Re: Reactive power: was Re: MOT help

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

Original poster: "Terry Fritz" <twftesla-at-qwest-dot-net>

Hi Jason,

I am not sure I can explain this with an E-mail.  It takes a lot of arm
waving :-))

But basically, reactive components (capacitors and inductors) do not
dissipate energy (heat) but they store energy and they do draw current.  If
you place a 100uF motor run cap across the AC line, it draws:

Rc = 1 / (2 x pi x 60 x 100e-6) = 26.52 ohms

120 / 26.52 = 4.52 amps

It does not get hot but the current is there and it will blow a 3 amps
fuse.  If you add a 20 ohm resistor to this in parallel, you will draw 6
amps through it too.  The total current is SQRT (4.52^2 + 6^2) = 7.51 amps.
 If you add an inductor, it operates in the opposite way as the capacitor
and it will reduce the current by nullifying the capacitors current.

From a sine wave point of view, the resistor draws current at zero degrees,
the inductor draws current 90 degrees latter and the cap draws current 90
degrees sooner.

When we add PFC caps, we are nullifying the NSTs inductance with the cap to
remove this unneeded current.

It could get unbelievably complex from here, especially in a TC where the
spark gap effects screw up the nice steady state sine wave equations for
such things, but that is the basic idea.

Cheers,

	Terry


At 06:28 PM 10/27/2001 +0100, you wrote:
>> 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
>