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Homework Help: EECS: Find i_0 in a circuit with 4 resistors and 1 indep. current source

  1. Oct 2, 2006 #1
    (a) Using nodal analysis, determine [itex]i_0[/itex] in the circuit.

    (b) Now use linearity to find [itex]i_0[/itex].


    ch4prob4.jpg


    Work so far:

    [tex]i_0\,=\,\frac{v_1\,-\,0}{6\,\ohm}\,=\,\frac{v_1}{6\,\ohm}[/tex]

    KCL@[itex]v_1[/itex]: [tex]\frac{v_1\,-\,0}{3\,\ohm}\,+\,\frac{v_1\,-\,0}{6\,\ohm}\,+\,\frac{v_1\,-\,v_2}{2\,\ohm}\,=\,0[/tex]

    KCL@[itex]v_2[/itex]: [tex]\frac{v_2\,-\,v_1}{2\,\ohm}\,+\,\frac{v_2\,-\,0}{4\,\ohm}\,+\,9A\,=\,0[/tex]

    [tex]v_1\,-\,\frac{1}{2}\,v_2\,=\,0[/tex]

    [tex]-\,\frac{1}{2}\,v_1\,+\,\frac{3}{4}\,v_2\,=\,-9[/tex]

    Whe I put these two equaitons into a 3 X 2 coefficient matrix, I get

    [tex]v_1\,=\,-\,9V[/tex]
    [tex]v_2\,=\,-\,18V[/tex]

    Then I plug into the first equation:

    [tex]i_0\,=\,\frac{v_1}{6\,\ohm}\,=\,\frac{(-\,9V)}{6\,\ohm}\,=\,-\,\frac{3}{2}A[/tex]

    Is this the correct value for [itex]i_0[/itex] or have I got the sign wrong?
     
  2. jcsd
  3. Oct 2, 2006 #2
    You are repeating the same mistake here as you did in the previous exercise.
     
  4. Oct 3, 2006 #3
    What is that, can you explain?
     
  5. Oct 3, 2006 #4

    berkeman

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    Staff: Mentor

    The sign on the 9A is wrong. The other two terms in that KCL are for currents *out* of the node. So since the 9A is into the node, it has to be -9A in that equation. Makes sense?
     
  6. Oct 3, 2006 #5
    OIC! There has to be a minus in there somewhere if all of the terms are on one side and it's equal to zero. Whoops!

    Using your correction:

    KCL @ [itex]v_2[/itex]: [tex]\frac{v_2\,-\,v_1}{2\,\ohm}\,+\,\frac{v_2\,-\,0}{4\,\ohm}\,-\,9A\,=\,0[/tex]

    [tex]-\,\frac{1}{2}\,v_1\,+\,\frac{3}{4}\,v_2\,=\,9[/tex]

    Which switches the signs on the [itex]v_1[/itex] and [itex]v_2[/itex]!

    [tex]i_0\,=\,\frac{v_1}{6\,\ohm}\,=\,\frac{(9V)}{6\,\ohm}\,=\,\frac{3}{2}A[/tex]
     
  7. Oct 4, 2006 #6
    Yes, that's much better. And be careful when you formulate the KCLs next time.
     
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