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Impedance of practical Capacitor

  1. Dec 9, 2013 #1
    1. The problem statement, all variables and given/known data
    A practical capacitor can be modeled by an ideal capacitor in parallel with a resistor.Find the impedance of practical capacitor at the radian frequency ω=377rad/s.Known C1=0.1 x 10^-6F R1=1 *10^6.



    2. Relevant equations

    1/Z=1/R1 + jCω

    3. The attempt at a solutionI am first determining the Z, by adding the impedance of resistor and the capacitor in parallel.
    And I am getting Z1(377)=10^6/(1+j37.7)= 2.6516*10^4 ∠-1.519 However book solutions gives ∠-1.5443.I would like to get some help about where my mistake is,or how should I approach the question. I know that finding the angle we should do θ=arctan(y/x) where y is the imaginary part and x is the real part.Hope I was clear and did not complicate it.
     
  2. jcsd
  3. Dec 9, 2013 #2

    Simon Bridge

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    Please show your reasoning and working.
     
  4. Dec 9, 2013 #3
    R//C Z=R//1/jωC = R1/(1+jωC1R1) inputing numbers I get =10^6/(1+j37.7) multiplying by 1+j37.7's conjugate I got 703.09-26506.54j , tried to find the length got the 26515.86 , finding the angle theta=arctan(-26506.54/703.09) doesnt give me same as book result. I get -1.519 book got -1.5443
     
  5. Dec 9, 2013 #4
    Looking at my procedure everything looks fine.I am wondering if book has done any mistake,or there is something else I should do, which I have missed,and it is making me do the mistake.
     
  6. Dec 9, 2013 #5

    gneill

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    Your numbers all look fine except for the result of the arctan(). When I enter that argument and take the arctan I get the book's result.
     
  7. Dec 9, 2013 #6
    You're saying that arctan(26506.54/703.09) gives 1.554 , either my two calculators are wrong or I am doing something wrong.
     
  8. Dec 9, 2013 #7

    gneill

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    That's what I'm saying :smile:
     
  9. Dec 9, 2013 #8
    Okay then, what about this one 101.92+j90.38,what do you get as your ∠? The book answer is ∠0.723 and about this one I have no clue how did they get it
     
  10. Dec 9, 2013 #9

    gneill

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    The numbers are probably rounded somewhat. I get 0.725 radians as the angle.
     
  11. Dec 9, 2013 #10

    mfb

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    Multiple threads are confusing... it looks like the error was here:
    I have no idea how you get -1.519 in degree mode... well does not matter.
     
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