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What is the peak current for this circuit?

  1. Jun 13, 2014 #1
    1. The problem statement, all variables and given/known data

    An RLC circuit has a sinusoidal voltage supplied to it at 632 kHz with a peak voltage of 748 kΩ; a 25 kΩ resistance; a 16 µF capacitance; and a 30 H inductance. What is the peak current for this circuit?
    A) 30 μA
    B) 26 μA
    C) 6.3 μA
    D) 11 μA



    2. Relevant equations

    Xc= 1/(2∏ƒC)
    Z = √R^2 + (Xl-Xc)^2
    Xl= 2∏ƒL


    3. The attempt at a solution

    Irms = Vrms / √R^2 + [(2∏ƒL) - (1/(2∏ƒC))]^2

    when i plug everything in I am getting 4.7 x 10^-8 which is nowhere correct. i'm not sure if i'm doing something wrong in relation to the theory aspect when making the variable equation to plug everything into; but it seems right so i'm a little confused
     
  2. jcsd
  3. Jun 13, 2014 #2

    NascentOxygen

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    Staff: Mentor

    We can't check your working when you don't show it.

    But I note that the question asks for peak current, and you show a formula with RMS. This may not necessarily be the problem, though.

    You assumed the given elements are all to be connected in series. Was this stated?
     
  4. Jun 13, 2014 #3
    it doesn't say if its in series or parallel.
    so i was just using the generalized RLC circuit equations.

    (748,000)/ √(25000)^2 + [{2∏*(632,000)} - (1/(2∏*632,000* 16x10^-6))^2]
    that's the work, its a lot to take in.
    that gives the Irms .

    i tried taking the voltage and using ohm's law but that also doesn't work so i'm really confused.
     
  5. Jun 13, 2014 #4

    ehild

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    Homework Helper
    Gold Member

    Check the problem wording: The peak voltage can not be given in kΩ. Also, the inductance looks too large.

    If you divide the peak voltage with the impedance, you get peak current. Dividing rms voltage with the impedance results in rms current.

    ehild
     
  6. Jun 13, 2014 #5
    yeah i think that's just a mistake in the problem that I was given, because i double checked and it has voltage in ohms which is weird.

    but the peak voltage/impedance works i think. thanks for the help!
     
  7. Jun 13, 2014 #6
    aw man okay, i did that and got 26 but its not right. i don't know what I'm missing.
     
  8. Jun 13, 2014 #7

    NascentOxygen

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    Staff: Mentor

    OK, I missed the voltage being expressed in kΩ. In that case, you need a whole different set of obtuse formulae to solve it. :smile:

    In other words, it's anyone's guess what the question should be. Do they mean V, or kV? And the same holds for the other data.

    You could try working backwards from each answer, since there is consistency there in the units, and see what you end up with. Maybe try various permutations of units, e.g., Hz or kHz, H or mH, etc.

    Good luck!
     
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