Question 16.13: Figure 16–17 shows a balanced Δ-connected load in parallel w......

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 \pmb{3Z_{Y}} 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 \pmb{3Z_{Y}} is in parallel with phase impedance \pmb{Z_{Δ}}. Combining these parallel impedances as \pmb{Z_{Δ}||3Z_{Y}} 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, Z_{YEQ} is the phase impedance of a balanced Y-connected load that is equivalent to the two parallel loads in Figure 16–17.