AC Circuit SuperPosition

Discussion in 'Homework Help' started by Digit0001, Jun 6, 2011.

  1. Digit0001

    Thread Starter Member

    Mar 28, 2010
    89
    0
    Hi
    DSC00068.JPG
    I am having trouble getting the answer for the second part which used the current source. I have used nodal analysis like the one below, but still my calculated answer is incorrect

    where
    -j2 => 0.2 F
    10j => 10j

    Vo/8 + V0/-j2 + V0/10j = 4

    P.S
     
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    It should be

    Vo/8+j2xVo+Vo/10j=4
     
  3. Digit0001

    Thread Starter Member

    Mar 28, 2010
    89
    0
    if j2Vo then Vo/10j = -10jVo?
     
  4. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    What you are forgetting is that

    Z_inductor=jωL

    Z_capacitor=1/(jωC)

    For C=0.2F at ω=10

    Z_C=1/(j2)=-j0.5

    For L=1H at ω=10

    Z_L=j10

    Therefore ...

    1/(Z_C)=1/(-0.5j)=j2

    1/(Z_L)=1/(10j)=-0.1j
     
  5. Digit0001

    Thread Starter Member

    Mar 28, 2010
    89
    0
    oh ok i see
     
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