# Re: Current Limiting and Impedence

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• Subject: Re: Current Limiting and Impedence
• From: "Tesla list" <tesla@xxxxxxxxxx>
• Date: Wed, 27 Apr 2005 11:00:46 -0600
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• Resent-date: Wed, 27 Apr 2005 11:02:12 -0600 (MDT)
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`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```