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!
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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