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Re: transformer measured L, etc.



Original poster: "Jim Lux by way of Terry Fritz <teslalist-at-qwest-dot-net>" <jimlux-at-earthlink-dot-net>

Loss is only due to the resistive component, not the entire impedance...
The current in an unloaded MOT will be quite high, but out of phase with the
voltage, so the active power (loss) is quite low.

----- Original Message -----
From: "Tesla list" <tesla-at-pupman-dot-com>
To: <tesla-at-pupman-dot-com>
Sent: Tuesday, February 04, 2003 6:37 AM
Subject: Re: transformer measured L, etc.


 > Original poster: "Matthew Smith by way of Terry Fritz
<teslalist-at-qwest-dot-net>" <matt-at-kbc-dot-net.au>
 >
 > Hi All
 >
 > >Therefore: N2/N1 = sqrt(L2/L1)
 >
 > Yep that's what I was getting at in my original post; whereas my books say
 > that I should be using Z in place of L in Jim' equation (above), L works
 > "near enough for Jazz" for me, and makes a real nice, quick way of getting
 > the turns ratio.
 >
 > Still can't get over those MOT's having a calculated primary loss of
 > 600W!  (Calculated as VV/Zp).
 >
 > Cheers
 >
 > M
 >
 > --
 > Matthew Smith
 > IT Consultant - KBC, South Australia
 > http://www.kbc-dot-net.au
 >
 >
 >
 >