RE: I've lost my k. Can someone help me find it?

["Tesla list" <tesla-at-pupman-dot-com>]

Original poster: "John H. Couture by way of Terry Fritz <twftesla-at-qwest-dot-net>" <couturejh-at-mgte-dot-com>


Paul -

The results of K factor tests of one of my coils. I calc a K of .1720
compared to .1708 using your calcc method. About 0.7 percent difference.
  F1 = 425
  F2 = 505
  Fr = 465

  K = (F2-F1)/Fr = (505-425)/465 = .1720

  x = F2/F1 = 505/425 = 1.88235
  K = (x^2-1)/(x^2+1) = .4119/2.4119 = .1708

John Couture

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-----Original Message-----
From: Tesla list [mailto:tesla-at-pupman-dot-com]
Sent: Tuesday, September 24, 2002 4:34 PM
To: tesla-at-pupman-dot-com
Subject: Re: I've lost my k. Can someone help me find it?


Original poster: "Paul Nicholson by way of Terry Fritz <twftesla-at-qwest-dot-net>"
<paul-at-abelian.demon.co.uk>

> set me straight on how to calculate K given a pri/sec assembly's
> lower (F1)  and upper (F2) resonant frequency?

The two frequencies are
  F1 = Fres/sqrt(1+k)
for the lower, and
  F2 = Fres/sqrt(1-k)
for the upper, for high Q coils where Fres is the common
resonant frequency of the uncoupled coils.

So k is (x^2-1)/(x^2+1) where x is the ratio of the two
resonant frequencies, F2/F1.

Earlier answers were confusing in that they were either
wrong, or referred to a completely different situation.
--
Paul Nicholson
--