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Node method analysis with dependent current source annoying me

  1. Jun 19, 2006 #1
    I've been trying to solve the exercise 4 below,but with no avail.I've worked through the equations,which I think that are okay, but I'm not sure because the result doesn't match with the one that is provided.Could you help me and check if those equations are okay?Also,how do I work with dependent sources like this one(current dependent) in this circuit?I need any help possible urgently!
    Thanks in advance for the help!

    Attached Files:

  2. jcsd
  3. Jun 19, 2006 #2

    Sorry for the late post...Here are the equations for the nodes and for the dependent current source:

    Node 1 : (v1-20)/2 + v1/20 + + (v1-v2)/20

    Node 2 : (v2-v1)/5 + v2/10 + 8I0


    Could this be okay?Any help is appreciated!:rolleyes:
  4. Jun 21, 2006 #3


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    Staff: Mentor

    Boy, attachments sure are taking a long time to get approved lately in PF. Esmeco, can you post a link to the attachment? Like, post it someplace else and give us a link.
  5. Jun 22, 2006 #4
  6. Jun 22, 2006 #5
    When writing node equations I always do the following:

    [tex] \sum I_{in}= \sum I_{out} [/tex]

    and define a current as:
    [tex] \frac{V_{tail}-V_{head}}{R}=I [/tex]

    V_tail ----------- /\/\/ -------->--- V_head

    So V_tail would be the voltage on one side of the resistor (with respect to ground), and V_head would be the voltage on the other side (with respect to ground). The current is defined to travel from tail to head.

    This helps me keep the polarity straight.

    So for the first node ([itex] V_1 [/itex]) you have defined all the currents to be leaving, thus:

    We have ZERO current in, and THREE currents out:

    The node equation is:
    [tex] \frac{V_1-20}{2}+\frac{V_1}{20}+\frac{V_1-V_2}{5}=0[/tex]

    Notice that you have:
    (v1-20)/2 + v1/20 + + (v1-v2)/20

    Can you go from there?

    Also, on a side note (but related). After writing the node equations, I collect terms, and put them into a matrix. Then simply solve the matrix and you get your results. The resultant matrix for this question would be:
    [tex] \left(\begin{array}{ccc} \frac{1}{2}+\frac{1}{20}+\frac{1}{5}& \frac{-1}{5}& 0 \\ \frac{1}{5}&-\frac{-1}{5}-\frac{1}{10}&-8 \\ \frac{1}{5} & \frac{-1}{5} & -1 \end{array} \right) \left( \begin{array}{c} V_1 \\ V_2 \\ I_0 \end{array} \right) = \left( \begin{array}{c} \frac{20}{2} \\ 0 \\ 0 \end{array} \right) [/tex]
    Last edited: Jun 22, 2006
  7. Jun 22, 2006 #6
    Thanks for the reply!Actually,I mistyped the (v1-v2)/5 for (v1-v2/20).What I really have doubts in is in the node 2 equations...Is my equation of node 2 right?If not,could someone please correct me?I don't understand much about matrixes...
    Last edited: Jun 22, 2006
  8. Jun 22, 2006 #7


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    Staff: Mentor

    Be sure to make them equations, which requires an equal sign and two sides. Like this:

    Your corrected first equation and your second equation match the circuit. Now you need to solve the simultaneous equations to get the two voltages and the current. You can use matrices to help speed this up as FrogPad suggested, or you can just do it more manually with several steps of algebra. Show us how you go from these equations to the answers in the problem page.
  9. Jun 22, 2006 #8
    Well,I tried to substitute the values of v1 and v2 on the equations,so that the final result on the equation side should equal 0.If I substitute the values for v1 and v2 on the current equation the value matches the one on the answers page but if I substitute the values of v1 and v2 on the node 2 equation it doesn't equal 0,so i thought that should be something wrong with my node 2 equation...Also,shouldn't I0=(v1-v2)/5 since the current is flowing from left to right?
    Last edited: Jun 22, 2006
  10. Jun 22, 2006 #9
    Post your final nodal equations, and post the steps you are taking to solve them.
  11. Jun 23, 2006 #10
    eq. 1:v1=(200 + 4v2)/15 and eq. 2: 14v1-13v2=0
    In equation 2 those were the values that turned out after I've changed the I0 equation from (v2-v1)/5 to (v1-v2)/5 .And it works,since the results are the same on the answers page...but it still doesn't equal 0 when I substitute the values on the equation.Also,I'm wondering now,why one doesn't put the v2/2 on eq. 2?Thanks for the all the help!:)
  12. Jun 23, 2006 #11


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    Staff: Mentor

    You get almost zero when you substitute the answers back into the equations 1 & 2. It's just rounding error that keeps it from being exactly zero. Carry more significant figures in the answers to get closer to zero. Good point about the polarity of your 3rd equation, BTW. I missed that.

    And you don't use V2/2 in equation 2 because you are summing the currents out of the node, and since there is a labelled current source in that leg, 8I0 is the current. The voltage across the 2 Ohm resistor is not V2, because you don't know what the voltage is across the current source.
  13. Jun 23, 2006 #12
    Thanks for claryfying me about the v2/2 part and thanks for being very helpful!:)
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