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A 3-phase, 3-wire, 240 volt, CBA system supplies a delta-connected load in which ZAB = 25 ∠90º, ZBC = 15 ∠30°, ZCA = 20 ∠0º ohms. Find the line currents and total power

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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.

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