有一平衡三相負載之線電壓為600 V，功率因數為0.6落後，消耗功率為360 kW。送電端經輸電線送電至負載，輸電線每相之阻抗值為0.015+j0.025 Ω，求送電端之線電壓？

正確答案629 V。計算步驟：P=360kW，pf=0.6滯後，sin φ=0.8；I_L=P/(√3×V_L×pf)=360000/(√3×600×0.6)=577.35A；相電壓V_R=600/√3=346.41V；令V_R為參考相量，I=(577.35∠−53.13°)A；V_S=V_R+I×Z_line=346.41+577.35∠−53.13°×(0.015+j0.025)；Z_line模=√(0.015²+0.025²)=0.02915Ω，角度=59.04°；I×Z_line=577.35×0.02915∠(−53.13°+59.04°)=16.83∠5.91°；實部=16.75，虛部=1.74；V_S=346.41+16.75+j1.74=363.16+j1.74；|V_S|=363.16V；線電壓=363.16×√3≈628.7≈629V。