived "easy bake" equations for calculating everything you would need, but if you dig down and dissect a lot of those equations you tend to find some Maxwells equation somewhere in there.

Contrary to popular belief engineers are mostly not math geniuses, they are just really good at looking up the right equation and quickly punching it into a calculator. Its simply the fastest way to get work done on a deadline.

If we engineers deliberately neglect the fundamentals, how can we criticize Lewin who is teaching them?
 

Offline radioactive

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #271 on: November 28, 2018, 01:04:27 am »
The explanation video promised by Prof. Walter Lewin.



I haven't made it all the way through this entire thread yet, but that response video was excellent.  I still need to watch his lecture #20.  This quick video should clear things up for anyone who watches it though.
 

Offline ogden

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #272 on: November 28, 2018, 02:25:35 am »
If we engineers deliberately neglect the fundamentals, how can we criticize Lewin who is teaching them?

I would not call high level abstraction as negligence. It is common sense. We criticize Dr.Lewin because he neglect fundamentals himself (read "Lewin’s Circuit Paradox" by Dr. Kirk T. McDonald - you'll see). I do not see anybody who is against Maxwell's equations or saying that Kirchhoff's law *always* hold. Those who do not agree (to "KVL for the birds") say that Kirchhoff’s loop equations apply to Lewin’s circuit.
 

Offline bsfeechannel

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #273 on: November 28, 2018, 02:37:35 am »
If we engineers deliberately neglect the fundamentals, how can we criticize Lewin who is teaching them?

I would not call high level abstraction as negligence. It is common sense. We criticize Dr.Lewin because he neglect fundamentals himself (read "Lewin’s Circuit Paradox" by Dr. Kirk T. McDonald - you'll see). I do not see anybody who is against Maxwell's equations or saying that Kirchhoff's law *always* hold. Those who do not agree (to "KVL for the birds") say that Kirchhoff’s loop equations apply to Lewin’s circuit.

So, you criticize Lewin because someone else wrote an article criticizing him? Not because you yourself master the fundamentals and is in a position to confront him? What kind of engineers do we want to be? Just a bunch of dilettantes ranting at random in forums? Let's shut up and do our homework. Thank you.
 

Offline ogden

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Re: Does Kirchhoff's Law Hold? Disagreeing with a Master
« Reply #274 on: November 28, 2018, 03:00:48 am »
We criticize Dr.Lewin because he neglect fundamentals himself (read "Lewin’s Circuit Paradox" by Dr. Kirk T. McDonald - you'll see). I do not see anybody who is against Maxwell's equations or saying that Kirchhoff's law *always* hold. Those who do not agree (to "KVL for the birds") say that Kirchhoff’s loop equations apply to Lewin’s circuit.

So, you criticize Lewin because someone else wrote an article criticizing him? Not because you yourself master the fundamentals and is in a position to confront him? What kind of engineers do we want to be?

I already provided my position here in this thread. That's why I just refer to article I agree to and do not repeat what is already said. Don't blame me if you did not read thread or do not remember what I did say or whatever.

Quote
Just a bunch of dilettantes ranting at random in forums? Let's shut up and do our homework. Thank you.

You better behave
 


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