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Calculate Values of L and C

by wayneinsane
Tags: values
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wayneinsane
#1
Mar22-09, 02:27 AM
P: 16
1. The problem statement, all variables and given/known data


An RLC circuit has a resonance frequency of 2000/pi hertz. When operating at angular frequency w > w0, reactance of inductance is 12 ohms and reactance of capacitance is 8 ohms. Calculate the values of L and C.
2. Relevant equations

XL = wL
XC = 1/wC


3. The attempt at a solution
welll, what i thouht was w = resonance frequency which is 2000/pi hertz... so since XL is given (12) and XC is given (8), then I plugged in the numbers and got this.

2000/pi = 636.6 hz...so, XL/w = L
12/636.6hz = .0188 H = L

and, 1/wXC = C which is 1/(636.6)(8) = .000196F = C

Correct or not? I feel like something is missing!
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Kruum
#2
Mar22-09, 04:54 AM
P: 220
Your answer isn't correct. There are a couple of things you're doing wrong. First of all, frequency isn't the same as angular frequency. And second, you're calculating the reactances of L and C using the resonance frequency, when values are given with higher frequency in the problem. You'll need three equations to solve this one, you've got two of them. Do you have an equation for resonance frequency or can you perhaps derive it with the knowledge you've got?
wayneinsane
#3
Mar22-09, 10:52 AM
P: 16
But isnt w the symbol for resonance frequency? and resonance frequency is given in the problem, along with the reactance of capaciatance and inductance...im not sure what other equation I would need..

Kruum
#4
Mar22-09, 12:07 PM
P: 220
Calculate Values of L and C

Quote Quote by wayneinsane View Post
But isnt w the symbol for resonance frequency? and resonance frequency is given in the problem, along with the reactance of capaciatance and inductance...im not sure what other equation I would need..
Nope. The w is the angular frequency (i.e. how fast sinusoidal signal goes through one cycle), the angular frequency for the resonance frequency is usually denoted by w0. There is link between angular frequency and frequency, though. [tex] \omega=2 \pi f[/tex]. When RLC circuit resonates, XL-XC=0. You can derive the third equation from that.
wayneinsane
#5
Mar22-09, 03:44 PM
P: 16
Hmm, this is starting to make sense... I just checked my notes and you are absolutely right. Wo is the resonance frequency...

so.... since w = 2 pi f, and XL = wL... can I do XL = 2 pi f L which would be XL/2 pi f = L, so 12/4000 which is 3 x 10-3...

then, for XC = 1/wC, can I do XC = 1/2pi f C, which would rearrange to C = 1/XC 2 pi f which would be 1/8x4000 which is 3.125x10-5...

correct?

BTW, thank's a lot!
Kruum
#6
Mar23-09, 06:06 AM
P: 220
Still not correct. Seems like my earlier point didn't go across the way I meant. This sentence "When operating at angular frequency w > w0, reactance of inductance is 12 ohms and reactance of capacitance is 8 ohms." says, that XL is 12 ohms and XC is 8 ohms at some unknown frequency, that is higher than the resonance frequency. So you can't use the resonance frequency to directly determine L and C, like you are now trying to. Solve XL-XC=0 as a function of L and C (i.e. isolate the w to one side and the rest to the other) to get the third equation. From there you can solve the frequency that makes XL 12 ohms and XC 8 ohms.


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