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*To*: tesla-at-pupman-dot-com*Subject*: Re: Directions for tesla coil research*From*: "Tesla list" <tesla-at-pupman-dot-com>*Date*: Mon, 06 May 2002 16:50:53 -0600*Resent-Date*: Mon, 6 May 2002 16:51:12 -0600*Resent-From*: tesla-at-pupman-dot-com*Resent-Message-ID*: <wXTY.A.kwG.Zjw18-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> > How does a bigger top load increase Q. if Q = Xl / R and Xl = > 2 * pi * f * l. > > from these equations, one would assume that if frequency goes down, due to a > shift in resonant frequency, then Q decrease because Xl decrease. I know > I'm missing something important here, right? > > Shaun Epp I think so. The discussion was about reducing the frequency by increasing the top loading, but you have the wrong expression for Q in this case. The Q is the parallel resistance divided by the reactance so, for a given resistance (same power in the streamers) the Q will increase. Your expression is for an inductor with a resistance in series with it. If you look at the equivalent circuit you will see that increasing the parallel resistance does lower the series resistance, with the same effect of increasing the Q a bit. For coils with a Q much over 10, as are all the ones under discussion here, the product of series resistance and parallel resistance equals the square of the reactance. Rp Rs = X^2 if I didn't make another typo. Ed Ed

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