Re: "De-coupling" coefficient?

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

Original poster: "Antonio Carlos M. de Queiroz by way of Terry Fritz <teslalist-at-qwest-dot-net>" <acmq-at-compuland-dot-com.br>

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?

No. This is the output voltage with an ideal voltage source at one
side and an open circuit at the other. Comes from the equivalence
between a "real" transformer and an ideal transformer loaded by two
inductors (see with a fixed-width font):

         k                    1:n
o-----+   +-----o    o--+---+   +--Lx--o
       |   |             |   )   (
       Lp  Ls       =    Lp  )   (
       |   |             |   )   (
o-----+   +-----o    o--+---+   +------o

n=k*sqrt(Ls/Lp)
Lx=Ls(1-k^2)

 > 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?

The current limiting occurs due to the equivalent inductance Lx in
series with the secondary.
With excitation by a perfect voltage source Vin at the primary side,
at frequency f (Hz), the short-circuit output current would be:
Imax = Vin*k*sqrt(Ls/Lp)/(2*pi*f*Ls*(1-k^2)), with a 90 degrees
phase shift.
With k=1, there is no current limiting. With k=0, no current.

Antonio Carlos M. de Queiroz