1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: Percentage Error Analysis

  1. Jan 21, 2009 #1


    User Avatar

    1. The problem statement, all variables and given/known data
    I have a resistor which has value written as [itex]5K\Omega[/itex] and I measured the value and it is [itex]4.8K\Omega[/itex]. I want to work out the error.

    2. The attempt at a solution
    I can use simple percentage calculation to get [itex]\frac{4800-5000}{5000} \times 100 = -4[/itex]% error but I've been told that derivative can also be used to calculate error, how would I go about doing that? :confused: Thanks in advance.
  2. jcsd
  3. Jan 21, 2009 #2
    Determining % error using a derivative requires a continuous function that is derivable. Since the data is discrete and only two points are given a derivative in this case cannot be taken.
  4. Jan 21, 2009 #3


    User Avatar
    Homework Helper

    I think you may be confusing percent error with expected error. As chrisk suggests, you need to have a theoretical function that depends continuously on various physical parameters, and then you take the derivative w.r.t. those parameters, and then multiply by the precision of those parameters (basically). This error estimation using derivatives is not a percent error, because it makes no assumption regarding what the result of the measurement should be.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook