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Basic circuit problem

  1. Sep 12, 2013 #1
    1. The problem statement, all variables and given/known data

    In the circuit below, VS = 65 V, R1 = 64 Ω, R2 = 190 Ω, R3 = 90 Ω, R4 = 290 Ω.
    Find:

    (a) The equivalent resistance (Req) at the dashed line
    (b) v1
    (c) v2
    (d) i4



    2. Relevant equations



    3. The attempt at a solution

    I found the equivalent resistance of the circuit (139.612 [itex]\Omega[/itex]) by collapsing the circuit step-by-step, but that didn't seem to work. Apparently this line is referring to something else, and I'm not sure what to do.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     

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  3. Sep 12, 2013 #2

    gneill

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    Staff: Mentor

    Can you show the steps you took to find the equivalent resistance? I'm not seeing the same result.
     
  4. Sep 12, 2013 #3
    First I combine R3 and R4 (290+90) which gives 380. Then I combine that with its parallel R2 (190) which gives 1/[(1/380)+(1/190)] = 126.66, then I add that to R1 (64) which gives 190.66, then combine it with its parallel (50), giving 39.612, which I then add to the last resistor, 100, to get 139.612 ohms. If I am making a mistake, I'd like to know where.
     
  5. Sep 12, 2013 #4

    gneill

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    Staff: Mentor

    Ah. That last resistor, 100 Ω, is also in parallel, not in series; You're want to look at the circuit from the point of view of where the dotted line is, not from where VS is connected.
     
  6. Sep 12, 2013 #5
    Okay, well I'm not sure how to handle it at that point then. I did everything the same as before, only with 100 in parallel, but that didn't seem to work, so I'm all out of ideas.
     
  7. Sep 12, 2013 #6

    gneill

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    Staff: Mentor

    What value did you get?
     
  8. Sep 12, 2013 #7
    28.373 ohms
     
  9. Sep 12, 2013 #8

    gneill

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    Staff: Mentor

    That's what I'm seeing, too. So...

    It's just possible that they want the equivalent resistance for the network to the left of the dotted line only. This would be a common step that one might do if one wanted to replace that piece of the circuit with a Thevenin equivalent for the source and network (you may or may not have learned about Thevenin equivalents yet).
     
  10. Sep 12, 2013 #9
    Definitely not, considering this is the homework for just day 3 of circuits 1.
     
  11. Sep 12, 2013 #10

    gneill

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    Staff: Mentor

    Well, I don't know what else to suggest. That's the only likely alternative interpretation of the question that springs to mind.
     
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