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Finding the unknown resistor using only an ohmmeter

  1. Feb 10, 2008 #1
    [SOLVED] Finding the unknown resistor using only an ohmmeter

    In a box there are 12 resistors, 11 of them have a resistance of exactly 1 ohm, and 1 of them doesn't. Using only an ohmmeter, what's the least measurements you can make to find out which one is the odd resistor and what's its value? (You can't count on luck!)

    Equations:
    For resistors in series: R(equivalent) = R1+R2+R3+...
    For resistors in parallel: 1/R(equivalent)=1/R1+1/R2+1/R3+...

    I got to 5 by deviding them up into 2 sets of 6, then measuring the resistance of each one (2 measurements). That would give me R. Then I would get 2 groups of 3 and measure one of them (3 measurements). If I get 3 ohm I know the odd resistor is in the other group. Then I measure the resistance of 2 of the 3 and finally find the odd one (5 measurements).

    I'm pretty sure there's a better way... :smile: and I think I should be playing with the series/parallel concept, but I couldn't find anything better than 5... Any ideas?
     
  2. jcsd
  3. Feb 10, 2008 #2
    Hmm, I came up with a similar method:

    1) Split into 2 piles of 6, and wire them up all in parallel. If you get 1/6 the resistor is not in the pile, and if you get something different then it is.

    2) Take the 6 and split it into 3 and 3. Wire them up in series. If you get 3 then the resistor is not in the pile, if you get something else then it is.

    3) Take two of the resistors and put them in series. If you get 2 then the resistor is the one you haven't hooked up, and you're done! If not you have to look between the two and find which one is the odd resistor.

    So possibly in 3, but definitely in 4.

    There may be a better way yet, but it's late and can't think as well. You definitely want to eliminate half of them, which requires one step, and once you have 6 you may be able to do something clever.
     
    Last edited: Feb 10, 2008
  4. Feb 10, 2008 #3
    You only need 3 measurements.

    • Split the resistors in 2 sets of 6 and connect each set in series. Measure one of the set. If you get 6 [itex]\Omega[/itex] the odd resistor is on the second set.
    • Take the set that has the odd resistor and split it in 2 sets of 3 and connect each set in series again. Measure one set. If you get 3 [itex]\Omega[/itex] the odd resistor is on the second set.
    • Take the set with the odd resistor and connect two resistors, say [itex]R_1,R_2[/itex] in series and the 3rd one [itex]R_3[/itex] parallel with the two ones. Then take one resistor from the other set, which you know that it's resistance is 1 [itex]\Omega[/itex] and connect it parallel with [itex]R_1[/itex] and measure the system. Now you have 3 possibilities: the odd resistor is [itex]R_1[/itex] or [itex]R_2[/itex] or [itex]R_3[/itex].
      Calculate the equivalent resistance of each possibility and you will see that they are different, since [itex]R_{odd}\neq 1 \Omega[/itex]. Depending of the result of the 3rd measurement you can decide which of [itex]R_1,R_2,R_3[/itex] is the odd resistor and find it's resistance.
     
  5. Feb 10, 2008 #4
    Thank you!!!!!!!! ^-^v
    How did you come up with that? (I'm just curious :) I've been thinking about this for ages....)
     
    Last edited: Feb 10, 2008
  6. Feb 4, 2009 #5
    Re: [SOLVED] Finding the unknown resistor using only an ohmmeter

    i dont understand....doesnt that mean you can get different values for R_1, R_2, and R_3? am i missing something simple?
     
  7. Apr 16, 2009 #6
    Re: [SOLVED] Finding the unknown resistor using only an ohmmeter

    I actually heard the answer was 1... haha I have no idea...
     
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