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Resistance of a given wire is obtained by measuring the current flowing in it and the voltage difference applied across it. If the percentage errors in the measurement of the current and the voltage difference are $3 \%$ each, then error in the value of resistance of the wire is
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$6 \%$
$6 \%$
$\mathrm{R}=\frac{\mathrm{V}}{\mathrm{i}}$
$\Rightarrow \quad\left|\frac{\Delta \mathrm{R}}{\mathrm{R}}\right|=\left|\frac{\Delta \mathrm{V}}{\mathrm{V}}\right|+\left|\frac{\Delta \mathrm{i}}{\mathrm{i}}\right|$
$\frac{\Delta \mathrm{V}}{\mathrm{V}} \times 100=3$
$\Rightarrow \quad \frac{\Delta \mathrm{V}}{\mathrm{V}}=0.03$
Similarly, $\frac{\Delta i}{i}=0.03$
Hence $\frac{\Delta \mathrm{R}}{\mathrm{R}}=0.06$
So percentage error is $\frac{\Delta \mathrm{R}}{\mathrm{R}} \times 100=6 \%$
$\Rightarrow \quad\left|\frac{\Delta \mathrm{R}}{\mathrm{R}}\right|=\left|\frac{\Delta \mathrm{V}}{\mathrm{V}}\right|+\left|\frac{\Delta \mathrm{i}}{\mathrm{i}}\right|$
$\frac{\Delta \mathrm{V}}{\mathrm{V}} \times 100=3$
$\Rightarrow \quad \frac{\Delta \mathrm{V}}{\mathrm{V}}=0.03$
Similarly, $\frac{\Delta i}{i}=0.03$
Hence $\frac{\Delta \mathrm{R}}{\mathrm{R}}=0.06$
So percentage error is $\frac{\Delta \mathrm{R}}{\mathrm{R}} \times 100=6 \%$
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