The phase currents in windings AB and BC are
80KVA/240V = 333A.
These currents combine to yield,
Ib = 333Ax1.732 = 577A for the 80 kva per phase.
Single phase =34 + (34+ 7) kva = 68 + 7 kva.
For winding CA, 68KVA of the single phase load may be treated as a 240V load and added to the 80KVA load to yield 148KVA of 240V load . Phase (not line) current is 148KVA/240 = 616A. To obtain the line current, Ia, we combine vectorially the phase current from winding AB.
Ia = 616A + 333A[cos(60) +jsin(60)] = 783A + j289A yields 835A magnitude.
The remaining 7KVA of single phase load adds 58.3A into node C. Then
Ic = 616A + 58.3A + 333A[cos(60) +jsin(60)] = 841A + j289A yields 890A magnitude.
Ia = 835A
Ib = 577A
Ic = 890A
