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RLC Circuits question

  1. Jan 22, 2012 #1
    A series RLC circuit is connected to a 5 V supply, the frequency of the supply is adjusted to give a maximum current of 11.9 mA at 2.5 kHz. The Q factor is 70. Determine the component values of the circuit.

    R= 5/.0119=420.2 Ohm
    Q = (1/R)*(sqrt L/C)
    70 = (1/420.2)*(sqrt L/C)
    70/(1/420.2)= 29414
    L/C = 865183396
    L = 865183396*C
    Resonate frequency
    =1/[(2pi)*(sqrt L*C)]
    2500=1/[(6.28)*(sqrt ((865183396*C)*C)]
    2500=.159/(29414*C)
    29414*C=.159/2500
    29414*C=.0000636
    C=.002uF
    L=865183396*C
    L=1.73 H

    Is this an ok way to calculate this? Does it appear correct?
     
    Last edited: Jan 22, 2012
  2. jcsd
  3. Jan 22, 2012 #2

    rude man

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    Your values for L and C are close but off just a bit: I got L = 1.87 Hy and C = 2.17nF

    I think you went about it the hard way: start with R as you did; realize that f = 2500 Hz so ω = 2pi*2500 rad/s; then use relation between f, L, R and Q to get L, then use relationship between L, C and ω to get C.
     
  4. Jan 23, 2012 #3
    Thanks, i agree i did think that. do you mind me asking what equation you transposed for L? All I can come up with is that involving capacitance
     
  5. Jan 23, 2012 #4

    rude man

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    Q = ωL/R so L = QR/ω.

    ω = 2πf and f = 2.5e3 Hz.

    Then LC = 1/ω2 to get C.
     
  6. Jan 24, 2012 #5
    Thanks rude man, I now can confirm I get the same as you
    L=1.872 Hy
    C=2.165 nF
     
  7. Jan 24, 2012 #6

    rude man

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    Yer' welcome!
     
  8. Sep 17, 2013 #7
    Hi people. Did you get this question correct?
     
  9. Sep 17, 2013 #8

    rude man

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    Yes. Look at the last two posts.
     
  10. Nov 14, 2013 #9
    How did you find the equation for C

    Hi rude man.

    I've been looking at this question for a while now and am yet to see how you found the equation LC =1/ω2


    I assume you found it from 2πfL/R = 1/R x (√L/c)
    But I cannot confirm this is the case.
    I tried to find the capacitance using L/QR2 but I get a different answer.

    Can you help me? The answer is there I know but I want to understand.
     
  11. Nov 14, 2013 #10

    rude man

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    A series R-L-C circuit has impedance √[R^2 + (wL - wC)^2] which is mimimized when wL = 1/wC. Minimum impedance occurs at the resonant frequency.
     
  12. Nov 20, 2013 #11
    Another way to complete

    Thanks! I understand where you're coming from now.

    I also managed to calculate the capacitor value using the resonance formula.

    Resonance -fo/r = 1 / 2π√LC

    ∴ C = (1/-fo/r)2 / ((2π)2 * L)

    C = (1/2500)2 / ((2)2 * 1.87)

    C = 2.167ηF
     
  13. Nov 20, 2013 #12

    rude man

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    Without actually checking your numbers you seem to be on-course.
     
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