RE: Current in the Coil -

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

Original poster: "David Thomson by way of Terry Fritz <twftesla-at-qwest-dot-net>" <dave-at-volantis-dot-org>

Thanks Ed.  So having the peak current in the primary known, and assuming
that I have a 1:40 primary to secondary wire length ratio, do I ballpark
calculate the secondary peak current as 1/40th of the primary?

Dave

-----Original Message-----
From: Tesla list [mailto:tesla-at-pupman-dot-com]
Sent: Friday, May 17, 2002 4:20 PM
To: tesla-at-pupman-dot-com
Subject: Re: Current in the Coil -


Original poster: "Ed Phillips by way of Terry Fritz <twftesla-at-qwest-dot-net>"
<evp-at-pacbell-dot-net>

Tesla list wrote:
>
> Original poster: "David Thomson by way of Terry Fritz
<twftesla-at-qwest-dot-net>" <dave-at-volantis-dot-org>
>
> Hi Dave L.,
>
> Thanks for the link.  That's quite an input, 2800A!  The peak current in
the
> Electrum secondary was 61A.  Do you know what kind of secondary coil
cooling
> mechanism they used, if at all?
>
> Dave T.

	It's easy to figure the peak current in the primary.  It's just the
peak capacitor voltage divided by the sqrt(L/C) of the primary.  For
example, if the primary capacitance is 0.01 ufd, the resonant frequency
is 200 kHz and the peak charging voltage is 15000, the peak current will
be about 188 amps.  The RMS current will be a lot less, depending on the
Q of the primary and the spark rate.

Ed