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Circuits and current

  1. May 17, 2010 #1
    Last edited by a moderator: Apr 25, 2017
  2. jcsd
  3. May 17, 2010 #2
    In the given figure, at the node where R3 and R4 are connected (let us call this as Nd), the total current flowing out of the node through R3 and R4 is (4-2mA=2mA). Now there are two other paths joining at Nd. So the current from these two paths in to the node Nd could be 4mA ( from R1 and R2 paths) and current from R5 path (let us call this as I5). Now applying KCL at node Nd, 4mA+I5=2mA , giving I5=-2mA, so I5 should flow from bottom to top.
     
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