Question 16.16: The line voltage at a Δ-connected load with ZΔ = 40 + j30 Ω ......
The line voltage at a Δ-connected load with Z_{Δ} = 40 + j30 Ω per phase is V_{L} = 2.4 kV (rms). Find the line and phase current phasors using ∠\pmb{V}_{AN} = 0° as the phase reference.
We first calculate the line current \pmb{I}_{A} in the equivalent Y-connected load. The phase voltage at the load is V_{P} = 2400/\sqrt{3} = 1386 V (rms) and the load impedance is Z_{Y} = Z_{Δ}/3 = 13.33 + j10 Ω. For the specified phase reference we have \pmb{V}_{AN} = 1386∠0° V (rms), and the phase A line current is calculated as
\pmb{I}_{A} = \frac{\pmb{V}_{AN}}{Z_{Y}}= \frac{1386∠0°}{|13.33 + j10|∠36.9°} = 83.1 ∠ – 36.9° A (rms)
The other two line currents lag \pmb{I}_{A} at – 120° intervals. For a positive phase sequence these currents are \pmb{I}_{B} = 83.1 ∠ – 156.9° A (rms) and \pmb{I}_{C} =83.1 ∠ – 276.9° A(rms). The magnitude of the phase current is I_{P} = 83.1/\sqrt{3}= 48 A (rms). For a positive phase sequence the phase current \pmb{I}_{AB} leads \pmb{I}_{A} by 30°; hence \pmb{I}_{AB} = 48 ∠ -6.9° A (rms) and the other two phase currents are \pmb{I}_{BC} = 48 ∠ – 126.9° A (rms) and \pmb{I}_{CA} = 48∠ – 246.9° A (rms).
Another approach is to calculate \pmb{I}_{AB} directly in the Δ-connected load. Using ∠\pmb{V}_{AN} = 0° as the phase reference, the appropriate line voltage is \pmb{V}_{AB} =V_{L}∠30° =2400 ∠30° V(rms). This voltage appears across the impedance Z_{Δ}; hence \pmb{I}_{AB} is found to be
\pmb{I}_{AB} = \frac{\pmb{V}_{AB}}{Z_{Δ}}=\frac{ 2400∠30°}{|40 + j30| ∠36.9°} = 48∠ – 6.9° A (rms)
which is the same as the result derived using the line current \pmb{I}_{A}.