# 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
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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 Well-Known 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 Well-Known 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.

(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 AAC Fanatic!

Jun 17, 2014
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Hi,

I did it mathematically this way...

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 Well-Known Member

May 20, 2015
1,980
597
djsfantasi
(L+M)*(L-M)/(2*(L-M))=(L+M)*(L-M)/2/(L-M) and not =(L+M)*(L-M)/(2/(L-M))