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*To*: tesla-at-pupman-dot-com*Subject*: Re: shunts/Magnetek NST*From*: "Tesla list" <tesla-at-pupman-dot-com>*Date*: Wed, 18 Apr 2001 07:09:31 -0600*Resent-Date*: Wed, 18 Apr 2001 07:42:31 -0600*Resent-From*: tesla-at-pupman-dot-com*Resent-Message-ID*: <UOrQi.A.H4D.EnZ36-at-poodle>*Resent-Sender*: tesla-request-at-pupman-dot-com

Original poster: "harvey norris by way of Terry Fritz <twftesla-at-uswest-dot-net>" <harvich-at-yahoo-dot-com> --- Tesla list <tesla-at-pupman-dot-com> wrote: > Original poster: "cougercat by way of Terry Fritz > <twftesla-at-uswest-dot-net>" <felix1063-at-home-dot-com> > > Hello Group, > > Does anybody have a photo of the transformer > "shunts" that can be removed for > greater power? > > Thanks > > --jeff Yes I would like to see those also. My basic question for the list is this, do ALL NST's have these shunts? Recently I tried to measure the inductance of the secondary of my 15,000 volt,30 ma Magnetek Jefferson NST's with a Wavetek LCR meter. The Highest inductance reading scale is 200 Henry, and it shows a overrated value on the highest scale reading, meaning the inductance is too high for the meter to measure. It does however make a reading of 20,000 ohms on the resistive reading. Discouraged I decided to calculate the value of inductance by finding the impedance that the secondary will have at 60 hz and delivering its rated 30 ma as a shorted secondary. At 15,000 volts and delivering the 30 ma across the short, the acting impedance of that secondary would be by Ohms Law extended to AC where V=IR then becomes V= IZ : 15000(v)=.03(A)*Z where Z then =15,000/.03= 500,000 ohms. Using the equation Z(impedance)=sq rt [X(L)^2 +R^2]= 500,000=sq rt [X(L)^2+ (20,000)^2] where then squaring both sides yeilds 2.5 *10^11= X(L)^2+ 4*10^8 showing that X(L)^2 =2.496*10^11 and therefore the sq rt of that value at ~499,600 ohms as the inductive reactance figure X(L) Since X(L)= 6.28 Freq(L), we can then divide the 499,600 by 6.28(60) to obtaing the L value as ~1326 Henry. So by the above calculations we can see that the 1326 henry secondary with 20,000 ohms resistance will permit a current of only ~ 30 ma if short circuited. This rated current that is delivered is already current limited by the acting impedance of that secondary. This is why I cannot understand what, why, and where the shunt exists. I conclude at this point in time that my Jefferson contains no shunts, and as I have said it is already current limited by the acting impedance of the secondary, based on the 20,000 ohms figure that has been measured. Terry had also posted figures for other NST's where the resistance of this NST secondary had the highest figure at 20,000 ohms. Is it possible that only the jeffersons function this way without the need for a shunt? Any comments on this shunt business, and potential errors in my calculations showing that no shunt exists on this transformer would certainly bring me to a better understanding of this matter. Perhaps I have made a silly error, and can stand to be corrected. Sincerely HDN ===== Binary Resonant System http://members3.boardhost-dot-com/teslafy/ __________________________________________________ Do You Yahoo!? Yahoo! Auctions - buy the things you want at great prices http://auctions.yahoo-dot-com/

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