# Re: pole pig question

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

At 07:34 PM 1/30/2003 -0500, you wrote:
>I will use the example of a 14.4 kV, 10 kva units. I know that the 14.4kv
>means, 14,400 volts out.

Actually 14,400vac "RMS" which is the equivalent voltage that would heat a
resistor the same as 14,400 volts DC.  Since the voltage may not be a
perfect sine wave (and in reality, never is) everyone prefers to use the
"RMS" thing.  Not real important in our case, but do remember that the peak
and firing voltage is 1.414 times the RMS voltage or about 20,000
volts!  That is real important to us ;-))

>But, what does the 10kva mean?

"VA" is Volts multiplied ba Amps.  In the case of DC, that is simply the
same as watts.  But in AC, the current and voltage sine waves may be in or
out of phase (always somewhere in between...), so the term "watts" gets
real fishy.  So they go with the worst case number which they call
"VA".  That is the maximum RMS voltage multiplied be the maximum RMS
current.  "Real power (watts)" will always be less or perhaps equal in a
perfect case.  By using "VA" they can never specify a transformer that is
too small.  It is usually way too big ;-))

In general:

Watts = Vrms x Irms x COS(the angle between voltage and current (radians)
(a number between zero and one))

VA = Vrms x Irms x 1 (worst case)

>I think it is amps, but what does the kva mean?

The "k" is kilo or x1000.  Big transformers are usually rounded to the
nearest 1000 VA ;-)

A 10kVA tranny at 14400 Vrms should only put out a maximum current of:

power / voltage = current

10000 / 14400 = 0.69444 Arms or 694.4 mA  (a darn big NST ;-)))

>Also, when we use pole pigs for tesla coils, we have to use some sort of
>current limiting. If you use a variac, won't that change the voltage going
>in, thus changing the output power of the pole pig?

Yes, think of the ballast inductor as a current limiting resistor that
dissipates no heat.  The "resistance" is:

Xr = 2 x pi x f x L

So a 100mH inductor on the AC line has an equivalent resistance (the real
term used is "reactance" (still measured in ohms)) of:

2 x pi x 60 x 0.1 = 37.7 ohms.  So, if you put 240Vac across it you will
have 6.366 amps (be careful of saturation if it is iron core!).

>How do you figure out how much power is really coming out of the pig to
>find out how many caps you need?

That can be messy ;-))  beyond me this moment, but get this stuff first
;-))  It is really one of those Joules going in, Joules stored, Joules
going out things....

>One more question, on the pigs, their are 3 inputs (well, i guess outputs
>in the wild) and 2 outputs (inputs in the wild), since their are 3 inputs,
>which two do you use for 2 phase current?

Line at zero degrees, neutral or ground, line at 180 degrees...  Basically,
120, 0, -120...  Time to start thinking about AC sine waves being x degrees
different in relation to each other...   Your reaching the next level of
your understanding now :-))))  And, starting to ask the "messy" questions :o)))

Cheers,

Terry

>Thanks