I need to solve the following equation to get I = V/((R^2+(Xl-Xc)^2)^1/2).
-I[(1/c - w^2L)sin(Φ)-Rwcos(Φ)]-wV=0
we know that tan(Φ) = (Xl-Xc)/R
Xc = 1/wc
Xl = wL
The Attempt at a Solution
I[(1/c - w^2 L)sin(Φ)-Rwcos(Φ)]=wV
I[sin(Φ)/c - sin(Φ)w^2L - Rwcos(Φ)] = wV
I[sin(Φ)/wc - sin(Φ)wL - Rcos(Φ)] = V
By using a helping right hand triangle, I know that sin(Φ)=(Xl-Xc)/((R^2+(Xl-Xc)^2)^1/2) and cos(Φ)=R/((R^2+(Xl-Xc)^2)^1/2)...
So I won't have to type this over and over, I'll assume the hypotnuse of the triangle is h = ((R^2+(Xl-Xc)^2)^1/2). Now when I plug in the values I have...
I[(Xl-Xc)/h (Xc) - (Xl-Xc)/h (Xl) - R^2/h]=V
I[(XcXl-Xc^2)/h - (Xl^2-XcXl)/h - R^2/h]=V
I = (Vh)/(XcXl-Xc^2-Xl^2+XcXl-R^2) plug in for h now from above...
I = [V ((R^2+(Xl-Xc)^2)^1/2)] / (XcXl-Xc^2-Xl^2+XcXl-R^2)
This is where I get stuck... I don't know what to do here... This could be better to post in the Mathematics forum. If someone could help me out, it would be greatly appreciated...