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A current carrying conductor obeys Ohm's law $(V=R I)$. If the current passing through the conductor is $I=(5 \pm 0.2)$ A and voltage developed is $V=(60 \pm 6) \mathrm{V}$, then find the percentage of error is resistance, $R$
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Verified Answer
The correct answer is:
14
Given, current passing through the conductor,
$I=(5 \pm 0.2) \mathrm{A}, \Delta I=0.2 \mathrm{~A}$
Voltage developed,
$\begin{array}{rlrl}
& & V & =(60 \pm 6) \mathrm{V} \\
\Rightarrow \quad & \Delta V & =6 \mathrm{~V}
\end{array}$
By Ohm's law,
$\begin{aligned}
V & =R I \\
R & =\frac{V}{I} \\
\frac{\Delta R}{R} & =\frac{\Delta V}{V}+\frac{\Delta I}{I} \\
& =\frac{6}{60}+\frac{0.2}{5} \\
& =0.1+0.04 \\
\frac{\Delta R}{R} & =0.14 \\
\therefore \quad \frac{\Delta R}{R} \times 100 & =0.14 \times 100=14 \%
\end{aligned}$
$I=(5 \pm 0.2) \mathrm{A}, \Delta I=0.2 \mathrm{~A}$
Voltage developed,
$\begin{array}{rlrl}
& & V & =(60 \pm 6) \mathrm{V} \\
\Rightarrow \quad & \Delta V & =6 \mathrm{~V}
\end{array}$
By Ohm's law,
$\begin{aligned}
V & =R I \\
R & =\frac{V}{I} \\
\frac{\Delta R}{R} & =\frac{\Delta V}{V}+\frac{\Delta I}{I} \\
& =\frac{6}{60}+\frac{0.2}{5} \\
& =0.1+0.04 \\
\frac{\Delta R}{R} & =0.14 \\
\therefore \quad \frac{\Delta R}{R} \times 100 & =0.14 \times 100=14 \%
\end{aligned}$
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