[SI-LIST] Re: Can L12 ever exceed L1 or L2 ??
- From: "Andrew Ingraham" <a.ingraham@xxxxxxxx>
- To: <si-list@xxxxxxxxxxxxx>
- Date: Sat, 29 Oct 2005 12:11:13 -0400
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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