Is MIT Prof. Lewin wrong about Kirchhoff's law?

cabraham

I'll draw a pic later. I've been tidying up the house today because I have guests arriving tonight. Been quite busy. Tonight or tomorrow, I'll have an illustrative pic. But here is a quick comment.

The pic you showed has the 0.1 ohm sec Cu resistance & the xfmr sec inductor depicted as separate lumped parameters. You earlier argued that the reality is distributed parameters including emf sources.

The emf in the xfmr case is forced to be roughly the value of CVS exciting the primary. Ee & Ei will oppose each other, but the CVS provides whatever primary current is needed to force a flux density of 1.5 tesla, resulting in 120V rms terminal voltage. But the voltage along the Cu path is 1.0V, & along the heater path is 119V. Why?

I'll have a pic later. For now please refer to the Dr. Lewin paper I recently posted. At the Cu-heater boundary, we have differing resistances. But the current in the Cu & the heater are the same, being connected in series.

So per Ohm's Law, OL. Jcu = sigma_cu*Ecu, & Jhtr + sigma_htr*Ehtr. But we know that Jcu = Jhtr = J. So that sigma_cu*Ecu = J = sigma_htr*Ehtr. We know that the sigma values differ, as Cu has much lower resistivity. But the CVS driving the xfmr primary forces an emf of 120V rms. We know this. This emf is distributed, i.e. 1.0 volt/turn. But why is there 1.0V across the Cu inside the wire, & 119V across the heater?

What if the sec winding was made with high resistance wire, so that both the Cu sec & the heater were 6.0 ohms each? There would be no difference in terminal voltage regardless of path. It would be 60V across the Cu & 60V across the heater. The 120V emf, forced by the CVS at the xfmr primary is distributed around the loop, & since the resistances are identical for each path, no Ec exists, i.e. the E field in each section is due to emf only, no static charges are present.

But if sec Cu is 0.10 ohm, heater is 11.9 ohm, we still have 120V emf for the sec loop regardless of resistance values. But due to the differing resistances, there is a charge build-up at each interface. Refer to Dr. Lewin's diagram. One boundary (between Cu & heater) has an accumulation of "+ve" charge, the other boundary "-ve" charge.

These discrete charges have their own E field, Ec. The Ec has a polarity as follows. In the Cu, Ec is oriented so as to subtract from Ee. In the heater Ec adds to Ee. So the loop emf stays at 120V because this Ec field has zero curl. Going around the loop integrating Ec only results in a rise & a drop of equal magnitude. Around the loop, Ee integrates to 120V, & Ec to zero.

In the Cu, Ee integrates to half (assuming symmetry) which is 60V, & Ec to -59V. The net voltage in the Cu is the difference of 1.0V. But in the heater, Ee integrates again to half, or 60V, but Ec integrates to +59V. The net voltage in the heater is the sum of 119V.

Meanwhile, the CVS is oblivious to Ec. Take the 2 cases I detailed, 6.0 ohm Cu sec with 6.0 ohm heater, & the other case being 0.1/11.9 ohm resp. The net load is 12 ohm, 10A, either way, likewise for the 120V emf. The CVS & the process of induction are oblivious to the exact resistance distribution across the secondary. In both cases, the emf is 120V, & the CVS, the core flux, etc. is unaware of how the 120V is distributed between the Cu resistance & load resistance.

Nature fixes this with the accumulation of static charges at the interface between the 2 media. This charge field has its own Ec field. The Ee type of field has curl. Around a loop it has a non-zero integral & non-zero emf. But the Ec field is non-curly, having zero net emf around the loop. In other words the CVS is oblivious to Ec. Ec has zero curl & hence it cannot "mess up" the 120V emf, mandated by laws of induction & the nature of a CVS.

But the Ec field exactly fixes the voltage distribution while preserving Ohm's Law, Faraday, Ampere, conservation of energy, Lenz, etc. By definition, the voltage from xfmr sec terminals a to b, is the integral of the composite E field, whose parts are Ee, Ei, counter-counter Ee provided by the CVS, as well as the all-important Ec.

Do NOT neglect Ec. It does not show up in the induction laws, & has no interaction w/ the CVS. It merely distributes the 120V emf in accordance w/ the resistances preserving Ohm's law.

J = sigma*E, so that if J is the same for 2 regions, which it is, & E were the same, then we would have a paradox. How can J & E both be the same while the sigma values differ for each region. The answer is that if we combine Ee & Ei & the CVS Ee all into a composite Ee', that value of Ee' is uniform. It would integrate to the same value in Cu & heater. Hence we get a paradox. The energy per unit charge to transport charge through copper is but 1.0V, vs. 119V for heater.

Ec fixes all that. Symmetry would result in 60V in each section. Ec provides -59V in the Cu & +59V in the heater. All laws are upheld.

Not so intuitive is it? Great minds stumble with this concept. It takes a lot of thought to sort it out. Prof. Lewin is spot on dead right!!!. Cheers.

Claude

stevenb

I second that comment.

I'll add one of my own too. Thanks to the great mind of Faraday, we don't need a great mind, nor should we stumble, when we study this subject. We need only trust in his Law. And, even if we want to get into these nitty gritty details, how much easier even that will be if we start with an acceptance that Faraday's Law trumps our intuition.

cabraham

Faraday was inded a great mind. Lest we forget, he had a high school education, the son of a blacksmith. He read books on his own, employed in a book store. Books were costly & hard to obtain in those days.

Faraday's & all other verified laws trump our intuition 24/7/365, i.e Ohm, Ampere, energy, Kirchoff current/voltage, etc. Relying on intuition is the most dangerous think I know of. The problem w/ intuition is that it seems so logical to us. But we only factor in the truths we are aware of, omitting details we are unaware of, but are pertinent nonetheless. Regarding this Prof. Lewin problem, the fact that static charges accumulate near the boundary region of the 2 media, & form another E field, Ec, that is conservative, & adds/subtracts w/ the Ee field, is well --- who'd have thunk it! But the presence of said E field dispels any paradox & upholds OL, CEL, FL, LL, LFL, AL, etc.

Science is amazing!

Claude

yungman

I'll read yours and Levin's tonight. I have to admit, I am a lot more open mind reading your's or any reference materials or text books. I just find it hard to read anything from Levin anymore. But I'll read it with as open a mind as possible.

BTW, I pull out your two posts and started another thread. Please comment on that one, I just found that very interesting and start thinking more, I am not stating any opinion, just want to listen to people.

yungman

I read the lecture notes. So what Levin said is the E only concentrate in the part of the loop that is of highest resistance and the rest of the loop that make up of wires ( very low resistance) has very little electric field. That he use the example of the two half of the loop make of R1 and R2 where R2>R1 and show less voltage drop (itex] V=\int E \cdot d \vec l [/itex] across R1 ( left side ) than right side of the loop.

I understand all the formulas he gave, but when come the part of the charges build up at the junction, I don't follow at all and there is no formulas to support this.


So basically what Levin claimed is in an open circuits, there is no voltage across the two terminals ( A and B) on the secondary of the transformer. Voltage only created when you put a resistor across the terminals. So there is no equivalent voltage source in the transformer or matter of fact in Levin's experiment.

Has this gone through peer review? Any other reputable people support this? Can you sight some other articles to support this or IEEE papers? Or in AIP papers?

yungman

You are basically repeating Levin's notes. So you claim that the reason you can measure 119V across the terminal is only because of the charge build up at the junction that make up 118V and the whole secondary only drop 1V? I don't get the charge buildup at the junction at all. Is there any experiment proof on this? Or is this his theory? Is there others with authority in this field support this?


This is pretty much the bottom line of the arguement. I sure have never seen anything like this in books and I don't see any proof or formulas to support this that the charge make up the voltage. How come nobody else want to join in the discussion. I am sure most of your guys has better theory background than me!!!!

hikaru1221

Yungman, don't refuse the fact that the argument comes straight from J = E*sigma. And that formula certainly is somewhere in Cheng's or Griffiths'. That's the theoretical base. Most books don't touch on this kind of phenomenon I believe, but that doesn't mean it doesn't exist.

This IEEE paper's abstract, though barely related to this discussion, points out that "Different surface treatments have further been performed in order to study the influence of space charge accumulation in the boundary layer", and evidences the existence of the charge accumulation at the boundary of 2 different media.
http://ieeexplore.ieee.org/xpl/freea...rnumber=522993
I know you may say it exists but is small & negligible. You can check the theory yourself, or stay happy with your own theory.

I also did provide you a document & an experiment video from some EE professors of MIT in support of Lewin's explanation a long time ago, but you probably didn't bother looking at it. This is the document & the video:
http://ocw.mit.edu/courses/electrica...emonstrations/
The download link of the document is at "PDF" at chapter 10 when you scroll down - see example 1 from page 2 to 4 of the document. Below that, at section 10.0.1, there is a downloadable video, which demonstrates the experiment of the example 1 of the document. And I don't think that these professors don't know what IEEE is.
(that doesn't mean I favor MIT. MIT OCW is the only place where I can find the supporting document)

This is not true. Lewin didn't conclude that, and even the theory won't end up at such conclusion. Ec (static E field) can be built up everywhere, even in the air, the same as Ei (induced E field). For an open circuit, we can regard the gap as a resistor of very high resistance. Extrapolating from the conclusion of Lewin, which is that total voltage across the resistor of higher resistance is higher, we can see that this air resistor takes all the voltage.

Voltage is created with total E field. The induced E field Ei is independent of the resistor. The static E field Ec is dependent on Ei and the resistor. Without resistor and thus Ec, there is still voltage by Ei. With resistor and thus Ec, there is also voltage, but with different distribution.

P.S.: To ellaborate on the point of J = E*sigma, that's not the only thing to apply. You will need Gauss law. Applying J = E*sigma, you arrive at that total E at 2 different media are different, if J is the same. Applying Gauss law, you arrive at that at the boundary surface, there is charge accumulation.

yungman

Thanks

I have to take a look at chapter 10 first.

Alan