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Voltage Division - AC cct.

  1. Feb 26, 2013 #1
    1. The problem statement, all variables and given/known data
    Hi,
    We have this cut. :

    Screen%20Shot%202013-02-26%20at%204.24.05%20PM.png

    and in the image below, I10 is calculated by using current division :

    Screen%20Shot%202013-02-26%20at%204.24.30%20PM.png

    3. The attempt at a solution

    I tried to find it using voltage division but the answer is wrong :
    Vx(of the upper node) = 100 * [tex]\frac{j5}{4+j5}[/tex] = 60.98 + j48.78 V

    => I10 = [tex]\frac{Vx}{10 - j5}[/tex] = 2.93 + 6.34 = 6.98 [itex]\ 65.2[/itex]

    which is wrong, WHY?

    thanks
     
  2. jcsd
  3. Feb 26, 2013 #2

    gneill

    User Avatar

    Staff: Mentor

    Your voltage division isn't taking into account the impedance of the 10Ω and -5j capacitor branch that's also connected at the Vx node. You might try applying nodal analysis to find Vx...
     
    Last edited: Feb 26, 2013
  4. Feb 26, 2013 #3
    oh
    thank you gneill
     
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