Two resistance are measured in Ohm and is&nbs | vedclass

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Question

$\frac{1}{R_{e q}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}$

$\frac{\mathrm{d} \mathrm{R}_{\mathrm{eq}}}{\mathrm{R}_{\mathrm{eq}}^{2}}=\frac{\mathrm{dR}_{1}

Vedclass · Accepted Answer

$1.33$

Vedclass · Answer

$\frac{1}{R_{e q}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}$

$\frac{\mathrm{d} \mathrm{R}_{\mathrm{eq}}}{\mathrm{R}_{\mathrm{eq}}^{2}}=\frac{\mathrm{dR}_{1}}{\mathrm{R}_{1}^{2}}+\frac{\mathrm{d} \mathrm{R}_{2}}{\mathrm{R}_{2}^{2}}$

$\Rightarrow \frac{\mathrm{d} R_{e q}}{R_{e q}}=R_{e q}\left\{\frac{d R_{1}}{R_{1}} 	imes \frac{1}{R_{1}}+\frac{d R_{2}}{R_{2}} 	imes \frac{1}{R_{2}}ight\}$

$\%$ error $=2\left\{\frac{1}{3}+\frac{2}{6}ight\}$

$=\frac{4}{3} \%=1.33 \%$

Two resistance are measured in $Ohm$ and is 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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