# Resistance of metal is given by $R=\large\frac{V}{I}$. $V$ is potential difference, $I$ is current. In a circuit $V=(8 \pm 0.5)v$ and current is $I=(2 \pm 0.2)A$. what is value of resistance with percentage error?

$(a)\;4 \Omega \pm 16.25 \% \quad (b)\;(4 \pm 0.7)\Omega \quad (c)\;4 \Omega \pm 0.7\%\quad (d)\;4 \Omega \pm 7\%$

Resistance $=\large\frac{8}{2}$$=4 \Omega % error in R = % error in V + % error in I. \qquad\qquad\qquad=\large\frac{.5}{8}$$\times 100+ \large\frac{.2}{2}$$\times 100$
$\qquad\qquad\qquad=16.25\%$
Hence the $\%$ of error is $16.25 \%$
Therefore, the correct answer is $(A)\;4 \Omega \pm 16.25 \%$
edited Mar 12, 2014