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Finding R, L, C of a tuned RLC circuit


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!



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:


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.


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!

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