Two resistance are measured in ohm and is given as:-
$R_{1} = 3 Ω \neq 1$
$R_{2} = 6 Ω \neq 2$
When they are connectrd in parallel, the percentage error in equivalent resistance is

3.33 %

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4.5 %

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0.67 %

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1.33 %

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Solution

The correct option is B $3.33 %$
$Given: R_{1} = 3 Ω \pm 0 \cdot 1 R_{2} = 6 Ω \pm 0.2$

$\begin{matrix} For parallel \frac{1}{R_{e q}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} = \frac{1}{3} + \frac{1}{6} = \frac{2 + 1}{6} = \frac{3}{6} = \frac{1}{2} \end{matrix}$

$\begin{matrix} R_{e q} = 2 Ω {Differentiating eq}^{n} (i) both side - \frac{d R_{e q}}{R_{e q}^{2}} = - \frac{d R_{1}}{R_{1}^{2}} - \frac{d R_{2}}{R_{2}^{2}} \end{matrix}$

$\begin{matrix} Error is always added \begin{matrix} \frac{d R_{e q}}{R_{e q}^{2}} & = \frac{d R_{1}}{R_{1}^{2}} + \frac{d R_{2}}{R_{2}^{2}} \frac{d R_{e q}}{R_{e q}} & = (\frac{d R_{1}}{R_{1}^{2}} + \frac{d R_{2}}{R_{2}^{2}}) R_{0} V = (\frac{0.1}{3^{2}} + \frac{0.2}{6^{2}}) 2 = (\frac{0.1}{9} + \frac{0.2}{36}) 2 \end{matrix} \end{matrix}$

$\begin{matrix} \frac{d Req x}{Reg} \times 100 % & = {(\frac{4 + 0.2}{36})}^{2} \times 100 % = \frac{0.6}{36} \times 2 \times 100 = \frac{100}{3} = 3.33 % \end{matrix}$

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