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Homework Help: Equivalent Resistance Problem

  1. May 7, 2010 #1
    1. The problem statement, all variables and given/known data

    All resistors have the same value, 'R'



    2. Relevant equations

    I don't think mesh/loop analysis or nodal analysis is applicable here, I think I'm just having a hard time seeing how the series-parallel connections work.


    3. The attempt at a solution

    I know the top left two are in parallel and then they come back together, and I know all the corners are in series, it's just the middle box that throws me.
     

    Attached Files:

    Last edited: May 7, 2010
  2. jcsd
  3. May 7, 2010 #2
    My approach would be to redraw the circuit so it's a bit easier to see what's going on.

    You are right that mesh and nodal don't apply- you have no need to find voltages or currents.

    Now, lastly, why do you say the top left are in parallel and the rest are in series? Step through that thought process, it might help you. I'm not saying you're wrong, but I'm not saying you're right either- just walk through the process.
     
  4. May 7, 2010 #3

    tiny-tim

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    Homework Helper

    Welcome to PF!

    Hi axleboy57! Welcome to PF! :smile:

    I don't normally do electricity questions unless they're really easy :rolleyes:

    can someone tell me whether the centre "diamond" is supposed to have zero resistance, because if it is, why don't the right-hand sides of the "diamond" short-out the four right-hand resistances? :confused:
     
  5. May 7, 2010 #4
    Re: Welcome to PF!

    Yes, it has zero resistance (the diamond) but I don't want to say more lest I give away the answer. Think about when a source is applied, where is current going to flow?

    This is a very common type of problem for intro courses to electronics. It's basically designed to force the student to think about parallel and series resistances (though, honestly, it's really just meant to confuse more than anything). I can PM you the solution, tiny-tim, so you can see what to do and I can explain what I can't in this thread if you want me to?
     
  6. May 7, 2010 #5

    berkeman

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    Re: Welcome to PF!

    Good hint, TT :wink:
     
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