When the voltage and current in a conductor are measured as $(100\pm4) V$ and $(5\pm0.2)A$ , then the percentage of error in the calculation of resistance is

Answer (1)

Given,

$V=(100\pm4)V,I=(5\pm0.2)A$

As, $\displaystyle R=\frac{V}{I}$ ,

$\displaystyle\begin{aligned}\frac {\Delta R} {R} \times 100&=\left(\frac {\Delta V} {V} +\frac {\Delta I} {I} \right)\times 100\\&=\left(\frac {4} {100} +\frac {0.2} {5} \right)\times 100\\&=4\%+4\%=8\%\end{aligned}$