Question 16.17: A balanced three-phase source with VL = 200 V (rms) feeds a ......
A balanced three-phase source with V_{L} = 200 V (rms) feeds a Δ-connected load with Z_{Δ} = 12 + j6 Ω per phase through a three-wire line with Z_{W} = 0.1 + j0.55 Ω per phase. Find the line current and phase current phasors using ∠ \pmb{I}_{A} = 0° as the phase reference.
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 ∠ \pmb{I}_{A} as the phase reference means \pmb{I}_{A} = 23.9∠0° A (rms) and the other two line currents lag \pmb{I}_{A} at – 120° intervals. For a positive phase sequence these currents are \pmb{I}_{B} = 23.9 ∠ – 120° A(rms) and \pmb{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 \pmb{I}_{AB} leads \pmb{I}_{A} by 30°; hence \pmb{I}_{AB} =13.8∠+30° A (rms) and the other two phase currents are \pmb{I}_{BC} =13.8∠ – 90° A (rms) and \pmb{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 \pmb{I}_{A} in the equivalent Y-Y circuit.