Z=20%=3467991=cs508=3687 3467/3687°9221=115636.87°92 (2)求Z:设隅接的相电流为/a 7500 P2 =1295A P2 CoSy 380 3×=×0.88 380 lZ2F 3/295:1694202=c082836° Z2=1694/2836°92 闪四 西南交通大学西南交通大学 = = 34.67W 1 1 1 P P I U Z = = ° - cos 0.8 36.87 1 j1 （2）求Z2：设Y接的相电流为IP2 A U P I P P 12.95 0.88 3 380 3 7500 3 cos 2 2 2 2 = ´ ´ = = j \Z2 = 16.94 28.36° W = = 12.95 »16.94W 3 380 | | 2 2 2 P P I U Z = = ° - cos 0.88 28.36 1 j2 \ Z1 = 34.67 36.87 ° W Z1 ¢ =11.56 36.87° W