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Complex Numbers added as impedances in parallel

  1. Feb 18, 2012 #1
    1. I have been asked to add 1/z1+1/z2+1/z3=Y. When z1=2+j2 and z2=1+j5 and z3=j6.


    3. I have basically treated them like a normal resistors in parallel equation using 1+j0 and dividing them individually and then adding the product to get Y=0.29-j0.55. Is this the right way to go about this ?
     
  2. jcsd
  3. Feb 18, 2012 #2
    The answer you arrived at looks correct but I don't follow your explaination of how you arrived there.

    Usually what you do in these problems with division by complex numbers is to multiply by 1.
    By that I mean, for term you do the following;

    [itex]\frac{1}{z} = \frac{1}{z} \frac{z^*}{z^*} = \frac{z^*}{|z|^2}[/itex]

    Then it becomes a pretty simple vector equation.
     
  4. Feb 18, 2012 #3
    Hi,
    thank you so much for your reply.
    Do you mean multiplying by 1 that you are multiplying by the conjugate of the value of z ? Sorry I don't quite understand your method ?
    I got to my answer by making 1/2+j2 into 1+j0/2+j2 then multiplying both values by the conjugate of 2+j2 which is 2-j2, this gave me a value of 0.25-j0.25, I repeated this for z2 and z3 then added those values to gain Y.
    Thanks again
    Kyle
     
  5. Feb 18, 2012 #4
    Yes, I believe we are going the same thing here, in your notation what I did was

    [itex]\frac{1}{a+j\ b} = \frac{1}{a+j\ b} \frac{a-j\ b}{a-j\ b} = \frac{a-j\ b}{(a+j \ b)(a- j\ b)} = \frac{a- j\ b}{a^2 + b^2}[/itex]

    This is all correct because [itex]\frac{a-j\ b}{a-j\ b} = 1[/itex].

    Then you have an equation of the form
    [itex](a + jb) + (c + jd) + (e +jf) = (a+c+e) + j(b+d+f) = Y[/itex]

    We were doing the same thing, I just didn't understand your notation
     
  6. Feb 18, 2012 #5
    Got it, thank you so much for your help :)
     
  7. Feb 18, 2012 #6
    No problem buddy :D
     
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