📚 A-Level Physics: a-level-physics-jun-18-insert3 Experimental Investigation | A-Level 物理：2018年6月插页3 实验探究
In many A-Level Physics examinations, an insert provides raw experimental data for students to analyse. This article uses a typical June 2018 Insert 3 style dataset to guide you through a full investigation: determining the resistivity of a metal wire by measuring how its resistance changes with length. We will cover data processing, graph plotting, gradient analysis, uncertainty propagation and evaluation of the experiment. Mastering these steps is essential for scoring top marks in practical-based questions.
在许多A-Level物理考试中，插页会提供原始实验数据供学生分析。本文利用一份典型的2018年6月插页3风格的数据集，带你完成一次完整的实验探究：通过测量金属丝电阻随长度的变化来测定其电阻率。我们将涵盖数据处理、图像绘制、斜率分析、不确定度传递以及实验评估。掌握这些步骤对于在实验类题目中取得高分至关重要。
1. Overview and Aim | 实验概述与目的
ngth L, given by R = ρL/A, where A is the cross-sectional area. By measuring R for different lengths and plotting a graph of R against L, the gradient equals ρ/A. Measuring the diameter d allows calculation of A = πd²/4, after which ρ can be found.
本实验的目的是测定一段未知金属丝的电阻率ρ。均匀导线的电阻R与其长度L成正比，关系式为R = ρL/A，其中A是横截面积。通过测量不同长度对应的R，并绘制R-L图像，其斜率等于ρ/A。测量直径d后可算出A = πd²/4，从而求出ρ。
2. Equipment Used | 所用器材
An unknown metal wire mounted on a metre ruler, with a sliding contact to vary the effective length.
安装在米尺上的未知金属丝，带有滑动触头以改变有效长度。
Digital ohmmeter (or a voltmeter and ammeter to calculate R = V/I).
数字欧姆表（或伏特表和安培表，通过R = V/I计算出电阻）。
Micrometer screw gauge, resolution 0.01 mm, to measure the wire diameter.
千分尺（螺旋测微器），分辨力0.01 mm，用于测量导线直径。
Connecting leads and a switch (if using a power supply).
连接导线和开关（若使用电源）。
Metre ruler to confirm the length settings.
米尺，用于确认长度设置。
3. Data Collection: Length and Resistance | 数据收集：长度与电阻
d 1.000 m. At each length, the resistance R is recorded directly from the ohmmeter. To minimise heating effects, the current is kept small and readings are taken quickly. The diameter of the wire is measured at several points along its length using the micrometer; six readings are taken and averaged to reduce random error.
滑动触头分别设置在长度L = 0.200 m、0.400 m、0.600 m、0.800 m和1.000 m处。在每个长度下，直接从欧姆表记录电阻值R。为了尽量减小热效应，电流保持很小并快速读数。使用千分尺沿导线多个位置测量直径；共取6个读数并求平均值，以减小随机误差。
4. Raw Data Table | 原始数据表
| Length L / m | Resistance R / Ω |
|---|---|
| 0.200 | 1.05 |
| 0.400 | 2.12 |
| 0.600 | 3.18 |
| 0.800 | 4.22 |
| 1.000 | 5.30 |
ndard uncertainty from the spread is estimated as ±0.004 mm. Thus d = (2.74 ± 0.04) × 10⁻⁴ m in SI units.
直径读数（mm）：0.27, 0.28, 0.27, 0.28, 0.27, 0.28。平均值为0.275 mm，剔除一个明显异常值后修约为0.274 mm。由离散程度估得的标准不确定度为±0.004 mm。因此以国际单位制表示为 d = (2.74 ± 0.04) × 10⁻⁴ m。
5. Plotting the Resistance–Length Graph | 绘制电阻–长度图像
ical axis is resistance R / Ω. Error bars on R are small (typically ±0.01 Ω from the meter' nd it is clear that the line passes very close to the origin, as expected from R = (ρ/A)L.
在坐标纸上以合适的标度画出这五个数据点。横轴为长度 L / m，纵轴为电阻 R / Ω。R 的误差棒很小（通常来自仪表精度，约±0.01 Ω），并在图中标出。过这些点作一条最佳拟合直线，可以清楚地看到该直线非常接近原点，这与 R = (ρ/A)L 的预期一致。
6. Determining the Gradient and Its Uncertainty | 确定斜率及其不确定度
5.30 - 1.05) / (1.000 - 0.200) = 4.25 / 0.800 = 5.3125 Ω/m. To estimate the uncertainty, a 'worst acceptable' maximum gradient line (through 0.200, 1.00 and 1.000, 5.35) gives mmax 0.800 = 5.4375 Ω/m. A minimum gradient line (through 0.200, 1.10 and 1.000, 5.25) gives mmin ) / 0.800 = 4.15 / 0.800 = 5.1875 Ω/m. The absolute uncertainty in the gradient is Δm = (mmax - mmin) / 2 = (5.4375 - 5.1875) / 2 = 0.125 Ω/m.
最佳拟合直线近似通过(0.200, 1.05)和(1.000, 5.30)两点，斜率为 (5.30 − 1.05)/(1.000 − 0.200) = 4.25/0.800 = 5.3125 Ω/m。为估计不确定度，作一条“最差可接受”的最大斜率线（通过0.200, 1.00与1.000, 5.35），得到 mmax = (5.35 − 1.00)/0.800 = 5.4375 Ω/m。再作最小斜率线（通过0.200, 1.10与1.000, 5.25），得到 mmin = (5.25 − 1.10)/0.800 = 5.1875 Ω/m。斜率的绝对不确定度为 Δm = (mmax − mmin)/2 = (5.4375 − 5.1875)/2 = 0.125 Ω/m。
± 0.13 Ω/m (rounding appropriately). The percentage uncertainty is (0.125 / 5.3125) × 100% ≈ 2.35%.
因此斜率为 m = 5.31 ± 0.13 Ω/m（合理修约）。百分不确定度为 (0.125/5.3125)×100% ≈ 2.35%。
7. Calculating the Cross-sectional Area | 计算横截面积
The cross-sectional area A is given by A = πd²/4. Using d = 2.74 × 10⁻⁴ m, we compute:
A = π × (2.74 × 10⁻⁴)² / 4 = π × 7.5076 × 10⁻⁸ / 4 ≈ 5.896 × 10⁻⁸ m²
Rounding to three significant figures gives A = 5.90 × 10⁻⁸ m².
横截面积A由A = πd²/4给出。代入d = 2.74 × 10⁻⁴ m，计算得：
A = π × (2.74 × 10⁻⁴)² / 4 = π × 7.5076 × 10⁻⁸ / 4 ≈ 5.896 × 10⁻⁸ m²
修约至三位有效数字为A = 5.90 × 10⁻⁸ m²。
The percentage uncertainty in diameter is (0.004 / 0.274) × 100% = 1.46%. Since A ∝ d², the percentage uncertainty in A is 2 × 1.46% = 2.92%.
直径的百分不确定度为(0.004/0.274)×100% = 1.46%。由于A∝d²，A的百分不确定度为2×1.46% = 2.92%。
8. Computing the Resistivity | 计算电阻率
the relationship R = (ρ/A)L, the gradient m = ρ/A. Therefore the resistivity is ρ = m × A.
ρ = 5.3125 × 5.90 × 10⁻⁸ = 3.134 × 10⁻⁷ Ω·m
e is typical for some alloys like nichrome, indicating the wire is likely not pure copper.
由关系式R = (ρ/A)L，斜率m = ρ/A。因此电阻率为ρ = m × A。
ρ = 5.3125 × 5.90 × 10⁻⁸ = 3.134 × 10⁻⁷ Ω·m
修约至两位有效数字得ρ = 3.1 × 10⁻⁷ Ω·m。这个数值接近某些合金（如镍铬合金），说明该导线很可能不是纯铜。
9. Combining Uncertainties to Find the Uncertainty in Resistivity | 组合不确定度以求出电阻率的不确定度
The resistivity depends on both the gradient and the area. Since the two quantities are independent, the percentage uncertainties add in quadrature when multiplication or division is involved. However, a simpler worst-case approach adds relative uncertainties linearly:
Δρ/ρ = Δm/m + 2Δd/d
Using the values: Δm/m = 2.35% and 2Δd/d = 2.92%, the total relative uncertainty ≈ 2.35% + 2.92% = 5.27%.
电阻率取决于斜率和面积两者。由于这两个量相互独立，当涉及乘除运算时，百分不确定度应按平方相加开方；但这里采用更保守的线性相加方式进行最差情况估算：
Δρ/ρ = Δm/m + 2Δd/d
代入数值：Δm/m = 2.35%，2Δd/d = 2.92%，总相对不确定度 ≈ 2.35% + 2.92% = 5.27%。
Rounded to one significant figure, Δρ = 0.2 × 10⁻⁷ Ω·m. The final result is expressed as:
ρ = (3.1 ± 0.2) × 10⁻⁷ Ω·m
因此绝对不确定度为Δρ = 0.0527 × 3.13 × 10⁻⁷ ≈ 0.165 × 10⁻⁷ Ω·m。修约至一位有效数字得Δρ = 0.2 × 10⁻⁷ Ω·m。最终结果表示为：
ρ = (3.1 ± 0.2) × 10⁻⁷ Ω·m
10. Key Sources of Error | 主要误差来源
Heating effect: even with a small current, the wire can warm up, increasing its resistance slightly. Taking readings quickly and allowing cooling between measurements reduces this.
热效应：即使电流很小，导线仍可能升温，导致电阻略有增大。快速读数并在各次测量之间让导线冷却，可减少这一影响。
Zero error on the ohmmeter or micrometer: all instruments should be checked for zero offset before use.
欧姆表或千分尺的零位误差：使用前应检查所有仪器的零点偏移。
Contact resistance: poor connection at the sliding contact can introduce extra resistance. Ensuring clean, firm contacts minimises this error.
接触电阻：滑动触头处的不良连接会引入额外电阻。保证触点清洁并接触良好，可尽量减小该误差。
Parallax error when reading the ruler and micrometer: the eye must be directly in line with the scale to avoid misreading.
读数时的视差：读米尺和千分尺时，眼睛必须正对刻度，以避免误读。
Diameter is not perfectly uniform: taking several readings along the wire accounts for irregularities, but a mean value still has uncertainty.
直径并非完全均匀：沿导线多处取读数可考虑不均匀性，但平均值仍存在不确定度。
11. Evaluation of the Experiment | 实验评估
Overall the experiment yields a consistent dataset, with the R–L graph showing a clear straight line passing close to the origin. The percentage uncertainty in the final resistivity (about 5%) is acceptable for a school laboratory. The largest contributor to the uncertainty is the diameter measurement, because it is squared in A, doubling its relative uncertainty. Using a micrometer with a smaller resolution, or taking many more diameter readings, would improve precision. Repeating the experiment with a longer wire and more data points would also refine the gradient determination.
总的来说，该实验获得了一组一致的数据，R–L图像显示清晰的过原点附近直线。最终电阻率的百分不确定度（约5%）在中学实验室条件下是可以接受的。不确定度的最大来源是直径测量，因为它在A中被平方，使得相对不确定度加倍。使用更高分辨力的千分尺，或取更多次直径读数，能提高精度。换用更长的导线并增加数据点进行重复实验，也可细化斜率的测定。
Furthermore, using a four-terminal (Kelvin) method to eliminate lead and contact resistance would make the resistance data more reliable. The absence of a systematic shift in the intercept (close to zero) suggests that systematic errors are small.
此外，采用四端（开尔文）法消除引线和接触电阻，可使电阻数据更可靠。截距接近零，表明系统误差很小。
12. Conclusion | 结论
0⁻⁷ Ω·m. The investigation demonstrates key practical skills: handling raw data, drawing a linear graph, extracting gradient with uncertainty, propagating errors through calculations, and critically evaluating the procedure. These are precisely the competencies examined in A-Level Physics Insert 3 tasks and similar practical-themed questions.
未知金属丝的电阻率被测定为(3.1 ± 0.2) × 10⁻⁷ Ω·m。本实验探究展示了关键的实验技能：处理原始数据、绘制线性图像、求取斜率及不确定度、在计算中传递误差，以及批判性地评估实验步骤。这些正是A-Level物理插页3及同类实验主题题目所考查的核心能力。
Published by TutorHao | Physics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图