# Phase angle? L-C series

1. Dec 20, 2009

### SAT2400

Phase angle?? L-C series

1. The problem statement, all variables and given/known data
In an L-C series circuit with very small resistance, the voltage drops across inductor and capacitor are 100V and 120V,. What's the phase angle?

2. Relevant equations
tantheta= Xl -Xc)/ R

3. The attempt at a solution

According to book's explanation, tantheta= (Xl -Xc)/ R is infinity..!??

Why?? and the answer is -90degree...

2. Dec 20, 2009

### ehild

Re: Phase angle?? L-C series

XL-XC is negative as the voltage across the capacitor is higher than across the inductor. So the tangent is negative: the angle itself is negative, as its cosine (which is proportional to R) is positive. R is very small, so the tangent is very large, tends to -infinity as R tends to zero. What is the tangent of -90 degree?

ehild

3. Dec 20, 2009

### SAT2400

Re: Phase angle?? L-C series

as its cosine is positive?!? what do you mean??

tan(-90) is ERR: DOMAIN on my calculator lol...

4. Dec 20, 2009

### ehild

Re: Phase angle?? L-C series

Have you seen a tangent function plotted out? how does it look very close to 90 degree?

The limit is infinite, if you calculate the tangent of angles closer and closer to 90 degree with your calculator. Try.

The real part of the complex impedance is R; it can not be negative.

The imaginary part is called reactance and denoted by X. For a series circuit, X=XL-XC.

If Z is the magnitude of the complex impedance, and theta is its phase angle, then R=Z*cos(theta), X=Z*sin(theta).

X/R= Z*sin(theta)/(Z*cos(theta))= sin(theta)/cos(theta)=tan (theta)

As R >=0, the cosine of the phase angle is positive: The angle must be between -90degree and 90 degree.

As tan(theta)=X/R, you can calculate the angle as the arctangent of X/R if R is finite. If it is zeo, the angle is either 90 of -90 degree.

ehild