What Is the Phase Angle in a Low Resistance L-C Series Circuit?

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Homework Help Overview

The discussion revolves around determining the phase angle in a low resistance L-C series circuit, where the voltage drops across the inductor and capacitor are given. Participants are exploring the implications of the phase angle in relation to the circuit's reactance and resistance.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to understand why the tangent of the phase angle approaches infinity when resistance is very small. There is a focus on the relationship between the voltages across the inductor and capacitor and how this affects the phase angle. Questions are raised about the behavior of the tangent function near -90 degrees and the implications of a zero resistance.

Discussion Status

There is an active exchange of ideas regarding the mathematical relationships involved in the circuit, particularly concerning the tangent function and its limits. Some participants are questioning the definitions and implications of the phase angle in the context of the circuit's parameters, while others are providing insights into the behavior of the tangent function as it approaches critical angles.

Contextual Notes

Participants are navigating the complexities of the phase angle in relation to very small resistance values, which raises questions about the assumptions made in the problem setup. The discussion reflects a mix of interpretations regarding the mathematical properties of the tangent function and the physical implications in the circuit.

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Phase angle?? L-C series

Homework Statement


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?


Homework Equations


tantheta= Xl -Xc)/ R


The Attempt at a Solution



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

Why?? and the answer is -90degree...
Please explain!:)
 
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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
 


ehild said:
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

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

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


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
 

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