Re: Current Limiting and Impedence

["Tesla list" <tesla@xxxxxxxxxx>]

Original poster: "Mark Dunn" <mdunn@xxxxxxxxxxxx>


Gerry:

Thanks for all your help.  Now if you can bear with me and help me solve
the original problem with the math(BTW <> is "not equal to")....

I have (2) inductors in series.  My current limiter and my transformer.
Mains is 120V/60 Hz/1 ph.

The current limiter is 15 mH, X = 5.65 Ohms, R = 0.3 Ohms.
The transformer is Z = 1.2 Ohms, R = 0.3 Ohms, so X = 1.16 Ohms

So System Impedence Z = 6.82 Ohms

This gives I = 120/6.82 = 17.6 Amps and is near exactly what I am
getting.

The measured voltage between the inductor and transformer is about 90 to
95 volts.  Have gotten as high a 100 volts.  This is the voltage drop
across the transformer.  This is inconsistent with the above values and
in fact seems reversed from what the math suggests.  I have double
checked it.

For more confusion, lets compute power:

True power = R*I^2 = 0.6*17.6^2 = 186 watts
Reactive power = X*I^2 = 6.81*17.6^2 = 2109 var
Apparent power = 2117 VA
Power Factor: 186/2117 = .08 (makes no sense)

I know I have 120*17.6 = 2112 VA input and approx 7500*.2=1500 watts
output from the transformer secondary. Thus I have a power factor of
approximately .71.  I get 48 to 60" sparks from the coil/toroid so I've
got to have near that kind of power.

I believe that I am improperly accounting for the power that is
transferred from transformer primary to secondary in the above
equations.  It's as if some of the reactive power "created" by the X of
the inductor is actually true power for the transformer.

I tested the current limiter and don't think it is saturating.
(BTW - I'm a "math guy" so I can handle the vector addition and complex
variables no problem.  But struggling with setting up equations and how
to apply them.)

Thanks again for your help.
Mark