Reactance

Discussion in 'Homework Help' started by Robert.Adams, Mar 31, 2010.

  1. Robert.Adams

    Thread Starter Active Member

    Feb 16, 2010
    112
    5
    I have an exemption test for Circuits in 2 days and I've been doing practice problems.

    I came across this and am very confused how the reactance switches signs in the simplify step.

    Can someone explain this to me or is it a typo on that site (Like I hope it is.)?

    Thanks.
     
  2. hondabones

    Active Member

    Sep 29, 2009
    123
    1
    Did you mean the impedance?

    2k - 250 = +1750
     
  3. Robert.Adams

    Thread Starter Active Member

    Feb 16, 2010
    112
    5
    No, I mean just the reactance part of the impedance in the step that converts to milli instead of kilo. It becomes 1/3m + 1m - j0.57m

    The signs on the DC resistance stayed positive and the reactance due to the inductor/cap flipped (It does come out as the right magnitude).

    EDIT: Oh Snap! I got it...I think. It is (1 < 0)/(1750 < 90) which makes it become -90.
     
  4. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    I'm guessing your question is how does one get .....

    \frac{1}{j1750}=-j0.571m

    Are we agreed that 'm' just stands for 10^{-3}

    So

    \frac{1}{1750}=0.571 X 10^{-3}=0.571m

    Also remember that for the complex operator 'j'

    \frac{1}{j}=-j

    combining these two concepts

    \frac{1}{j1750}=-j0.571m

    Or you do it like you did in your EDIT .....
     
    Last edited: Apr 1, 2010
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