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Use the node-voltage to find v1 and v2 in the circuit below

  1. Sep 17, 2012 #1
    Use the node-voltage to find v1 and v2 in the circuit below(see picture)

    4.8)
    http://s716.photobucket.com/albums/ww168/Pitoraq/?action=view&current=DSC_0204.jpg

    http://s716.photobucket.com/albums/ww168/Pitoraq/?action=view&current=DSC_0206.jpg

    4.26)

    http://s716.photobucket.com/albums/ww168/Pitoraq/?action=view&current=DSC_0205.jpg

    http://s716.photobucket.com/albums/ww168/Pitoraq/?action=view&current=DSC_0207.jpg



    2. Relevant equations

    4.8)

    v1 + (v1-v2)/8 = 6 (node 1)
    1 + v2/120 = (v2-v1)/8 (node 2)

    4.26

    1/R(eq) = 1/150 + 1/75 = 30 ohm

    2 - ((-v1-25)/50) + v1/150 + v1/75 = 0

    3. The attempt at a solution

    4.8)
    After combining node 1 and node 2 i got v = 146 v which is wrong

    what about the r=80 resistor ?

    The answer should be 120v,96v
    4.26)

    I got v1 to be -37.5 v and p = (-37.5)*2 = -75 W

    The answer should be

    a) -37.5 V,75W
    b) -37.5 v, 75W
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
    Last edited: Sep 17, 2012
  2. jcsd
  3. Sep 17, 2012 #2

    gneill

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    Staff: Mentor

    Re: Usa the node-voltage to find v1 and v2 in the circuit below

    Check carefully the implied direction of the currents in your Node 2 equation.
     
  4. Sep 17, 2012 #3
    Re: Usa the node-voltage to find v1 and v2 in the circuit below

    But i cant see anything wrong with the current direction
     
  5. Sep 17, 2012 #4

    gneill

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    Staff: Mentor

    Re: Usa the node-voltage to find v1 and v2 in the circuit below

    Your equation is:

    1 + v2/120 = (v2-v1)/8 (node 2)

    On the left you have currents flowing out of node 2. On the right you also have a current flowing out of node 2. If it's on the right, and it must balance the currents on the left, it should be flowing INTO node two.

    You might find it more straightforward to write the sum of all currents on one side, setting the sum to zero. That way the individual current directions may be more obvious.

    One other thing: what happened to the contribution of the 80 Ohm resistor?
     
  6. Sep 17, 2012 #5
    Re: Usa the node-voltage to find v1 and v2 in the circuit below

    Yes. I started with writing
    1 + v2/120 - (v2-v1)/8 = 0
    But im not sure about the 80 ohm resistor
     
  7. Sep 17, 2012 #6

    gneill

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    Re: Usa the node-voltage to find v1 and v2 in the circuit below

    Okay, but since you are summing all the currents LEAVING the node, the sign on the last term should be positive, too.
    Add a term for it that is just like the one for the 120 Ohm resistor.
     
  8. Sep 18, 2012 #7
    From the first node equation, should the 80 resistor be added to that to ?
     
  9. Sep 18, 2012 #8
    Re: Usa the node-voltage to find v1 and v2 in the circuit below

    I have done that:

    1 + v2/120 -(v2-v1)/8 = 0

    1 + v2/120 leaves the node 2, while (v2-v1)/8 enters node 2.

    Should i add the resistors: 1/R = 1/80 + 1/120 , R = 48 ohm ?

    -6 + v1/40 + (v1-v2)/8 = 0 (node 1)
    1 + v2/48 - (v2-v1)/8 = 0 (node 2)

    Are the equations right ?
     
    Last edited: Sep 18, 2012
  10. Sep 18, 2012 #9

    gneill

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    Staff: Mentor

    Nope. The 80Ω path "belongs" to the second node; it directly connects to the second node, whereas there is the 8Ω resistor between it and the first node.

    attachment.php?attachmentid=50973&stc=1&d=1347973052.gif

    Each path through which current can flow into (or out of) a given node needs to have a term in the node's equation.

    Out of habit I tend to assume that all currents, except explicitly defined ones such as fixed current sources, always flow out of a given node, and just bang down the equation accordingly all on the left hand side. The math will automatically take care of the details of the "true" current directions since its the relative potentials that define them anyways, and they start out as unknowns.

    So, can you now take another stab at writing the two node equations?
     

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  11. Sep 18, 2012 #10
    I got it to be:
    -6 + v1/40 + (v1-v2)/8 = 0 (node 1)
    1 + v2/120 - v2/80 - (v2-v1)/8 = 0 (node 2)
    Is this correct ?
     
  12. Sep 18, 2012 #11

    gneill

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    Staff: Mentor

    The node 1 equation looks fine.

    You've got some sign issues with the node 2 equation. For example, why would the signs for the 80Ω and 120Ω paths be different? Surely current must flow in the same direction through the two?
     
  13. Sep 18, 2012 #12
    Then it must be:

    1 + v2/120 + v2/80 -(v2-v1)/8 = 0

    Should 80 ohm be possitive ?
     
  14. Sep 18, 2012 #13

    gneill

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    Staff: Mentor

    ALL of the terms should be positive if you're summing them as outgoing currents.
     
  15. Sep 18, 2012 #14
    Shouldn't (v2-v1)/8 be negative since 6A goes into node 2 ? Furthermore shouldn't the sum of all be zero ?
     
  16. Sep 18, 2012 #15

    gneill

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    Currently the potentials V1 and V2 are unknowns. Regardless of whether you know by intuition that current will flow from the V1 node to the V2 node, you can rely on the math to take care of that detail. By writing that current as (V2 - V1)/8, it will automatically evolve to have V1 > V2, so that the term's actual sign will take care of itself. This means that you don't have to rely on intuition when a complex circuit doesn't make it obvious.

    So, if you always write the expressions assuming that the current node has a higher potential than the surrounding ones, and that all currents are outgoing (or incoming --- your choice), then the math will take care of the rest. Less thinking = fewer missteps.
     
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