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!