Supernode

Discussion in 'Homework Help' started by lukepfister, Sep 16, 2007.

  1. lukepfister

    Thread Starter New Member

    Sep 16, 2007
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    [​IMG]

    Use the node voltage method to find how much power the 2 A source extracts from the circuit.



    Okay, I figure I will find the power dissipated at each resistor, and use that to find how much the 2A source is extracting from the circuit.

    I redrew the circuit so I could use the supernode a little easier. I chose the bottom node to be my reference node, and the supernode would fall between node's 2&3. HOWEVER, I am very, very weak with this concept of a supernode, so hopefully you can steer me in the right direction.

    At node one, we have
    \frac{V_1}{50} + \frac {V_1-V_2}{1}  - 2 = 0

    And the supernode equation, I came up with...
    \frac{V_2-V_1}{1} + \frac{V_3}{4} = 0

    Now, I have no idea what to write as my third equation. There are no dependent sources to use to create a constraint. Where can I use the 45 V? Should I set my supernode equation equal to it?
     
  2. hgmjr

    Moderator

    Jan 28, 2005
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    One strategy would be to analyze the power dissipated by the circuit with the 2A source removed. You could then put the 2A current source back in and use Norton's Theorem to convert it and the 50 ohm resistor to an equivalent voltage source and a source resistance. Then recompute the power dissipated by this configuration. The difference should then be the power extracted by the 2A source.

    hgmjr
     
  3. lukepfister

    Thread Starter New Member

    Sep 16, 2007
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    Similar to superposition? I am not familiar with Norton's theorem. I will go look it up...in the meanwhile, can you see any way to do this problem using node-voltage method? I'm afraid that when our teacher says to do it one way...we do it that way. That being said, I would love to have another method to look at it and solve it, so your insight is MUCH appriciated.
     
  4. hgmjr

    Moderator

    Jan 28, 2005
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  5. Eduard Munteanu

    Active Member

    Sep 1, 2007
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    Norton's theorem is like Thevenin's theorem ending up with a current source instead of a voltage source.
     
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