Calculating Uncertainties with linear combinations

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Homework Statement



I am having trouble determining the error for a set of linear equations that represent a simple circuit with two voltage sources. I have found two possible uncertainties by solving using substitution, detailed below.

The circuit is shown below:
http://img233.imageshack.us/img233/9165/circuit1small.jpg [Broken]



Homework Equations



I am solving the circuit using loop analysis and the two equations found are:

1. I1R1 + R2(I1+I2) = V1
2. I2R3 + R2(I1+I2) = V2



The Attempt at a Solution




Solving 2. for I2 and subbing into 1. gives:
I1 = 25.8 mA +- 17.2%

Solving by back subbing into 2. gives:
I2 = 128.5 mA +- 6.1%

or subbing I1 into 1. gives:
I2 = 128.5 mA +- 25.8%


So which one is correct 25% or 6%?
 
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Answers and Replies

  • #2
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What rule are you using?
 
  • #3
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What rule are you using?
I'm sorry, I do not understand what you mean by rule. As far as I can tell I have not violated any rules?
I'm really just curious to know if anyone has an idea why the uncertainty would be different depending on which equation is used to solve for I2.
 
  • #4
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I mean that there are different "rules" for calculating uncertainties whether you are adding, or multiplying, etc. There is a "master" rule that goes as

given function f(x,y,...)
[tex]\sigma_f = \sqrt{(\frac{\partial f}{\partial x} \sigma_x)^2 + (\frac{\partial f}{\partial y} \sigma_y)^2+...}[/tex]
 
  • #5
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I mean that there are different "rules" for calculating uncertainties whether you are adding, or multiplying, etc. There is a "master" rule that goes as

given function f(x,y,...)
[tex]\sigma_f = \sqrt{(\frac{\partial f}{\partial x} \sigma_x)^2 + (\frac{\partial f}{\partial y} \sigma_y)^2+...}[/tex]
Oh interesting, I have not seen the Partial method of calculating uncertainties.
We are using the standard method most people learn in first year physics, addition/subtraction is adding absolute error; multiplication/division is adding of relative errors.

I showed it to my prof relatively quickly and even she was dumbfounded as to why the uncertainties are different.
 
  • #6
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Well, they should be the same. Maybe we can figure it out. I don't know the value and uncertainty of R2 though.

How did you calculate this uncertainty if all you've covered is addition/subtraction? This is a problem for mathematica.
 
  • #7
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R1= 370+- 10 Ohm
R2= 100+- 3 Ohm
R3= 220+- 6 Ohm
V1= 25V+- 2%
V2= 45V+- 2%


The uncertainty that I ended with was a result of manually calculating each operation with each absolute or relative uncertainty. Our lab fell on a holiday last week, so the prof assigned this 'simple' question as a replacement, I suppose to keep us busy.

The uncertainty for each component is a given in the question.
 
Last edited:
  • #8
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I hate to say it, but your values are probably just not right. According to matlab, using the master rule, the uncertainty in I1 is

[tex]\Delta I_1 = \sqrt{\Delta V1^2 + \Delta V2^2 + (\frac{V2R1+V1R3)R3}{R1R3+R1R2+R2R3}\Delta R2)^2 + (\frac{V1-R2V2/(R3+R2)}{R1+R2-R1^2/(R1+R2)}\Delta R1)^2 + (\frac{V2R1+R2V2-V1R2)R3}{R1R3+R1R2+R2R3)^2}\Delta R3)^2}[/tex]
 
  • #9
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It is entirely likely that I made an error even though I checked the work multiple times to be sure; but, I did end with the same current for both equations.

Is there another name for the Master rule? I cannot seem to find much on it.
 

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