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Calculating Rth error?

  1. Dec 16, 2014 #1
    1. The problem statement, all variables and given/known data
    Hi Guys,

    I am doing a number of questions on finding Rth and can across this example. When I looked at the answers I had a different one to them but I actually think they may be wrong.

    Could you guys please take a look?

    2. Relevant equations


    3. The attempt at a solution
    So they have Rth found (once voltages are short circuited) as Rth = 8*2/8+2 = 1.6 Ohms but that implies that one of the 2ohm resistors is it series with the 6 ohm resistor but I don't think it is. I think that both 2ohm resistors are in series and then those are in parallel with the 6 ohm so Rth is 2.4 ohms.
    I may be wrong but what do you guys think?

    Thanks
     

    Attached Files:

  2. jcsd
  3. Dec 16, 2014 #2

    gneill

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    Staff: Mentor

    As you say, the text solution is incorrect. But your result is also incorrect, likely due to a calculation issue. Check your calculation.
     
  4. Dec 16, 2014 #3

    mfb

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    2016 Award

    Staff: Mentor

    What is Rth?
    That would be ##\frac{8\cdot2}{8}+2##, I guess you mean 8*2/(8+2). Where does that come from?
    Right.
    Only if you remove all power sources, but then the network becomes pointless.
     
  5. Dec 16, 2014 #4

    gneill

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    Staff: Mentor

    GBA13 is looking for the Thevenin resistance of the network as seen at the open terminals ( ##V_{th}## ).
     
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