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Alternating Currents

  1. Jun 21, 2015 #1
    1. The problem statement, all variables and given/known data

    A 10 ohms resistor, a 12 microFarad capacitor and a 28 mH inductor are connected in series with a 170 V generator. A.) At what frequency is the rms current maximum? B.) What is the maximum value of the rms current?
    2. Relevant equations

    A.) Fo = 1 / 2pi sqrtLC
    b.) Irms = Vrms / 2pi fL

    3. The attempt at a solution
    This is a lecture thag i'm trying to study. I know how to get for A. But when I try inputting the variables, i get the wrong answer for B. It says its 17A. I've been getting 3.52. Can somebody help me please?
     
  2. jcsd
  3. Jun 21, 2015 #2

    mfb

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    Please show how you got 3.52 A, otherwise it is hard to understand what went wrong.
     
  4. Jun 21, 2015 #3

    Hesch

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    That's not correct ( the capacitor is not included ).

    The impedance: Z = R + jωL + 1/(jωC).
    ( ω = 2πf )

    I = V / Z
     
  5. Jun 21, 2015 #4
    Using the formuka for B
    Irms = Vrms / 2pi fL
    = 170V / 2pi (274.57Hz) (28×10^-3H)
     
  6. Jun 21, 2015 #5
    So my professor was wrong? :0
     
  7. Jun 21, 2015 #6

    Hesch

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    I don't know what your professor has told you. :)
     
  8. Jun 21, 2015 #7
    Hahaha
    that's what she used in that number though. Can you tell me what's wrong with the formula and how to really use the right one? I'm really sorry for asking much but I'm quite stuck....
     
  9. Jun 21, 2015 #8

    Hesch

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    Well, The "formula" as for Z is the right one.

    To find the current, you must find the absolute value for Z.

    1/(jωC) = -j/(ωC) →
    Z = R + j(ωL - 1/(ωC) )

    Obvious the minimum value for |Z| is found when (ωL - 1/(ωC)) = 0. You have already found ω0 in (A), where (ω0L - 1/(ω0C)) = 0.

    So |Z|min = R + j(ω0L - 1/(ω0C) ) = R →

    Imax = V / R
     
  10. Jun 21, 2015 #9

    rude man

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    Hint: at f = f0 there is no reactance, meaning L and C reactances cancel each other out. What's left?
     
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