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Parallel LR impedance

  1. May 4, 2017 #1
    1. The problem statement, all variables and given/known data

    Find the complex parallel impedance of a 20 Ω resistor and 110 mH inductor from a 50 Hz supply.

    2. Relevant equations

    Z = R (XL2 / R2+XL2) + j XL (R2 / XL2 + R2)

    3. The attempt at a solution

    R = 20 Ω
    XL = j34.6 Ω

    Z = 20 ( j34.62 / j34.62 + 202) + j34.6 ( 202 / j34.62 + 202) Ω

    Z = 20 ( -1197 / 400 - 1197 ) + j34.6 ( 400 / 400 - 1197 ) Ω

    Z = 30 - j17 Ω

    Apparently this is wrong but I can't find where I have gone wrong.

    Thanks
     
  2. jcsd
  3. May 4, 2017 #2

    BvU

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    Hi,

    Not clear where your relevant formula comes from.
    Why not simply $${1\over Z} = {1\over R} + {1\over j\omega L}\ \ ?$$
     
  4. May 4, 2017 #3

    BvU

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    Correction: I confirm your equation, so it's in a later step. Check out what XL should be.
     
  5. May 4, 2017 #4
    I calculated XL at 50 Hz and got j34.6 Ω

    Then I tried to calculate XL2 as follows:

    XL2 = (j34.6)2 = -1197

    Are you suggesting -1197 is incorrect?

    Thanks for your response.
     
  6. May 4, 2017 #5

    BvU

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    I am indeed. The complex impedance is ##{\bf j}\omega L## but the reactance XL is a real number.
     
  7. May 4, 2017 #6
    I've been sat trying to interpret what you said and getting nowhere unfortunately. 34.6 is a real number? 34.62 = 1197
     
  8. May 5, 2017 #7

    BvU

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    So XL2 is 1197, not -1197. Indeed.

    And: be clearer with the brackets when you write down an expression.
     
  9. May 5, 2017 #8
    Right I understand what you are saying and thanks for the help.
     
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