[例]已知： i1(t)= 5cos(314t+ 60°) A i2(t)=10sin(314t+ 60°)A i3(t)= – 7cos(314t+ 60°) A
+j
I1m 60° +1 –30°
写出振幅相量，绘相量图。解： I =5/60° A 1m i2(t) =10sin(314t+60°) =10cos(314t – 30° )
I3m
–120°
I2m
i3(t) = – 7cos(314t+ 60°)
=7cos(314t – 120° )A
I3m=7/﹣120° A I2m=10/﹣30° A 用复数表示正弦量的方法称相量表示法