Answer
$0.059 ohm$
Work Step by Step
Write the differential form such as:
$dR=\dfrac{\partial R}{\partial R_1} \times dR_1 + \dfrac{\partial R}{\partial R_2} \times dR_2+ \dfrac{\partial R}{\partial R_3} \times dR_3$
$\triangle R=\dfrac{\partial R}{\partial R_1} \times \triangle R_1 + \dfrac{\partial R}{\partial R_2} \times \triangle R_2+ \dfrac{\partial R}{\partial R_3} \times \triangle R_3$
Need to plug the values.
$\triangle R=[\dfrac{(11.7647)^2}{(25)^2}] (0.125)+ [\dfrac{(11.7647)^2}{(40)^2}] (0.2)+ [\dfrac{(11.7647)^2}{(50)^2}] (0.25)=(138.408) (0.0002+0.000125+0.0001)$
Thus, we get $dR =\dfrac{1}{17} \approx 0.059$ ohm
