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RLC Circuit Find inductance and capacitance

  1. Feb 27, 2011 #1
    The energy of an RLC circuit decreases by 1.00% during each oscillation when R=2.00 ohms. If this resistance is removed, the resulting LC circuit oscillates at a frequency of 1.00 kHz. Find the values of inductance and capacitance.
     
  2. jcsd
  3. Feb 27, 2011 #2

    gneill

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    Do you have a strategy? What do you know? What equations are relevant? Where's your attempt?
     
  4. Feb 27, 2011 #3
    w=1/(LC)^(1/2)
    f=w/2pi = 1/(2pi(LC)^(1/2)) = 1.00kHz

    some how this is supposed to relate to the equation for a damped object on a spring.

    L(d^2Q/dt^2) + R(dQ/dt) + Q/C = 0 <---> m(d^2x/dt^2) + b(dx/dt) + kx = 0

    Other than this I really have no idea...
     
  5. Feb 27, 2011 #4

    gneill

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    Okay, it may be a bit simpler than you think.

    From what you have written you can determine the value of ωo. Next determine the Q of the circuit. You're told that the energy decreases by 1% each cycle, so what is the Q? (hint: Q is energy stored / energy dissipated per cycle).
     
  6. Feb 27, 2011 #5
    So ωo = 2pif = (2pi)1.0 khz

    and the resistance would have something to do with Q?
     
  7. Feb 27, 2011 #6

    gneill

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    Yes, the resistance is where energy is dissipated. But in this case you're given specific information about how the energy is lost (per cycle of oscillation). You can determine the Q from that.
     
  8. Feb 27, 2011 #7
    Q= 2pif x (energy stored / energy dissipated per cycle)
    = ωo(energy stored / energy dissipated per cycle)
    = ωo(0.01)
    ?
     
  9. Feb 27, 2011 #8

    SammyS

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    Problem statement: The circuit loses (dissipates) 1.00% of its energy during each cycle.

    The Quality Factor, Q0, is the ratio: (energy stored)/(energy dissipated) for each cycle.

    What is (energy stored)/(1.00% of energy stored) ?
     
  10. Feb 27, 2011 #9
    oh, so Qo= 100
     
  11. Feb 27, 2011 #10

    gneill

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    It's simply energy stored/energy lost for a given cycle. You're told that 1% of the energy is lost per cycle. Imagine that there happens to be 100 units of energy (you don't care what the units are) that begin a cycle. A 1% loss represents 1 unit of energy. So Q = 100/1 = 100.

    Now, there are expressions for the natural frequency ωo and Q for RLC circuits. These involve the circuit components R, L, and C (naturally). Since you have R, with the expressions for ωo and Q you can solve for L and C. The tricky thing is trying to decide whether its a parallel RLC circuit or a series RLC circuit, because the expression for Q is different for each.

    What formulas have you learned for ωo and Q for RLC circuits?
     
  12. Feb 27, 2011 #11
    ωo = 1/(LC)^1/2 -> L = ((1/ωo)^2)/C

    Q = (1/R)(L/C)^(1/2) -> L = C(QR)^2

    -> C=1/QRωo = 7.96x10^(-9) Farads

    -> L=Q^2(R^2)(C) = 3.184

    I think this looks right!

    Thanks a bunch, really appreciate it!
     
    Last edited: Feb 27, 2011
  13. Feb 27, 2011 #12

    gneill

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    Watch your orders of magnitude. I put the capacitance in the ~1μF range, and the inductance around 30 mH.
     
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