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(D) The equivalent resistance $R_{eq}$ for two resistors in parallel is given by $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}$.First,calculate $R

Vedclass · Accepted Answer

$1.33$

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(D) The equivalent resistance $R_{eq}$ for two resistors in parallel is given by $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}$.First,calculate $R_{eq}$: $R_{eq} = \frac{R_1 R_2}{R_1 + R_2} = \frac{3 	imes 6}{3 + 6} = \frac{18}{9} = 2 \Omega$.Differentiating the expression $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}$,we get $-\frac{dR_{eq}}{R_{eq}^2} = -\frac{dR_1}{R_1^2} - \frac{dR_2}{R_2^2}$,which simplifies to $\frac{dR_{eq}}{R_{eq}^2} = \frac{dR_1}{R_1^2} + \frac{dR_2}{R_2^2}$.Multiplying by $R_{eq}$,we get $\frac{dR_{eq}}{R_{eq}} = \frac{R_{eq}}{R_1} \left( \frac{dR_1}{R_1} ight) + \frac{R_{eq}}{R_2} \left( \frac{dR_2}{R_2} ight)$.Given $\frac{dR_1}{R_1} = 1\%$ and $\frac{dR_2}{R_2} = 2\%$,we substitute the values:$\frac{dR_{eq}}{R_{eq}} = \frac{2}{3} (1\%) + \frac{2}{6} (2\%) = \frac{2}{3}\% + \frac{2}{3}\% = \frac{4}{3}\%$.$\frac{dR_{eq}}{R_{eq}} = 1.33\%$.

Two resistances are measured in $\Omega$ and are given as $R_1 = 3 \Omega \pm 1\%$ and $R_2 = 6 \Omega \pm 2\%$. When they are connected in parallel,the percentage error in equivalent resistance is .......... $\%$

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