Interesting postulate. Please give me a technical reference, and not from an amateur radio publication.
Hard to find it in a non-ham publication, because any Engineering text would
leave it as an exercise for the reader.
But try this: Terman* gives the inductance of two coupled coils in parallel as
Ltot = (L1 * L2 - M * M ) / ( L1 + L2 -/+ 2M )
(we use the minus sign in the last term because the fields are aiding)
where L1 and L2 are the coils and M is the mutual inductance. In the case
in question, L1 = L2, and M = k * the inductance of either winding, where k
is the coupling coefficient.
Unfortunately the formula reduces to 0 / 0 when k = 1 and M = L, but
if you solve for the limit as k -> 1 (or M -> L) you'll find that this reduces
to L * ( 1 + k ) / 2, so when k = 1, Ltot = L.
Or you can simply solve the original equation for M = 0.90 L and for M = 0.99 L
and draw your own conclusions about how close Ltot will be to L.
*
Terman, Fredrick Emmons, Radio Engineers' Handbook, First Edition,
McGraw-Hill, 1943, pp 64-65While it approaches the value of L, it does indeed "blow up" if one arrives at a k of 1. Terman doesn't do a good job of explaining himself. Perhaps a coupling coefficient of 1 cannot be realistically achieved. So, in reality, L1 and L2 in parallel will always be something less that 1. Which will yield a total inductance of something less than L, for n identical windings with the same sense in parallel. BTW, the induced voltage in L2 from a test current in L1 opposes forced current flow, hence the subtraction of M. Just the opposite if the winding direction were opposite for L2.
