Re: MOT current limiting shunts

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

Original poster: "Jim Lux" <jimlux-at-earthlink-dot-net> 

Choose L so that the reactance (Xl = 2 * pi * f * L) is suitable to limit
the current.  You'll have to use the complex form of Ohm's law,  I = E /
(R+jX) >>
     I = E * (R-jX)/(R^2+X^2)

Actually, all you care about is the magnitude of I -> sqr(real^2 + imag^2)

And, in practice X is very much larger than R, so you can ignore R
Ilimit = E/X
X = E/Ilimit = (2*pi*f*L) >rearranging>> L = E/(Ilimit * 2 * pi *f)

Handy number to remember in 60Hz land -> 2*pi*60 = 377


AS to shunts...

An ideal transformer has two windings with the flux from one entirely
coupled to the other.  In a real transformer, not all the flux is coupled:
some of the flux goes through one winding only, called leakage flux.  The
equivalent circuit for such a real transformer is an inductor in series with
the winding of the ideal transformer.  If you want a transformer with some
inductance in series with the winding (i.e. ballast L), then you can
deliberately create more leakage flux by "shunting" the magnetic flux around
the winding.


From: "Tesla list" <tesla-at-pupman-dot-com>
To: <tesla-at-pupman-dot-com>
Sent: Sunday, January 11, 2004 7:22 PM
Subject: MOT current limiting shunts


 > Original poster: John <fireba8104-at-yahoo-dot-com>
 >
 >
 > Hello all,
 >
 > A question for the ages, at least in this hobby. Is there any formula for
 > finding L of current limiting shunts in Mots ,or any other transformer for
 > that matter? Now, to clear up a bit of confession, are the shunts
 > equivalent to a inductor in parallel with a winding, as the name shunt
 > implies, or are they equivalent to series inductance, as what would be
 > desirable for a neon xformer in order to limit voltage after an arc is
 > started? Finally, after I've killed your love for the written word, to
what
 > winding is ohm's law applied?
 >
 > Thanks for at least reading this somewhat long post,
 >
 > John
 >
 >