📚 AS Physics Insert June 2022 Experimental Investigation | AS物理实验探究（2022年6月插入部分）
In the June 2022 AS Physics examination, the insert provided students with a structured experimental scenario designed to test practical skills such as planning, data handling, and evaluation of uncertainties. This article unpacks a typical investigation that mirrors the style of that insert – determining the resistivity of a metal wire using a voltmeter–ammeter method. Whether you are preparing for Paper 3 or strengthening your understanding of core practicals, mastering this experiment is essential for top marks.
在2022年6月AS物理考试中，插入材料为学生提供了一个结构化的实验场景，旨在考查规划、数据处理和不确定性评估等实践技能。本文深入剖析一个与该插入风格相似的典型探究——使用伏安法测定金属丝的电阻率。无论你是在备考Paper 3还是巩固核心实验理解，掌握这个实验对于取得高分至关重要。
1. Connecting the Insert to the Syllabus | 将插入材料与考纲联系起来
e investigation of how the resistance of a component varies with length, cross-sectional area, or temperature. By focusing on a metal wire at constant temperature, we can extract its resistivity – a fundamental material property that links directly to the equation R = ρL/A.
AQA AS物理2022年6月的插入材料通常会包含一个电路图、一张原始测量数据表以及用于分析的图像。一个常见的主题是探究元件的电阻如何随长度、横截面积或温度变化。通过关注恒温下的金属丝，我们可以提取其电阻率——这是一种直接关联公式R = ρL/A的基本材料属性。
The key syllabus requirements assessed include: applying the resistivity equation, calculating cross-sectional area from diameter, drawing a line of best fit, determining gradient, and combining percentage uncertainties. This article walks through every step, using specimen data typical of an exam insert.
考纲中考查的关键要求包括：应用电阻率公式、由直径计算横截面积、绘制最佳拟合线、确定斜率以及合成百分比不确定度。本文使用考试插入材料中常见的示例数据，逐步讲解每一步。
2. Equipment and Setup | 仪器与装置
To obtain reliable data, a simple circuit is built with a power supply, an ammeter in series, and a voltmeter in parallel across the test wire. A metre ruler is used to measure the length of the wire, and a micrometer screw gauge measures its diameter at several points along the wire. A switch and a variable resistor (rheostat) allow the current to be adjusted without altering the length during a single set of readings.
为了获取可靠数据，搭建一个简单的电路：电源、串联的电流表以及并联在待测金属丝两端的电压表。使用米尺测量金属丝的长度，使用千分尺在金属丝的多处测量其直径。一个开关和一个可变电阻（变阻器）使得在单组读数过程中无需改变长度即可调节电流。
- Power supply (1.5 V to 6 V DC) | 电源（1.5 V至6 V直流）
- Digital ammeter (0–1 A, resolution 0.01 A) | 数字电流表（0–1 A，分辨力0.01 A）
- Digital voltmeter (0–10 V, resolution 0.01 V) | 数字电压表（0–10 V，分辨力0.01 V）
- Nichrome or constantan wire (diameter ~0.25 mm) | 镍铬或康铜丝（直径约0.25 mm）
- Micrometer screw gauge (resolution 0.01 mm) | 千分尺（分辨力0.01 mm）
- Metre ruler (resolution 1 mm) | 米尺（分辨力1 mm）
- Connecting leads and crocodile clips | 连接导线和鳄鱼夹
3. Procedure | 实验步骤
The wire is taped gently to a metre ruler and connected into the circuit using crocodile clips placed exactly at the zero mark and at the chosen length L. The initial length is set to 1.000 m. The circuit is switched on, and the rheostat is adjusted to give a small current (e.g. 0.20 A) to minimise heating. The voltmeter reading V and ammeter reading I are recorded. The process is repeated for at least six different lengths, reducing L in steps of 0.100 m while keeping the current roughly constant by adjusting the rheostat each time.
将金属丝轻轻固定在米尺上，并使用鳄鱼夹将其接入电路，鳄鱼夹精确放置在零刻度线和选定的长度L处。初始长度设为1.000 m。接通电路，调节变阻器使电流较小（例如0.20 A）以尽量减少发热。记录电压表读数V和电流表读数I。对至少六个不同的长度重复上述过程，每次将L减少0.100 m，同时通过调节变阻器使电流大致保持恒定。
lculated to reduce random error. The cross-sectional area A is then found using A = πd²/4.
关键是要使用千分尺在三个不同方向以及沿金属丝五个不同位置测量其直径d。计算平均直径以减少随机误差。然后利用公式A = πd²/4求出横截面积A。
Safety note: Even at low voltage, hot wires can cause burns. Touch the wire only after switching off.
安全提示：即使在低电压下，发热的导线也可能引起灼伤。务必在关断电源后再触碰导线。
4. Data Collection | 数据收集
e current was maintained at I = 0.20 A throughout to keep the wire’s temperature constant.
以下是一组示例读数，类似于插入材料中常给出的数据表。整个过程中电流保持在I = 0.20 A，以确保金属丝温度恒定。
| Length L / m | Voltage V / V | Current I / A | Resistance R = V/I / Ω |
|---|---|---|---|
| 1.000 | 1.62 | 0.20 | 8.10 |
| 0.900 | 1.46 | 0.20 | 7.30 |
| 0.800 | 1.30 | 0.20 | 6.50 |
| 0.700 | 1.14 | 0.20 | 5.70 |
| 0.600 | 0.98 | 0.20 | 4.90 |
| 0.500 | 0.81 | 0.20 | 4.05 |
asurements: 0.25 mm, 0.24 mm, 0.26 mm, 0.25 mm, 0.25 mm (mean d = 0.25 mm = 2.5 × 10⁻⁴ m).
直径测量值：0.25 mm、0.24 mm、0.26 mm、0.25 mm、0.25 mm（平均值 d = 0.25 mm = 2.5 × 10⁻⁴ m）。
5. Calculating Resistance | 计算电阻
nce the current I is constant, the resistance values are directly proportional to the voltage. The table already includes the computed resistances. These are plotted on a graph of R (y-axis) against L (x-axis).
对于每个长度，利用欧姆定律R = V/I计算电阻。由于电流I恒定，电阻值与电压成正比。上表已包含计算出的电阻值。将这些数据绘制在R（y轴）对L（x轴）的图中。
The graph should be a straight line passing through the origin, confirming that R ∝ L for a wire of uniform cross-sectional area and constant temperature. The gradient of the line is equal to ρ/A. A student would draw a best-fit line, not simply connect points.
图像应是一条通过原点的直线，这证实了对于横截面积均匀且温度恒定的金属丝，R ∝ L。该直线的斜率等于ρ/A。学生应绘制最佳拟合线，而不是简单地连接各点。
6. Determining Resistivity | 测定电阻率
ng two widely spaced points on the best-fit line: (1.000 m, 8.10 Ω) and (0.500 m, 4.05 Ω).
电阻率ρ可从斜率求得。使用上述数据，斜率 = ΔR/ΔL。在最佳拟合线上选取两个间隔较远的点：(1.000 m, 8.10 Ω) 和 (0.500 m, 4.05 Ω)。
Gradient = (8.10 - 4.05) / (1.000 - 0.500) = 4.05 / 0.500 = 8.10 Ω m⁻¹
sectional area A = πd²/4 = π × (2.5 × 10⁻⁴ m)² / 4 = π × 6.25 × 10⁻⁸ / 4 ≈ 4.91 × 10⁻⁸ m².
横截面积 A = πd²/4 = π × (2.5 × 10⁻⁴ m)² / 4 = π × 6.25 × 10⁻⁸ / 4 ≈ 4.91 × 10⁻⁸ m²。
0 × 4.91 × 10⁻⁸ = 3.98 × 10⁻⁷ Ω m. This value is typical for nichrome wire (accepted value ~ 1.0 × 10⁻⁶ Ω m; the slight difference implies possible systematic error).
由于斜率 = ρ/A，因此 ρ = 斜率 × A = 8.10 × 4.91 × 10⁻⁸ = 3.98 × 10⁻⁷ Ω m。该值对于镍铬合金丝来说是典型的（公认值约为1.0 × 10⁻⁶ Ω m；细微差异表明可能存在系统误差）。
7. Uncertainty Analysis | 不确定性分析
Uncertainty calculations are central to AS Physics practical questions. The percentage uncertainty in resistivity comes from combining the uncertainties in the gradient and in the area. The area uncertainty itself derives from the diameter measurement: since A ∝ d², the percentage uncertainty in A is twice the percentage uncertainty in d.
不确定度计算是AS物理实验题的核心。电阻率的百分比不确定度由斜率和面积的不确定度合成而来。面积的不确定度本身源于直径测量：因为A ∝ d²，所以A的百分比不确定度是d的百分比不确定度的两倍。
ter is 0.25 mm. The absolute uncertainty in d is taken as ±0.01 mm (the instrument limit).
假设千分尺的分辨力为±0.01 mm，平均直径为0.25 mm。d的绝对不确定度取作±0.01 mm（仪器极限）。
%U(d) = (0.01 / 0.25) × 100% = 4.0%
%U(A) = 2 × 4.0% = 8.0%
spread is 0.80, so the absolute uncertainty is approximately half the spread: ±0.40 Ω m⁻¹.
可以通过绘制最陡和最浅可接受的最佳拟合线来估算斜率的不确定度。假设它们得出的斜率分别为8.50和7.70 Ω m⁻¹。跨度为0.80，因此绝对不确定度约等于跨度的一半：±0.40 Ω m⁻¹。
%U(gradient) = (0.40 / 8.10) × 100% ≈ 4.9%
Total percentage uncertainty in ρ = %U(gradient) + %U(A) = 4.9% + 8.0% = 12.9%. The absolute uncertainty in ρ is 0.129 × 3.98 × 10⁻⁷ ≈ 0.51 × 10⁻⁷ Ω m.
ρ的总百分比不确定度 = %U(斜率) + %U(A) = 4.9% + 8.0% = 12.9%。ρ的绝对不确定度为0.129 × 3.98 × 10⁻⁷ ≈ 0.51 × 10⁻⁷ Ω m。
d as ρ = (4.0 ± 0.5) × 10⁻⁷ Ω m, which is consistent with the accepted order of magnitude.
因此最终结果表示为 ρ = (4.0 ± 0.5) × 10⁻⁷ Ω m，这在数量级上与公认值一致。
8. Sources of Error | 误差来源
The dominant sources of error in this experiment are:
- Zero error of the micrometer – if the micrometer does not read zero when fully closed, all diameter readings must be corrected. Failure to do so introduces a systematic error.
- Heating effect of the current – even at 0.20 A, the wire’s temperature may rise, increasing its resistance. This causes the R–L graph to deviate from linearity at higher currents.
- Contact resistance – crocodile clips may not make perfect contact with the wire, adding a small fixed resistance that shifts the graph upward but does not change the gradient.
- Length measurement – ion of the clip contact point may be uncertain by ±1 mm, giving a small random error in L.
- Non-uniformity of the wire – diameter may vary along the length, making a single mean area insufficient.
本实验的主要误差来源有：
- 千分尺的零误差——如果千分尺完全闭合时读数不为零，所有直径读数都必须修正。未修正会引入系统误差。
- 电流的热效应——即使电流为0.20 A，金属丝的温度仍可能升高，导致其电阻增大。这会使R–L图像在较大电流时偏离线性。
- 接触电阻——鳄鱼夹可能与金属丝接触不良，产生一个小的固定电阻，使图像上移但不改变斜率。
- 长度测量——夹子接触点的确切位置可能存在±1 mm的不确定度，给L带来小的随机误差。
- 金属丝的非均匀性——沿长度方向的直径可能变化，使得单一的平均横截面积不够精确。
9. Improvements and Refinements | 改进与完善
To minimise these errors, several refinements can be implemented:
- Use a jockey and a metre bridge instead of crocodile clips to eliminate contact resistance and to measure length more precisely.
- Employ a low-resistance rheostat and a very low current (e.g. 0.10 A) to reduce heating, but ensure voltmeter sensitivity is adequate.
- Take diameter readings at 10 different positions and at two orientations at each position, then use the mean.
- Check the micrometer for zero error before and after measurements, subtract any zero reading.
- For length, use a travelling microscope to measure the exact distance between the potential contacts.
为尽量减少这些误差，可采取以下改进措施：
- 使用滑动触头和米桥代替鳄鱼夹，以消除接触电阻并更精确地测量长度。
- 采用低电阻变阻器和非常小的电流（例如0.10 A）以减少发热，但需确保电压表的灵敏度足够。
- 在10个不同位置并在每个位置的两个方向读取直径，然后取平均值。
- 在测量前后检查千分尺的零误差，并扣除任何零读数。
- 对于长度，使用游动显微镜来精确测量电势触点之间的距离。
10. Exam-Style Data Interpretation | 考试型数据解读
In an exam, an insert may present a graph of R against L already plotted, or ask you to complete a table and draw conclusions. A typical question might be:
“ ameter of 4.0 × 10⁻⁴ m. Calculate the resistivity of the metal. State an appropriate unit.”
在考试中，插入材料可能直接呈现一张已绘制的R对L图像，或者要求你完成表格并得出结论。典型的问题可能是：
“图像的斜率为7.95 Ω m⁻¹。金属丝的直径为4.0 × 10⁻⁴ m。计算该金属的电阻率。给出适当的单位。”
0⁻⁴)²/4 = 1.26 × 10⁻⁷ m². ρ = 7.95 × 1.26 × 10⁻⁷ = 1.00 × 10⁻⁶ Ω m. The unit is ohm metre.
解答：A = π(4.0 × 10⁻⁴)²/4 = 1.26 × 10⁻⁷ m²。ρ = 7.95 × 1.26 × 10⁻⁷ = 1.00 × 10⁻⁶ Ω m。单位为欧姆米。
Another common task is to discuss whether the wire obeys Ohm’s law. Since the graph is a straight line through the origin for constant temperature, it does. If the line curves at higher currents, the wire is being heated and Ohm’s law no longer holds.
另一个常见任务是讨论该金属丝是否遵循欧姆定律。由于在恒温条件下图像为过原点的直线，因此它遵循欧姆定律。如果直线在较大电流处弯曲，则说明金属丝被加热，欧姆定律不再成立。
11. Conclusion | 结论
The June 2022 AS Physics insert exercise reinforces fundamental skills: using a micrometer, calculating cross-sectional area, plotting and interpreting a linear graph, extracting gradient, and combining uncertainties. These competencies are transferable to many other core practicals, from determining acceleration due to gravity to investigating Young’s modulus. A systematic approach – coupled with a clear understanding of the underlying physics – is the key to securing full marks in practical-based questions.
2022年6月AS物理插入练习强化了基本技能：使用千分尺、计算横截面积、绘制并解释线性图像、提取斜率以及合成不确定度。这些能力可迁移到许多其他核心实验中，从测定重力加速度到探究杨氏模量。系统化的方法——加上对基础物理的清晰理解——是在实践类问题中取得满分的关键。
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