Question 16.19: In Figure 16–23 the load is Y-connected with a phase impedan......

In this example the line current I_{L} and phase impedance Z_{Y} are given, hence the complex power delivered to the load is

S_{L} = 3I^2_{L}Z_{Y} = 3(10)^2 (15 + j6)

= 4.5 + j1.8 kVA

The apparent power delivered to any balanced load can be written as |S_{L}| = \sqrt{3}V_{L}I_{L}. Solving for the line voltage gives

V_{L} =\frac{|S_{L}|}{\sqrt{3}I_{L}}=\frac{|4.5  ×  10^3  +  j1.8  ×  10^3|}{\sqrt{3}  ×  10}= 280  V (rms)

Again, no phasors are needed to solve this problem.