Re: "De-coupling" coefficient?

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

Original poster: "Malcolm Watts by way of Terry Fritz <teslalist-at-qwest-dot-net>" <m.j.watts-at-massey.ac.nz>

Hi Jolyon,

On 17 Jun 2003, at 12:14, Tesla list wrote:

 > Original poster: "Jolyon Vater Cox by way of Terry Fritz 
<teslalist-at-qwest-dot-net>" <jolyon-at-vatercox.freeserve.co.uk>
 >
 > There is a coefficient of coupling, k, which is essentially a measure of
 > the proportion of total magnetic flux which "cuts" both the primary and
 > secondary windings.
 >
 > Looking at the equation for secondary voltage Vout = Vin*sqrt(Ls/Lp)*k,
 > where k is unity in the case of an ideal transformer
 > would it not be possible  to use same formula with a different coefficient
 > say, l, to calculate the discrepancy in secondary voltage of a real
 > transformer due to the imperfect coupling -in which instance 1 would
 > represent the proportion of total flux that does NOT cut both primary and
 > secondary windings i.e. the "leakage" flux.
 >
 > In both instances,is Ls/Lp the mutual inductance?

No. Where did that equation come from? It cannot be correct because
it doesn't take loading into account. Even if k is less than 1, the
relation between inductances and voltages will hold with no loading.

A familiar equation relates k to Ls and Lp thus: k = M/SQRT(Lp.Ls)
where M is the mutual inductance.

Malcolm