Question

I need help solving this problem...
If R is the total resistance of three resistors, connected in parallel, with resistances r1 r2 r3 then

1/R = 1/R1 + 1/R2 + 1/R3

R1= 25 ohms R2 = 40 ohms R3= 50 ohms ,

possible error of 0.5 % in each case

Estimate the max error in the calculated value of R

please show the steps in solving this because its a type of a final exam problem (accdng to my prof.)

Answer 1

\[\frac{1}{R}=\frac{1}{25\Omega}+\frac{1}{40\Omega}+\frac{1}{50\Omega}=\frac{8+5+4}{200\Omega}=\frac{17}{200\Omega}\quad\Rightarrow\quad R=\frac{200}{17}\Omega\]\[R=\frac{R_1R_2R_3}{R_1R_2+R_2R_3+R_3R_1}\]\[\frac{\partial R}{\partial R_1}=\frac{R_2R_3(R_1R_2+R_2R_3+R_3R_1)-R_1R_2R_3(R_2+R_3)}{(R_1R_2+R_2R_3+R_3R_1)^2}=\]\[=\left(\frac{R_2R_3}{R_1R_2+R_2R_3+R_3R_1}\right)^2\]\[\frac{1}{R}\frac{\partial R}{\partial R_1}\Delta R_1=\frac{R_2R_3}{R_1R_2+R_2R_3+R_3R_1}\delta R_1\]\[\frac{1}{R}\frac{\partial R}{\partial R_2}\Delta R_2=\frac{R_3R_1}{R_1R_2+R_2R_3+R_3R_1}\delta R_1\]\[\frac{1}{R}\frac{\partial R}{\partial R_3}\Delta R_3=\frac{R_1R_2}{R_1R_2+R_2R_3+R_3R_1}\delta R_3\]\[\delta R=\frac{1}{R}\frac{\partial R}{\partial R_1}\Delta R_1+\frac{1}{R}\frac{\partial R}{\partial R_2}\Delta R_2+\frac{1}{R}\frac{\partial R}{\partial R_3}\Delta R_3\]\[\delta R_1=\delta R_2=\delta R_3=0.5\%\quad\Rightarrow\quad\delta R=0.5\%\]