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Re: Reactive power: was Re: MOT help



Original poster: "Jason by way of Terry Fritz <twftesla-at-qwest-dot-net>" <jasonp-at-btinternet-dot-com>

John,

I wasnt actually talking about PFC - this reactive power draw occurs even on
transformers that have a PFC factor ov very close to 100%.....

Jason
----- Original Message -----
From: "Tesla list" <tesla-at-pupman-dot-com>
To: <tesla-at-pupman-dot-com>
Sent: Monday, October 29, 2001 3:54 AM
Subject: RE: Reactive power: was Re: MOT help


> Original poster: "by way of Terry Fritz <twftesla-at-qwest-dot-net>"
<couturejh-at-telocity-dot-com>
>
>
> Jason, All -
>
> There is also an explanation for this problem on the following site. Click
> on
>
>      http://www.miramar.sdccd.cc.ca.us/faculty/jcouture/tesla/index.asp
>
> Click on 5. Power Factor Correction.
>
> John Couture
>
>
> ----------------------------
>
>
> -----Original Message-----
> From: Tesla list [mailto:tesla-at-pupman-dot-com]
> Sent: Saturday, October 27, 2001 11:43 AM
> To: tesla-at-pupman-dot-com
> Subject: Reactive power: was Re: MOT help
>
>
> Original poster: "Jason by way of Terry Fritz <twftesla-at-qwest-dot-net>"
> <jasonp-at-btinternet-dot-com>
>
> > Now someone correct me if I'm wrong, but in order to see how many amps
you
> are
> > actually supplying your transformer with, you just see how many amps it
> (the
> > ballast) would normally draw. I don't know for sure about non-current
> limited
> > ballast, like another MOT, but I would guess somewhere around 10-12 A.
>
> I am not sure about this one. I have been told and also found out for
myself
> that reactive power will increase the current draw by a good few amps.
When
> you use a purely resistive load then yes, this 'short and test' approach
> would work very well. However when you are using an inductive or
capacitive
> circuit (like a TC) then the current drawn through the transformer can be
> bumped up further. I am not entirely sure how this works, and someone
please
> set me straight if I am wrong...
>
> The capacitor or inductor when in the charging stage (i.e. high current
> draw) are actually negative in relation to ground, assuming an HV+ =>
> Inductor/Capacitor => Ground connection. This means that the potential
> across the cap or inductor to ground is actually higher than the potential
> from the transfomer directly to ground. In this way the virtual resistance
> stays the same, allowing more current to be drawn due to V=IR.
>
> I think........... :)
>
> Jason
>
>
>
>
>
>