Uncertainty in parallel resistors.

Question

I need help with one of my study guide questions. We learned about uncertainty in class but am not sure how to attack this problem:

Could someone walk me through this example? Any help will be appreciated!

Answer 1

Parallel resistance is \begin{equation} R_p = \frac{R_1 R_2}{R_1+R_2} \end{equation} Uncertainty can be calculated by using differentials and a linear model. \begin{equation} dR_p = \frac{\partial R_p}{\partial R_1}dR_1 + \frac{\partial R_p}{\partial R_2} dR_2= \frac{R_2^2}{(R_1+R_2)^2}dR_1+\frac{R_1^2}{(R_1+R_2)^2}dR_2 \end{equation}

Hence, the total max uncertainty for $dR_1=dR_2=dR$, \begin{equation} dR_p = \frac{dR}{2}=0.05 \end{equation}

The total resistance is $1\pm0.05$. So it is not better to use a single resistor of value $1\pm0.1$

Answer 2

Another method :

The maximum resistance of the two R. in pll. is $(2+0.1)/2 = 1.05$

The minimum resistance of the two R. in pll. is $(2-0.1)/2 = 0.95$

So, the resistance is between 0.95 and 1.05

Comparison to the single resistor which resistance is between 0.9 an 1.1 shows that it is not better.

Note : the method shown by L__ is more general and better on the scolar viewpoint.