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Ac analysis with dependent source

  1. Mar 15, 2009 #1
    1. The problem statement, all variables and given/known data
    Determine I1

    independent AC voltage source 10 cos 10^5t
    inductor 60mH
    capacitor 5nF
    dependent voltage controlled voltage source 3Vx



    2. Relevant equations
    KVL & KCL


    3. The attempt at a solution

    [tex]
    \begin{array}{l}
    - 7.071 + 6000I_1 - j6000I_3 = 0 \\
    I_3 = I_1 - I_2 \\
    - 7.071 + 6000I_1 - j6000(I_1 - I_2 ) = 0 \\
    j6000(I_1 - I_2 ) + 3v_x - j2000 = 0 \\
    v_x = 6000I_1 \\
    - 7.071 + 6000I_1 - j6000(I_1 - I_2 ) = 0 \\
    j6000(I_1 - I_2 ) + 3(6000I_1 ) - j2000 = 0 \\
    24000I_1 = 7.071 + j2000 \\
    I_1 = (2.94625 \times 10^{ - 4} ) + j0.083333 \\
    I_1 = 0.0833\angle 89.79^o \\
    \end{array}
    [/tex]

    The answer for I1 is given as 0.33mA @ 171.9 degrees

    Could someone help me here please, gone over this for a couple hours still stumped.
    Cheers
     

    Attached Files:

  2. jcsd
  3. Mar 15, 2009 #2
    Quickly going through your attempt, all I can spot is that your complex value for the independent supply is incorrect. cos(x)=sin(x+pi/2)...

    One sure-fire way to get it is to use mesh method.
     
  4. Mar 15, 2009 #3
    If I convert to the frequency domain and have my voltage as my zero phase reference it shouldn't matter should it? ie voltage 7.071 V @ 0degrees

    I thought what I tried was mesh loop analysis.
     
  5. Mar 16, 2009 #4
    Yes, of course you can calculate it as cos as the reference. But in that case you need to start playing with the signs a bit more due to the definition of inductor and capacitor.

    And by mesh method I meant using the matrix form. You only get 2x2 matrix, which is easy to deal with.

    Edit: Going more in-depth to your calculations, looks like your sign convention differs from mine. Can you explain the method how you got rows 1 and 4? Especially in row 4 why do you only use the impedance of the capcitor or is it a typo? And is the voltage source given as 10cos(10^5t), because in the picture the frequency is 15.9kHz, which would give w=31831pi?
     
    Last edited: Mar 16, 2009
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