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Resistance equivalence and circuit current

  1. Mar 13, 2016 #1
    1. The problem statement, all variables and given/known data
    a) find the equivalent resistance of the circuit below b) each current in the circuit c) the potential difference across each resistor d the power dissipated by each resistor.

    upload_2016-3-13_17-43-6.png

    2. Relevant equations
    V=IR
    R=(1/R)^-1

    3. The attempt at a solution

    Not sure if I did the resistance equivalency correctly. I combined R2 and R6 in series and then combined R2, R3, and R5 as in parallel. I then treated R1, R2/3/5/6 and R4 as in series. Also, need help going back and finding the current. Please advise.

    upload_2016-3-13_17-43-29.png
     
  2. jcsd
  3. Mar 13, 2016 #2

    berkeman

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    Staff: Mentor

    That's not making sense to me...

    Start at the outside (far right side), and start doing parallel and series combinations. Start with R4 and R5 and try again? :smile:
     
  4. Mar 13, 2016 #3
    If the current can go to one resistor or another, but not both, then the resistors are not in series. See if you can redraw the circuit in a less-complicated looking manner by redrawing the diagonal portion to look more like a typical rectangular circuit.
     
  5. Mar 13, 2016 #4
    I see R3(bottom right) and R5 being in parallel. If I combine those they would be in series with R2 (top right). Is that a good start?
     
  6. Mar 13, 2016 #5
    Correct! Notice that R3 and R5 share both of their nodes. That's how you would spot that.
     
  7. Mar 13, 2016 #6
    Ok. Maybe I'm making a mountain out of a mole hill! I saw something online that showed R5(R2+R6)/R2+R5+R6. So, R2 and R6 are in series with each other but Parallel with R5. Do you agree? I Think that's what was throwing me for a loop. Wasn't sure which way to go with it.
     
  8. Mar 13, 2016 #7
    No, R2 and R6 are not in series with each other as the circuit is written. Eventually, however, you'll find new series and parallel combinations as you continue to combine your resistors. Your parallel combination of R3 and R5 are in series with R2. What can you combine next?
     
  9. Mar 13, 2016 #8
    So, combine R3 and R5 as in parallel. Then The sums R1R2 and R4R1, respectively, which puts them in series with R6. Combine the remaining as in series?
     
  10. Mar 14, 2016 #9

    CWatters

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    I struggle to read your hand writing but R3 and R5 in parallel looks correct. As far as I can see R1 and R2 are not in series so summing them appears incorrect.

    As I see it the starting sequence should be

    Step 1... (R3//R5)
    Step 2... (R3//R5) + R2
    Step 3... (R3//R5) + R2) // R6

    where // means "in parallel with"
     
  11. Mar 14, 2016 #10

    OmCheeto

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    2016 Award

    I agree with axmls, the original diagram makes my head hurt, and it should be redrawn.
    I did it by pretending all the resistors were connected with rubber bands, disconnecting the negative end of the terminal, and stretching everything to the right.
    Triple check that all the paths are correct.
    Then add in a wire, blue, to complete the circuit.

    2016.03.14.hw.problem.861942.jpg

    This, visually, makes it much easier for me to see what's going on.
     
  12. Mar 14, 2016 #11
    I agree. THANK YOU!!!!
     
  13. Mar 14, 2016 #12

    OmCheeto

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    Please let us know what you get as an answer, as...

    rubber.bandy.method.png
     
  14. Mar 14, 2016 #13
    15.4 ohm's
     
  15. Mar 14, 2016 #14

    OmCheeto

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    That's what I get.
    :partytime:
     
  16. Mar 14, 2016 #15
    Graicas my friend!
     
  17. Mar 14, 2016 #16

    SammyS

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    Conductive rubber, I hope !
     
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