📚 Analysing the AQA Unit 3 EMPA Insert January 2019: A Guide to Experimental Investigations | AQA 2019年1月Unit 3 实验插入资料分析：实验探究指南
In AQA A-Level Physics, Unit 3 (EMPA) challenges students to analyse experimental data provided in a separate insert booklet. The January 2019 paper featured a classic investigation: determining the resistivity of a metal wire. This article will guide you through the typical content of such an insert, unpacking the experimental setup, measurement techniques, data processing, and uncertainty evaluation required for top marks. You will learn how to extract the maximum information from the given data, perform necessary calculations, and present a logical evaluation of errors and limitations.
在AQA A-Level物理中，Unit 3 (EMPA) 要求学生分析一份单独的插入资料册中提供的实验数据。2019年1月试题包含一项经典探究：测定金属丝的电阻率。本文将引导你了解此类插入资料的典型内容，深入解析实验设置、测量技巧、数据处理以及获得高分所需的不确定度评估。你将学到如何从给定数据中提取最多信息、进行必要的计算，并对误差与局限性进行合乎逻辑的评估。
1. Understanding the Insert Document | 理解插入资料文档
The insert typically begins with a brief description of the experiment's aim, a list of apparatus, and a labeled diagram. For Jan 19, students were provided with a circuit to measure current and voltage across a length of constantan wire, along with a micrometer screw gauge and a metre ruler. You must first identify the independent variable (length of wire), the dependent variable (resistance, derived from V/I), and the control variables (temperature, wire diameter). The insert also includes pre-recorded repeated readings of the wire diameter and a single value for the length between crocodile clips.
插入资料通常以简要的实验目的说明、仪器列表和带标注的示意图开始。在1月19日的试题中，学生获得的资料包括一个测量康铜丝电流和电压的电路，以及一把千分尺和一把米尺。你必须首先识别自变量（导线长度）、因变量（电阻，由 V/I 得出）和控制变量（温度、导线直径）。资料中还包含预先记录的对导线直径的多次重复读数以及一个鳄鱼夹间长度的单次测量值。
2. Initial Observations: The Circuit Assembly | 初步观察：电路组装
The insert diagram showed a power supply, a rheostat used as a potential divider, an ammeter in series with the test wire, and a voltmeter in parallel across the wire. Recognising this arrangement is crucial: the potential divider allows you to vary the p.d. smoothly from zero, reducing heating effects. You should note that the ammeter must be connected to measure the current flowing through the wire only, and the voltmeter leads should be attached exactly at the crocodile clips to avoid measuring the resistance of the connecting leads. The insert explicitly reminded candidates to record the room temperature and keep the current low to minimise joule heating.
插入资料中的示意图展示了一个电源、一个作为分压器使用的变阻器、一个与待测导线串联的电流表以及一个并联在导线两端的电压表。识别这种连接方式至关重要：分压器可使电压从零开始平稳变化，从而减小热效应。你应当注意，电流表必须仅测量通过导线的电流，电压表的引线应恰好连接在鳄鱼夹处，以避免测量连接导线的电阻。插入资料明确提醒考生记录室温并保持电流较小，以尽量减少焦耳热效应。
3. Mastering Micrometer Readings: Wire Diameter | 掌握千分尺读数：导线直径
m, 0.276 mm, and 0.272 mm. The mean diameter d is calculated as (0.274 + 0.276 + 0.272) ÷ 3 = 0.274 mm. The precision of the micrometer is 0.001 mm, so the resolution uncertainty is ±0.001 mm. However, the spread of the readings (0.276 − 0.272 = 0.004 mm) suggests a larger random uncertainty. Using the range/2 method, the absolute uncertainty in d is ±0.002 mm. Therefore, the reported value is d = 0.274 ± 0.002 mm. Always convert to metres before area calculations: d = 2.74 × 10⁻⁴ m.
插入资料给出了康铜丝直径的三次千分尺原始读数，分别在导线不同位置测得：0.274 mm、0.276 mm 和 0.272 mm。平均直径 d 计算为 (0.274 + 0.276 + 0.272) ÷ 3 = 0.274 mm。千分尺的仪器精确度为 0.001 mm，因此读数分辨率不确定度为 ±0.001 mm。然而，读数的分散程度（0.276 − 0.272 = 0.004 mm）表明随机不确定度更大。使用半极差法，d 的绝对不确定度为 ±0.002 mm。因此，报告值为 d = 0.274 ± 0.002 mm。进行面积计算前务必转换为米：d = 2.74 × 10⁻⁴ m。
4. Uncertainty in Length and Repeated Measurements | 长度测量与重复测量的不确定度
, but alignment errors at both ends give an absolute uncertainty of ±0.002 m. Although only a single length is used, the experiment is designed around varying voltage to find resistance at fixed L. A more sophisticated investigation would vary L, but the Jan 19 insert focused on resistance determination from an I– . You must later propagate the percentage uncertainty in L, which is (0.002 ÷ 0.800) × 100% = 0.25%, into the final resistivity.
插入资料指出，使用米尺设定鳄鱼夹间导线的长度 L 为 0.800 m。米尺的最小分度为 1 mm，但两端的对准误差使得绝对不确定度为 ±0.002 m。尽管只使用了一个长度值，该实验通过改变电压来测定固定 L 下的电阻。更复杂的探究会改变 L，但 2019 年 1 月的插入资料侧重于通过 I–V 特性曲线测定电阻。你必须随后将 L 的百分比不确定度 (0.002 ÷ 0.800) × 100% = 0.25% 传递到最终的电阻率中。
5. Designing the Data Table for I-V Characteristics | 设计I-V特性数据表格
d. I will reconstruct a typical set: (0.50, 0.12), (1.00, 0.24), (1.50, 0.36), (2.00, 0.48), (2.50, 0.60). You must add a column for the calculated resistance R = V/I, and the insert often asks for the absolute uncertainty for each instrument. The voltmeter used had a resolution of 0.05 V, so the absolute uncertainty in V is ±0.025 V. The ammeter had a resolution of 0.01 A, giving an uncertainty of ±0.005 A. This table will be the basis for plotting a graph of V against I.
| V / V | I / A | R = V/I (Ω) |
|---|---|---|
| 0.50 ± 0.03 | 0.12 ± 0.005 | 4.17 |
| 1.00 ± 0.03 | 0.24 ± 0.005 | 4.17 |
| 1.50 ± 0.03 | 0.36 ± 0.005 | 4.17 |
| 2.00 ± 0.03 | 0.48 ± 0.005 | 4.17 |
| 2.50 ± 0.03 | 0.60 ± 0.005 | 4.17 |
d. I will reconstruct a typical set: (0.50, 0.12), (1.00, 0.24), (1.50, 0.36), (2.00, 0.48), (2.50, 0.60). You must add a column for the calculated resistance R = V/I, and the insert often asks for the absolute uncertainty for each instrument. The voltmeter used had a resolution of 0.05 V, so the absolute uncertainty in V is ±0.025 V. The ammeter had a resolution of 0.01 A, giving an uncertainty of ±0.005 A. This table will be the basis for plotting a graph of V against I.
一个结构良好的表格至关重要。插入资料提供了五组电压 V（单位V）和电流 I（单位A）的读数，随着电压逐渐增加测得。我将重构一组典型数据：(0.50, 0.12)、(1.00, 0.24)、(1.50, 0.36)、(2.00, 0.48)、(2.50, 0.60)。你需要添加新的一列用于计算电阻 R = V/I，并且插入资料通常要求给出每台仪器的绝对不确定度。所用电压表的分辨率为 0.05 V，因此 V 的绝对不确定度为 ±0.025 V。电流表的分辨率为 0.01 A，故不确定度为 ±0.005 A。此表格将作为绘制 V-I 图线的基础。
6. Performing Linear Regression: Finding the Gradient | 执行线性回归：求斜率
best-fit line gives the resistance R. Using the given points, a perfect fit line passes through (0,0) and (0.60, 2.50), giving gradient = (2.50 − 0) ÷ (0.60 − 0) = 4.1667 Ω. To determine uncertainty in R, draw worst-fit lines by adding the absolute uncertainty bars: V ± 0.03 V, I ± 0.005 A. The steepest reasonable line might give a gradient Rmax = (2.53) ÷ (0.595) = 4.252 Ω, and the shallowest line Rmin = (2.47) ÷ (0.605) = 4.083 Ω. The absolute uncertainty in R is (Rmax − Rmin 4.083)/2 = 0.0845 Ω. So R = 4.167 ± 0.085 Ω. Percentage uncertainty = (0.085/4.167) × 100% ≈ 2.0%.
绘制 V（纵轴）对 I（横轴）的图线应得到一条通过原点的直线，证实欧姆定律。最佳拟合线的斜率即为电阻 R。使用给定的数据点，一条完美拟合线通过 (0,0) 和 (0.60, 2.50)，可得斜率 = (2.50 − 0) ÷ (0.60 − 0) = 4.1667 Ω。为确定 R 的不确定度，需通过添加绝对误差棒来绘制最差拟合线：V ± 0.03 V，I ± 0.005 A。最陡的合理直线可能给出斜率 Rmax = (2.53) ÷ (0.595) = 4.252 Ω，最平缓的直线 Rmin = (2.47) ÷ (0.605) = 4.083 Ω。R 的绝对不确定度为 (Rmax − Rmin)/2 = (4.252 − 4.083)/2 = 0.0845 Ω。因此 R = 4.167 ± 0.085 Ω。百分比不确定度 = (0.085/4.167) × 100% ≈ 2.0%。
7. Propagating Percentage Uncertainties | 合成百分比不确定度
⁻⁴)² = 5.90 × 10⁻⁸ m². The percentage uncertainty in A is twice the percentage uncertainty in d because A ∝ d². From earlier, %Δd = (0.002 mm/0.274 mm) × 100% = 0.730%. Hence %ΔA = 2 × 0.730% = 1.46%. %ΔL = 0.25% as calculated. %ΔR = 2.0%. Therefore, the total percentage uncertainty in ρ is %Δρ = %ΔR + %ΔA + %ΔL = 2.0% + 1.46% + 0.25% = 3.71% (rounded to 3.7% or 4% for practical reporting).
电阻率 ρ 由 ρ = RA/L = (πd²R)/(4L) 计算得出。首先计算横截面积 A = π(d/2)²。代入 d = 2.74 × 10⁻⁴ m，得 A = π(1.37 × 10⁻⁴)² = 5.90 × 10⁻⁸ m²。由于 A ∝ d²，A 的百分比不确定度是 d 的百分比不确定度的两倍。依据前文，%Δd = (0.002 mm/0.274 mm) × 100% = 0.730%，因此 %ΔA = 2 × 0.730% = 1.46%。%ΔL = 0.25%（已计算），%ΔR = 2.0%。故 ρ 的总百分比不确定度为 %Δρ = %ΔR + %ΔA + %ΔL = 2.0% + 1.46% + 0.25% = 3.71%（在实际报告中可四舍五入为 3.7% 或 4%）。
8. Calculating Resistivity from the Gradient | 由斜率计算电阻率
ily found: ρ_max = (4.252 × 5.90×10⁻⁸) / 0.800 = 3.136 × 10⁻⁷ Ω·m; ρ_min = (4.083 × 5.90×10⁻⁸) / 0.800 = 3.011 × 10⁻⁷ Ω·m. The absolute uncertainty Δρ ≈ (ρ_max − ρ_min)/2 = (3.136 − 3.011)/2 × 10⁻⁷ = 0.0625 × 10⁻⁷ Ω·m. Hence the final result is ρ = (3.07 ± 0.06) × 10⁻⁷ Ω·m. This agrees well with the accepted value for constantan (around 4.9 × 10⁻⁷ Ω·m) within uncertainty, but note that constantan’s exact resistivity depends on its composition; our slightly lower value hints at possible systematic errors.
现在计算标称值：ρ = (R × A) / L = (4.167 Ω × 5.90 × 10⁻⁸ m²) ÷ 0.800 m = 3.074 × 10⁻⁷ Ω·m。利用绝对不确定度，可轻松求出 ρ 的范围：ρ_max = (4.252 × 5.90×10⁻⁸) / 0.800 = 3.136 × 10⁻⁷ Ω·m；ρ_min = (4.083 × 5.90×10⁻⁸) / 0.800 = 3.011 × 10⁻⁷ Ω·m。绝对不确定度 Δρ ≈ (ρ_max − ρ_min)/2 = (3.136 − 3.011)/2 × 10⁻⁷ = 0.0625 × 10⁻⁷ Ω·m。因此最终结果为 ρ = (3.07 ± 0.06) × 10⁻⁷ Ω·m。在不确定度范围内，该值与康铜的公认值（约 4.9 × 10⁻⁷ Ω·m）基本一致，但请注意康铜的精确电阻率取决于其成分；我们略低的结果暗示可能存在系统误差。
9. Evaluating Systematic vs. Random Errors | 评估系统误差与随机误差
Systematic errors may arise from a zero error on the micrometer, incorrect calibration of the voltmeter, or the temperature rise of the wire despite low currents. The insert may have asked students to describe how to check the micrometer's zero reading – by closing the jaws and reading the barrel. If a zero error exists, all diameter readings must be corrected. Random errors are evident in the scatter of the resistance values if repeated, but here the given data is highly consistent. A key evaluation point is the assumption that the wire’s diameter is uniform; measuring at several positions and finding a small spread reduces this uncertainty. Additionally, using a micrometer with a digital readout rather than an analogue scale could reduce reading error.
系统误差可能来源于千分尺的零误差、电压表校准不准或即使电流较低但导线温度的升高。插入资料可能要求学生描述如何检查千分尺的零位读数——通过闭合测量面并读取套管刻度。如果存在零误差，所有直径读数都必须修正。随机误差在多次重复电阻值时会表现为数据的离散，但在此给出的数据高度一致。一个关键的评估要点是假设导线直径均匀；在多个位置测量并发现较小离散度可降低该不确定度。此外，使用数字显示的千分尺而非模拟刻度尺可减小读数误差。
10. The Final Practical Write-Up: Conclusion | 最终实验报告：结论
tate whether the agreement is satisfactory and discuss the main sources of error. For the Jan 19 insert, a model answer would read: “ etermined to be (3.07 ± 0.06) × 10⁻⁷ Ω·m, with a percentage uncertainty of approximately 2%. This is 37% antan or a zero error in the micrometer that was not accounted for. The linearity of the I–V graph confirms that temperature effects were negligible.” Such a structured evaluation meets the criteria for the practical skills assessment.
你的结论必须引用带有绝对不确定度和百分比不确定度的电阻率，并与参考值进行比较。说明该吻合度是否令人满意，并讨论主要的误差来源。对于 2019 年 1 月的插入资料，一份标准答案可能为：“测得康铜丝的电阻率为 (3.07 ± 0.06) × 10⁻⁷ Ω·m，百分比不确定度约为 2%。这比公认值 4.9 × 10⁻⁷ Ω·m 低 37%，表明存在显著的系统误差，最可能的原因是导线非纯康铜或存在未予修正的千分尺零误差。I–V 图的线性关系证实温度效应可忽略不计。”这种结构化的评估符合实验技能评估的标准。
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