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Need help finding equivalent resistance in a circuit.

  1. Sep 25, 2011 #1
    1. The problem statement, all variables and given/known data

    nTqaP.png

    2. Relevant equations

    delta wye transformations?

    3. The attempt at a solution

    PM9Qm.png
    I feel like I should do delta wye transforms here, but that doesn't really lead me anywhere. Am I missing how I should do the problem?
     
  2. jcsd
  3. Sep 25, 2011 #2
    Perhaps if you treated the two pairs of 3 kΩ resistors as two 6 kΩ resistors, you would realize the next step.
     
  4. Sep 25, 2011 #3
    Oh, wait... treat them as 6 kΩ resistors and then they become resistors in parallel?

    In that case, does the whole half above A become equivalent to 4.8 kΩ? If so, then the whole circuit would have an equivalent resistance of 14.4 kΩ? Is that right? Or did I take your hint the wrong way....
     
  5. Sep 25, 2011 #4
    Your thinking and answer was wrong.

    What is each 6 kΩ in parallel with?

    Where is A?
     
  6. Sep 25, 2011 #5
    So I figured the following:

    wQWD1.png

    Treat the boxed resistors as 6 kΩ and then treat those to be in parallel with the 4 kΩ resistors under them.

    Then we can treat the two pieces on the top (the two sets of parallel resistors) to be parallel with each other and be equivalent to 4.8 kΩ.

    The region I'm thinking of as being equivalent to 4.8 kΩ is shown in the circle below.

    QcQkN.png

    Then, we can just treat the 4.8 kΩ to be in series with the other four 3 kΩ resistors to get an answer of (4.8 kΩ + 12 kΩ) 16.8 kΩ.

    Can you please guide me as to where I messed up? Thank you so much for your help!
     
  7. Sep 25, 2011 #6
    You're supposed to be finding the resistance between A and B. You have to keep the A node!
     
  8. Sep 25, 2011 #7
  9. Sep 25, 2011 #8
    That's correct.
     
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