Mutually coupled inductors in parallel

Discussion in 'General Electronics Chat' started by montepionte, May 29, 2016.

  1. montepionte

    Thread Starter New Member

    May 29, 2016
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    Hey guys.

    Can anyone explain how do you use, for k = 1 and equal inductance values (L1=L2=M), the parallel coupled inductors equation, which is:

    Leq = [(L1*L2)-(M)^2] / [(L1+L2-(2*M))]


    I'm asking that because everyone just says that, for the conditions I described above, the equivalent inductance Leq comes out as L1=L2=M, but, if I substitute it, say for L1=L2=M=1, the numerator becomes zero, thus Leq = 0.

    Thanks.
     
  2. Kermit2

    AAC Fanatic!

    Feb 5, 2010
    3,785
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    But k=1 is theoretical.

    Theory is NEVER anything more than a tool for understanding. Reality will not allow the outcome you tortured out of theory
     
  3. Bordodynov

    Active Member

    May 20, 2015
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    L1=L2=L
    Leq=(L^2-M^2)/(2*(L-M)=(L+M)*(L-M)/2/(L-M)=(L+M)/2
    M-->L ==> Leq=L
     
    kubeek likes this.
  4. djsfantasi

    AAC Fanatic!

    Apr 11, 2010
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    I think there are a couple of typos, but I get the intent. It still bothers me that (L-M)=0, and is in the denominator but the logic is there and I can see the answer.

    * there is a missing parentheses in the second line and did you mean to divide by (L-M) in the third line?
     
  5. Bordodynov

    Active Member

    May 20, 2015
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    I used a simplified mathematical notation.
    Leq=limit(L^2-M^2)/(2*(L-M)=(L+M)*(L-M)/2/(L-M)=(L+M)/2 for M-->L
    limit(L^2-M^2)/(2*(L-M)=limit((L+M)*(L-M)/2/(L-M))=limit(L+M)*limit((L-M)/2/(L-M))=2*L/2=L
    I used limit(x/x)=1 x-->0, x=L-M
     
  6. djsfantasi

    AAC Fanatic!

    Apr 11, 2010
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    I'll try once more, and then I'll give up.

    Your second line which starts
    (L+M)*(L-M)/2/(L-M)
    Is NOT equal to (L+M)/2
    By my manipulations, it is equal to
    ((L+M)*(L-M)^2)/2
    Somehow, the denominator:
    (2*(L-M))
    got rewritten as
    (2/(L-M))
    between the first and second lines.

    The second line should start:
    ((L+M)*(L-M))/(2*(L-M)) which is equal to (L+M)/2
     
  7. MrAl

    Well-Known Member

    Jun 17, 2014
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    Hi,

    I did it mathematically this way...

    Start with:
    Leq=(L1*L2-M^2)/(L2+L1-2*M)

    Replace L2 with L1 because L2=L1 we get:
    Leq=(L1^2-M^2)/(2*L1-2*M)

    Simplify, we get:
    Leq=(M+L1)/2

    Replace M with L1 we get:
    Leq=(L1+L1)/2

    Simplify, we get:
    Leq=L1

    Note the squares and variables in the numerator in:
    Leq=(L1^2-M^2)/(2*L1-2*M)

    changes the way we interpret the denominator, because it's not in the right form yet. But if you prefer to take the limit, you can take the limit as M approaches L1 and get:
    Leq=L1

    again :)
     
  8. Bordodynov

    Active Member

    May 20, 2015
    637
    188
    djsfantasi
    (L+M)*(L-M)/(2*(L-M))=(L+M)*(L-M)/2/(L-M) and not =(L+M)*(L-M)/(2/(L-M))
     
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