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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
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The correct answer is:
4.33
4.33
Equivalent resistance in parallel combination is given by
$\begin{aligned}
& \frac{1}{R_{\text {eq }}}=\frac{1}{R_1}+\frac{1}{R_2} \\
& \Rightarrow R_{\text {eq }}=\frac{R_1 R_2}{R_1+R_2}=\frac{10 \times 15}{10+15}=6
\end{aligned}$
Differentiating both sides, we get
$\begin{aligned} & \frac{\Delta R_{\text {eq }}}{\mathrm{R}_{\mathrm{eq}}^2}=\frac{\Delta \mathrm{R}_1}{\mathrm{R}_1^2}+\frac{\Delta \mathrm{R}_2}{\mathrm{R}_2^2} \\ & \Rightarrow \frac{\Delta \mathrm{R}_{\mathrm{eq}}}{\mathrm{R}_{\mathrm{eq}}}=\left(\frac{\Delta \mathrm{R}_1}{\mathrm{R}_1^2}+\frac{\Delta \mathrm{R}_2}{\mathrm{R}_2^2}\right) \mathrm{R}_{\mathrm{eq}} \\ & =\left(\frac{0.5}{100}+\frac{0.5}{225}\right) \cdot 6=\left(\frac{6 \times 0.5}{25}\right)\left(\frac{1}{4}+\frac{1}{9}\right)=\frac{13}{300} \\ & \frac{\Delta \mathrm{R}_{\mathrm{eq}}}{\mathrm{R}_{\mathrm{eq}}} \times 100=\frac{13}{3}=4.33 \%\end{aligned}$
$\begin{aligned}
& \frac{1}{R_{\text {eq }}}=\frac{1}{R_1}+\frac{1}{R_2} \\
& \Rightarrow R_{\text {eq }}=\frac{R_1 R_2}{R_1+R_2}=\frac{10 \times 15}{10+15}=6
\end{aligned}$
Differentiating both sides, we get
$\begin{aligned} & \frac{\Delta R_{\text {eq }}}{\mathrm{R}_{\mathrm{eq}}^2}=\frac{\Delta \mathrm{R}_1}{\mathrm{R}_1^2}+\frac{\Delta \mathrm{R}_2}{\mathrm{R}_2^2} \\ & \Rightarrow \frac{\Delta \mathrm{R}_{\mathrm{eq}}}{\mathrm{R}_{\mathrm{eq}}}=\left(\frac{\Delta \mathrm{R}_1}{\mathrm{R}_1^2}+\frac{\Delta \mathrm{R}_2}{\mathrm{R}_2^2}\right) \mathrm{R}_{\mathrm{eq}} \\ & =\left(\frac{0.5}{100}+\frac{0.5}{225}\right) \cdot 6=\left(\frac{6 \times 0.5}{25}\right)\left(\frac{1}{4}+\frac{1}{9}\right)=\frac{13}{300} \\ & \frac{\Delta \mathrm{R}_{\mathrm{eq}}}{\mathrm{R}_{\mathrm{eq}}} \times 100=\frac{13}{3}=4.33 \%\end{aligned}$
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