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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
$6.33$
$2.33$
$4.33$
$5.33$
Solution
$\frac{1}{ R }=\frac{1}{ R _1}+\frac{1}{ R _2}$
Differentiating both sides, we get
$\frac{\Delta R }{ R ^2}=\frac{\Delta R _1}{ R _1^2}+\frac{\Delta R _2}{ R _2^2}\left[ R =\frac{ R _1 R _2}{ R _1+ R _2}=\frac{10 \times 15}{10+15}=6\right]$
$\Rightarrow \frac{\Delta R }{ R }=\left(\frac{\Delta R _1}{ R _1^2}+\frac{\Delta R _2}{ R _2^2}\right) R$
$=\left(\frac{0.5}{100}+\frac{0.5}{225}\right) 6$
$=\left(\frac{6 \times 0.5}{25}\right)\left(\frac{1}{4}+\frac{1}{9}\right)=\frac{13}{300}$
$\frac{\Delta R }{ R } \times 100=\frac{13}{3}=4.33 \%$
