Two resistances are given as \(R _1=(10 \pm 0.5)\,\Omega\) and \(R_2=(15 \pm 0.5)\, \Omega\). The percentage error in the measurement of equivalent resistance when they are connected in parallel is

Differentiating both sides, we get

\(\frac{\Delta R }{ R ^2}=\frac{\Delta R _1}{ R _1^2}+\frac{\Delta R _2}{ R _2^2}\left[ R =\frac{ R _1 R _2}{ R _1+ R _2}=\frac{10 \times 15}{10+15}=6\right]\)

\(\Rightarrow \frac{\Delta R }{ R }=\left(\frac{\Delta R _1}{ R _1^2}+\frac{\Delta R _2}{ R _2^2}\right) R\)

\(=\left(\frac{0.5}{100}+\frac{0.5}{225}\right) 6\)

\(=\left(\frac{6 \times 0.5}{25}\right)\left(\frac{1}{4}+\frac{1}{9}\right)=\frac{13}{300}\)

\(\frac{\Delta R }{ R } \times 100=\frac{13}{3}=4.33 \%\)