# 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>
>
> 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