Propagation of uncertainty problem

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Homework Help Overview

The problem involves calculating the equivalent resistance of a circuit with multiple resistors and determining the associated uncertainty in that value. The resistors have specified values and tolerances, which contribute to the uncertainty calculation.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to calculate the equivalent resistance and its uncertainty using formulas for propagation of uncertainty for both parallel and series resistors. Some participants question the correctness of the formulas used and whether the approach is overly complicated.

Discussion Status

The discussion is ongoing, with some participants confirming the original poster's calculations while others suggest that the provided answer may be incorrect. There is a suggestion to consider using absolute uncertainties instead of relative ones, indicating a potential shift in approach.

Contextual Notes

The problem includes specific tolerances for each resistor, which are relevant to the uncertainty calculations. The original poster expresses confusion regarding the discrepancy between their calculated uncertainty and the answer provided in the problem statement.

libelec
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Homework Statement



Find the equivalent resistance viewed by A and B and its equivalent uncertainty:
http://img707.imageshack.us/img707/5040/dadadadaz.png

R1 = 10ohm, 5% tolerance.
R2 = 2ohm, 1% tolerance.
R3 = 5ohm, 5% tolerance.
R4 = 15ohm, 1% tolerance.

The Attempt at a Solution



The Req = (1/R1 + 1/R2)-1 + (1/R3 + 1/R4)-1 = 65/12ohm = 5,41ohm.

Now, for the uncertainty, I separated the problem in two: first, find RA (10/6ohm), the equivalent resistance of the first two resistors and RB (15/4ohm), the equivalent resistance of the second two resistors, and then add them.

I was given a formula for propagation of uncertainty for parallel resistors and for series resistors, based in the formula for a function f(x1, x2,..., xk): [tex]\mu[/tex]f = [tex]\sum[/tex]|(xk/f)*(f'xk)|*[tex]\mu[/tex]xk:

For two parallel resistors: [tex]\mu[/tex] = (R2/(R1+R2))*[tex]\mu[/tex]R1 + (R1/(R1+R2))*[tex]\mu[/tex]R2

For two series resistors: [tex]\mu[/tex] = (R1/(R1+R2))*[tex]\mu[/tex]R1 + (R2/(R1+R2))*[tex]\mu[/tex]R2

If I do the calculus, I get that the total uncertainty is 3,28%: the uncertainty of A is 2/(10+2)*0.05 + 10/(10+2)*0.01 = 1/60, and the one of B is 15/(15+5)*0.05 + 5/(15+5)*0.01 = 1/25. Then the total uncertainty is (10/6)/(65/12)*1/60 + (15/4)/(65/12)*1/25 = 0.0328 = 3.28%.

But the answer given by the problem says it has to be 8.7%.

I'd like to know if I'm using the formulas right, or if I should be doing something with the uncertainties instead of that.

Thanks.
 
Last edited by a moderator:
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Anybody?
 
I got the same result as you.

ehild
 
Then the answer I was given is wrong? Or is there another method I could have used to reach that answer?

Thanks.
 
Last edited:
Sometimes wrong answers are given. Your calculation was correct but a bit overcomplicated.

Unless the function is a product or fractions of its variables, calculate with absolute uncertainties instead of relative ones.

[tex]\Delta f = \Sigma (|\frac{\partial f }{\partial x_i}|\Delta x_i)[/tex]

Divide the absolute error with the value of f if relative uncertainty is needed.

ehild
 
Last edited:

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