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Resistance in a series / parallel?

  1. Feb 5, 2013 #1
    1. The problem statement, all variables and given/known data

    Find the equivalent resistance between points A and B shown in the figure(Figure 1) . Consider R1= 2.2 , R2= 1.0 , R3 = 3.6 , R4= 5.0 , R5= 4.0 , and R6= 7.6 .

    http://i.imgur.com/EBJmSvk.png

    2. Relevant equations

    For resistors in parallel, 1 / Req = 1/r1 + 1/r2 + 1/r3...etc
    for resistors in series, Req= R1 + R2 + R3...etc



    3. The attempt at a solution

    So I think R3, R4 and R5 are parallel

    1/3.6 + 1/5 + 1/4 = .72

    and then I think we can combine R6, R1 and R2 into one since they are parallel also.

    1 / 2.2 + 1/1 + 1/7.6 = 1.58

    Now add up .72 and 1.58 since they are in series? = 2.3

    Would appreciate any help, this is pretty confusing to me lol
     
  2. jcsd
  3. Feb 5, 2013 #2
    Check this again.
     
  4. Feb 5, 2013 #3
    Ahhh I'm sorry I did a typo. I meant parallel
     
  5. Feb 5, 2013 #4
    Did you get the correct answer now?
     
  6. Feb 5, 2013 #5
    nope :( I'm not sure what I'm doing wrong
     
  7. Feb 5, 2013 #6
    0.72 is reciprocal of equivalent resistance of R3, R4 and R5. The equivalent of these three resistors is in series with R6.
     
  8. Feb 5, 2013 #7

    Ok I was just thinking that, so its actually.

    .72 = 1 / Req

    Req = 1.39?

    Then we add that to R6 in a series? so

    1.39 + 7.6 = 8.99

    Now we add R6 to R2 and R1 which are parallel.

    So... 1/2.2 + 1 + 1/8.99 = 1 / Req

    = 1.24 = 1 / Req -----> 1 / 1.24 = .8?
     
  9. Feb 5, 2013 #8
    I haven't checked out your calculations but your steps look fine to me.
     
  10. Feb 5, 2013 #9
    ARG.....It's not .8...what the f.....heck
     
  11. Feb 5, 2013 #10
    Calculation mistake. 1/Req=1.56
     
  12. Feb 5, 2013 #11
    Finally!! Got it right, thanks so much for your help!
     
  13. Feb 5, 2013 #12

    haruspex

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    Try that step again (btw, 8.97 is more accurate)
     
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