Question

When the voltage and current in a conductor are measured as $(100 \pm 4)$ V and $(5 \pm 0.2)$ A, then the percentage of error in the calculation of resistance is

• A. $8\%$
• B. $4\%$
• C. $20\%$
• D. $10\%$
• E. $6\%$

Hint:

For $R = V/I$, percentage errors always add.

Answer 1

The Correct Option is A

Concept: Error propagation in division
Resistance: \[ R = \frac{V}{I} \] For division: \[ \frac{\Delta R}{R} = \frac{\Delta V}{V} + \frac{\Delta I}{I} \] ---

Step 1: Calculate fractional error in voltage \[ \frac{\Delta V}{V} = \frac{4}{100} = 0.04 = 4\% \] ---

Step 2: Calculate fractional error in current \[ \frac{\Delta I}{I} = \frac{0.2}{5} = 0.04 = 4\% \] ---

Step 3: Add errors \[ \frac{\Delta R}{R} = 4\% + 4\% = 8\% \] But this is incorrect because error in denominator contributes differently. Actually for quotient: \[ \frac{\Delta R}{R} = \frac{\Delta V}{V} + \frac{\Delta I}{I} \] So: \[ = 4\% + 4\% = 8\% \] But we must consider significant propagation properly: More precise: \[ R = \frac{100}{5} = 20 \] Absolute error: \[ \Delta R = R \left(\frac{\Delta V}{V} + \frac{\Delta I}{I}\right) = 20 \times (0.04 + 0.04) = 20 \times 0.08 = 1.6 \] Percentage error: \[ \frac{1.6}{20} \times 100 = 8\% \] Correction: The correct answer should be $8\%$. --- Final Answer: \[ \boxed{8\%} \]