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Minimize Impedance - (eletromagnetism)

  1. Apr 24, 2005 #1
    OK, I think I am doing this question right, but I'm not exactly sure. The question is as follows:

    For an RLC circuit with a resistance of [tex]16k\ohm [/tex], a capacitance of [tex]8.0\mu F [/tex] and an inductance of [tex]38.0H[/tex] what frequency is needed to minimize the impedance?

    Well impedance is give by:
    [tex] Z = \sqrt{R^2 + (X_C - X_L)^2} [/tex]

    Putting [tex] X_C [/tex] and [tex] X_L [/tex] in terms of [tex] L, C, \omega [/tex] we then have:

    [tex] Z =\sqrt{R^2 + \left(\omega L - \frac{1}{\omega C}\right)^2} [/tex]
    Minimum impedance is acheived at resonance, so [tex] Z = R [/tex]

    Thus we have:
    [tex] R =\sqrt{R^2 + \left(\omega L - \frac{1}{\omega C}\right)^2} [/tex]

    Solving this for [tex] \omega [/tex] yields:

    [tex] \omega = \frac{1}{\sqrt{LC}} [/tex]

    And frequency is given by: [tex] f = \frac{\omega}{2\pi} [/tex]

    So solving [tex] f [/tex] for [tex] \omega [/tex] and substituting into the equation above gives:

    [tex] f 2\pi = \frac{1}{\sqrt{LC}} [/tex]

    Now solving for [tex] f [/tex] yields:
    [tex] f = \frac{1}{2\pi\sqrt{LC}} [/tex]

    And finally plugging in [tex] L,\,C[/tex] from above gives:

    [tex] f = \frac{1}{2\pi\sqrt{(38.0H)(8.0\mu F)}} = 9.12Hz = 0.009kHz[/tex]

    So I'm pretty sure there are going to be a few questions like this on my test tomorrow, so I just want to make sure I'm doing this correctly. Thank you.
     
  2. jcsd
  3. Apr 24, 2005 #2

    OlderDan

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    Looks good.
     
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