Physics

# When a current of $(2.5 0\pm 0.5)A$ flows through a wire, it develops a potential difference of $(20\pm 1)V$. The resistance of the wire is :

$(8 \pm 2)\Omega$

##### SOLUTION
Given that current of  $2.50 \pm 0.5$ A flows across a wire of potential difference $20\pm 1 V$.
$V=IR$ or $R=\dfrac{V}{I}=\dfrac{20 }{2.5}=8 \Omega$

$\Rightarrow \dfrac{\Delta R}{R}=\dfrac{\Delta V}{V}+\dfrac{\Delta I}{I}$

$=\dfrac{1}{20}+\dfrac{0.5}{2.5}\\=0.05+0.2\\=0.25$

$\Delta R =0.25 \times 8 =2$
Now, we have resistance with limits of error $= 8\pm 2 \Omega$

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Single Correct Medium Published on 18th 08, 2020
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