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In the middle of an AC analysis problem, I am stuck - please help!

  1. Apr 18, 2007 #1
    OK, I have this complex number equation:

    [tex]5\,V\,=\,\left[\left(j2\,+\,1\right)\,\left(1000\,-\,j10000\right)\,+\,\left(200\,+\,j2000\right)\right]\,i_L[/tex]

    Now I try to solve for [itex]i_L[/itex]:

    [tex]i_L\,=\,\frac{5\,V}{\left(j2\,+\,1\right)\,\left(1000\,-\,j10000\right)\,+\,\left(200\,+\,j2000\right)}[/tex]

    [tex]i_L\,=\,\frac{5\,V}{-20000j^2\,-\,6000j\,+\,1200}[/tex]

    Since [itex]j^2[/itex] is just -1:

    [tex]i_L\,=\,\frac{5\,V}{21200\,-\,6000j}[/tex]

    And since [itex]\frac{1}{j}[/itex] = -j, the final complex numbered answer I get is:

    [tex]0.0002358\,+\,0.00083333j[/tex]

    However, this is incorrect! I have the answer for the problem, step-by-step, given by the prof. and I have double checked the answer using the Symbulator for the TI-89.

    I should be getting:

    [tex]0.00021836\,+\,0.0000618j[/tex]

    What am I doing wrong?:frown:
     
  2. jcsd
  3. Apr 18, 2007 #2

    cristo

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    What've you done to get from (1) to (2)? I would first divide top and bottom by 5 to get
    [tex]i_L\,=\,\frac{V}{4240-1200j}[/tex]

    Now, you need to get this into the form V(x+jy), so I would multiply top and bottom by the complex conjugate of the denominator

    [tex]i_L\,=\,\frac{V}{4240-1200j}\cdot\frac{4240+1200j}{4240+1200j}[/tex]

    Now expand out the denominator, and you should obtain a purely real number, and the result will follow from dividing the real and imaginary parts of the numerator by the denominator.
    That's what I get too!
     
    Last edited: Apr 18, 2007
  4. Apr 18, 2007 #3
    Wow, thanks!

    Why does the way I was trying not work?

    [tex]\frac{1}{j}\,=\,-j[/tex] <----- Right?

    [tex]\frac{5}{21200}\,=\,0.0002358[/tex]

    and for the complex part:

    [tex]\frac{5}{-6000}\m\left(-j\right)\,=\,0.000833j[/tex]

    That is where I get the answer I was getting (wrong):

    [tex]0.0002358\,+\,0.000833j[/tex]
     
    Last edited: Apr 18, 2007
  5. Apr 18, 2007 #4

    cristo

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    Because you made the error of assuming that [tex]\frac{a}{b+cj} = \frac{a}{b}+\frac{a}{cj}[/tex]. This is not true (in general) and you cannot split up a fraction like this.

    The trick for solving such a question is to always multiply the fraction by the conjugate of the denominator-- this makes the denominator real, and so the fraction becomes one of the form [tex]\frac{a+bj}{c}[/tex] which of course can be simplified to obtain [tex] \frac{a}{c}+\frac{b}{c}j[/tex]
     
  6. Apr 18, 2007 #5
    Sweet, thanks again! Now I understand.
     
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