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Determining resistance for unknown circuit

  1. Feb 10, 2016 #1
    1. The problem statement, all variables and given/known data
    This was an experiment we did in my lab.
    You're given a black box with four resistors inside, with resistances 20, 51, 100, and 200 ( [itex]k\Omega[/itex]). The box has three plugs on it, labeled A, B, and C. You take your multimeter and plug into two jacks at a time and measure the resistance across the terminals.
    You get $$R_{AB} = 60$$ $$R_{AC} = 139$$ $$R_{BC}=119$$ Note that these are the actual numbers I measured so there may be a slight discrepancy between theoretical values and observed values.
    Anyways, the goal is to try to draw the complete circuit in the box with the four resistances.

    2. Relevant equations
    Parallel: $$\frac{1}{R_{eq}} = \frac{1}{R_1} + . . . + \frac{1}{R_n}$$
    Series: $$R_{eq} = R_1 + ... + R_n $$
    3. The attempt at a solution
    I've tried doing a ton of different possible circuit combinations between A, B, C. Examples include straight lines (A to B, B to C, or A to C, C to B) as well as closed loops (A, B, C). I guess my real question is does anyone know a way to approach this systematically, because the guess work is not coming anywhere for the problem. For the closed loop, I wrote out the system of equations governing the relationship between the equivalent resistances, and used mathematica to solve that system, but nothing came out that matched with the given resistances. If anyone could provide any insight at all, it would be much appreciated.
    Regards.
     
  2. jcsd
  3. Feb 10, 2016 #2

    haruspex

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    It's not exactly methodical, but I think I have solved it.
    Start with the smallest reading. Is it possible that the path from A to B includes a resistor that has no parallel element? Is it possible that every resistor in that path has a parallel element? That should get you down to very few possibilities for that path. Then see how you can add any resistors left over, if any, without changing the AB reading.
     
  4. Feb 10, 2016 #3

    BvU

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    No systematic approach found. Sherlock Holmes way seems indicated:

    none of the three resistances follows from a single component
    none of the three resistances < 20 ##k\Omega##
    none of the three resistances > 200 ##k\Omega##
    so there are parallel and series combinations involved (well, we knew that, didn't we?).​

    What did help me enormously is a simple table of the ... possible parallel combinations of two components. None of these appears as a measured resistance, so further parallel or series is required.

    (you could also make a table for idem series, but that's easier to keep in mind without much calculating)

    Then see if and how you can make RAB (Haru hint) -- this should probably take only one more component, otherwise you run out of options later on
    The last component should then allow you to make the two others (RAC and RBC) and there is almost no alternative to how it should be connected.

    Elementary, my dear Watson !

    And as a colleague experimental physicist :rolleyes:, I should add: my compliments for the quality of your measurements !
    I enjoyed this exercise, so if you are also in a good mood (once you had the eureka experience) you might compliment teacher with this challenging experiment !
     
  5. Feb 10, 2016 #4
    Haha delightful. Thank you both, I'll try to figure it out with your advice and update if I've figured it out
     
  6. Feb 10, 2016 #5
    After trying it myself, I was able to get a circuit which matched two of the terminal resistances precisely (less than 1% error) while the last terminal resistance is at about 8% error. I was wondering if either of you had anything closer?

    Also, thanks to your help, I was able to determine the circuit for the same problem with different data (which I did not post) and had all three terminal values at less than 1% error.
     
  7. Feb 10, 2016 #6

    BvU

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    I had ##1 \ k\Omega## max deviation, hence the compliment for the measurements ....
    Want to show your steps ?
     
  8. Feb 10, 2016 #7
    Yes I'll post it when I get the chance
     
  9. Feb 10, 2016 #8
    Ok I finally think I've got it. So you have a center point that forks three ways. One of those forks you put 51 and 200 in parallel, so 40 equivalent. Another fork you put 20. The last fork you put the 100. Connecting any two ends of the forks, you get 20+40, 20+100, 40+100 which is 60,120, and 140. All within 1 kohm. Thanks BvU :)
     
  10. Feb 10, 2016 #9

    BvU

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    Elementary! Well done !
     
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