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Simplifying a resistance circuit

  1. Nov 3, 2014 #1
    1. The problem statement, all variables and given/known data
    Hi guys,

    I am trying to find the Thevein resistance of a circuit, I'm only stuck on finding Rth. Can you help?

    I have included a screen shot.

    2. Relevant equations


    3. The attempt at a solution
    Well I thought that I could combine the two bottom 1 ohms in series, the 4 and 12 in parallel and then combine the three remaining for a total of 10 ohms. But I am starting to think that the two 1 ohm resistors don't actually share the same to nodes so I am now a bit stuck! Any help would be greatly appreciated.
     

    Attached Files:

  2. jcsd
  3. Nov 3, 2014 #2

    gneill

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    Staff: Mentor

    The two 1 Ohm resistors do indeed share the same nodes; That makes them parallel connected, not serial.
     
  4. Nov 3, 2014 #3
    Opps.. my mistake! I'm a little bit confused though, I can see that the two 1ohm obviously share the same bottom node, but I'm not really sure how the top ones are the same, I see they are connected by a wire but as the nodes are also connected to other resistors I thought they wouldn't be in parallel..

    So then that makes then that means that their combined resistance is 1 ohm. So then the 4 and 12 are is parallel for a resistance of 3 ohms, making the total resistance 9 ohms?

    Thanks for the help!
     
  5. Nov 3, 2014 #4

    gneill

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    Staff: Mentor

    As long as two components share two nodes they are parallel-connected regardless of what other components may also connect to those nodes. It's series connections that are finicky that way -- components that are in series must be the "sole proprietors" of the node that they share.
    Ah, so close. Two 1 Ohm resistors in parallel is how many Ohms?
     
  6. Nov 3, 2014 #5
    Oh thanks! makes much more sense!

    Once again, Opps! 1/1 + 1/1 = 2 so its 1/2. So the total resistance is 8.5 Ohms! :)
     
  7. Nov 3, 2014 #6

    gneill

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    Staff: Mentor

    Ta da! Yup. :smile:
     
  8. Nov 3, 2014 #7
    Yay! Thanks very much, you've been really helpful! :)
     
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