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AC Steady State Analysis

  1. Sep 28, 2014 #1
    1. The problem statement, all variables and given/known data

    Steif.ch06.p41_1.jpg
    Find the steady state expressions in terms of Acos(wt+theta) for nodes v1 and v2.

    2. Relevant equations

    Ohm's law
    KCL
    Node Analysis

    3. The attempt at a solution
    Convert to phasors and find impedances-> R1= 50 Ohms R2= 30 Ohms
    C1= -j2500=1/j*4e-4 Ohms C2= -j2000=1/j5e-4 Ohms
    L1= j5 Ohms L2= j10 Ohms

    I'm not getting an acceptable answer when I use Wolfram to solve the systems produced. I believe the error is that I need to combine the impedances of the capacitor and the inductor coming off of node 2, but I'm not sure since there is another node there. It's quite possible there are multiple errors on my part.
     

    Attached Files:

    Last edited: Sep 28, 2014
  2. jcsd
  3. Sep 28, 2014 #2

    NascentOxygen

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    Staff: Mentor

    Hi butleRonius. http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif [Broken]

    There is no restriction on what impedance can comprise each branch; it doesn't have to be a single element. In your eqn (2) you need to replace 1/(j3E4) by 1/(z) where z is the series impedance of the capacitor and inductor.

    So there's a little sub-exercise for you, determining that series impedance before using it in eqn (2).
     
    Last edited by a moderator: May 7, 2017
  4. Sep 28, 2014 #3
    so -j2000 + j10 = -j1990

    Sub that in: v2/-j1990

    Solve using nodal?

    I actually tried combining as (Zc+ZL) || Zr but that turned into a nightmare

    Thanks.
     
  5. Sep 28, 2014 #4
    Solved. Thank you.
     
  6. Sep 28, 2014 #5

    NascentOxygen

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    Staff: Mentor

    Sure, go ahead.
    A little bit of work involved, yes, but it simplifies the circuit to one with no nodes so the subsequent analysis is shorter. Swings and roundabouts ....
     
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