Question

When a current of $\left(2.5 \pm0.5\right)$ A flows through a wire, it develops a c of $\left(20\pm1\right)V,$ the resistance of wire is

• A. $\left(8 \pm2\right) \Omega$
• B. $\left(8 \pm1.6\right) \Omega$
• C. $\left(8 \pm1.5\right) \Omega$
• D. $\left(8 \pm3\right) \Omega$

Answer 1

Answer: (A)

Here,
Current, $I = (2.5 ± 0.5) \,A$
Potential difference, $V = (20 ± 1)V$
Resistance, $R=\frac{V}{I}=\frac{20\,V}{2.5\,A}=8\,\Omega$
$\frac{\Delta R}{R}=\pm\left(\frac{\Delta V}{V}+\frac{\Delta I}{I}\right)$
$=\pm\left(\frac{1}{20}+\frac{0.5}{2.5}\right)$
$\pm\left(0.05+0.2\right) =\pm\left(0.25\right)$
$\Delta R=\pm\left(0.25\right)\left(8\right)=\pm2\,\Omega$
$\therefore $ Resistance of the wire is $\left(8 \pm2\right)\,\Omega$.