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 :


ANSWER

$$(8 \pm 2)\Omega $$


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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