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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].


    Work so far:


    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]



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


    Then I plug into the first equation:


    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


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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]


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

  7. Oct 4, 2006 #6
    Yes, that's much better. And be careful when you formulate the KCLs next time.
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