66 amps at 460 at 480?

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(460/480)*66
Why not (480/460)*66?

Do we know enough to give the correct answer? Zog's formula would work if this were a constant-power load. Mine would work if it were a constant-impedance load. So which is it? Doesn't that depend on the type of welder? :-?
 
I would just ignore the 20 volts difference just like we do with 460 volt motors. You will never hit peak input amps unless the operator is maxing out the welder which is in my experience very unusual and once you figure in the duty cycle it really will not matter.
 
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