# Propagation of uncertainty problem

1. May 8, 2010

### libelec

1. The problem statement, all variables and given/known data

Find the equivalent resistance viewed by A and B and its equivalent uncertainty:

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

3. 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): $$\mu$$f = $$\sum$$|(xk/f)*(f'xk)|*$$\mu$$xk:

For two parallel resistors: $$\mu$$ = (R2/(R1+R2))*$$\mu$$R1 + (R1/(R1+R2))*$$\mu$$R2

For two series resistors: $$\mu$$ = (R1/(R1+R2))*$$\mu$$R1 + (R2/(R1+R2))*$$\mu$$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: May 4, 2017
2. May 9, 2010

Anybody?

3. May 9, 2010

### ehild

I got the same result as you.

ehild

4. May 9, 2010

### libelec

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

Thanks.

Last edited: May 9, 2010
5. May 10, 2010

### ehild

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.

$$\Delta f = \Sigma (|\frac{\partial f }{\partial x_i}|\Delta x_i)$$

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

ehild

Last edited: May 10, 2010