Question

已知正弦电流i1=70.7sin（314t-30°）A，i2=60sin（314t+60°）A。求正弦电流i1，i2的和

Answer 1

i1+i2=70.7sin（314t-30°）+60sin（314t+60°）=70.7[sin(314t)cos30°-cos(314t)sin30°]+60[sin（314t）cos60°+cos（314t）sin60°]=70.7[sin(314t)√3/2-cos(314t)/2]+60[sin（314t）/2+cos（314t）√3/2]=70.7[sin(314t)√3/2-cos(314t)/2]+60[sin（314t）/2+cos（314t）√3/2]=35.35[sin(314t)√3-cos(314t)]+30[sin（314t）+cos（314t）√3]=(35.35√3+30)sin(314t)+(30√3-35.35)cos(314t)=(35.35√3+30)sin(314t)+(30√3-35.35)cos(314t)=91.2sin(314t)+16.6cos(314t)

Answer 2