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Phasor circuit analysis

  1. Apr 8, 2015 #1
    1. The problem statement, all variables and given/known data
    Selection_028.png
    (Note, my answer below is also wrong. I've attempted this problem for about 1 hour now and I can't figure it out.)

    2. Relevant equations
    KCL (Kirchoffs Current Law)
    Z for the capacitor, Z = 1/(jwC)



    3. The attempt at a solution

    I'm using KCL at V1 with the following convention: +ve for current flowing out of the node, -ve for current flowing in

    If the bottom of our circuit is the reference:
    V2 = Vs

    KCL at Node 1
    -i1(t) + i2(t) + V1/23 - (Vs-V1/(-40j) = 0

    Now I convert all the current and voltage functions to phasors.
    i1(t) = 0.2(60d)
    i2(t) = 0.1(-90d)
    Vs(t) = 10(-180d)

    Subbing in to above...

    -0.2(60d) + 0.1(-90d) + V1/23 - (10(-180d))/(-40j) + V1/(-40j) = 0

    V1( 1/23 + 1/(-40j) ) = 0.2(60d) - 0.1(-90d) + 10(-180d) / (-40j)

    V1 = (0.1 + 0.1732j + 0.1j - 10/(40j)) / (1/23 + 1/(-40j))

    V1 = 6.93 + 8.05j

    Converting this to phasor form,

    V=10.622(40.3d)

    But this is also wrong. I can't figure out what I'm doing wrong.
     
    Last edited: Apr 8, 2015
  2. jcsd
  3. Apr 8, 2015 #2

    berkeman

    User Avatar

    Staff: Mentor

    It looks like you are double-converting the signs. Do you really want to invert those phase angles? (It's been a long time since I used phasors, so maybe there is a reason).
     
  4. Apr 8, 2015 #3
    Right, I think you are right. cosx = sin(x+90)

    I'm going to try again with i2(t) = 0.1(0180d) = 0.1(0d) and Vs(t) = 10(90d)

    -i1(t) + i2(t) + V1/23 - (Vs-V1/(-40j) = 0
    -0.1(0d) + 0.2(60d) + V1/23 - 10(90d)/(-40j) + V1/(-40j) = 0

    V1(1/23 + 1/(-40j)) = 0.1(0d) - 0.2(60d) + 10(90d) / (-40j)

    V1 = -6.04 - 0.51 J

    Yep. It is correct. Thank you, I really need to be more careful when calculating the little things :)
     
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