I dont understand how to calculate percentage uncertainty!?by mutineer123 Tags: percentage error, uncertainity 

#1
Mar3012, 09:33 PM

P: 93

I know in chem we calculate uncertainties by smallest division/the reading X 100. But in physics i am clueless. For instance heres an example
The power loss P in a resistor is calculated using the formula P = V^2/R. The uncertainty in the potential difference V is 3% and the uncertainty in the resistance R is 2%. What is the uncertainty in P? (this is not homework) I just want to know how to get sums like these. 



#2
Mar3112, 10:43 AM

P: 84

mutineer123,
if your quantity is a function of other variables then you first calculate the absolute error and then divide by the measured value to get the % error. At all points in the calculation you must use the absolute error not the %. The method is to find the error relating to each variable (R and V in your example) these are squared and the squares added, finialy take the square root. This will give you the absolute error in the power. To find the contribution from each of the variables you differentiate the function with respect to the variable and multiply by the absolute error of that variable. This is the quantity you square as mentioned above. See section 4 of http://www.phas.ubc.ca/~phys209/erro...sis/errors.pdf Regards sam 



#3
Apr112, 03:50 PM

P: 210

On my course we are taught that when quantities are multiplied or divided the % errors are added together.
It was shown like this....imagine V=100 +or3%. This means that V could be as large as 103 or4 as small as 97. V^2 could be as large as (103x103) = 10609 which is pretty much 6% larger than (100x100) Also V^2 could be as small as (97x97)= 9409 which is also pretty much 6% smaller than (100x100). So error in V^2 is +or6%. You can do this exercise to include the 2% error in R and you will find that the overall % error is pretty much 8% (3 + 3 + 2) Our teacher did several examples like this and it was convincing. 



#4
Apr112, 06:09 PM

HW Helper
P: 6,925

I dont understand how to calculate percentage uncertainty!?P_min = (.97 V)^{2} / (1.02 R) ~= 0.922451 ~= 1  0.077549 P_max = (1.03 V)^{2} / (.98 R) ~= 1.082551 P_nonerror = (1.00 V)^{2} / (1.00 R) = 1.000000 P_avg_with_error ~= 1.002501 ± 8.005 % This is excessive precision since the orginal values only show percentages (assume 3 digits of precision), but generally high precision calculations are done for worst case until a final answer is produced. On a side note, if you're looking for distribution for the uncertainty, it's complicated by the fact that a 2% resistor could have been selected from a large batch of resistors where all the resistors with 1% or less error were removed and labeled as 1% resistors, so you have a bell curve with an empty center. If the 2% resistors were removed with the rest of the resistors to be labeled as greater than 2% resistors, then you'd have a bell curve only with the region betwen +/ 1% to 2%, empty center, empty edges. 



#5
Apr212, 02:16 PM

P: 210

Our method of simply adding %s when you have quantities multiplied or divided gave the same value....8%




#6
Apr212, 05:46 PM

HW Helper
P: 6,925





#7
Apr312, 12:42 PM

Mentor
P: 10,791

In general, the error depends on the correlation of the individual errors from V and R (and the distribution of the errors, but that is often close to a Gaussian distribution).
Without correlation, you shouldn't add the percentages, but the squares of them. The 6% error for V^2 are fine, but the total error is then given by [tex]\sqrt{6^2+2^2}\%=\sqrt{40}\%=6.3\%[/tex] The difference to the "worst case" scenario of rcgldr comes from the fact that a larger voltage can come together with a larger resistance at the same time, too: (1.03U)^2/(1.02R) =~ 1.04 U^2/R 



#8
Apr312, 02:15 PM

HW Helper
P: 6,925





#9
Apr312, 08:50 PM

P: 93

Okay Thanks guys, Il apply this knowledge to the questions Il solve, and if I have any difficulty Il get back to the thread. Again Thanks alot




#10
Apr312, 09:08 PM

Math
Emeritus
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Thanks
PF Gold
P: 38,879

Suppose X has error dX, Y has error dY. If Z= X+ Y then dZ= dX+ dY while if Z= XY, dZ= XdY+ YdX so, dividing by Z= XY, [tex]\frac{dZ}{Z}= \frac{XdY}{XY}+ \frac{YdX}{XY}= \frac{dY}{Y}+ \frac{dZ}{Z}[/tex] 



#11
Apr412, 01:09 AM

P: 210

I know that when quantities are added the absolute errors must be added.
The total error is then converted to a % (relative?) error if the quantity is part of a multiplication such as working out an area or volume from lengths. Then adding the %s seems to give the overall error for the number of significant figures involved 



#12
Apr412, 08:10 AM

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P: 10,791





#13
Apr412, 03:20 PM

P: 210

I am getting confused by so many techniques for dealing with errors, the different techniques do not seem to make much difference.
If adding % errors give 8% total error and doing the square root calculatio give 6.4% total error how will this show in the significant figures of an actual answer?....if at all 



#14
Apr512, 08:30 AM

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P: 10,791

With just two errors, the difference is not so significant. But imagine 4 different uncertainties of 5% each. Adding the uncertainties linearly gives 20% uncertainty, adding them in quadrature gives 10% total uncertainty.
100+10 and 100+5 (as an example) are really different measurements in a scientific context. 


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