Question 16.13: Figure 16–17 shows a balanced Δ-connected load in parallel w......
Figure 16–17 shows a balanced Δ-connected load in parallel with a balanced Y-connected load. The two-phase impedances are ZΔ = 120 + j40 Ω and ZY = 50 + j30 Ω. Find the phase impedance of an equivalent Y-connected load.
S T E P 1 We first convert the Y-connected load in Figure 16–17 into an equivalent Δ-connected load using Eq. (16–22). The conversion yields phase impedances of 3ZY and produces the circuit configuration in Figure 16–18(a).
Z_{Y} = \frac{Z_{Δ}}{3} (16–22)
S T E P 2 Each phase impedance 3ZY is in parallel with phase impedance ZΔ. Combining these parallel impedances as ZΔ || 3ZY produces the equivalent Δ-connected load in Figure 16–18(b), where
Z_{ΔEQ} = Z_{Δ} || 3Z_{Y} = \frac{Z_{Δ} × 3Z_{Y}}{Z_{Δ} × 3Z_{Y}}
= \frac{(120 + j40)(150 + j90)}{120 + j40 + 150 + j90}
= 67.6 + j29.7 Ω
S T E P 3 We use Eq. (16–22) again, this time to convert the load in Figure 16–18(b) into the equivalent Y-connected load in Figure 16–18(c), where
Z_{YEQ} = \frac{Z_{ΔEQ}}{3} = \frac{67.6 + j29.7}{3}
= 22.5 + j9.9 Ω
In sum, ZYEQ is the phase impedance of a balanced Y-connected load that is equivalent to the two parallel loads in Figure 16–17.
