Question about Magnetic Circuits in a Transformer Core

  • #1
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Main Question or Discussion Point

Hello community,
I can't find an explanation for something and I thought I'd come and ask you.
I'm studying for a test I have tomorrow and I came across a problem of a three legged steel core with 400 turns on the center leg, the magnetization curve is given, the first part of the problem asks you to find the current required to produce a certain flux density in the central leg of the core.
So what I did was to first find the magnetic flux across the leg.
Φcenter=B.Across
Then because the two outer legs are symmetric, I found their magnetic flux by dividing the one coming from the center leg, and found the flux density on each outer leg with this information.
B=Φouter/(Across
But then, when trying to find the MMF, for me, it would be the MMF required to produce that magnetizing intensity on the center leg plus the magnetizing intensity on each outer leg so what I did was:
Ftotal=H*lcenter+H*louter*2
but checking the Instructor's manual, it only takes in consideration the magnetizing intensity for the center and one leg (one magnetic loop):
Ftotal=H*lcenter+H*louter
I guess it's because of the analogy with electric circuits when they are in parallel, but I'm not sure, so I was wondering why is it we only take one of the outer legs in consideration, instead of the two of them to find the total MMF.
Can anybody enlighten me? I'd appreciate it! Thanks in advance!

The problem is 1.13 in Chapman Electric Machinery
https://imgur.com/a/PD8A7xQ
 

Answers and Replies

  • #2
Charles Link
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The way I would work this problem is to write ## \oint H \cdot dl=NI ## for the loop that goes through the center and around the right side, and also the same equation for a loop that goes through the center and around the left side. You can assign ## H_1 ## to the center, ## H_2 ## to the right, and ## H_3 ## to the left. (## H_2 ## and ## H_3 ## are the values of ## H ## in the integral once the path of the loop leaves the center leg). Meanwhile, by symmetry ## H_2=H_3 ##. Since the thickness is 8 cm everywhere it simplifies the ## B ## computation: You can work with ## B ## instead of flux ## \Phi ##. Let ## B_1 ## be the magnetic field in the center leg. ## B_1=B_2 +B_3 ## by flux conservation. Writing ## B_1=\mu H_1 ##, ##B_2=\mu H_2 ## and ## B_3=\mu H_3 ## makes it simply an algebraic exercise. I'll leave the calculation to you, and see if it agrees with the answer that they give. ## \\ ## Upon further study of it, I see that all of the ## \mu's ## are not equal. The center ## \mu ## will be different from the ## \mu's ## in the outer legs. I think the problem may require having an equation for ## \mu ## as a function of ## H ## for a complete solution, but I am going to need to study it further. ## \\ ## Additional editing: I studied their solution to part (a) and it makes sense. ## \\ ## And I think I see what your question is: The way the equation ## \oint H \cdot dl=NI ## works is you just take the integral around a single loop in the transformer. ## \\ ## It might interest you that the equation ## \oint H \cdot dl =NI ## is actually a form of Ampere's law ## \oint B \cdot dl=\mu_o I ## when magnetic materials are involved. Beginning with the equation ## \nabla \times B=\mu_o J_{total} ## where ## J_{total}=J_{free}+J_m ##, (## J_{free} ## is current density of any currents in conductors, and ## J_m ## is magnetization current density to to variations in the magnetization ## M ##), and also writing ## B=\mu_o H+M ##, we can take the curl of both sides of this last equation, and we get ## \nabla \times B=\mu_o \nabla \times H+\nabla \times M ##. An in-depth analysis shows that ## \nabla \times M=\mu_o J_m ## so that ## \nabla \times H=J_{free} ##. Integrating over an area ## dA ## and applying Stokes' theorem to this result and including a number of turns factor gives the result ## \oint H \cdot dl =NI ##. ## \\ ## This last part is some theoretical detail, but you might find it of interest that the application here with the MMF formula is really just an extension of the same concept of using loop integrals for calculating magnetic fields using Ampere's law. The equation ## \oint H \cdot dl=NI ## is a modified form of the integral form of Ampere's law that is used when magnetic materials are involved.
 
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  • #3
jim hardy
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I guess it's because of the analogy with electric circuits when they are in parallel, but I'm not sure,
You're exactly right.
.
 
  • #4
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You're exactly right.
.
Yes, I realized this after some more reading. I had the test on Thursday and there even was a problem that involved infinite permeability (zero reluctance) and that part of the core behaved like a "short circuit". I'm amazed at how Kirchhoff and Ohm's Laws apply so satisfactorily with magnetism as well.
 
  • #5
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I think the problem may require having an equation for ## \mu ## as a function of ## H ## for a complete solution, but I am going to need to study it further. ## \\ ## Additional editing: I studied their solution to part (a) and it makes sense. ## \\ ## And I think I see what your question is: The way the equation ## \oint H \cdot dl=NI ## works is you just take the integral around a single loop in the transformer. ## \\ ## It might interest you that the equation ## \oint H \cdot dl =NI ## is actually a form of Ampere's law ## \oint B \cdot dl=\mu_o I ## when magnetic materials are involved.
This was sort of another -more theoretical- way to put it, but it actually was a very good way to understand why magnetic circuits are analyzed this way taking only one magnetic loop in consideration. Thank you.

PD. I admire your knowledge in magnetism.
 
  • #6
jim hardy
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I'm amazed at how Kirchhoff and Ohm's Laws apply so satisfactorily with magnetism as well.
They work fine for most of what you'll need to do.
Reluctance is a function of B and that's why the B-H graph is a curve not a straight line.
We usually handle that graphically .
Just remain aware we do basic magnetic design work with approximations, graphs and tables that let us reduce the math to algebra.

Maxwell's equations are elegant
if perhaps overkill for an introduction to electric machinery. .
but will become part of your repertoire.

Glad it "clicked" for you.
 

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