Question

Given $R_1=5.0{+}_0.2\Omega s,$ and $R_2=10.0{+}_0.1\Omega$. What is the total resistance in parallel with possible % error?

Answer 1

$R_1=5.0+0.2\Omega$

$R_2=10.0+0.1\Omega$

$R=\frac{R_1{R}_2}{R_1+R_2}=\frac{5\times 10}{5+10}=\frac{50}{15}=3.33\Omega$

$\frac{1}{R_p}=\frac{1}{R_1}+\frac{1}{R_2}\Rightarrow \frac{{\Delta R}_p}{R_p^2}=\frac{{\Delta R}_1}{R_1^2}+\frac{{\Delta R}_2}{R_2^2}$

${\Delta R}_p={\left(3.33\right)}^2[\frac{0.2}{{\left(5\right)}^2}+\frac{0.1}{{\left(10\right)}^2}]$

$={\left(3.33\right)}^2[\frac{0.2}{25}+\frac{0.1}{100}]$

${\Delta R}_p={\left(3.33\right)}^2\times 0.009$

${\Delta R}_p=0.099$

$R=3.33\pm 0.099\Omega$

% error $=\frac{\Delta R}{R}\times 100=\frac{0.099}{3.33}\times 100=2.997$%