SOLUTION: Hi i'm stuck with the following problem can anyone help If i1=0.02sin(ωt) and i2 =0.032cos(ωt-π/3) Determine an expression for output current i=i1+i2 in the

Algebra ->  Trigonometry-basics -> SOLUTION: Hi i'm stuck with the following problem can anyone help If i1=0.02sin(ωt) and i2 =0.032cos(ωt-π/3) Determine an expression for output current i=i1+i2 in the      Log On


   

Question 22523: Hi i'm stuck with the following problem can anyone help
If i1=0.02sin(ωt) and i2 =0.032cos(ωt-π/3)
Determine an expression for output current i=i1+i2 in the form R sin(ωt+ά)
and thus state the its amplitude and phase angle
Answer by venugopalramana(3286) About Me  (Show Source):
You can put this solution on YOUR website!
I1 = 0.02Sin(wt)
I2 = 0.032Cos(wt-pi/3) = 0.032{Cos(wt)Sin(pi/3)+Sin(wt)Cos(pi/3)} =
I2 = 0.032{0.866Cos(wt)+0.5Sin(wt)} = 0.028Cos(wt)+0.016Sin(wt)
I1+I2 = 0.02Sin(wt)+0.028Cos(wt)+0.016Sin(wt)
= 0.036Sin(wt)+0.028Cos(wt)
NOW MULTIPLY AND DIVIDE WITH SQUARE ROOT OF (0.036^2+0.028^2)= 0.0456
LET Sin(a)=0.028/0.0456…SO THAT…Cos(a)=0.036/0.0456…HENCE WE GET
I1+I2 = Sin(wt)Cos(a)+Cos(a)Sin(wt) = Sin(wt+a)…..WITH A BEING ARC SIN (0.028/0.0456)
OR …Sin(a) = (0.028/0.0456)