Guaranteed accuracy and limiting error: A 0–25 A ammeter has ±1% of full-scale accuracy. If it reads 5 A, what is the limiting error expressed as a percentage of the measured value?
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A2%
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B2.5%
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C4%
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D5%
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E1%
Answer
Correct Answer: 5%
Explanation
Introduction / Context:Guaranteed accuracy is often specified as a percentage of full-scale (FS). When the reading is far below full-scale, the same absolute error corresponds to a larger percentage of the reading. Understanding this distinction is crucial for correct error reporting and for deciding appropriate instrument ranges.
Given Data / Assumptions:
- Ammeter range: 0–25 A.
- Accuracy: ±1% of full-scale.
- Measured current: 5 A.
Concept / Approach:
Absolute error at any reading = (accuracy% of FS) * FS. Limiting error as a percentage of the measured value = (absolute error / reading) * 100%. This converts a full-scale-based tolerance to a reading-based percentage, which is commonly requested in test reports.
Step-by-Step Solution:
Compute absolute error: ±1% of 25 A = ±0.25 A.Convert to percentage of reading: (0.25 / 5) * 100% = 5%.Hence the limiting error at 5 A is ±5% of the indicated value.Verification / Alternative check:
If the meter read 25 A, the same ±0.25 A would be ±1% of reading, matching the specification. Lower readings inflate the percent-of-reading error, which is a key reason to select the smallest meter range that safely covers the expected value.
Why Other Options Are Wrong:
- 2%, 2.5%, 4%: underestimate the impact of applying a full-scale accuracy to a sub-range reading.
- 1%: confuses accuracy-of-FS with accuracy-of-reading.
Common Pitfalls:
- Reporting “±1%” at all readings without converting to percentage of the actual reading.
- Mixing up full-scale and of-reading specifications.
Final Answer:
5%