Question 16.16: The line voltage at a Δ-connected load (see Figure 16–22(a))......
The line voltage at a Δ-connected load (see Figure 16–22(a)) with ZΔ = 40 + j30 Ω per phase is VL = 2.4 kV(rms). Find the line and phase current phasors using ∠VAN = 0° as the phase reference. Verify your results using Multisim. Assume 60 Hz.
We first calculate the line current IA 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 ZY = ZΔ/3 = 13.33 + j10 Ω. For the specified phase reference we have VAN = 1386 ∠ 0° V(rms), and the phase A line current is calculated as
\textbf{I}_{A} = \frac{\textbf{V}_{AN}}{Z_{Y}} = \frac{1386 ∠ 0°}{|13.33 + j10| ∠ 36.9°} = 83.1 ∠ −36.9° A(rms)
The other two line currents lag IA at −120° intervals. For a positive phase sequence, these currents are IB = 83.1 ∠ −156.9° A(rms) and IC = 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 IAB leads IA by 30°; hence IAB = 48 ∠ −6.9° A(rms) and the other two phase currents are IBC = 48 ∠ −126.9° A(rms) and ICA = 48 ∠ −246.9° A(rms).
Another approach is to calculate IAB directly in the Δ-connected load. Using ∠VAN = 0° as the phase reference, the appropriate line voltage is VAB = VL∠30° = 2400∠30° V(rms). This voltage appears across the impedance ZΔ; hence IAB is found to be
\textbf{I}_{AB} = \frac{\textbf{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 IA.
Multisim allows the Δ-circuit to be readily analyzed. Draw the circuit as shown in Figure 16–22(b). Note that the impedances need to be converted into the time domain. Hence, the load inductance is j30 = j2π × 60L or L = 79.6 mH. Ask Multisim to perform a Single Frequency AC Analysis at 60 Hz. The desired line current is the current exiting (not entering) the sources VAN, VBN, and VCN. Hence, add new expressions: −I(V1), −I(V2), and −I(V3). The phase currents are simply the currents through each of the elements in each branch of the delta load. We selected the current through the load resistors: I(R4), −I(R5), and I(R6). Grapher View returned the table shown in Figure 16–22(c). These results are the same as those calculated earlier.
