1. The problem statement, all variables and given/known data Find the equivalent resistance viewed by A and B and its equivalent uncertainty: http://img707.imageshack.us/img707/5040/dadadadaz.png [Broken] 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): [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.