# How uncertainties affect values

If I calculate the equivalent resistance of a circuit, for example:

$$1/R_{eq} = 1/R_1 + 1/R_2 + 1/R_3 = 1/1472 + 1/3260 + 1/5580 \Rightarrow R_{eq} = 858.22\,\Omega$$

And then calculate its uncertainty:

$$\Delta R_{eq} = |\delta 1/\delta R_1| \times \Delta R_1 + |\delta 1/\delta R_2| \times \Delta R_2 + |\delta 1/\delta R_3| \times \Delta R_3 = 0.001\,\Omega$$

Will the uncertainty affect the value $(858.22)$, given that it only has 2 decimal numbers?

• I get $\Delta R_{eq}=1$, so $R=858\pm 1$. Commented Oct 15, 2013 at 23:15
• if the uncertainties are uncorrelated, you should be adding the squares and taking the square root. Commented Oct 16, 2013 at 0:57

As @NowIGetToLearnWhatAHeadIs notes in a comment, the uncertainties in the resistors are almost certainly uncorrelated, so you want to add them in quadrature (NB: pdf). So that would give you $$(\Delta(1/R_{eq}))^2 = (\delta(1/R_1))^2 + (\delta(1/R_2))^2 + (\delta(1/R_3))^2$$
In addition, $\delta(1/R_1) \neq \delta(1)/\delta(R_1) = 0$. The general form for the uncertainty of a power is $\delta(x^n) = |nx^{n-1}|\delta(x)$ (see Taylor, Introduction to Error Analysis for a derivation). You can use that here with $n=1$.
To answer your title question directly, an uncertainty of $\Delta R_{eq} = 0.001 \Omega$ will not noticeably impact a value of $R_{eq} = 858.22 \Omega$. Any measurement you made to measure that $R_{eq}$ would not be affected by an uncertainty that small. However, I strongly suspect (without working things out myself) that your uncertainty is much larger than that.