Re: High Voltage Output

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

Original poster: "Malcolm Watts" <m.j.watts-at-massey.ac.nz> 

Hi Gerry,

On 6 Jun 2004, at 11:09, Tesla list wrote:

 > Original poster: "Gerry Reynolds" <gerryreynolds-at-earthlink-dot-net>
 >
 > Hi Antonio,
 >
 > You are right, my intuition was that inductance was proportional to
 > the number of turns, which was wrong.  I went back to the text books
 > to find out why the N^2 relationship and basically, if one uses the
 > definition of inductance L=flux/current and looks at the inductance of
 > one turn, that inductance will be proportional to the number of turns
 > since the total flux will be proportional to the number of turns.  So
 > the total inductance is proportional to the this single turn times the
 > number of turns, hence the N^2.  There will be a proportionality
 > constant since not all turns are perfectly coupled into this one turn.

The proportionality is built into Wheeler's formula in the form of
n.h in the denominator where h is the length of the winding and n is
some constant depending on the length units being used.

Malcolm

 > Gerry R.
 >
 >  > Original poster: "Antonio Carlos M. de Queiroz" <acmdq-at-uol-dot-com.br>
 >  > > Tesla list wrote: >  > >  > Original poster: "Gerry Reynolds"
 >  <gerryreynolds-at-earthlink-dot-net> >  > >  > Hi Antonio, >  > >  > Would
 >  this be the sqrt of the turns ratio?? > > No. The turns ratio,
 >  because for coils with identical geometries the > inductance is
 >  proportional to the square of the number of turns, and > for a
 >  transformer with high coupling coefficient, the mutual inductance >
 >  is M=sqrt(L1*L2). > The actual voltage gain of a transformer is
 >  A=M/L1, that for this > value of M is equal to sqrt(L2/L1), that is
 >  equal to n2/n1 if the > geometries of the coils are identical. > >
 >  Antonio Carlos M. de Queiroz > >
 >
 >
 >