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Resonate Frequency of Parallel RLC Circuit

  1. Apr 17, 2012 #1
    1. The problem statement, all variables and given/known data

    Determine the resonant frequency of the circuit in Fig 14.28 (See attached)


    2. Relevant equations

    [tex]\begin{array}{l}
    {\rm{How is }} \to {\omega _0}0.1 - \frac{{2{\omega _0}}}{{4 + 4\omega _0^2}} \\
    {\rm{derived from }} \to j{\omega _0}0.1 + \frac{{2 - j{\omega _0}}}{{4 + 4{\omega ^2}}} \\
    \end{array}[/tex]

    I'm sure this is not difficult but I just can't see it.

    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Apr 17, 2012 #2

    NascentOxygen

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    Staff: Mentor

    You consider just the two terms with a "j" in them. I've corrected where you omitted a "2".
     
  4. Apr 17, 2012 #3
    Still can't see it.
     
  5. Apr 17, 2012 #4

    NascentOxygen

    User Avatar

    Staff: Mentor

    One of the terms with a j in it is jω0.1

    What is the other term with a j in it?
     
  6. Apr 17, 2012 #5
    The other term is -j2w. I still cant see how this helps.
     
  7. Apr 17, 2012 #6

    NascentOxygen

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    Staff: Mentor

    That's only the numerator; what is its denominator?
     
  8. Apr 17, 2012 #7
    How's this? I new it was simple and the answer was already sitting there. It felt a bit like pulling teeth.

    Thanks for that.

    [tex]\begin{array}{l}
    \begin{array}{*{20}{c}}
    {{\rm{How is}} \to {\omega _0}0.1 - \frac{{2{\omega _0}}}{{4 + 4\omega _0^2}}} \\
    {{\rm{derived from}} \to j\omega 0.1 + \frac{{2 - 2j\omega }}{{4 + 4{\omega ^2}}}} \\
    \end{array} \\
    {\rm{Answer:}} \\
    {\rm{At resonance }}Im(Y){\rm{ = 0:}} \\
    j{\omega _0}0.1 - \frac{{j{\omega _0}2}}{{4 + 4\omega _0^2}} = 0 \to j\left( {{\omega _0}0.1 - \frac{{{\omega _0}2}}{{4 + 4\omega _0^2}}} \right) = 0 \\
    {\omega _0}0.1 - \frac{{{\omega _0}2}}{{4 + 4\omega _0^2}} = 0 \\
    \end{array}[/tex]
     
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