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[Circuits] Computing the Thevenin Equivalent #2

  1. Jan 21, 2014 #1
    1. The problem statement, all variables and given/known data

    h_1390340433_1355490_7cf39d8773.png

    2. Relevant equations



    3. The attempt at a solution

    I am trying to find the equivalent resistance for part b.

    First, we turn off independent sources. We open circuit the 2A source and we close circuit the 30V source. We're left with two resistors. Are they in series or parallel and why? I think it is in series because there's no other elements other than the two resistors.
     
  2. jcsd
  3. Jan 21, 2014 #2

    berkeman

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    Staff: Mentor

    Slide the 60 Ohm resistor down and to the right, into the position where the 30V source used to be. Now do the resistors look like they are in series or parallel, with respect to the two probe ponts 1 & 2?
     
  4. Jan 21, 2014 #3
    They look like they're in parallel but I can also make them look like they're in series by sliding the 60 ohm resistor even further past the (2) node.
     
  5. Jan 21, 2014 #4

    berkeman

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    Staff: Mentor

    But then you've changed the question. The position of the 1 & 2 nodes is given to be where they are with respect to the two resistors. They are indeed in parallel. :smile:
     
  6. Jan 21, 2014 #5
    But I removed the load so (1) and (2) aren't connected to anything.

    As a side-question, what if we ignore (1) and (2), would the resistors be in series or parallel?
     
  7. Jan 21, 2014 #6

    berkeman

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    Staff: Mentor

    I don't understand what that means.

    It all depends. You have to specifiy in series or parallel with respect to what. That circuit reduces to just the two resistors connected nose-to-tail by wires. If that's all there is, they are in series to make a total of 90 Ohms going around the loop.
     
  8. Jan 21, 2014 #7
    I think I get it. Thanks.
     
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