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*To*: tesla-at-pupman-dot-com*Subject*: Re: High Voltage Output*From*: "Tesla list" <tesla-at-pupman-dot-com>*Date*: Sun, 06 Jun 2004 11:09:00 -0600*Resent-Date*: Sun, 6 Jun 2004 11:13:17 -0600*Resent-From*: tesla-at-pupman-dot-com*Resent-Message-ID*: <auN7dC.A.IjE.fC1wAB-at-poodle>*Resent-Sender*: tesla-request-at-pupman-dot-com

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. 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 > >

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