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Node-voltage method problem, to find voltage.

  1. Sep 24, 2012 #1
    1. The problem statement, all variables and given/known data

    Use the node-voltage method to find vo and the power delivered by the current source in the circuit in figure(Figure 1) if the v = 30V and the i = 1.6A.
    #1: Use the node a as the reference node to find vo
    #2: Use the node b as the reference node to find vo
    #3: Find the power delivered by the 1.6A current source

    I am stuck on the first one at the moment.



    http://imageshack.us/a/img152/2156/eeehomework04p4.png [Broken]Left is the Given with my adjustment notes, right is modified result.

    2. Relevant equations
    Kirchoff's Volt Law
    Kirchoff's Current Law
    Ohm's Law
    Node-voltage method
    supernode & mesh analysis

    3. The attempt at a solution
    In the pic above you can see I combined the series resistances, 20+55=75. Then I combined the parallel resistors, 150 and the new 75 to get (150*75)/(150+75)=50. This problem instructs me to use node a as the reference, so I put the ground/ref. there for #1.

    This is when I setup my node-voltage method equation.
    node b: 1.6 + (1/50)Vb + (1/50)(Vb - 30) = 0

    This was the only equation I was able to make, and it gives me 0 for Vb which is wrong... I'm probably handling the voltage source incorrectly in the problem.. Is the modified picture correct? If so, did I make an error with my equation for Vb, or should I have a 2nd equation from node C?
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Sep 24, 2012 #2

    gneill

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    Staff: Mentor

    Your node b equation looks okay. Perhaps an algebra problem while solving for Vb?
     
  4. Sep 24, 2012 #3
    yup, that was the problem, after correcting it, I got -25 for Vb
     
  5. Sep 24, 2012 #4

    gneill

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    Staff: Mentor

    Yup. That looks fine :smile:
     
  6. Sep 24, 2012 #5
    for the 2nd part, with node b as reference, will I get 2 equations this time or just 1 again? I'm getting:
    node a: -1.6 + (1/50)Va + (1/50)(Va-30) = 0
     
  7. Sep 24, 2012 #6

    gneill

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    Staff: Mentor

    Two things. First, note the polarity of Va (how it's "measured" as per the schematic). The Va you've written into your equation has the opposite polarity. Second, note that the voltage source is increasing the potential between node a and the reference node b; you've written it as decreasing the potential difference.

    EDIT: Never mind! I was thinking that you were solving for Vo, but you've in fact written a correct equation for the node potential Va with respect to node b except for the slip with the voltage source. You can translate Va to Vo by noting their relative polarities on the schematic.
     
    Last edited: Sep 24, 2012
  8. Sep 24, 2012 #7
    so to correct my eq for that....

    -1.6 - (1/50)Va + (1/50)(Va + 30) = 0

    That should adjust to the polarity, and the Voltage gain from the source.
     
  9. Sep 24, 2012 #8

    gneill

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    Staff: Mentor

    D'oh! See my edit above.
     
  10. Sep 24, 2012 #9
    For the third portion, I thought I should be summing up the power across the resistors, I know Vo is -25V so that's the one, and the V over the other resistor should be Vb-30, which is -55V

    So, 25^2/50 + 55^2/50 = 73W ... This isn't correct, am I calculating it wrong or am I missing an additional source of power?
     
  11. Sep 24, 2012 #10

    gneill

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    Staff: Mentor

    The voltage source is another possible source (or sink) for power.

    If you have the potential across the current source, and if you know its current, then you can determine the power it delvers or sinks directly. P = I*V. It's delivering power if the current emerges from its more positive terminal.
     
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