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*To*: tesla-at-pupman-dot-com*Subject*: Re: Current in the Coil -*From*: "Tesla list" <tesla-at-pupman-dot-com>*Date*: Sat, 18 May 2002 12:48:04 -0600*Resent-Date*: Sat, 18 May 2002 12:49:24 -0600*Resent-From*: tesla-at-pupman-dot-com*Resent-Message-ID*: <7sRHiB.A.gAB.VIq58-at-poodle>*Resent-Sender*: tesla-request-at-pupman-dot-com

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> > > 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 Not that simple. The determining factor is the ratio of secondary inductance to primary inductance, or the ratio of secondary capacitance to primary capacitance. The inductance varies as the square of the number of turns plus some geometric factors, so no simple ratio using wire lengths would work unless the primary and secondary had the same size and shape. In that case the ratio would be as you state. I find it easier to think in terms of primary and secondary capacitance. If there is 100% energy transfer from the primary capacitor to the secondary capacitor, the ratio of secondary current to primary current is sqrt(Cs/Cp). Since Lp x Cp = Ls x Cs, the current ratio is also sqrt(Lp/Ls). Normal practice is pick a primary capacitor, build the secondary and add some top loading, and then to select a primary inductance for resonance. It's much easier to adjust the number of turns on the primary than to fix the number of turns and fiddle with the capacitance. Ed

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