Question

For two resistor $R_1$ and $R_2$, connected in parallel, find the relative error in their equivalent resistance, if $R_1=(50\pm 2)\Omega$ and $R_2=(100\pm 3)\Omega$

Answer 1

$\text{Two resistors}$, $R_{1}and{R}_2$ $\text{are connected in parallel.}$

We know,

$\frac{1}{R_e}=\frac{1}{R_1}+\frac{1}{R_2}$

Then $R_e=\frac{R_1{R}_2}{R_1+R_2}$

$=\frac{50\times 100}{50+100}=\frac{5000}{150}=\frac{100}{3}$

Now, Parallel Connection error:

$=\frac{R_1^2\left(dB\right)+R_2^2\left(dA\right)}{(R_1+R_2{)}^2}$

$=\frac{{50}^2\left(3\right)+{100}^2\left(2\right)}{{150}^2}$

$=\frac{7500+20000}{150\times 150}$

$=\frac{11}{9}$

$\text{Relative Error}$ $=\frac{\left(11/9\right)}{\left(100/3\right)}$$=0.03666$