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Impedance network

  1. Oct 29, 2006 #1
    Hi all, I'm stuck on this part of a problem. I have gotten the net impedance between two terminals A and B. The impedance I got was

    [tex] Z=8.39+2.22i [/tex]

    Now the question is...

    The network is equivalent a to a resistor and an energy storage element connected in parallel. Find their values.

    I thought this was going to be easy then I realized that it was parallel not in series. So then I solved the general equations for the impedance of a resistor and a capacitor and a resitor and inductor in parallel. This is what I got:

    Note: w is the angular frequency (the given frequency is 60Hz)
    R//C
    [tex]\frac{R-R^2wci}{R^2w^2c^2+1}[/tex]

    R//L
    [tex]\frac{RwL(wL+Ri)}{R^2+w^2L^2}[/tex]

    Then I figured that i could equate the Real and Imaginary parts of the impedance together and then use the magintude of the impedance to solve for R or the energy storage element. I can't seem to get that to work out. Is this the right track?
     
    Last edited: Oct 29, 2006
  2. jcsd
  3. Oct 29, 2006 #2
    This is certainly the right track. Let me see what I can do...
     
  4. Oct 29, 2006 #3
    I think it's the right track, you just need to get the [itex] j [/itex] part separated.

    [tex] Z_C = \frac{1}{j \omega C} [/tex]
    [tex] Z_L = j \omega L [/tex]

    [tex] R || Z_C = \frac{R \left(\frac{1}{j \omega C}\right) }{R+\frac{1}{j \omega C}} = \frac{R}{R j \omega C + 1} [/tex]

    Now if you multiply the top and bottom by [itex] Rj \omega C - 1 [/itex]

    [tex] \frac{R}{R j \omega C + 1 } \frac{(Rj \omega C -1)}{(Rj \omega C -1)} = \frac{R}{C^2R^2 \omega^2 + 1}- \frac{CR^2\omega}{C^2R^2 \omega^2 + 1}j[/tex]

    Now obviously this expression will not allow you to get, [itex] Z=8.39+2.22i [/itex].

    So if you do the same for the inductor you should be good. I got,
    [tex] R \approx 9 \Omega [/tex]
    [tex] L \approx 90 mH [/tex]
     
  5. Oct 29, 2006 #4
    I think you are suppose to multipy the top and bottom by the complex conjugate which is [itex] -Rj \omega C + 1 [/itex] rather than [itex] Rj \omega C - 1 [/itex]
     
    Last edited: Oct 29, 2006
  6. Oct 29, 2006 #5
    I'm pretty sure it will come out to the same thing. Using the complex conjugate will save some time factoring out the negatives though.

    EDIT:
    Since,
    [tex] R j \omega C - 1 [/tex]
    [tex] -R j \omega C + 1 [/tex]

    [tex] R j \omega C -1 = -1(-R j \omega C + 1) [/tex]
    So you are still multiplying by the complex conjugate.
     
    Last edited: Oct 29, 2006
  7. Oct 29, 2006 #6
    thanks guys Ill keep working on, I figured the cap one looked like it wouldn't work.
     
  8. Oct 29, 2006 #7
    I got the 9 ohms and the 90mH as well. I ended up with a couple of ugly quadratic equations and then used my TI-89 to solve them. Thanks for all the help guys!
     
  9. Oct 29, 2006 #8
    used my 89 to solve them also ;)
     
  10. Nov 17, 2006 #9
    You can do this finding admittance first as Y=1/Z=1/(8.39+j2.22) = 0.1114-j0.0295.
    Next observe that, if you have R and L or C in parallel, then the admiitance has real part as 1/R and imaginary part as -j/Lw, that is
    Y=(1/R) -j(1/Lw) with w=120pi.
    Therefore, you get 1/R= 0.1114 giving R=8.97 and 1/Lw=.0295 thus L=90mH. There is no need to solve quadratic equations.

    xxxx
     
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