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Finding R, L, C of a tuned RLC circuit

okraina

New Member
The question is: Tuned Series RLC circuit is connected to 25mV source has Bandwidth of 10kHz and f1 (lower half power frequency) = 600kHz. Find R, L, C

First I calculated resonance frequency = 600kHz + (10kHz/2) = 605kHz and f2 = 610kHz.

After that I calculate Quality Factor Q = fr / BW = (605kHz / 10 kHz) = 60.5.

My teacher tell me that I should solve the following equations:
Q = Xl / R at fr (A),
Xl - Xc = R at f2 (B),
and Xc - Xl = R at f1 (C) . I am stuck on this step.

First I equate (B) and (C) which gives:

(2pi)*(610K)*L - (1/(2*pi*(610K)*C)) = R = (1/(2*pi*(600K)*C)) - (2*pi)*(610K)*L)

2*pi*L (610K + 600K ) = (1/ (2*pi*C) ) * ( ( 1 / (610K ) ) + ( 1 / ( 600K) ) ) . Moving L and C to LHS:

LC = ( 1210K / (366000K ) ( 1 / ( (2pi)^(2) ( 1210K) ) ) ~= 6.92 * 10 ^(-11)

For equation (A): I rearrange to get R = ((2*pi) * (605K) ) / (60.5)
further rearranging I get (62.83K) * L = R.

Now I don't know how to continue this question to eliminate any of the three variables.

Please let me know if you have a way of solving this question. I have attached my worked solution to this post.

Thank you!
 

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Ylli

Active Member
Math is not by best subject, but I don't think you can find a unique set of values. You need the reactance of the L and the C to be the same at 605 KHz, and the the equivalent shunt resistance 60.5x the reactance of either. Could be Xl = Xc = 10 ohms, and R = 605 ohms, or could be Xl = Xc = 10,000 ohms, R = 605K ohms.
 
Last edited:

crutschow

Well-Known Member
Most Helpful Member
That question does not have a unique answer for the component values.
You need to know one of the component values for that.
But you can find the relative value of all three.
 

okraina

New Member
Thanks for the responses. I don't know why this problem is in the back of the book if you are unable to solve it; but in either case it was a good exercise in problem solving!
 

Ratchit

Well-Known Member
okraina
Your teacher is wrong. You have 3 equations and 3 unknowns, but equations B and C are useless for finding a solution because they both contain all three variables and therefore are equivalent. You need to know another variable like Xl, Xc. or R to complete the solution. Ratch
 

MrAl

Well-Known Member
Most Helpful Member
Hello,

Yes it appears that the question given makes it sound like there is just one solution (because of the voltage specification) but if we dont know what something else is we can only solve it symbolically. You may want to do that, but it is a little questionable why your professor gave you a particular voltage input when the voltage cancels when computing the upper, lower, or center frequency. Perhaps there is something that was left out of this question that was supposed to be in there.

Also interesting, if the LOWER frequency is given as 600kHz exactly and the bandwidth is 10kHz then the center frequency is not exactly 605kHz although that is somewhat intuitive. The actual center frequency is about 0.003 percent different than that. That's a small difference however so 605kHz is probably acceptable.
The reason for this is because the center frequency is not exactly in the center of the upper and lower frequencies mathematically although we often think of it that way because the difference is so small.
If you do a full calculation of any center frequency and upper and lower frequencies you will find this to be true for any analog filter that has a 2nd order bandpass type response.

Just to note, you can reduce the L and C to just L or just C using a well known property of a series RLC circuit. That helps to solve these problems, but will still not lead to a single numerical result.
 

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