A-Level · 15 September 2026 · 7 min
Percentage uncertainty in Physics: worked examples and common mistakes
Percentage uncertainty compares an estimated absolute uncertainty with the size of a measurement: percentage uncertainty = (absolute uncertainty ÷ magnitude of the measured value) × 100%. Keep both quantities in the same units before dividing.
The calculation is simple once the uncertainty has been chosen appropriately. These original examples explain that choice and the school-level combination rules used in Edexcel International A-Level Physics. Follow the uncertainty stated in a question and the conventions of your own board; different experiments can require different estimates.
Absolute uncertainty and percentage uncertainty are different
Suppose a measured length is L = (40.0 ± 0.5) cm. The absolute uncertainty is 0.5 cm. The percentage uncertainty is (0.5/40.0) × 100% = 1.25%, which could be reported as 1.3% to two significant figures.
For a length of 10.0 cm with the same absolute uncertainty, the percentage is 5.0%. The ruler has not changed, but the uncertainty is a larger fraction of the shorter measurement. This explains why measuring a larger suitable length can reduce relative uncertainty.
Changing units does not change the percentage: (0.005 m/0.400 m) × 100% is still 1.25%. Mixing 0.5 cm with 0.400 m without conversion gives a meaningless ratio. Percentage uncertainty is also not the same as percentage difference from an accepted value.
Use the spread of repeat readings thoughtfully
Three repeated times for the same event are 1.42 s, 1.46 s and 1.50 s. Their mean is 1.46 s, and half the range is (1.50 − 1.42)/2 = 0.04 s. Using half-range as the estimated uncertainty gives (1.46 ± 0.04) s and approximately 2.7% uncertainty.
That estimate describes the observed spread under this method. It does not prove that a stopwatch has no zero offset or that the same event was identified correctly each time. Repeated identical readings also do not mean zero uncertainty: the instrument and the method still have limitations.
For a single instrument reading, first use any uncertainty supplied in the question. Edexcel’s specification uses half the instrument resolution in its single-reading examples. Do not turn that into a claim that every real measurement’s total uncertainty is half a display increment; timing, alignment and calibration may contribute too.
A length or extension may involve two readings
Suppose an extension is found by subtracting two ruler positions: x₁ = (12.0 ± 0.1) cm and x₂ = (18.0 ± 0.1) cm. The extension is 6.0 cm. Using the simple maximum-uncertainty addition rule, the absolute uncertainties add: 0.1 + 0.1 = 0.2 cm.
The extension is therefore (6.0 ± 0.2) cm, with percentage uncertainty (0.2/6.0) × 100% ≈ 3.3%. Subtracting the uncertainties would incorrectly give zero. The two readings can shift in directions that make their difference less certain.
This example assumes the stated uncertainties contribute separately under the school-level rule. A common systematic offset shared by both readings may cancel in a difference; more advanced uncertainty analysis treats such correlations explicitly. Read what the question says about the measurement rather than adding every possible source automatically.
Combining uncertainties: multiply, divide and powers
For Edexcel Unit 6 calculations using the simple maximum-uncertainty approach, add percentage uncertainties when independent measured quantities are multiplied or divided. For a power, multiply the percentage uncertainty by the magnitude of that power. These are approximations for small relative uncertainties, not universal statistical rules.
For example, take V = (6.0 ± 0.1) V and I = (0.40 ± 0.02) A. The calculated resistance is R = V/I = 15 Ω. The voltage uncertainty is about 1.67%; the current uncertainty is 5.0%. Adding them gives approximately 6.7% in R, or 1.0 Ω. A suitable result is R = (15 ± 1) Ω.
The current measurement contributes more to this estimate. Improving only the voltage reading would therefore leave the larger contribution largely unchanged. This is how uncertainty calculations help choose a practical improvement, rather than merely add another number to a report.
Why diameter uncertainty doubles when calculating area
A circular wire has diameter d = (0.50 ± 0.01) mm. Its diameter uncertainty is 2%. Because A = πd²/4, the area uncertainty is approximately 2 × 2% = 4%. The exact constants π and 4 do not introduce measurement uncertainty in this model.
The area itself is π × (0.50 mm)²/4 ≈ 0.196 mm², or 1.96 × 10⁻⁷ m². Convert area using the square of the length conversion factor: 1 mm² = 10⁻⁶ m². Rounding appropriately gives about 0.20 mm² with a 4% estimated uncertainty.
If you first calculate radius by dividing diameter by 2, its absolute uncertainty also halves, so its percentage uncertainty remains 2%. Do not halve the percentage uncertainty and then square the radius: that would underestimate the uncertainty in the area.
Check your method with two short questions
Question 1: a time is (25.0 ± 0.2) s. What is its percentage uncertainty? Answer: (0.2/25.0) × 100% = 0.8%.
Question 2: a quantity is proportional to L²/t. The percentage uncertainties in independent measurements of L and t are 1.5% and 2.0%. What is the approximate combined percentage uncertainty using the rules above? Answer: 2 × 1.5% + 2.0% = 5.0%.
Before finishing any question, check that you used the appropriate estimate, kept units consistent and combined absolute or percentage uncertainties according to the operation. A negative measured coordinate can still have a positive uncertainty; use its magnitude in a relative calculation. Relative uncertainty is undefined for a measured value of zero, so quote an appropriate absolute uncertainty instead.