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>

Malcolm,
My mistake -the RHS of the equation was missing a primary voltage
term -without which it is of course, complete nonsense.

It should have been

Vsec = Vprim.k.sqrt(Lsec/Lprim)

Sorry for any  confusion.

Jolyon


----- Original Message -----
From: "Tesla list" <tesla-at-pupman-dot-com>
To: <tesla-at-pupman-dot-com>
Sent: Monday, June 30, 2003 12:16 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>
 >
 > On 29 Jun 2003, at 9:44, Tesla list wrote:
 >
 >  > 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?
 >
 > It never holds. The LHS describes a voltage and the RHS is
 > dimensionless.
 >
 >  > 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?
 >
 > k doesn't change if the geometry of the transformer isn't changed.
 >
 > Malcolm
 >
 >  > Jolyon
 >  > ----- Original Message -----
 >  > From: "Tesla list" <tesla-at-pupman-dot-com>
 >  > To: <Tesla-at-pupman-dot-com>
 >  > 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
 >  >  >
 >  >  >
 >  >  >
 >  >
 >  >
 >  >
 >
 >
 >