Solve for Z

Discussion in 'Homework Help' started by davidand, Jun 24, 2005.

  1. davidand

    Thread Starter Active Member

    Jun 2, 2005
    43
    0
    I'm trying to solve this for Zt and It
    do I add R1 and R2 for Rt

    do I subtract Xc-XL =x
    then add Rt squared + X squared
    then take sqt root of result to equal Z

    Z = 5.4 ohms

    E / Z = I = 5.57A
     
  2. kinyo

    Member

    Jun 6, 2005
    13
    0
    no, doing so would mean R2 and Xc are in series which they are not
     
  3. davidand

    Thread Starter Active Member

    Jun 2, 2005
    43
    0
    Should I solve for R1 and XL then R2 and Xc
     
  4. celect

    Member

    May 31, 2005
    18
    0

    I believe Z t = 3ohms + 6ohms + (2ohms) (8ohms)/2ohms+J8
     
  5. Semyazza

    Member

    Jun 25, 2005
    12
    0
    3@0 + 6@90 + [(2@0 * 8@-90) / (2@0 + 8@-90)] = Ztotal = 7.376@48.56

    7.376@48.56 / 30@0 = Itotal = 245.88mA @48.556



    OR............

    [(2 * -j8)/ (2 + -j8)] + j6 + 3 = Ztotal

    Ztotal / 30 = Itotal
     
  6. Semyazza

    Member

    Jun 25, 2005
    12
    0
    To correct myself......

    3@0 + 6@90 + [(2@0 * 8@-90) / (2@0 + 8@-90)] = Ztotal = 7.376@48.56

    30@0 / 7.376@48.56 = Itotal = 4.07 @ -48.56
    OR............

    [(2 * -j8)/ (2 + -j8)] + j6 + 3 = Ztotal

    30 / Ztotal = Itotal
     
  7. celect

    Member

    May 31, 2005
    18
    0
    What chapter of circuit analysis should I reference to learn how to solve this
     
  8. davidand

    Thread Starter Active Member

    Jun 2, 2005
    43
    0
    here is a problem I'm working on now with the answer.

    I dont understand how you got Z t = 7.376<48.56
     
  9. Semyazza

    Member

    Jun 25, 2005
    12
    0

    Celect: I just looked at the book and I would read Volume II Chapters 2,3,4,5 you can skip chapter 2 if you already are familiar with complex numbers (it dosn't hurt to review though). Chapter 5 puts everything all together so it can be skipped but its nice to know (and important as far as AC circuits go).

    Davidand:
    A resistor is always at 0˚ an inductor at 90˚(or j) and a capacitor at -90˚(or -j). R2 and Xc are in parallel because two different currents pass through them (Components are in series if the same current passes through them otherwise they are in parallel). R1 and XL are in series with the parallel combination so they are added.

    (R2 * Xc) / (R2 + Xc) + R1 + XL = Ztotal


    If you have any other questions just ask.
     
  10. davidand

    Thread Starter Active Member

    Jun 2, 2005
    43
    0
    If I plug my numbers in above equation I get 10.6 not 7.376

    (2*8) / (2+8) + 3 + 6 =
     
  11. Semyazza

    Member

    Jun 25, 2005
    12
    0
    When dealing with AC circuits, inductors, and capacitors in terms of impedance you are adding vectors. When adding vectors you can not simply add, multiply, and subtract the way you are doing, you must add, multiply, subtract vectors. http://en.wikipedia.org/wiki/Phasor_%28electronics%29 is a good starting place.
     
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