Question

Given $R_1=5.0\pm 0.2\Omega ,R_2=10.0\pm 0.1\Omega .$ What is total resistance in parallel with possible percentage error?

• A. $15\Omega \pm 2\%$
• B. $3.3\Omega \pm 3\%$
• C. $15\Omega \pm 3\%$
• D. $3.3\Omega \pm 7\%$

Answer 1

Answer: (B)

In parallel,
$R_p=\frac{R_1{R}_2}{R_1+R_2}$ $=\frac{5.0\times 10.0}{5.0+10.0}$ $=\frac{50}{15}=3.3\Omega$
Also,
$R_P=3.3\Omega$
$\frac{1}{R_p}=\frac{1}{R_1}+\frac{1}{R_2}$
Differentiating
$⟹$ $\frac{\Delta R_p}{R_p^2}=\frac{\Delta R_1}{R_1^2}+\frac{\Delta R_2}{R_2^2}$
$\frac{\Delta R_p}{R_p}=\frac{50}{15}\left(\frac{0.2}{25}+\frac{0.1}{100}\right)$
%error = $\frac{50}{15}\left(\frac{0.8+0.1}{100}\right)\times 100=3\%$