[SI-LIST] Re: Can L12 ever exceed L1 or L2 ??

Andy, you've humbled me on that one.  Insert N^2 into the equation and it 
holds true:

For L2 = N^2*L1:

L1 > N^2*L12

But, I'm still not so sure about all those college professors.  You can 
never be too sure about their political leanings.

Regards,


Steve.
At 12:11 PM 10/29/2005 -0400, Andrew Ingraham wrote:
>Imagine the theoretical case where you have two coils "occupying the same
>plane" such that all the flux lines of one also go through the other.  Or
>that they are wound on the same hi-mu toroid.  No leakage inductance.
>
>Now suppose the two coils have a different number of windings, and therefore
>different self inductances.
>
>Since the mutual inductance L12=L21 cannot exceed the smaller of the two
>self inductances, this means k will have to be significantly less than 1.
>
>In simple circuit theory we tend to think of the coupling coefficient k as a
>measure of the degree to which the two coils are coupled.  k=1 implies
>perfect coupling.  In this case, despite the fact that there is perfect
>coupling, k<1.
>
>Where is the flaw in this somewhat over-simplified way of looking at coupled
>inductors?
>
>Perhaps |k|<=1 is allowed only when the two inductors have equal self
>inductances, and perfect coupling requies a smaller upper-bound on k when
>they aren't equal?
>
>There's a lot of published circuit theory that allows Lm > L1, even uses it
>in examples.  But whoever said college professors know what they're talking
>about?
>
>Regards,
>Andy
>
>
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