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Significant figures

  1. Nov 4, 2012 #1
    Can you give me the answer to the right significant number please???
    I got R_L(max) = 0.0022 [ohm] and R_L(min) = 0.3 [ohm] but I have a feeling that they're not correct. Someone please check.

    R_L(max) = |(5 - 2.4)/(3*(-400) + 40))|

    R_L(min) = (5 - 0.4)/(16 - (-1.6))
     
  2. jcsd
  3. Nov 4, 2012 #2

    mfb

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    Assuming 5 and 16 are fixed constants without uncertainty, you could quote R_L(min) as 0.26, as 5-0.4=4.6 and 16+1.6 = 17.6.

    R_L(max) looks fine.
     
  4. Nov 4, 2012 #3
    Actually this is the actual question. It's about finding the external load resistance of an open collector of standard TTL.

    Here's the picture so you can everything clearly. So can you check if I am doing it right?
     

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  5. Nov 4, 2012 #4

    mfb

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    You cannot answer this based on your equations.

    Significant figures can be a handy tool to estimate the uncertainty of a result. If you need a proper analysis, use actual uncertainties and not significant figures.
     
  6. Nov 4, 2012 #5
    But that's the formula given to find the max/min resistance. So how am I supposed to calculate them?
     
  7. Nov 4, 2012 #6

    mfb

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    As you did, with the improvement given in post 2 if you know that 5 and 16 are exact. If you do not know that, 0.3 is fine as result.
     
  8. Nov 4, 2012 #7
    Did I get the significant figure right?


    I know that when we take account of the significant figures, 5-2.4 would be 3 because there is no decimal value after five so it becomes unknown and we can't just make it 0 and minus 4. But I am not sure about this calculation though, because I don't know if I have to calculate everything normally first and then think about the s.g. when I get the final answer OR taking in account of the s.g. for every step of the calculation. I hope I am clear with my doubts here.
     
  9. Nov 4, 2012 #8

    mfb

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    The second method is better, the first can lead to problems in some types of calculations.
     
  10. Nov 4, 2012 #9
    So I would have to calculate 5-2.4 as 3??Then that means my final answer would only have 1 significant figure. So do I have to do that?
     
  11. Nov 4, 2012 #10

    mfb

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    If that value 5 is known to one significant figure only, yes.
     
  12. Nov 4, 2012 #11
    As you can see in the formula, that value of 5 is the source voltage. 5 is obviously one significant figure, right?
     
  13. Nov 4, 2012 #12

    mfb

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    Probably, and if you don't know anything else about it, just assume that it is 5 with an uncertainty of ~1.
     
  14. Nov 4, 2012 #13
    So that would mean the significant figure is 1. Correct?
     
  15. Nov 4, 2012 #14

    mfb

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    As I said before, yes.
     
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