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Impedance of a circuit

  1. Aug 28, 2015 #1
    1. The problem statement, all variables and given/known data
    The impedance of the circuit shown is:

    q13-1.jpg
    A. 41.1 Ω

    B. 100 Ω

    C. 173 Ω

    D. 187 Ω

    E. 241 Ω


    2. Relevant equations
    Z= sqrt (R^2 + [Xinduct - Xcapac]^2)
    Wd = 1/sqrt(L * C)
    Xinduct = WdL
    Xcapac = 1/WdC


    3. The attempt at a solution

    I think my issues come with the 2nd to 4th equations.
    The answer I get is 100 ohms, which i get as wrong, I follow the below logic
    Wd = 1/sqrt (0.0002*0.5)
    Wd=100
    Xinduct = 0.02
    Xcapac = 0.02
    thus Z = R which is 100 ohms.

    Where have I gone wrong here?
     
  2. jcsd
  3. Aug 28, 2015 #2

    BvU

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    Hi Jimbo,

    You want to work with the complex impedance, because there is a phase involved.
    So ##Z_L = j\omega L ## and ##Z_C = {1\over j\omega C}##.

    Are you comfortable with that approach ?
     
  4. Aug 28, 2015 #3
    Is Jw here the same as the angular frequency wd?

    The textbook chapter that this refers to says nothing about complex impedance as opposed to regular impedance, so I am not sure I get what you mean?
     
  5. Aug 28, 2015 #4

    BvU

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    ##\omega## is the angular frequency allright, but not the LC resonance frequency ##\omega_r## (*)
    It is the driving frequency in the circuit under consideration, so in this case ##50 \times 2\pi = 100 \pi ## rad/s.


    (*) your wd, for which ##\ |Z_{LC}| = 0\ ## from $$\ j\omega L + {1\over j\omega C} = j\omega L\; \left ( 1 + {1\over j^2\omega^2 LC }\right ) = j\omega L\; \left ( 1 - {1\over \omega^2 LC }\right )\ = 0 $$ if ## \omega^2 LC = 1##, a frequency of ##100/2\pi## Hz.
     
  6. Aug 28, 2015 #5
    zrlc.gif
    I have a feeling this picture tells me all I need?

    w = 2pi *50
    Xc = 15.92
    XL = 157.08

    Thus Z = 173, thus C.

    Is that correct reasoning?
     
  7. Aug 28, 2015 #6

    BvU

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    Yes. But now I am afraid you have no idea where the ##X_L - X_C## comes from ...
     
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