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Homework Help: Changing values of R in a RLC series circuit

  1. Sep 28, 2010 #1
    1. The problem statement, all variables and given/known data
    A television channel is assigned the frequency range from 54 MHz to 60 MHz. A series RLC tuning circuit in a TV receiver resonates in the middle of this frequency range. The circuit uses a 16 pF capacitor. Inductor is .487 micro Henries.
    In order to function properly, the current throughout the frequency range must be at least 50% of the current at the resonance frequency. What is the minimum possible value of the circuit's resistance?

    2. Relevant equations

    3. The attempt at a solution
    I'm kinda lost on this one.
    I tried setting .5(V/Z1)=(V/Z2) and solving for R.
    Z1 is using the 57MHz and Z2 is using 54 or 60 MHz, neither gave me a correct answer.

    I meant to put "finding a value" instead of "changing values of," sorry.
  2. jcsd
  3. Sep 29, 2010 #2
    It seems like your set up is correct. What is the expression you get for [tex] \frac{| Z_1 |}{| Z_2 |} [/tex]?

    Though it doesn't seem like you need the info I wrote up below, it might be worth checking out:

    Your expression for the magnitude of the impedance of the circuit is correct, but is more instructive written in this way:

    [tex] | Z(w) | = \sqrt{ R^2 + ( \frac{ ( \frac{\omega}{\omega_o} )^2 - 1 }{ \omega C })^2 } [/tex]

    Where [tex] \omega_o = \frac{1}{\sqrt{LC}}[/tex] , or the angular frequency at which [tex] | Z | [/tex] is a minimum. Notice is only depends on L and C, not R. This is correspondingly the frequency of maximum current.

    When you plug in the values for L and C, you find that the resonant frequency ([tex] \frac{\omega_o}{2 \pi} [/tex] ) is very close to 57 MHz, right in the middle of the range given in the problem.

    Your task is to find the value R such that the ratios of the currents at the edge of the frequency range are at least 50% of the current at the resonant frequency. Hope this helps.
    Last edited: Sep 29, 2010
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