Series R, L, and C

Discussion in 'Homework Help' started by DRKM, Apr 21, 2010.

  1. DRKM

    Thread Starter New Member

    Apr 21, 2010
    8
    0
    Hi,

    I am currently working on a project to show how RLC series circuits can be solved using complex numbers. I have worked through http://www.allaboutcircuits.com/vol_2/chpt_5/2.html# and understand the process but when calculating for my figures I cannot find where the Ztotal value comes from. I get the adding of Zr+Zl+Zc, but it is the value of 1.5437kΩ /_ -80.680°

    If anyone can help me out with how this value comes about that would be great.
     
  2. c_omalley2002

    Member

    Mar 18, 2010
    31
    3
    I would suggest going back here: http://www.allaboutcircuits.com/vol_2/chpt_2/5.html

    They are simply adding the values in rectangular form, because it's easier that way. Then converting back to polar to express the answer. you take the square root of (250^2 + -1.5233k^2) which equals 1.5437k that's your polar magnitude. Then take the arctangent(-1.5233k/250) to get the angle. Which equals -80.680°
     
    DRKM likes this.
  3. DRKM

    Thread Starter New Member

    Apr 21, 2010
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    0
    Thanks for the response that's answered my question.
     
  4. DRKM

    Thread Starter New Member

    Apr 21, 2010
    8
    0
    Well, my original question was answered but I have another question regarding applying ohms law to the remaining portion of http://www.allaboutcircuits.com/vol_2/chpt_5/2.html# where ohms law has to be applied. I am applying various methods to try and find how the Itotal has been calulated from Etotal and Ztotal in complex form. Using ohms law I=E/Z.

    Am I on the right lines in thinking I need to multiply the complex numbers together or is there another format using other numbers.

    All help is greatly appreciated
     
  5. c_omalley2002

    Member

    Mar 18, 2010
    31
    3
    sorry, I forgot to subscribe to this thread, and I haven't been on here in a while. If you reference the online AAC Book here: http://www.allaboutcircuits.com/vol_2/chpt_2/6.html
    you will see how to properly divide complex numbers.


    Quote:
    "Division of polar-form complex numbers is also easy: simply divide the polar magnitude of the first complex number by the polar magnitude of the second complex number to arrive at the polar magnitude of the quotient, and subtract the angle of the second complex number from the angle of the first complex number to arrive at the angle of the quotient"

    So you would take 120 @0° / 1.5437kΩ @-80.680°.
    Divide 120 / 1.5437k which give you 77.34mA
    subtract 0° - -80.680°= 80.680°

    hope that helps
     
    Last edited: May 6, 2010
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