RE: Reactive power: was Re: MOT help

["Tesla list" <tesla-at-pupman-dot-com>]

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