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Using Thevenins therom to solve a simple circuits

  1. Sep 7, 2011 #1
    Using Thevenins therom to solve a "simple" circuits

    1. The problem statement, all variables and given/known data

    The problems I am stuck on is problem two, figure 1 and figure 2. It's attached to this post.

    2. Relevant equations



    3. The attempt at a solution

    My initial attempt at solving for Rthev for figure 1 was that Rthev=[itex]\frac{1}{4}[/itex]+[itex]\frac{1}{14}[/itex] = [itex]\frac{9}{28}[/itex] = [itex]\frac{28}{9}[/itex].

    However, on the answers it says that the 12K is shorted out since the source is connected at the same points, so it became Rthev=[itex]\frac{1}{4}[/itex]+[itex]\frac{1}{2}[/itex]=[itex]\frac{3}{4}[/itex]=[itex]\frac{4}{3}[/itex]. At first this made sense, but when I tried doing the second question, the 10K wasn't shorted out when solving for the Rthev.

    Could someone please clarify why the 12K was shorted, but the 10K wasn't?
     

    Attached Files:

  2. jcsd
  3. Sep 7, 2011 #2

    NascentOxygen

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    Staff: Mentor

    Re: Using Thevenins therom to solve a "simple" circuits

    In determining the Thevenin equivalent circuit,

    (i) voltage sources are replaced by their internal series resistance, which for an ideal voltage source is zero ohms (i.e., a short circuit).

    (ii) current sources are replaced by their internal series resistance, which for an ideal current source is infinite ohms (i.e., an open circuit).

    Fig 2 involves a constant current source.
     
  4. Sep 8, 2011 #3
    Re: Using Thevenins therom to solve a "simple" circuits

    Got it, thanks a lot!!
     
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