1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

How to Calculate Resistors in Parallel Uncertainty

  1. Mar 5, 2016 #1
    1. The problem statement, all variables and given/known data

    We are told that resistor 1 has a resistance of 4700Ω ± 5% and resistor 2 has a resistance of 6800Ω ± 5% and the equivalent series and parallel resistances are to be calculated with an estimate of uncertainty.

    2. Relevant equations

    RSeries = R1 + R2
    RParallel = R1 × R2 / R1 + R2

    Uncertainties can simply be added if their measured numbers are added or subtracted.

    The formula we were given for multiplying or dividing uncertainty is the following:

    If z = x ⋅ y or z = x / y
    Then

    Δz=z(√(Δx/x)2+(Δy/y)2) (The square root covers everything in the equation except for z)

    3. The attempt at a solution

    I believe that I am doing the math correctly, but my the method doesn't really appear to correspond to anything else I have seen.

    I have done the following:

    RSeries = 4700 + 6800 =11500Ω
    Uncertainty = 235 + 340 = 575Ω (That's just 5% of each number)

    I feel that my method might be going wrong here:

    R1 × R2 = 4700 × 6800 = 31,960,000
    Uncertainty ≅ 2,259,913.273 (Using the uncertainty multiplication/division formula)

    The bottom half of the parallel equation has already been calculated.

    To get the equivalent resistance and uncertainty, I used the multiplication/division formula again.

    RParallel = 31,960,000 / 11500 ≅ 2,779Ω
    Uncertainty = 2779(√(2,259,913.273 / 31,960,000 )2 + (575 / 11500)2)

    Which gives ~241Ω.

    Is 2779 ± 241Ω a correct and acceptable way to display this answer or have I done something wrong here? Any feedback would be appreciated. :smile:
     
  2. jcsd
  3. Mar 5, 2016 #2

    berkeman

    User Avatar

    Staff: Mentor

    Welcome to the PF.

    You can easily check your answers by substituting the max values for each resistor in the equations for the series and parallel combinations, and then substitute in the minimum values and re-calculate. Then you can see what the % difference is (+/-%) between the max and min combinations...
     
  4. Mar 5, 2016 #3
    Hi. Thanks for the reply. Doing the following suggests that the uncertainty is ~139Ω.

    Looking online, I found a formula that corresponds to this answer.

    http://physics.stackexchange.com/qu...quivalent-resistance-of-resistors-in-parallel

    I would therefore have to think that I should apply this formula instead. Thank you for guiding me to the answer. :oldsmile: Having said that, I don't know what's wrong with what I did above. My two best guesses are that you can't use it for both a product and a quotient, because the uncertainty gets compounded somehow, or alternatively, the formula is wrong.
     
  5. Mar 5, 2016 #4

    berkeman

    User Avatar

    Staff: Mentor

    Yeah, I'd go with the formula that you found online. The parallel combination calculation involves both products and quotients. I'm no expert in uncertainty, but my simple method usually works for me. :smile:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted