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RLC circuit impedance

  1. Apr 7, 2016 #1
    1. The problem statement, all variables and given/known data
    A series RLC circuit consists of a 60.0 Ω resistor, a 2.30 mH inductor, and a 690 nF capacitor. It is connected to an oscillator with a peak voltage of5.80 V .

    Part A
    Determine the impedance at frequency 3000 Hz.
    Part B
    Determine the peak current at frequency 3000 Hz.
    Part C
    Determine phase angle at frequency 3000 Hz



    2. Relevant equations
    Z=sqrt[R2+(Xl-Xc)2]
    Xc1/wC
    Xl=wL



    3. The attempt at a solution

    I found Xc=1/(2pi*3000Hz) =0.76886
    and XL2pi*3000=43.354

    so Z=73.5765

    As you might have guessed, this is not the correct answer.

    What am I doing wrong? I suspect it has something to do with how I converted Hz to Rad/s

    How would I go about parts B and C too?
     
  2. jcsd
  3. Apr 7, 2016 #2

    NascentOxygen

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    Staff: Mentor

    There should be only one "-" sign in your expression for Z.

    Did you involve the value of C in your calculation for XC?
     
  4. Apr 7, 2016 #3
    That was a typo on my part. I used the correct equation when I worked it out.
     
  5. Apr 7, 2016 #4

    NascentOxygen

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    Staff: Mentor

    You have the decimal incorrectly placed in your answer for XC.
     
  6. Apr 7, 2016 #5
    I converted nF to F wrong.... Thanks you!
     
  7. Apr 7, 2016 #6
    Now I'm working on a later part that involves finding impedance at 5000Hz
    Xc=1/(2pi*5000z) =46.13
    and XL=2pi*5000*L=72.26

    so Z=54.013 but this is somehow wrong even though the exact same process worked for 3000 and 4000 Hz
     
  8. Apr 8, 2016 #7

    gneill

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    Staff: Mentor

    Your individual reactance values look fine. Must be a calculator/finger interface issue :smile:

    Can you try again? If it still doesn't look right, give us a breakdown of the calculation step by step.
     
  9. Apr 8, 2016 #8
    XL=2pi(5000)(0.0023)=72.2566
    XC=1/(2pi5000*(6.9*10^-7))=46.1318

    Z=sqrt[(60^2)-(72.2566-46.1318)^2]=54.0138
     
  10. Apr 8, 2016 #9

    gneill

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    Staff: Mentor

    Ah. Why have you used a minus sign between the two terms within the square root?
     
  11. Apr 8, 2016 #10
    I keep doing that. I don't know why I keep doing that
     
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