AC Analysis Problem

Discussion in 'Homework Help' started by Digit0001, May 10, 2011.

  1. Digit0001

    Thread Starter Member

    Mar 28, 2010
    89
    0
    Hi
    Can someone tell me what i have done wrong in the following question:

    View attachment tut6_qns.doc

    Equation 1

    Vo=10V

    Equation 2

    Z = j4*j6/4j+j6 = -24/j10

    Vo= (-j2.4)/(j4+6-j2.4) * 12 = -j48/(j1.6+6) = 11.95+j7.4689

    Equation 3

    2H -> j4
    1/12F -> j6

    Z = (6*j6)/(6+j6) = -6j+6

    Vo = (j6)/(j4+6)*4 = (j24)/(4j+6) * (j4-6)/(j4-6) = (-96-144j)/(-16-36)
    = 1.816-j2.769


    P.S
     
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    The capacitive impedance in the case of equation 2 is -j4Ω not +j4Ω.

    So the parallel branch impedance is -j4*j6/(j6-j4)=-j12Ω

    Remember Zcap=1/(j*ω*C)=-j/(ω*C)
     
  3. Digit0001

    Thread Starter Member

    Mar 28, 2010
    89
    0
    for equation 2 i get

    12j/(12j+6) * 12 = 1.60.8j
    in phasor = 1.788/26.6 degrees
     
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