如图所示电路中，已知i1=3sin（314t-30°）A，i2=4sin（314t+60°）A，则

Question

如图所示电路中，已知i1=3sin（314t-30°）A，i2=4sin（314t+60°）A，则安培表A的示数为?

Answer 1

i1=3sin（314t-30°）A=3sin(314t)√3/2-3cos(314t)/2
i2=4sin（314t+60°）A=4sin(314t)/2+4cos(314t)√3/2
i3=(2+3√3/2)sin(314t)+(2√3-3/2)cos(314t)=4.598sin(314t)+1.964cos(314t)=
5(4.598/5*sin(314t)+1.964/5*cos(314t))=5sin(314t+23)