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Expressing equivalent impedance

  1. May 16, 2013 #1
    1. The problem statement, all variables and given/known data


    Determine the equivalent impedance of the following network:


    [Broken]

    (My bad, the element at the far right where the j2ohm is was supposed to be an inductor, fixed now. The numbers are correct too.)

    2. Relevant equations

    R(series) = R1 + R2?

    R(parallel) = [ (1/R1) + (1/R2) ]-1

    except with impedance now

    3. The attempt at a solution

    Not sure if I can do algebra with the imaginary components...

    but the right branch would simplify to

    (4Ω - j4Ω/3)

    Then this is in parallel with j4Ω, simplifying to (12Ω + j4Ω + j12Ω)/(j4Ω(12Ω - j4Ω)).


    All I want to know, is if this is really the right track? Can I really algebraically manipulate the imaginary components like this (maybe I Just need to review algebra now)? Any other tips would be helpful, thanks.
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. May 16, 2013 #2

    Simon Bridge

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    ... just treat then as you would a variable - except that you have extra simplifications: j.j=-1 and 1/j = -j.

    At the end you want to express it as z=a+jb - which may mean rationalizing the denominator.

    But you are basically on the right track.
     
  4. May 16, 2013 #3

    SammyS

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    Yes, this should work fine.

    Is that element at the far right a resistor or an inductor ?

    Added in Edit:

    I see that you have now revised the image to show it as an inductor.
     

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    Last edited: May 16, 2013
  5. May 16, 2013 #4
    Thank you.

    Although, what happens when you end up having to multiply (Ω * jΩ)? Or you shouldn't have to do this? Because acting on what you said, I ended up getting (continued from the work in my last post):

    (4-j3Ω)/(12Ω-j4Ω) all to the -1 power

    then 12Ω - j4Ω/4 - j3Ω


    and rationalizing the denominator I get

    (12Ω - j4Ω)(4 - j3Ω) which is where I ask the above question.


    Zeq would then be

    [ (12Ω - j4Ω)((4 - j3Ω) ] + 2Ω
     
  6. May 16, 2013 #5

    Simon Bridge

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    ##\omega \cdot j\omega = j\omega^2## ... pretty much as you'd expect.

    ... but I think you need to distinguish between the frequency symbol ##\omega## and the symbol for the unit ##\Omega## - or it will get confusing.

    i.e. A capacitor with impedance "##^-j4\Omega##" - makes sense if the cap-omega on the diagram is intended to be the unit - but your last post gives me pause to wonder if it's a mistake - like the inductor on the far right given a resistor symbol.

    You can use w or ω for frequency if you don't want to use LaTeX.
     
  7. May 16, 2013 #6
    No, not a mistake, the "-j4Ω" is correctly there. The only incorrect thing about the problem statement was the resistor in place of an inductor Sammy pointed out there.


    This might sound dumb but can I ask, where you are bringing in frequency / the lower-case omega in to this from though? I meant this was going to mean getting "Ohms squared" in the equation, unless I put down the complex numbers / units wrong somehow for simplifying the impedances.


    And doing what you say again and actually continuing trying to simplify & foiling from my last post would give:


    Zeq = (48Ω -j36Ω2 -j16Ω + 12Ω2) + 2Ω

    so

    Zeq = 50Ω -j36Ω2 -j16Ω + 12Ω2


    Seems all algebra now, but is there any way to ideally simplify this more?
     
  8. May 16, 2013 #7

    gneill

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    There should be no squared Ohms in the results; everything should shake out to Ohms. Perhaps something has gone awry during your simplification steps (perhaps the rationalization operations?).

    Frequency doesn't enter into the problem as given since all of the impedances are specified as constants.
     
  9. May 16, 2013 #8

    Simon Bridge

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    For a reactive component, the impedance depends on the frequency. Inductive impedance is directly proportional to the frequency and capacitative impedance is inversely proportional to frequency. See the contradiction?

    From your last reply - the ##\Omega## are units and should not appear as a variable in your equations.
    Please spell out how you are using the ##\Omega## symbol when you next reply.
     
  10. May 18, 2013 #9
    Okay, I made some mistakes in the math before here so I'm going to start over again here:

    The right-most elements are in parallel so

    [ -1/4Ωj + 1/2Ωj ]-1 = ( 1/4Ωj )2 = 4Ωj
    , which would then be in series with 4Ω so

    combined would be 4Ω + 4Ωj so at this point would look like....:


    [Broken]

    (and I'll leave the "resistor" there this time, just for reference)


    So trying to combine the right two parallels again:

    = [ 1/(4Ω+4Ωj) + 1/4Ωj ]-1

    = [ (4Ωj+4Ωj+4Ω)/4Ωj*(4Ω+4Ωj) ]-1

    = [ (8Ωj+4Ω)/4Ωj*(4Ω+4Ωj) ]-1

    = 4Ωj*(4Ω+4Ωj)/(8Ωj+4Ω)



    But how do I simplify this now?


    I would want to factor 4Ωj from the top, so I think that would be

    4Ωj(2 - j)/4Ωj(4Ω +4Ωj)

    but not sure.


    The final answer not simplified from that point would be (or would have been)

    =[ 4Ωj*(4Ω+4Ωj)/(8Ωj+4Ω) ] + 2Ω
     
    Last edited by a moderator: May 6, 2017
  11. May 18, 2013 #10

    gneill

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    The idea is to clear the imaginary term from the denominator, yielding a canonical "a + jb" form of result. Look up: rationalization of complex numbers.
     
  12. May 18, 2013 #11
    Ugh, so you multiply by the OPPOSITE conjugate.... thanks! That would mean simplify:




    [ 4Ωj*(4Ω+4Ωj)/(8Ωj+4Ω) ] * (4Ω - 8Ωj)/(4Ω - 8Ωj)





    but would that mean I get Ω2 again if I want to really foil say ie the denominator

    (8Ωj + 4Ω)(-8Ωj + 4Ω)

    when I multiply the 4Ω by 4Ω?


    Or was I just supposed to forget about even mentioning the "Ω" here and just worry about the "j"?
     
  13. May 18, 2013 #12

    gneill

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    If you carry out the multiplications and divisions diligently, the units will reduce to simple Ω's.

    Consider the units of the factors involved:

    $$\frac{Ω \cdot Ω}{Ω} \cdot \frac{Ω}{Ω} = \frac{Ω^2}{Ω} \cdot \frac{Ω}{Ω}$$

    How does that reduce?

    You never ignore units when you multiply or divide. The units are associated with the values and the mathematical operations apply to all parts of the expressions.
     
  14. May 19, 2013 #13
    I see now, and I almost got it, I think. So if you get stuff like j3 it "cancels" down to 1 power of j because j*j = -1?

    If that happens you have to switch the signs too for each coefficient, right?



    After simplifying the two parallel resistors above with your help I got:



    (-64Ω3/80Ω2) + (192Ω3/80Ω2)j

    =(-4/5)Ω + (12/5)Ωj


    So by adding the series 2Ω in, my final answer would just be


    Zeq = (6/5)Ω +(12/5)Ωj



    But still not sure if it's right.
     
  15. May 19, 2013 #14

    gneill

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    The imaginary term looks okay, but the real term does not.
     
  16. May 19, 2013 #15
    Ok, I looked over it again and got:


    (+64Ω3/80Ω2) + (192Ω3/80Ω2)j

    =(+4/5)Ω + (12/5)Ωj


    So adding 2Ω again:


    Zeq = (14/5)Ω +(12/5)Ωj
     
  17. May 20, 2013 #16

    gneill

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    Yup, that looks good.
     
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