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Resistors ratio

  1. Jan 6, 2013 #1
    1. The problem statement, all variables and given/known data
    Part of an elecrial circuit for the olympic score board consisted of a single battery V and several resistors having on of two values. R or 2R. They are hooked up in the confriguration as shown. Calculate the ratio of the two currents shown Ib/Ia
    http://s1302.beta.photobucket.com/u...pg.html?&_suid=135753198581409194345584751531

    3. The attempt at a solution
    So for Ib i did

    Rt = R + 1/2R
    Rt = 2R^2 /2R+ 1/2R
    Rt = 2R^2 + 1 /2R

    Ib = V/2R^2 + 1 /2R
    Ib = V/4R^3 + 2R

    for Ia
    Rp = 1/2R + 1/2R + 1/2R +1/2R +1/2R +1/2R
    Rp = 6/2R

    Rs = R + R + R + R + R
    Rs = 5R

    Rt = 5R + 6/2R
    Rt = 10R^2 + 6 / 2R

    Ia = V / 10R^2 + 6/2R
    Ia= V/ 20R^3 + 12R

    Ib/Ia= V/4R^3 + 2R / V/ 20R^3 + 12R
    = 20R^3 + 12R/4R^3 + 2R
    = 2R (10R^2 + 6) / 2R (2R^2 + 1)
    = (10R^2 + 6) /(2R^2 + 1)

    i don't know what to do next
     
  2. jcsd
  3. Jan 7, 2013 #2

    ehild

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    Homework Helper
    Gold Member

    What is Rt? Your last equation is dimensionally incorrect.

    ehild
     
  4. Jan 7, 2013 #3
    oh sorry i re-did this this way now:
    Ib= V/R
    Rtot = R + 2R
    Rtot = 3R
    Ib = V/3R

    for Ia = V/R
    Rseries = 5R
    Rparallel = 6/2R
    Rtot = 10R^2 + 6 / 2R
    Ia = V/20R^3 + 12R

    right?
     
  5. Jan 7, 2013 #4

    gneill

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    Staff: Mentor

    It is unclear what resistors you're combining and how. It appears that whatever approach you're taking it's not producing the results you want; the units in your expressions don't match (they are dimensionally incorrect as ehild mentioned).

    Looking at the diagram there are very few candidates for parallel or series connected components for direct simplification purposes. So you'll either have to make a brute-force kirchhoff law attack (mesh or nodal) or find a clever approach employing symmetry if you can spot it :wink:

    Hint: consider just the last three resistors at the right end. What equivalent resistance do they 'present' to the rest of the circuit?
     
  6. Jan 7, 2013 #5

    phinds

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    Uh ... I'm wondering if we are looking at the same circuit? The value of Ib is truly trivial and Ia isn't much harder

    EDIT: and I mean just sort of by looking at it ... there is NO sophisticated analysis needed.
     
  7. Jan 7, 2013 #6

    gneill

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    Staff: Mentor

    I'm looking at this circuit:

    attachment.php?attachmentid=54538&stc=1&d=1357613607.gif

    The symmetry of the "repeated cells" allows for a quick solution for the current ratio, otherwise it could be a bit of a slog.
     

    Attached Files:

  8. Jan 7, 2013 #7

    phinds

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    Exactly. That's why I was puzzled by your original statement.
     
  9. Jan 7, 2013 #8

    gneill

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    Staff: Mentor

    :smile: I was looking at the OP's attempts and it didn't appear that he'd spotted the short cut, which prompted my post; I wanted to find out more about his approach, and see if he could spot the short cut.
     
  10. Jan 8, 2013 #9

    phinds

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    Ahha ... thus the wink.
     
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