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FM Radio receiver in RLC circuit with .200 microH Inductor

  1. Apr 18, 2013 #1
    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?
     
  2. jcsd
  3. Apr 18, 2013 #2

    gneill

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    Staff: Mentor

    This assumption is suspect. The components are in series so they won't have the same potential across all three.

    Can you write expressions for the impedance at each frequency? Leave R as a variable, but plug in the known reactances at each frequency. If you assume some source voltage E, what's an expression for the peak current given impedance Z?
     
  4. Apr 18, 2013 #3

    Z = √(R2 + (XL-XC)2)

    XC,1 = 1/ (2π(104.3*106) (11.64 * 10-12)) = 131.094 Ω

    XL,1 = 2π (104.3*106)(.2*10-6) = 131.067 Ω

    XC,2 = 1/ (2π(103.9*106) (11.64 * 10-12)) = 131.599 Ω

    XL,2 = 2π (103.9*106)(.2*10-6) = 130.565 Ω

    I would think that XC,n = XL,n and the only reason why they're not equal right now is because of the slight rounding from a, so:

    Z = √(R2 + (XL-XC)2) = R at both frequencies

    Is that a wrong assumption to make?

    As for the current with source voltage E, I0 = E/Z, yes?
     
    Last edited: Apr 18, 2013
  5. Apr 19, 2013 #4

    gneill

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    Staff: Mentor

    Nope! Take a close look at the formulas for inductive and capacitive reactance. They are different because their values vary differently with respect to frequency. Only at resonance do they have the same magnitude (in fact, that's a defining characteristic of resonance).
    Yup. Bad assumption. There will be two different values of Z. Only at resonance will Z = R because the reactances exactly cancel then.
    Yes. Write expressions for the two values of current that are of interest.
     
  6. Apr 19, 2013 #5
    Thank you for your help! I figured it out last night. Stupid mistake on my part; I should've realized that XL =/= XC when the frequency changes.

    Anyway, here is the correct method:

    L = .200 μH
    C = 11.64 pF ≈ 11.6 pF (This can actually give you an entirely different answer)
    R = Some resistance we have yet to find
    Z = Impedance
    I0 = ε/Z, where some power supply with voltage ε is hooked up to the radio.
    Let all those with subscript 1 be that of 104.3 MHz, all those with subscript 2 be that of 103.9 MHz

    The radio should stay the same, so the inductance, capacitance, and resistance should all stay the same, regardless of frequency. We know that I0,2 = .10% I0,1, so we have this:

    ε/Z2 = .001 ε/Z1 <---- .001 because .10% = .001

    .001 Z2 = Z1
    .001 √(R2 + (XL,2 - XC,2)2) = √(R2 + (XL,1 - XC,1)2)

    We know that at 104.3, XC = XL
    ∴ XC,1 = XL,1 → XL,1 - XC,1 = 0 Ω

    So now we have:

    .001 √(R2 + (XL,2 - XC,2)2) = √(R2)

    Square both sides, and we have:

    10-6 (R2 + (XL,2 - XC,2)2) = R2
    R2 + (XL,2 - XC,2)2 = 106 R2

    106 - 1 is practically 106, so:

    (XL,2 - XC,2)2 = 106 R2

    Plug numbers in with C = 11.6 pF (!!), and you should get:
    (1.487)2 = 106 R2 → 1.487 = 103 R

    → R = 1.487 * 10-3 = 1.49 * 10-3 Ω
     
  7. Apr 19, 2013 #6

    gneill

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    Staff: Mentor

    Yep. Looks good.
     
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