Re: "De-coupling" coefficient?

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

Original poster: "Jolyon Vater Cox by way of Terry Fritz <teslalist-at-qwest-dot-net>" <jolyon-at-vatercox.freeserve.co.uk>

Will the equation Vs = k.sqrt(Ls/Lp) hold when the transformer is loaded?

What is implication of this (if any) in the example of a current-limited
transformer (I understand all practical transformers are in fact
current-limited to some extent by the % "regulation" factor)
Does k change with loading,  for if a current-limited transformer delivers
its max current when the secondary voltage approaches zero volts (i.e. a
short-circuit condition)
and the above equation was correct, must not the coupling coefficient be
less at higher currents to explain the concommitant decrease in secondary
voltage with loading assuming of course, that resistive losses in the
winding are discounted?

Jolyon
----- Original Message -----
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Sent: Wednesday, June 18, 2003 1:41 AM
Subject: Re: "De-coupling" coefficient?


 > 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
 >
 >
 >