Question 16.17: A balanced three-phase source with VL = 200 V(rms) feeds a Δ......

The phase impedance of the equivalent Y-connected load Z_Y = Z_Δ/3 = 4 + j2 Ω. The phase voltage magnitude at the source is V_{P} = 200/\sqrt{3} = 115.5  V(rms). In each phase, the voltage V_P appears across the series combination of Z_W + Z_Y; hence the line current magnitude is

I_{L} = \frac{V_{P}}{|Z_{W}  +  Z_{Y}|} = \frac{115.5}{|4.1  +  j2.55|} = 23.9  A(rms)

Using ∠I_A as the phase reference means I_A = 23.9 ∠ 0° A(rms) and the other two line currents lag I_A at −120° intervals. For a positive phase sequence, these currents are I_B = 23.9 ∠ −120° A(rms) and I_C = 23.9 ∠ −240° A(rms). The phase current magnitude is I_{P} = 23.9/\sqrt{3} = 13.8  A(rms). The phase current I_AB leads I_A by 30°; hence I_AB = 13.8 ∠ +30° A(rms) and the other two phase currents are I_BC = 13.8 ∠ −90° A(rms) and I_CA = 13.7 ∠ −210° A(rms).
Again, we can easily get the phase currents in the Δ-connected load by first finding the line current I_A in the equivalent Y-Y circuit.