The resistance R = frac{V}{I} where V = (200 | vedclass

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(A) Given,$R = \frac{V}{I}$.According to the rules of error analysis for division,the relative error in $R$ is given by:$\frac{\Delta R}{R} = \frac{\D

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$3.5$

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(A) Given,$R = \frac{V}{I}$.According to the rules of error analysis for division,the relative error in $R$ is given by:$\frac{\Delta R}{R} = \frac{\Delta V}{V} + \frac{\Delta I}{I}$Substituting the given values $V = 200 \ V$,$\Delta V = 5 \ V$,$I = 20 \ A$,and $\Delta I = 0.2 \ A$:$\frac{\Delta R}{R} = \frac{5}{200} + \frac{0.2}{20}$$\frac{\Delta R}{R} = 0.025 + 0.01 = 0.035$To find the percentage error,multiply by $100$:$	ext{Percentage error} = \frac{\Delta R}{R} 	imes 100 = 0.035 	imes 100 = 3.5 \%$

The resistance $R = \frac{V}{I}$ where $V = (200 \pm 5) \ V$ and $I = (20 \pm 0.2) \ A$. The percentage error in the measurement of $R$ is: (in $\%$)

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