Superposition Principle / Double Current Divider

Discussion in 'Homework Help' started by fiolvb, Jan 18, 2016.

  1. fiolvb

    Thread Starter New Member

    Jan 18, 2016
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    Hey guys, I'm trying to figure out how to use the superposition principle in this circuit to get the current through R4.. I know the basics of superposition
    So I first open-circuited the current source and ended up with a total resistance that couldn't help me find current i4...
    Then I also tried to short-circuit the voltage source ending up with a total resistance given by r1 in parallel with a series-connection between r2 in parallel with r4 and r3... if this is right, what should I do to get current I4?
    upload_2016-1-18_15-20-2.png
    Thanks in advance!
     
  2. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Have you try redrawn your circuit.
    For example if we remove the current source we get this circuit:
    -2.PNG
    And finding I4 for this circuit is a trivial task.
     
    Last edited: Jan 18, 2016
  3. fiolvb

    Thread Starter New Member

    Jan 18, 2016
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    Yeah from that point I would join the other three resistors in a parallel and then series connection into one piece, so I would have R_123 in series with R4, but I feel that's wrong, isn't it?
    Or could I apply KCL to that node? How do I get the other 2 currents then?
     
  4. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    No, the this is the right approach.
     
  5. WBahn

    Moderator

    Mar 31, 2012
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    But those three resistors are not in parallel. R1 and R3 are in series and that combination is in parallel with R2.

    You can use any circuit analysis technique you want (that is allowed by the instructor). But reducing the circuit to an effective resistance such that the current through that resistance also happens to be the current that you are seeking is probably the easiest way to go.
     
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