Engineering How Do You Calculate Reactance in a Parallel RLC Circuit?

[DethRose]

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

 

[Delta]

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.

 

[DethRose]

so are you saying that answer that i came up with is right?

 

[DethRose]

i know how to calculate everything else...but of course if the reactance is wrong so are all of the other values haha

 

[Delta]

The formulas are good, but I didn't get 0.32. What did you get for XL.

 

[DethRose]

i got 157 ohms

 

[DethRose]

anyone know how to figure this out?

 

[DethRose]

my answers are Xc=318 ohms and Xl=157 ohms

 

[Delta]

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.

 

[DethRose]

no i think it was just a marking error...thanks

 

[Bryonfire031]

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

 

[vk6kro]

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.