How Do You Calculate Reactance in a Parallel RLC Circuit?

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Discussion Overview

The discussion revolves around calculating the reactance of components in a parallel RLC circuit, specifically focusing on a circuit with a 500-ohm resistor, a 0.01 microfarad capacitor, and a 0.5 mH inductor at a frequency of 50 kHz. Participants are attempting to determine the individual reactances (Xc and Xl) and the overall circuit reactance.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant presents their calculations for capacitive reactance (Xc) and inductive reactance (Xl) but receives feedback suggesting their results may be incorrect.
  • Another participant questions whether the initial calculations are for individual reactances or the overall reactance of the circuit.
  • Multiple participants provide their own calculations for Xc and Xl, with varying results, indicating potential discrepancies in the calculations.
  • A later post suggests that the question may imply a different circuit configuration, potentially affecting the interpretation of the reactance calculations.

Areas of Agreement / Disagreement

Participants express differing results for the reactance calculations, and there is no consensus on the correct values or the interpretation of the question regarding overall circuit reactance.

Contextual Notes

There are indications of possible misunderstandings regarding the circuit configuration and the components involved, as well as discrepancies in the values calculated for Xc and Xl. The discussion does not resolve these issues.

Who May Find This Useful

Students and individuals interested in electronics, particularly those studying RLC circuits and reactance calculations.

DethRose
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Hey

im in first year electronics and have a problem with a parallel rlc circuit

im trying to find the reactance (Xc and Xl) for a circuit that has a 500 ohm resister and a .01 microfarad capacitor in parallel with a 500 ohm resistor and a .5 mH inductor. The frequency of the circuit is 50 KhZ

here are my calculations

xc= 1/(2)(pie)(50khz)(.01microfarads)
= 0.32

xl= (2)(pie)(50khz)(.0005H)

but i had them both marked wrong...help with this would be very much appreciated
 
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First thing, are you sure you got the right answer there.

Secondly, now that you have the reactance of the individual components, I suspect that the question is asking for the overall reactance.
 
so are you saying that answer that i came up with is right?
 
i know how to calculate everything else...but of course if the reactance is wrong so are all of the other values haha
 
The formulas are good, but I didn't get 0.32. What did you get for XL.
 
i got 157 ohms
 
anyone know how to figure this out?
 
my answers are Xc=318 ohms and Xl=157 ohms
 
That's what I get based on the information. Is that all the question asks for though, or is it asking for the circuit reactance.
 
  • #10
no i think it was just a marking error...thanks
 
  • #11
Parallel circuit consiting a 50Khz Power Supply with a 500 ohm resistor, capacitor with a .1microF value, and a .5mH inductor

-Use the Capacitive reactance formula XC= 1/2*3.14*Frequency*Capacitor : XC= 1/2*3.14*50KHZ (50,000Hz)*.1microF (.0000001F)= 31.8 Ohms
XC=31.8ohms

-Capacitive inductance formula XL=2*3.14*F*L : XL=2*3.14*50KHZ*.5mH= 157 ohms
XL=157ohms
 
  • #12
Hi Bryonfire031, and welcome.

The posts above were posted in 2005 (dates are on the left of the post) and neither poster has logged on since 2007.

However, the question seemed to imply that there was a 500 ohm resistor in series with the inductor (as well as the first 500 ohms ) and it was asking for the total impedance, so it may not be a surprise that the poster was marked wrong.

The capacitor was 0.01 µF.
 

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