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Re: Tesla's gotta lotta Gauss



fairly easy to calculate maximum possible peak current...
Do it with energy.. Energy in C == Energy in L  so CV^2 = LI^2... I = V *
SQRT(C/L)
This is IDEAL with no losses and no "pumping" of the C with pulses at just
the right rate..
Taking your coil as an example"
.01 uF at 350 kHz means that L is about 21 uH.  C/L is .01/21 or 4.76E-4..
Sqrt of that is .0218

Your peak voltage is 15*1.4 = 21 kV... Times the sqrt of C/L you get about
458 amps...

-----Original Message-----
From: Tesla List <tesla-at-pupman-dot-com>
To: tesla-at-pupman-dot-com <tesla-at-pupman-dot-com>
Date: Wednesday, May 24, 2000 7:50 PM
Subject: Re: Tesla's gotta lotta Gauss


>Original Poster: "John Williams" <jwilliams-at-edm-dot-net>
>
>I once calculated the peak current in the primary of a coil
>that ran off a 15 kv 60 ma neon sign transformer with a
>0.01 mfd cap at around 350 khz...
>
>I forget what it was off hand, exactly but it was seriously
>high.  So I don't doubt that the flux field would be fairly
>serious too.
>
>>Original Poster: ANTarchimedes-at-aol-dot-com
>>
>>
>>Has anyone tried measuring the magnetic feilds of their coil using a
>>Gaussmeter?  I'll bet that a fair current is created... despite the lack
of
>>an iron core.
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