Frequency where reactence is zero

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Homework Help Overview

The problem involves determining the frequency at which the reactances of a 58 mH inductor and an 84 µF capacitor are equal. The discussion centers around the equations for inductive and capacitive reactance.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants explore the relationship between the reactances by setting them equal to each other. There are attempts to isolate frequency in the equations and questions about the correctness of the equations used.

Discussion Status

Some participants have provided guidance on solving for frequency and have noted potential misunderstandings in the initial setup of the equations. There is an ongoing exploration of the correct forms of the reactance equations.

Contextual Notes

There appears to be some confusion regarding the equations for reactance, with participants questioning the definitions and forms presented in the original post.

kdrobey
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Homework Statement


At what frequency (in Hz) are the reactances of a 58 mH inductor and a 84 µF capacitor equal?


Homework Equations


Xl=1/2(Pi)fL
Xc=2(Pi)fC



3. The Attempt at a Solution [/b
F needs to be the same for both, so i got f by itself in both equations, then had them equal each other, giving me Xl2(Pi)L=1/2(Pi)XcC, but I am stuck from there
 
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kdrobey said:

Homework Statement


At what frequency (in Hz) are the reactances of a 58 mH inductor and a 84 µF capacitor equal?

Homework Equations


Xl=1/2(Pi)fL
Xc=2(Pi)fC
3. The Attempt at a Solution [/b
F needs to be the same for both, so i got f by itself in both equations, then had them equal each other, giving me Xl2(Pi)L=1/2(Pi)XcC, but I am stuck from there

I think you have misunderstood the question. The question wants to know for what frequencies are the reactances the same. In other words, find f such that Xc = Xl.
 
so if Xl=Xc, then 1/2(Pi)fl=2(Pi)fC? i tried that then got 1=(2pi fL)*(2pi fC). is that on the right track?
 
Yes, now just solve for f.

On a related note, wikipedia tells me
Xc = 1 / (2pi*f*C)
Xl = 2*pi*f*L
Which seems to be the opposite of what you have written, although it should still give the same answer.
 
kdrobey said:
so if Xl=Xc, then 1/2(Pi)fl=2(Pi)fC? i tried that then got 1=(2pi fL)*(2pi fC). is that on the right track?
Looks good thus far :approve:.
 
got it! thanks a lot
 
kdrobey said:
got it! thanks a lot
A pleasure :smile:.
 

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