Finding i(t) in a Circuit with Phasors: Introductory Exercise

[okh]

Homework Statement

In this circuit, find [itex]i(t)[/itex], knowing that [itex]v_s= 2 \cos \left(w x+\frac{\pi }{2}\right)[/itex], and that, at the source's frequency, [itex]X_C= -1 Ω[/itex] and [itex]X_L = 1 Ω[/itex].

Homework Equations

Basic phasors and dividers equations.
[itex]Z_C = jX_C[/itex]
[itex]Z_L = jX_L[/itex]

The Attempt at a Solution

I used dividers. The source divides between R and the parallel of C and the series of L and R.
[itex]
I=\frac{Z_C v_s \left(Z_L+R\right)}{\left(Z_L+R\right) \left(Z_C+Z_L+R\right) \left(\frac{Z_C \left(Z_L+R\right)}{Z_C+Z_L+R}+R\right)}[/itex]
Solving with [itex] R=1, Z_c=-j, Z_l=j, v_s=2j [/itex] I get [itex]i(t)=\frac{2}{\sqrt{5}}*cos(wt+0.46)[/itex], while the correct phase should be [itex]-2.68[/itex]. Basically I get the symmetrical cosine wave with respect to the x axis.

 

[gneill]

I agree with your result. It seems that they took the given voltage source phase ##\pi/2## to be a negative phase shift for some reason.

 

[okh]

Thank you. Yeah, that may be the reason.