The resistance R = VI where V = (200 5) V and I = (20 0.2) A. The percentage error in the measurement of R is:

The resistance R = VI where V = (200 5) V and I = (20 0.2) A. The… — Solution | JEE Physics

MCQEasyJEE 2024Significant Figures & Error Analysis

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

A
$3.5\%$
B
$7\%$
C
$3\%$
D
$5.5\%$

Answer

Correct answer:A

Step-by-step solution

the solution unavailable

Given: $R = \frac{V}{I}$ , $V = (200\pm 5)\,\text{V}$ , $I = (20\pm 0.2)\,\text{A}$ .

Find: Percentage error in $R$ .

Working could not be extracted from the solution. Using error propagation for division, the fractional errors add:

\frac{\Delta R}{R} = \frac{\Delta V}{V} + \frac{\Delta I}{I}

So,

\frac{\Delta R}{R}\times 100 = \left(\frac{5}{200} + \frac{0.2}{20}\right)\times 100

= (0.025 + 0.01)\times 100 = 3.5\%

Therefore, the correct option is A.

Common mistakes

Adding absolute errors $5$ and $0.2$ directly is incorrect because percentage or fractional errors must be combined for quantities related by division. First convert each uncertainty to fractional form, then add them.
Using subtraction of errors for $\frac{V}{I}$ is wrong. Although the variables are divided, the maximum fractional errors always add when propagating error in a quotient.

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Answer

Step-by-step solution

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