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Re: MOT charging reactor PS



Original poster: "Wells Campbell by way of Terry Fritz <twftesla-at-qwest-dot-net>" <wellscampbell-at-onebox-dot-com>

Jim lux wrote:

> The foregoing assumes that the inductor is a pure inductor and has
> no loss.
>  If you want to determine the loss, you'll need to a bit of math and
> use
> the source voltage (Vs) as well...  Consider the inductor as the sum
> of
> some resistance and inductance (Rx + Lx).  The math is left as an exercise
> for the reader...

Jim, 

thanks, I'll give it a try...

As for the loss in the inductor, could you use a constant DC to measure
the resistance of the inductor without any inductive effects?

-- 
Wells Campbell
wellscampbell-at-onebox-dot-com - email
(415) 430-2169 x3756 - voicemail/fax



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