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A question on norton method

  1. May 14, 2008 #1
  2. jcsd
  3. May 14, 2008 #2
    I is negative because you consider it from - to + (as you show in your paper).
     
  4. May 14, 2008 #3
    i know that KCL says that the sum of the currents that goes in
    equals the sum of the currents that goes out

    here i dont know what current goes into the node
    and what current goes out??
     
  5. May 16, 2008 #4
    And you don't need to. The sum of currents entering a node is equal to zero, the sum of currents leaving the node is also zero. Pick one and stay consistent.

    If you choose to sum the currents leaving the node, then by the passive sign convention you have (e-v)/R1 + e/R2 - I = 0. If you choose to sum the currents entering the node you have (v-e)/R1 + (0-e)/R2 + I = 0, which is exactly the same as the previous equation once you sort out the signs.

    So, to answer your question, you are summing the currents leaving the node. Since I is entering the node it follows that -I is leaving it.
     
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