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$R=\frac {V}{I}$

$V=(50\pm 2) V$

$I=(20 \pm 0.2 ) A $

% Error = ?

The resistance $R=\frac {V}{I}$, where $V=(50\pm 2) V$ and $I=(20 \pm 0.2 ) A $. The percentage error in R is 'x' %. The value of 'x' to the nearest integer is _ _ _ _ .

Solution

We have, $\frac{{|\Delta R|}}{R} = \frac{{|\Delta V|}}{V} + \frac{{|\Delta I|}}{I} = \frac{2}{{50}} + \frac{{0.2}}{{20}}$

$\therefore $ % Error = $\frac{{|\Delta R|}}{R} \times 100 = \frac{2}{{50}} \times 100 + \frac{{0.2}}{{20}} \times 100 = 4 + 1 = 5$

So, 'x' = 5

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