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Radio Receiver

  1. Jul 10, 2008 #1
    1. The problem statement, all variables and given/known data

    A radio receiver contains an LC circuit whose natural frequency of oscillation can be adjusted, or tuned, to match the frequency of the incoming radio waves. The adjustment is made by means of a variable capacitor. Suppose that the inductance of the circuit is 11.00 μH. What capacitance must the capacitor be adjusted to if the circuit is to span the 540.00 kHz range?


    2. Relevant equations

    [tex]\omega[/tex] = [tex]\sqrt{L/C}[/tex]

    3. The attempt at a solution

    [tex]\omega[/tex] = [tex]\sqrt{L/C}[/tex]
    [tex]\omega[/tex] = 540.00 kHz
    L = 11.00 microH
    C = what we're looking for

    I solved for C getting C = [tex]\frac{1}{\omega^2 L}[/tex]
    C = [tex]\frac{1}{(540000Hz)^2 * .000011 H}[/tex]
    C = [tex]\frac{1}{3.20E6}[/tex]
    C = 3.1176E-7 F

    Where am I going wrong?
     
  2. jcsd
  3. Jul 10, 2008 #2
    C = 1/3207600
     
  4. Jul 10, 2008 #3

    marcusl

    User Avatar
    Science Advisor
    Gold Member

    You are mixing up frequency with angular frequency. Remember that

    [tex]\omega=2\pi f[/tex]
     
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