Capacitive reactance current limiting is something you already understand.
[ I(peak) = Omega x Cap x Voltage(peak)] applies to pole pigs the same as
it does for NST.
Our system uses:
(10 Kva) pig.
30amp desired with 240V input yields 500ma (peak) current at the high
votage end.
14.4 kv x sqrt(2) yields approx. 20kv (peak).
Omega = 2pi x 60hz.
Solving yields approx .09uf, This is the value of our tank capacitor.
No more current can flow than is set by the equation above.
Of course you still need a ballast to handle the momentary shorts. But it
does not need to be as large as with systems using smaller tank capacitors.
Jared Dwarshuis, Larry Morris"
We all know Ohm's law but you've calculated the reactive charging
current with no spark discharges. Can't imagine it being as large as the
current with several sparks per cycle. The last sentence sure isn't
obvious! Larger capacitors will take larger charging current in general
and particularly when 'there's a whole lot of sparking going on'!
Ed