On 18/10/10 07:24, Owen wrote:
....
Only fairly basic AC circuit theory is needed to analyse the effect of
the shorted turns.

If you have a air cored solenoid inductor of n turns, and short m turns
at one end, you can treat that as two independent inductors of n-m and m
turns with some flux coupling factor k. The mutual inductance can be
calculated, and a T equivalent of Ln Lm-n Rn Rm-n M elements constructed
and solved. k of course depends on coil construction and n and m, a
value can be determined by measurement of the reactance of the
combination. (You might be surprised at how low k is.)

To give a mental arithmetic example...

Assuming ideal inductors for the moment...

Suppose you had an air cored inductor, that when you measure the
inductance of the first half of the inductor (other terminal open) you
get 10µH. You now measure the whole inductor and get 30µH. We can
calculate that M=5µH.

Now forming a T equivalent of the inductor with one half shorted,
L=10+5+(5//(10+5))=18.75µH. Notably, the current in the s/c is
3.75/15=25% of the current in the other section, so losses are about 6%
of that in the other section... not usually a big issue.

That is, if I got the maths right on the fly!

Now, real inductors have some distributed capacitance which changes this
as you approach the inductor's self resonance frequency.

This works ok because mutual inductance is lowish. Increase k by
introducing a magnetic core material for instance, and the situation
changes.

You might see the technique applied to powdered iron core inductors. Not
necessarily a good idea, but it 'works' for some because they are low
permeability powders and flux leakage is high (ie k is not nearly 1).

Owen

Owen wrote in news:vtKuo.256$tk4.180
@viwinnwfe02.internal.bigpond.com:

Well, I didn't get the maths right, there was a sign error in the formula
below. Here is what it should have read.

Suppose you had an air cored inductor, that when you measure the
inductance of the first half of the inductor (other terminal open) you
get 10µH. You now measure the whole inductor and get 30µH. We can
calculate that M=5µH.

Now forming a T equivalent of the inductor with one half shorted,
L=10-5+(5//(10-5))=7.5µH. Notably, the current in the s/c is
50% of the current in the other section, so losses are about 25%
of that in the other section... not usually a big issue.

Of course, the situation depends on the tapping point, and is much worse
when you short just one turn... but we don't usually do that. The example
has a fairly high coupling factor k (for an air cored coil), and losses
are lower for lower k.

I am working on a note that expands on this.

Owen