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Simple RLC Circuit

  1. Jul 3, 2006 #1
    Question:
    A parallel RLC circuit, which is driven by a variable frequency 10-A source, has the following parameters:
    [tex] R=500\Omega [/tex]
    [tex] L=0.5mH [/tex]
    [tex] C=20\muF [/tex]

    Find the resonant frequency, the [itex] Q [/itex], the average power dissipated at resonant frequency, the [itex] BW [/itex], and the average power dissipated at the half-power frequencies.

    Answer:
    [itex] Q [/itex], [itex] \omega_0 [/itex], and [itex] BW [/itex] are all straightfoward calculations.

    [tex] Q = 100 [/itex]
    [tex] BW = 100 \frac{rad}{sec}[/tex]
    [tex] \omega_0 = 10000\frac{rad}{sec} [/tex]

    The half power frequencies are:
    [tex] \omega_{hi} = 1005.01 [/tex]
    [tex] \omega_{lo} = 995.01 [/itex]

    I don't understand how to calculate the average power dissapated at resonant frequency, OR at the half power frequencies. If someone could give me a push in the right direction, that would be swell :redface:
     
    Last edited: Jul 3, 2006
  2. jcsd
  3. Jul 3, 2006 #2
    :smile: nm, figured it out. That was embarrassingly easy.
     
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