Two resistors R₁ = (100 ± 3) Ω and R₂ = (200 ± 4) Ω are connected in parallel. Choose the correct resistance of this parallel combination.

Here,

${\mathrm{R}}_1=(100\pm 3)\mathrm{\Omega },{\mathrm{R}}_2=(200\pm 4)\mathrm{\Omega }$

The equivalent resistance in parallel combination is

$\begin{array}{l}\frac{1}{{\mathrm{R}}_{\mathrm{p}}}=\frac{1}{{\mathrm{R}}_1}+\frac{1}{{\mathrm{R}}_2}\cdot \frac{1}{{\mathrm{R}}_{\mathrm{p}}}=\frac{1}{100}+\frac{1}{200}\\ =\frac{3}{200},{\mathrm{R}}_{\mathrm{p}}=\frac{200}{3}=66.7\mathrm{\Omega }\end{array}$

The error in equivalent resistance is given by

$\begin{array}{l}\frac{{\mathrm{\Delta R}}_{\mathrm{p}}}{{\mathrm{R}}_{\mathrm{\rho }}^2}=\frac{{\mathrm{\Delta R}}_1}{{\mathrm{R}}_1^2}+\frac{{\mathrm{\Delta R}}_2}{{\mathrm{R}}_2^2};{\mathrm{\Delta R}}_{\mathrm{p}}=\\ {\mathrm{\Delta R}}_1{\left(\frac{{\mathrm{R}}_{\mathrm{p}}}{{\mathrm{R}}_1}\right)}^2+{\mathrm{\Delta R}}_2{\left(\frac{{\mathrm{R}}_{\mathrm{p}}}{{\mathrm{R}}_2}\right)}^2\\ =3{\left(\frac{66.7}{100}\right)}^2+4{\left(\frac{66.7}{200}\right)}^2=1.8\mathrm{\Omega }\end{array}$

Hence, the equivalent resistance along with error in parallel combination is

$(66.7\pm 1.8)\mathrm{\Omega }$