Mutually coupled inductors in parallel

Thread Starter

montepionte

Joined May 29, 2016
1
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.
 

Kermit2

Joined Feb 5, 2010
4,162
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
 

djsfantasi

Joined Apr 11, 2010
9,155
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
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?
 

Bordodynov

Joined May 20, 2015
3,174
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
 

djsfantasi

Joined Apr 11, 2010
9,155
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
 

MrAl

Joined Jun 17, 2014
11,345
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 :)
 
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