Nodal analysis - w/ dependent current source

Discussion in 'Homework Help' started by castrogfx, Apr 13, 2012.

  1. castrogfx

    Thread Starter New Member

    Apr 13, 2012
    4
    0
    Hi,
    I think this problem is easy, but I can't do it.
    [​IMG]

    answer: 2 A

    ---
    Node A is on the left, node B is on the right.
    [​IMG]
    So,
    A)  i_2 + i_4 = i_1 + i_3
    B)  i_3 + i_7 = i_6 + i_5

    i_1=\frac{v_1}{5}, i_2=-3A, i_3=\frac{v_1 - v_2}{20}, i_4 = i_5 = 3 i_1 = \frac{3 v_1}{5}, i_6=\frac{v_2}{10}, i_7= 10 A

    A)  -3 + \frac{3 v_1}{5} = \frac{v_1}{5} + \frac{v_1 - v_2}{20}

    B) \frac{v_1 - v_2}{20} + 10 = \frac{v_2}{10} + \frac{3 v_1}{5}


    I have this system of linear equations:

    -60 = -7v_1 - v_2
    200 = 13v_1 + 3v_2

    v_1 = -2.5 V <br />
v_2 = 77.5 V


    i_1 =\frac{-2.5}{5} = -.5 A

    Thanks for the help!
     
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    The dependent source seems to be a CCVS despite having a symbol typical of a CCCS. However the answer of 2A is correct if it is a CCCS. I note you have I7 as 10A. Do you now see the error?
     
    Last edited: Apr 13, 2012
    castrogfx likes this.
  3. castrogfx

    Thread Starter New Member

    Apr 13, 2012
    4
    0
    I've tried this so many times that I copied wrong in a moment.
    Unbelievable.
    Thanks t_n_k for the help, I really appreciated it.
     
  4. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,957
    1,097
    castrogfx likes this.
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