# Faraday's Law vs Kirchhoff's Law !!

#### recca02

Joined Apr 2, 2007
1,214
IIRC (after a week?) The prof. said Kirchoff's law is a special case of Faraday's law.
Can anyone tell me how are they related?

Thanks a lot mik3, That was truly wonderful.

#### steveb

Joined Jul 3, 2008
2,436
Why?

All Prof Lewin has demonstrated is that an EMF and a Potential Difference are not two names for the same thing. They are in fact different animals.

....

However, if correctly stated, Kirchoffs Law can be successfully applied as I showed. Here I must differ from the good Prof.
There are two separate issues we should consider here.

First, what is the definition of Kirchoff's Voltage Law (KVL)? It is simply that the line integral of electric field around a closed loop is zero. This is the position that Prof. Lewin takes, and that is the correct statement. However, we often extend KVL for circuit analysis to include the effects of lumped inductance (as determined from Faraday's law). So, here is where you and Prof. Lewin have different opinions, and personally I do not object to this (although Prof. Lewin would vehemently) since it comes down to definitions, and I use KVL in that modified form all the time for circuits.

The second issue explains my opinion that you are missing the most important aspect to the experiment he demonstrated. Your modified version of KVL, which includes EMF from inductive components, allows us to have a convenient circuit equation. However, this KVL law does not allow us to explain the results of the experiment. The reason is that the physics of the experiment does not allow an equivelent circuit representation in which the new KVL law works. It can't work, because moving the meter physically, without changing the connection points, changes the measured voltage. Standard circuit theory does not allow this effect.

This issue can be demonstrated with any attempt to present an equivalent circuit that explains the results, as the meter is moved in space. You can't do it because only Faraday's Law (applied over the full geometry) can answer that question. Note that the Prof did the demonstration with the meter on the left and then on the right, but he did not talk about what happens if the meter is above or below the circuit. Each of these cases gives a different answer. The bottom line is that this situation is not a circuit problem obeying your modified KVL, but is a field problem obeying Faraday's law. If you say that KVL describes his experiment, then all you have done is modified the definition of KVL once again, and made it completely equivalent to Faraday's law, with all its implications that spatial positioning of components is important. This doesn't seem fair to Faraday who is credited with a great discovery.

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#### studiot

Joined Nov 9, 2007
4,998
So whose equation would you use if you had both the battery and the inductor in circuit?

You don't even need two resistors to get apparent anomalies. What happens if you repeat the experiment with only one resistor. First with the battery in circuit then without?
What is then the 'voltage' across the ends of the resistor? In one direction it is zero the other it is given by ohms law IR. Are we also denying Ohms law?

You haven't addressed my comments about either the differential or integral representation of this situation, or the energy viewpoint. Nor is there an answer to Recca.s question.

#### steveb

Joined Jul 3, 2008
2,436
So whose equation would you use if you had both the battery and the inductor in circuit?
I already answered that indirectly above. If the inductor is a lumped inductor and there is no magnetic flux change cutting the circuit, then I use your modified version of KVL. This is acutally Faraday's Law (applied to a circuit) and not KVL, but I just think of it as KVL when doing circuit equations.

You don't even need two resistors to get apparent anomalies. What happens if you repeat the experiment with only one resistor. First with the battery in circuit then without?
What is then the 'voltage' across the ends of the resistor? In one direction it is zero the other it is given by ohms law IR. Are we also denying Ohms law?.
Yes, one resistor would demonstrate the effect too. Without a battery, one meter would read zero volts and the other would not. But, this does not deny ohms law. The meter that reads zero volts on the resistor makes a loop around the time varying magnetic field; hence, the EMF from the changing field exactly cancels the voltage on the resistor. The other meter does not make a loop around the changing field, and so it reads the resistor voltage only. You can also think about it another way making a path through the wire. The meter that reads zero volts does not make a loop around the field but goes through the wire, which drops no voltage. The other meter is reading a voltage across the wire (which is strange) but this is due to the EMF from the fact that it makes a loop around the field change.

I'll leave it to you to do if for the case with the battery.

You haven't addressed my comments about either the differential or integral representation of this situation, or the energy viewpoint. Nor is there an answer to Recca.s question.
I'm not sure what I need to address about your comment about the differential representatons of the Laplace or Poisson equatons. Those equations are consistent with electrostatics only, and are not really relavent here.

I thought I did answer Recca's question. Strickly speaking, Kirchoff says that the integral of electric field around a closed loop is zero. I presumed that it was clear what Faraday's Law says, but to be clear: Faraday's Law says that the integral of electric field around a closed loop equals the negative of time rate of change of magnetic flux through an open surface bounded by the closed path.

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#### mik3

Joined Feb 4, 2008
4,843
I am happy you found that interesting. Well, I don't think this is very strange. It is like when you have two resistors connected in series with a coil and a changing magnetic field is nearby the coil. However, in this case the coil is just the wires and people think (at least initially or some never understand it) that wires cannot induce voltage across them like a coil.

#### studiot

Joined Nov 9, 2007
4,998
Steve I don't understand your obsession with voltmeters.

Any analysis technique worthy of the name should be able to predict the the voltages present at chosen points in a circuit, without attaching voltmeters, or flipping them about. So the situation of the voltmeter and its attachments forming part of a loop does not arise.

Of course both Faradays and Kirchoffs laws and sufficiently powerful to achieve this. I am not trying to detract in any way from either. Both have their applications and their limitations.

Incidentally I understand the generalised expression of Faradays Law, in integral form, to be

EMF = $$\oint$$(V x B) . dL - $$\int$$$$_{s}$$ $$\frac{dB}{dt}$$ .ds

Sorry I'm still struggling with TEX, this is as near as I can get. The db/dt should be a partial derivative.

#### studiot

Joined Nov 9, 2007
4,998
I'm not sure what I need to address about your comment about the differential representatons of the Laplace or Poisson equatons. Those equations are consistent with electrostatics only, and are not really relavent here.
Is it not true that when a wire moves relative to a magnetic field (or the field changes relative to the wire) an electric field, E appears in the wire such that.

E = V x B

And are we not trying to integrate this E around a closed loop in stating that

EMF = $$\oint$$ E.dL = $$\oint$$ (V x B) . dL

#### studiot

Joined Nov 9, 2007
4,998
Finally consider Prof Lewins's experiment with no resistors, just a wire loop.

We now have an EMF of 1 volt but zero potential difference in the system.

A pretty dramatic demonstration that EMF and PD are not the same thing, although both are measured in volts.

#### steveb

Joined Jul 3, 2008
2,436
Steve I don't understand your obsession with voltmeters.
I don't know what to say to a comment like that. I'm not obsessed with voltmeters. But the whole point of that lecture by Prof Lewin is the nonintuitiveness of the experimental results, which are revealed by the voltmeter measurements. The voltmeters are not being tricked. Those are the actual results that demonstrate that Faraday's Law works, and Kirchoff's Law doesn't. A voltmeter measures current through a high resistance placed in parallel. It is a competely valid circuit to consider.

By the way your expression of Faraday's Law is incorrect. You are supposed to take the line integral of the electric field. See the following link.

#### studiot

Joined Nov 9, 2007
4,998
By the way your expression of Faraday's Law is incorrect. You are supposed to take the line integral of the electric field. See the following link.
In what way incorrect?
The first line of your reference is exactly the same as my second term.

However this term only takes care of time aspect (i.e fields that vary in time but not space or with stationary conductors). This limitation is clearly stated in the next paragraph of your reference.

My first term takes care of spatial variations.

I would say that the important point from Prof Lewins lecture is the difference between Kirchoffs Law and Faradays.

Kirchoff doesn't care where the EMFs come from (chemical, magnetic, elctromechanical etc) he just equates their total to the pds.

Faraday doesn't care what happens to (or even if) any current flows just what EMF is generated.

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#### steveb

Joined Jul 3, 2008
2,436
In what way incorrect?
The first line of your reference is exactly the same as my second term.
OK, I see what you are saying, but most people don't use this obscure form, as it gets into issues of frames of reference. Anyway this is taking us off course. Getting back to the issue at hand.

In order to cure my obsession with voltmeters , I am providing an attachment with an example similar to what you suggested. A simple loop with no resistors. However, I modify this and show two cases in which we attach a resistor with both leads at the same point. In one case the restor makes the connection by wrapping around the central flux region, and in the other case it does not.

The important point is that in one case the current is zero, and in the other case it is not. Kirchoff's Law holds in one case and not the other. To try and salvage Krichoff's Law, you insert an artificial EMF source at a point in the circuit, and this will depend on the path of the resistor connections. So you will have a EMF source that changes based on the way wires move around. Note that you could move the wire right through the flux region and get very wild changes in resistor current.

Looking at it this way, I see your point of view, but hopefully you see Prof. Lewin's too. If you feel that Kirchoff's Law is describing this situation, then you are just redefining Kirchoff's Law and making it identical to Faraday's Law. I don't know how to make it any more simple than this.

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#### studiot

Joined Nov 9, 2007
4,998
#15 03-17-2009, 02:03 AM
steveb
Senior Member Join Date: Jul 2008
Posts: 328

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Quote:
Originally Posted by studiot
I does go to show what I am always banging on about - Make sure the conditions of validity apply before you use an equation. This is all too often overlooked.

Agreed. When dealing with a new situation, or when trying to answer a mystery, start with first principles (in this case Maxwell's Equations), and carefully derive the simplifications that are allowable based on the assumptions you are convinced you can make.
Steve, thank you for an interesting discussion so far, you have made me stop and think before I answer. So please don't take it that I am getting at you in any way, nothing could be further from the truth and I apologise if anything I have said has looked that way.

Now I don't have access to Kirchoff's original statements. Perhaps Dave would oblige?
However I learned my definitions from later editions of 'standard' textbooks that had been published and republished many times since their first editions.

My understanding is that K originally proposed two laws, one relating to meshes.

Kirchoff's law of nodes, which is basically the law of conservation of charge, now called KCL, and not in contention here.

Kirchoff's law of meshes which states that the sum of the EMFs equals the sum of the pds for any closed loop in the mesh. This is now presented as KVL and stated as $$\Sigma$$voltages = zero.

This is the crux of the matter, as your arrangement with the resistor tacked on violates the basic definition of a closed loop in a mesh. It is a fundamental requirement that each point is visited only once when tracing the loop. Figures of 8 are not allowed. This restriction is common to all normal mesh analysis theorems.

So, by definition I agree that KVL does not apply because the point of attachment of both ends of the resistor is visited more than once in one of the three loops in your diagram.

Yes I said three loops, one with the resistor, one without and one with a crossover.

This brings us to Faraday's law. If we regard your circuit as being three turns, with one turn shorted, then we can work. Otherwise F too has problems.

Incidentally my integral equation is not esoteric. When the first integral is zero (but not the second) we have the transformer equation; when the second is zero (but not the first) we have the motor/generator equation.
Both are in common use.

A more compact version states that

EMF = minus the time rate of change of flux from all causes

All I have done is account for the all causes more fully.

And thank you again Mik for promoting a most interesting discussion.

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#### Dave

Joined Nov 17, 2003
6,970
Now I don't have access to Kirchoff's original statements. Perhaps Dave would oblige?
I've had a quick look to see if I can get hold of the original paper he published as a student, "Ueber den Durchgang eines elektrischen Stromes durch eine Ebene, insbesonere durch eine kreisförmige", however I cannot get an original copy that I can translate.

However, I have read a couple of translations of subsets of the work and KVL, unlike KCL, is not explicitly derived or even given as a formal mathematical expression; it is merely a given statement of the algebraic sum of potential differences around a closed circuit being equal to zero with a worked example of a closed resistor network. It appears as though it is an assumed law without rigorous proof. I will continue to look for an original copy that I can translate myself.

Faraday's Law proves KVL which is essentially a special-case of the former.

Dave

#### steveb

Joined Jul 3, 2008
2,436
This brings us to Faraday's law. If we regard your circuit as being three turns, with one turn shorted, then we can work. Otherwise F too has problems.
I look forward to what further discussion comes out of this later.

I just wanted to mention here that Faraday's Law always holds true (classically speaking). This means that Faraday's Law will correctly describe the situation no matter how many loops, or how they are shorted. I think you know that, but your wording above seems to suggest that is not the case.

#### steveb

Joined Jul 3, 2008
2,436
I'm not sure if anyone will find this interesting, but this discussion got me to try a few examples similar to the loop problem we discussed. I've attached four simple loop examples that can be solved with Faraday's Law.

No math is needed, just logical reasoning.

Case 3 is probably the one that makes sense to most people since it is similar to a generator driving a resistor. But, the others are really challenging to think about. .. Or, at least I thought so.

I assume perfect conductors for the wires, and the thin insulator lets no connection through at all. I consider 4 paths in each example, and you need to get the same answer for each path. See if you get the same answers I did. I won't be surprised if I made a mistake somewhere. So, see if you catch one before I do.

EDIT: Just to clarify, no detailed knowledge of Calculus or field theory is needed to solve these problems. They can be considered like simple puzzles. Just consider all 4 possible loops in each of the 4 examples. Any loop than encloses the central flux change region must have voltages that add to 1 volt. Any loop that does not enclose the central flux region must have voltages that add to zero volts. You need to identify the voltage across the central coil, the resistor and the thin insulator. The right answer makes all 4 loops consistent with Faraday's Law.

EDIT: Updated PDF with color code

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#### studiot

Joined Nov 9, 2007
4,998
I'm not sure if anyone will find this interesting,
This amount of work deserves some response, surely someone else can also contribute?

I really don't see a problem with K's law, so long as you don't try to apply it to a loop where a point is visited twice (except beginning/end of course).

Kirchoff doesn't care where the EMFs come from (chemical, magnetic, elctromechanical etc) he just equates their total to the pds.
So using this

Case1

Ring is threaded by flux and has 1volt EMF, R=0 thus I must be ∞
1 Ohm loop is not threaded by flux thus EMF =0, I =0

Case2

Ring is threaded by flux and has 1volt EMF, R=0 thus I must be ∞
1 Ohm loop is threaded by flux thus EMF =1volt, R=1, thus I =1

Case3

Ring and I ohm section now form 1 long loop, threaded by flux as sketch3 thus
EMF = 1, R =1, I =1

Case4

Ring and I ohm section now form 1 long loop, not threaded by flux as sketch4 thus
EMF = 0, R =1, I =0

The interesting question, avoided by Prof Lewin, is what happens if the both the battery and the coil are included at the same time?

I suggest Faraday's law won't help here, but Kirchoff will as he simply adds the EMFs , without regard to their origin.

However Prof Lewin did show you cannot use Kirchoff's law of meshes to calculate the difference of potential in a potential field. But then a potential field is not a mesh.

It will take some time but I will try to present a fuller analysis and answer Recca's question, since no one else has actually done this.

#### steveb

Joined Jul 3, 2008
2,436
I suggest Faraday's law won't help here, but Kirchoff will as he simply adds the EMFs , without regard to their origin.
Studiot, thanks for your response. We seem to get the same answers.

I need to comment on the above statement you made. You have several times made a comment about Faraday's Law not working in certain cases. This is completely wrong. I'm not trying to give you a hard time about this, but Faraday's Law is one of the 4 Maxwell's Equations, and it always works. Faraday's law can easily handle voltage sources. A battery would generate voltage as the integral of electric field. Just take any path through the battery. Of course, if you know the battery voltage, then you just use that directly.

I must stress again. Faraday's Law is a fundamental Law. It's always valid. I'm quite sure that you are aware of this fact. Why would you want to question it? Am I mistaken in what you are trying to say here?

It will take some time but I will try to present a fuller analysis and answer Recca's question, since no one else has actually done this.
Also, I must stress that I attempted to answer Recca's question, although you may not agree with the answer. Faraday's Law is clearly shown in the PDF document and in the previous web link I provided, and also I described it in words. I would maintain that strickly speaking KVL amounts to the following.

The integral of electric field around a closed loop is zero.

Of course if you can produce historical evidence that shows this to not be true, I would be interested to see it. But, it seems to me that there are two possibilities that would refute the above statement.

1. Kirchoff first proved that the integral of electric field around a closed loop is equal to the negative of rate of change of flux through the area enclosed by the closed path. Thus, we have mistakenly given Faraday credit for Kirchoff's wonderful discovery. I doubt this is true; but, my god, if you uncover this, what a scandal we have!

2. Kirchoff just restated Faraday's Law, after Faraday (either knowingly or unknowingly). If this is true then we are just disagreeing about terminology and KVL is just a circuit representation of of Faraday's Law. This is possible, and I would certainly like to know about it if it is true.

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#### studiot

Joined Nov 9, 2007
4,998
OK so my terminology is a bit lax and I really meant not directly applicable. Of course Faraday's Law works where it is applicable.

I have no wish to detract from the greatness of any of these discoveries. By fuller analysis I meant the following.

Firstly there are a whole bunch of theorems, laws and equations honouring ( in no particular order) Maxwell, Gauss, Faraday, Kirchoff, Lenz, Lorenz, Laplace, Poisson, Coulomb, Ampere etc.
Many of these are equivalent statements in an alternative system because we can approach the math via differential calculus or integral calculus, vectors /scalars. In some cases some are similar but apply to different circumstances such as conductor / non conductor or conservative /non conservative fields etc.
Further we also have to consider both electric and magnetic effects and their inter-relation.
I was planning to try to rationalise these to show the relationships, including that between Faraday and Kirchoff.

My understanding of Faraday's law is that it is in differential form as we have both stated here. It is more far reaching than Kirchoff's Law as it connects electric and magnetic effects. Kirchoff's Law relates purely to electric effects. However the downside of this is that there must actually be magnetic flux to vary to yield the EMF.

If I hold a PP3 battery in my hand there is EMF, but no flux and therefore Faraday's Law is not applicable.

There is, of course, no circuit at this stage either.

If I now connect two identical batteries ( or carefully adjusted power supplies) in opposition in a circuit, perhaps including resistance, there is still no flux as there is no current.
Kirchoff's law can be applied to this situation as we can sum the opposing EMFs meaningfully.

#### steveb

Joined Jul 3, 2008
2,436
Firstly there are a whole bunch of theorems, laws and equations honouring ( in no particular order) Maxwell, Gauss, Faraday, Kirchoff, Lenz, Lorenz, Laplace, Poisson, Coulomb, Ampere etc.
Many of these are equivalent statements in an alternative system because we can approach the math via differential calculus or integral calculus, vectors /scalars. In some cases some are similar but apply to different circumstances such as conductor / non conductor or conservative /non conservative fields etc.
Further we also have to consider both electric and magnetic effects and their inter-relation.
I was planning to try to rationalise these to show the relationships, including that between Faraday and Kirchoff.

My understanding of Faraday's law is that it is in differential form as we have both stated here. It is more far reaching than Kirchoff's Law as it connects electric and magnetic effects. Kirchoff's Law relates purely to electric effects. However the downside of this is that there must actually be magnetic flux to vary to yield the EMF.
I think I found a good way to shed some light on this. After checking all of my books, each which provided a different interpretation of Kirchoff's Voltage Law (KVL), my thought was that we needed to get a translation of Kirchoff's original work to get his exact statement, but I was unable to do so. I found comments similar to Dave's statements that Kirchoff was talking about potential differences in resistor networks. This may in fact be true, but we need that paper to be sure about what he stated as a useful law/rule/condition.

However, this morning it occurred to me that it would make sense to consult Maxwell on this. So, I read through my copy of "A Treatise on Electricity and Magnetism", 3'rd edition, by James Clerk Maxwell. His translation/interpretation/understanding of both of Kirchoff's Laws is clearly shown there, and I made a PDF of the relevant section (art. 282a).

Maxwell states that in Kirchoff's conditions "the consideration of of the potential is avoided" and the voltage law states the following:

"In any complete circuit formed by the conductors the sum of the electromotive forces taken round the circuit is equal to the sum of the products of the current in each conductor multiplied by the resistance of that conductor."

This is clearly not in line with my stated opinion, but this is similar to your stated opinion, Studiot. You said, "Old fashioned versions of Kirchoff's law state: The sum of the EMF's = The sum of the voltage drops (or potential differences)". However, I think the statement by Maxwell is best/cleaner because it avoids any confusion about the whether a potential difference is itself a form of EMF. For example, is a capacitor voltage a voltage drop or an EMF?

Faraday's Law allows understanding/calculation of the EMF from batteries, inductors, capacitors, generators, etc, while Kirchoff does not care about how the EMF is generated. So, Kirchoff's Law does not rise to the level of a universal law. However, it seems that Kirchoff may have deliberately worded the statement to be consistent with Faraday's Law, which he must certainly have been aware of.

Of course, more historical research could be done on this subject, but I'm willing to accept Maxwell's judgement of this. I thank you Studiot for leading me to this information. I think this is important because there seems to be an inconsistency in the way Kirchoff's Laws are presented in books. There is something very simple and unambiguous about the Kirchoff's laws as restated by Maxwell.

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#### mik3

Joined Feb 4, 2008
4,843
Finally consider Prof Lewins's experiment with no resistors, just a wire loop.

We now have an EMF of 1 volt but zero potential difference in the system.

A pretty dramatic demonstration that EMF and PD are not the same thing, although both are measured in volts.
If you make the experiment with a wire loop then you will measure half of the EMF if you place the voltmeter across an arc which is half the circumference of the loop in length.

Wires have resistance too, they are not perfect conductors.