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Nodal analysis question

  1. Feb 17, 2014 #1
    nodalvxv.png

    How do I find the voltage at the node with the red dot using nodal analysis? The 2Ix voltage source to the left of the node with the red dot makes it confusing for me...

    Thanks
     
  2. jcsd
  3. Feb 17, 2014 #2

    Maylis

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    Gold Member

    First, I think you should find Ix in terms of the voltage at the red dot. Once you have Ix in terms of the node voltage, then you can use nodal analysis to find its value
     
  4. Feb 17, 2014 #3

    FOIWATER

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    Gold Member

    Do you have to use nodal that's sort of odd. Since the terminals are open, no current flows through the 5 ohm resistor and this is a simple series circuit
     
  5. Feb 18, 2014 #4
    Well my book uses mesh analysis. I was just wondering whether it is possible to use nodal analysis?
     
  6. Feb 18, 2014 #5

    The voltage at the red dot is 2Ix no?
     
  7. Feb 18, 2014 #6

    NascentOxygen

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    Staff: Mentor

    No. That's the voltage of just one of the sources in a branch between Red Point and ground.

    If you consider the other (shorter) branch between Red Point and ground, what is that branch voltage?
     
  8. Feb 18, 2014 #7
    Is this correct?

    nodalvfv.png
     
  9. Feb 18, 2014 #8
    Yes, you got that part right.
     
  10. Feb 18, 2014 #9
    Ok, so how would I apply nodal analysis now? I mean I can get (v-10)/3 but thats as far as I can get. I don't know how to treat the part to the right of node V.
     
  11. Feb 18, 2014 #10
    Treat it as one big node a supernode.

    Whats comes in must come out.
    (10 - V)/3Ω = ((V+ 2*Ix) - 0)/6Ω
    And
    Ix = (10 - V)/3Ω
     
    Last edited by a moderator: Sep 25, 2014
  12. Feb 18, 2014 #11
    The part in bold is the voltage that comes out of the red node rather than the blue node?

    Thanks
     
    Last edited by a moderator: Sep 25, 2014
  13. Feb 18, 2014 #12
    The part in the bold is a current that comes out of the red node. Or the current that leaves the positive terminal of a CCVS.
     
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