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