I = P/(E*1.732)
In 3 ph, the supply voltage would be 208V.
In practicality, we would find on the motor decal the FLA and multiply by 1.25 and if that includes enough current to match the LRA, fine, if not, then use the LRA. Otherwise, we would look in the table in the NEC with the motor's hp, voltage, and motor type.
Example: induction type motor, 208V 3 ph, 5 hp. Whatever the amp draw in the table is, that would get multiplied by at least 1.25 to ensure enough current to accomodate LRA or LRC (same difference. Got to have enough current to meet locked rotor, or complete standstill of the motor.) Usually, if you take the full load amperage and do the 125 % on that, it will be enough to meet LRA.
But those size fuses would still work, especially if they are standard trade sizes.
[This message has been edited by rws (edited 12-07-2006).]
