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 uncertatainty

  1. Sep 10, 2012 #1
    1. The problem statement, all variables and given/known data

    In a simple electrical circuit, the current in a resistor is measured as 2.5 ± 0.05 mA. The resistor is marked as having a value of 4.7 Ω ± 2%. If these values were used to calculate the power dissipated in the resistor, what would be the percentage uncertainty in the value obtained?

    2. Relevant equations


    3. The attempt at a solution
    please help me, what is the solution for this one? thanks

    i'm sorry, i post this in the advanced physics too, i'm confused where to post this.. hope you can help me...
  2. jcsd
  3. Sep 10, 2012 #2


    User Avatar
    Homework Helper

    I think that when multiplying two quantities, you merely add the percentage errors??

    In this case you are multiplying 3 things I x I x R as in P = I2R; so you would add all three percentage errors.
    You are given the percentage error in R, and can calculate the percentage error in I from the values given.

  4. Sep 10, 2012 #3
    Can you show me the solution? i'm confused... thanks peter...
  5. Sep 10, 2012 #4


    User Avatar
    Homework Helper

    If you wish to calculate the area of a rectangle with length 5.0 ±0.1 cm by 10.0 ± 0.1 cm, and give a percentage uncertainty, then

    A = l x w so Area = 50 cm2

    Now the percentage error.
    5.0 ± 0.1 means an uncertainty of 0.1 in 5 or 1 in 50 or 2%
    10.0 ± 0.1 mans an uncertainty of 1 in 100 or 1%

    SO the uncertainty in the answer is 3% [add them together]

    So Area is 50 cm2 ± 3%

    That is as close to the solution you seek I will give,
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook