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Nortonization and nodal analysis

  1. Jul 1, 2011 #1
    Hi, :)

    Please have a look on the following link:
    http://img836.imageshack.us/img836/1554/circuitmm.jpg

    1: I was trying to solve a circuit. I was successful except the part where I was trying to apply the nodal analysis (it's at the very end of the linked scan). Where am I going wrong? By nodal analysis I get Isc = 1/3 A which is wrong. Where did I go wrong?

    2: How would we proceed while applying nodal analysis if the 12V voltage source's polarities are reversed?

    It would be very nice of you if you could help me with the above queries. Thanks

    Cheers
     
  2. jcsd
  3. Jul 2, 2011 #2

    gneill

    User Avatar

    Staff: Mentor

    In your third line,

    2 + (V - 12)/4 - (V - 0)/16 = 0

    For the voltage source branch you've made the assumption that the current will be flowing into the node, but you've written the term as if it is flowing out of the node.

    It's generally easier to NOT make any assumptions about the current direction for branches that don't have fixed currents (like the one containing the 2A current source). Just assume that current is flowing out of the node for all such branches. So,

    2A - I - Isc = 0

    2A - (V - 12)/4 - (V - 0)/16 = 0

    proceed...
    The "V - 12" would become "V + 12".
     
  4. Jul 2, 2011 #3
    Many, many thanks, gneill.

    Cheers
     
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