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Induction and flux linkage clarification

  1. Jan 17, 2015 #1
    So if you have a single inductor then I assume that there is no flux linkage and that L = flux*N / i = L11
    and so that flux is total leakage flux. If you have two inductors then the flux leakage is any flux that isn't flux linkage (the self flux is flux leakage).

    (flux21 is the flux in 2 from 1, flux11 is the flux in 1 from 1 etc.)
    According to my book the coupling coefficients are:
    k1 = flux21/flux11
    k2 = flux12/flux22
    k = sqrt(k1*k2)
    And mutual inductance = L12 = L21 = k*sqrt(L11*L22)

    What I'm wondering is that how do you calculate flux12 or flux 21? And that value for inductance of:
    N2/reluctance
    is that the total inductance of L11 + L12 ?

    And if you have two inductors how do you calculate flux11 and flux22 and their respective linkages?

    If you have a transformer can you say inductance of side one is: L = NBA/i
    (where N is turns of that side, i is current of that side and BA is the flux going through that side)
    Is that the total inductance given to you by that formula, L11 + L12 and how do you calculate wich is wich flux, like what the leakage flux is?

    Cheers in advance!
     
  2. jcsd
  3. Jan 22, 2015 #2
    Dammit I was hoping to get some interesting feed back on this.
     
  4. Jan 29, 2015 #3

    Baluncore

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    Where you have two remote coils the mutual inductance will be low. Where you have a transformer the coupling will be close to one. The closest coupling between two inductors is when they share a common magnetic core.

    The inductance of L1 is the sum of the coupling of all it's geometrical parts. The same is true for L2. The mutual inductance between L1 and L2 can be solved algebraically for most simple geometries. You have not identified the geometry of your inductors or their relative positions. For that reason there is no simple answer to your question. Can you specify the geometric configuration of your two inductors?

    I would suggest you find a copy of “Inductance Calculations” by G.W.Grover. Use http://www.bookfinder.com/ to locate a low cost copy, or maybe a new Dover Publications reprint.
     
  5. Feb 2, 2015 #4
    Thanks for the reply.
    Ok I am concidering 3 different geometries, two are torroids, wound identically, from the top view and the third one is just bifilar coils, I suppose you could imagine a small magnetic core in the centre of the bifilar coils if you wanted:
    geometry 1and2.png
    geometry 3.GIF
    Say geometry 1 is on an angle of zero degrees and geometry 2 is offset by an angle of 180 degrees. I imagine that if they were 90 degrees offset there would be no mutual flux linkage as they would be perpendicular. I don't know how to calculate the coupling coefficients from the geometries but I imagine that geometry 1 would have more flux linkage than geometry 2 (as are functions of 'd') I woul guess it would be double the flux linkage of geometry 2.

    As for geometry 3, from what I have read (but don't really fully understand) that if you just took L to be the inductance of winding A or B, equal to:
    'N' (of A or B) squared / Reluctance
    Then this series layout adds to yeild a total inductance of Ltotal = (LA + LB)*M = (L + L)*2 = 4*L, because the mutual coupling coefficient is 2 (how? I don't know), which kind of makes sense, I suppose there would be no stray inductance, but still sounds a bit high, too good to be true, unless the reluctance was something really small, not sure what, maybe the centre diameter of the wire length with a permittivity of air or if there was a core in there.
    Thanks
     
  6. Feb 3, 2015 #5

    Baluncore

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    I assume your toroids have magnetic cores. The thing about a toroid is that the magnetic field is contained within the magnetic core. Geometry 1. If you stack two toroidal cores, each with it's own winding, then there will be very little coupling between the two windings. Geometry 2. The same containment applies, so there is little coupling.
    http://en.wikipedia.org/wiki/Toroid...tal_B_field_confinement_by_toroidal_inductors

    The third is a bifilar pancake coil. Because the windings are spread the coupling is not as good as you might expect. These can be complex to calculate, (see chapter 17 of Grover). There is also a correction needed because of the thickish insulation between the windings. When you connect the two filaments in series you simply double the number of turns which gives you four times the inductance of either filament by itself with the other open circuit.
     
  7. Feb 3, 2015 #6
     

    Attached Files:

  8. Feb 3, 2015 #7

    Baluncore

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    If two wires, joined at the ends, are wound together to form a spiral inductor, the inductance will be close to that of one wire only, while the resistance will be half of one wire only.
    In parallel you are making Litzendraht wire. http://en.wikipedia.org/wiki/Litz_wire
     
  9. Feb 3, 2015 #8
    Really good point! So what was the advantage of winding it out of two wires as opposed to more turns of one wire?

    P.S did you think much of that .pdf?
     
  10. Feb 3, 2015 #9

    Baluncore

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    It is the errata for Grover's book. I made the corrections to my copy a couple of years ago.

    I don't know where you found the bifilar pancake picture, or who drew it and why. You can wind a bifilar coil twice as fast as two mono-filar coils. Maybe they wanted a centre tapped coil to couple with another nearby coil.
     
  11. Feb 3, 2015 #10

    Baluncore

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    Take a look at this archive. https://archive.org/browse.php?field=subject&mediatype=texts&collection=NISTJournalofResearch

    Some of the papers written by or used by Grover in writing his book are listed below. I have renamed them but original filename is at end of new filename.
    They are available here; https://archive.org/details/NBSBulletin

    BBS Vol 2 no 2. Self—Inductance of Single—Layer Coils. Rosa. calcul216118719063131unse.pdf
    BBS Vol 4 no 1 Influence of Frequency on Coib influe416117819077676cohe.pdf
    BBS Vol 4 no 1 Self—inductance of Circles. Rosa,Cohen[1907]. onself414915919077575unse.pdf
    BBS Vol 4 no 2 B&W Inductance of finear conductors. Bureau of Standards, E.B.Rosa 1907.pdf
    BBS Vol 4 No 2 Inductance of Linear Conductors. Rosa[1907]. selfmu430134419088080unse.pdf
    BBS Vol 4 no 2. The Self and Mutual Inductances of Linear Conductors. Rosa[1907] selfmu430134419088080unse.pdf
    BBS Vol 5 no 1. Formulae and Tables for the Calculation of Mutual and Self—Inductance formulae5113219089393rosa.pdf
    BBS Vol 8 no 1 Formulas for Mutual and Self—Inductance formul812871912169169unse.pdf
    BBS Vol 8, No 1. Formulas for Self and Mutual Inductance, Sci169noerr.pdf
    BBS Vol 12 Effective Resistance and Inductance of Iron and Bimetalic Vlnres. Miller[1915]. eff122072671915252252mfll.pdf
    BBS Vol 14 Additions to Inductance Formulas add145375701919320320unse.pdf
    BBS Vol 17, part 2 Capacity Between Inductance Coils and the Ground. Breit[1921]. effectsofdistrib17521unse.pdf
    BBS Vol 18 Inductance of Polygonal Coib. Grover[1922] formulaetablesfo18737unse.pdf
    BBS Vol 19. Skin effect in solenoids, Hickman_Sci472.pdf
    BBS Vol 21, Inductance of a helix, Sci537.pdf
     
  12. Feb 3, 2015 #11
    Thank you very much for the papers, I'll follow them up tomorrow.

    As for were I got that picture, it was just from the wiki page (as you might have guessed, sorry, I remember your past advice) :
    http://en.wikipedia.org/wiki/Bifilar_coil
    It states "Some bifilars have adjacent coils in which the convolutions are arranged so that the potential difference is magnified (i.e., the current flows in same parallel direction). Others are wound so that the current flows in opposite directions. The magnetic field created by one winding is therefore equal and opposite to that created by the other, resulting in a net magnetic field of zero (i.e., neutralizing any negative effects in the coil). In electrical terms, this means that the self-inductance of the coil is zero." And I think the picture is from Tesla's patent.
    Which to mean indicates the design is for more than just the convenience of centre-tapping.

    Thanks!
     
  13. Feb 3, 2015 #12

    Baluncore

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    Back in Tesla's day, low value resistors were wire-wound. That usually meant they were also inductive. Inductance was not a problem with DC circuits, but it was with AC. By connecting the two filaments in series so the current flowed in both directions, most of the inductance could be cancelled.

    The pancake construction allows the W = I2R heat dissipated in the resistance to escape quickly.
     
  14. Feb 3, 2015 #13
    Interesting, ok well my main question was whether there was any special coupling interaction between: parallel-wound serise connected coils compared to a standard single coil with the same amount of turns. But from what you're saying there isn't and there is other reasons as to why the layout was is used.
    I do have two side questions though about what you raised about the number of turns: comparing a parallel-wound, series connected set of coils to a parallel-wound, parallel connected set of coils, what is it about the potential difference between turns that distinguishes each turn as being another turn? As opposed as to when they're parrallel and just appear to be different strands of the same turn?
    Furthermore, if you connect a bifilar set of coils from one connection to the other, does the value of 'N*I' remain the same with N and I changing in proportion, or would one set have a larger N*I?

    Thanks
     
  15. Feb 3, 2015 #14

    Baluncore

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    Where multiple filaments take the same path through a changing magnetic field, they will have identical voltages induced in them by the changing field. Consider two wires, each has a voltage V induced in them by the changing magnetic field. The magnetic field is proportional to the ampere * turns. Inductance obeys the relationship V = L * di/dt, therefore L = V * dt/di

    In parallel the voltage is once, but the current is twice. Lp = 1 * dt/2 ∝ 1 / 2 = 0.5
    In series the voltage is twice, but the current is once. Ls = 2 * dt/1 ∝ 2 / 1 = 2.0
    So series has 4 times the inductance of parallel, as is expected from N2 relationship.
     
  16. Feb 3, 2015 #15
    Ok, so NI will remain the same but the inductance will change. But what I meant when I asked:
    Irrespective of why there is moving charges flowing around the coils (magnetic induction or battery) there seems to need to be to be a potential difference between coils for them to count as different turns. So when you have them in parallel, they just look like a copy of eachother and don't count as extra turns (although as we discussed there is more current). I'm just wondering what it is about this potential difference that distinguishes between turns.
     
  17. Feb 3, 2015 #16

    Baluncore

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    For the same di/dt, terminal voltage is proportional to inductance. V = L * di/dt.
    If N turns are in series then the N voltages add, so the inductance increases by N times due to series connection.
    When in parallel, there is no change in voltage, but for the same ampere turns the inductor terminal current is N times greater.
     
  18. Feb 3, 2015 #17
    That's something to think about, are you saying that if there is a coil of N turns, throughout each N1 + N2 ...+ NN, there is a slightly different di/dt for each turn at that point as the current flows down the wire?
    So V = L1*di/dt +...+ LN*di/dt
    Assuming all the points of inductance for each turn are the same.
     
  19. Feb 3, 2015 #18

    Baluncore

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    No, current is the same in all turns and so is di/dt, if that was not the case it would be a transmission line. The inductor must be treated as an N x N matrix of mutual inductance between turns. The self inductance of the winding is the total.
    One thing about transformers and inductors is that they are optimised to operate close to a particular voltage per turn.
     
  20. Feb 5, 2015 #19
    Yes, I think I see what you're saying. And increasing the diameter of the wire, with another strand (in parallel) doesn't increase inductance, so in a way 'N' is really more of an arbitrary thing.
     
  21. Feb 5, 2015 #20

    Baluncore

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    It will actually lower the total inductance very slightly because the self inductance of the wire will be less. That is because the bigger wire will have parallel current filaments that are further apart, so there is less "self coupling". A single filament has greater self inductance than a tape or sheet.
     
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