Calculating Error in Voltage (V) from Errors in I & R

[mathlete]

I don't know if this more of a math question than a physics question, but here goes:

Let's say I have an equation, let's say V=IR. Let's also say I know $$\sigma I$$ and $$\sigma R$$, the possible errors in I and R, how can I calculate $$\sigma V$$, the error in voltage? Is there a way to generalize this for any equation?

 

[Cantari]

square root of the sum of the error over the actual values squared equals error of V over V.

√((∆I/I)2 + (∆R/R)2 ) = ∆V/V

so ∆V = (V)√((∆I/I)2 + (∆R/R)2 )

 

[HallsofIvy]

What you are saying, I think, is that the actual value of R lies between R-δR and R+ δR and that the actual value of I lies between I- δI and I+ δI.

The smallest possible value then of V is (R-δR)(I- δI)= RI- 2δRδI+ (δR)(δI) while the largerst possible value is (R+ δR)(I+ δI)= RI+ 2δRδI+ (δR)(δI).

Can you get the error in V from that?

 

[mathlete]

What you are saying, I think, is that the actual value of R lies between R-δR and R+ delta;R and that the actual value of I lies between I- δI and I+ δI.

The smallest possible value then of V is (R-δR)(I- δI)= RI- 2δRδI+ (δR)(δI) while the largerst possible value is (R+ δR)(I+ δI)= RI+ 2δRδI+ (δR)(δI).

Can you get the error in V from that?

Well I cheated a bit, I already KNEW the answer in the form Cantari gave it. I was wondering if that is the standard way to find error if you have an equation, by taking the square root of the sum of the errors of your components (I, R in this case) over the actual values squared?

 

[Cantari]

Yes that is the standard way. Can be used for multiplication or division. Would be the same form if you wanted to find the error in I if you only had the errors in V and R. I = V/R.

 

[jdstokes]

Yes that is the standard way. Can be used for multiplication or division. Would be the same form if you wanted to find the error in I if you only had the errors in V and R. I = V/R.

I did a lot of work with error propagation for a high school experiment earlier this year. Error analysis can be an extremely tedious task, especially when the problem involves many inter-related variables. The paper covers the error propagation in a complicated multistage chemical analysis. If you like I can email the pdf file to you or upload it to the net somewhere.

 

[Cantari]

The uncertainty equation I gave is derived from a taylor expansion, it is indeed a proper way.

 

[krab]

The formula given by Cantari is not correct in every case. It requires that the measurements of the 2 quantities to be multiplied (a) are independent, and (b) have random errors with normal distributions.

 

[HallsofIvy]

But the original problem said that you KNOW the errors (which I took to mean maximum possible error), not that the errors were random.

 

[mathlete]

OK, the equations i'll need it for aren't very complex. They're all independent and the errors have normal distribution. Thanks for your help everyone.