As explained in a 3-phase system has only two possible sequences : ABC and CBA. In the ABC sequence, the voltage of phase B lags behind voltage of phase A by 120º and that of phase C lags behind phase A voltage by 240º. In the CBA phase which can be written as A →C →B, voltage of C lags behind voltage A by 120º and that of B lags behind voltage A by 240º. Hence, the phase voltage which can be written as
EAB = E ∠ 0º ; EBC = E ∠ – 120º
and ECA = E ∠ – 240º or ECA = ∠120º
The circuit is shown in Fig..
Line current IA'A = IAB + IAC = IAB – ICA
= –j9.6 – (–6 – j10.4) = 6 + j0.08
Line current IB'B = IBC – IAB = j16 –(–j9.6) = j25.6 A
IC'C = ICA – IBC = (–6 – j10.4) – j 16 = (–5 – j26.4) A
Now, RAB = 0; RBC = 15 cos 30 = 13 Ω ; RCA = 20 Ω
Power WAB = 0; WBC = IBC2 RBC = 162 × 13 = 3328 W; WCA = ICA2 × RCA = 272 × 20 = 14,580 W.
Total Power = 3328 + 14580 = 17,908 W.