Ia=3kVA/120V=25A@0deg, Ib=4kVA/120V=33.3A@-120deg, Ic=5kVA/120V=41.7A@120deg. IA=Ia-Ic, IB=Ib-Ia, IC=Ic-Ib. IA=(25cos(0)+i25sin(0)) - (41.7cos(120)+i41.7sin(120))[\quote]
If the the j component of - 120 is negative your caculations would change.
Ia = Iab +Iac
Ia = 25/0 + 33cos/-120 + j33sin/-120
Ia = 25 + 33 x .5 + j33 x -.866
Ia = 25 + 16.6 - j28.85
Ia = 41.6 - j28.85 = 51 amps
Ib = Iab + Ibc
Bob, you assigned the wrong phase angle to Iba. It should be 60 degrees.
Iab = -Iba = -33.3A/-120 = 33.3A/60
The resultant phase angle is 35 deg. In a balanced system, it would be 30 deg.
The attachment shows the phasor diagram. The same method would be used at nodes B and C. Just pay attention to the "direction" of the currents. The approximate angles are:
30, -90, & -210 degrees,
