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IbB is equal to IaA but lagged by angle 120 degree, hence, IbB = 70 * (Cos(-140) + j Sin(-140)) = -53.623-j45.

This problem can be solved easily by applying KVL for loop (a-A-B-b) as it is without converting it into single phase values  where, IaA = 65.78-j23.94, IbB = -53.62-j45 ( IbBI = 70, angle = -140), Vab = (65.78-j23.94)*(5+j10) + 12500 -  (-53.62-j45)*(5+j10) Vab = 12886.47 + j1299.3 IVabI =...

I think this question may issued considering the permissions under article 440.41 (A) :"In case the motor controller is rated in horsepower but is without one or both of the foregoing current ratings, equivalent currents shall be determined from the ratings as follows. Table 430.248, Table...