1. The problem statement, all variables and given/known data The tuning circuit in an FM radio receiver is a series RLC circuit with a 0.200 µH inductor a. The receiver is tuned to a station at 104.3 MHz. What is the value of the capacitor int he tuning circuit? b. FM radio stations are assigned frequencies every 0.2 MHz, but two nearby statiosn cannot use adjacent frequencies. What is the maximum resistance the tuning circuit can have if the peak current at a frequency of 103.9 MHz, the closest frequency that can be used by a nearby station, is to be no more than 0.10% of the peak current at 104.3 MHz? The radio is still tuned to 104.3 MHz, and you can assume the two stations have equal strength. 2. Relevant equations ω = 1/ √(LC) = 2πf (f is for frequency, ω is for angular frequency, L inductance, C capacitance) Ip = Vp/R = Vp/Xc = Vp/XL Xc = 1/ (ωC) XL = ωL 3. The attempt at a solution I correctly found the answer to a as shown: ω = 1/ √(LC) = 2πf √(LC) = 1/(2πf) LC = 1/(2πf)2 --> C = 1/(4π2f2L) = 11.64 pF What I am stuck on is b. Here is my thought process so far: Ip = Vp/R = Vp/Xc = Vp/XL ∴ Vp/R = Vp/XL Since the peak current of the station at 103.9 MHz is no more than 0.10% that of the peak current of the station at 104.3 MHz, the max resistance occurs when the peak current is .10% that of the peak current of the station at 104.3 MHz, so: Vp/R = .001 * Vp/XL ---> .001 R = XL = ωL = 2πf * L Solve for R, and I get the wrong answer. The correct answer is apparently R = 1.49 * 10-3 Ω The reason why I used the inductance rather than the capacitance is because it seemed like a better number to use, since the capacitance was slightly rounded and the inductance was given. Am I missing something? Perhaps a minor detail in the problem that I may have overlooked?