📚 AS Physics Unit 4 June 2019: Experimental Investigation of Capacitor Discharge | AS物理单元4 2019年6月：电容放电实验探究
In the June 2019 AS Physics Unit 4 (WPH14/01) examination paper, a key practical investigation required students to design, carry out and analyse an experiment on the discharge of a capacitor through a known resistor. This classic experiment tests understanding of exponential decay, time constant, graphical representation and data linearisation, all of which are central to the electricity and fields topics. The question guided learners through the process of selecting appropriate apparatus, recording potential difference as a function of time, plotting suitable graphs and extracting the time constant to determine capacitance or resistance. This article breaks down every stage of that experimental inquiry, providing detailed explanations, model data, analysis techniques and common pitfalls to avoid – essential revision for any AS Physics candidate.
在2019年6月的AS物理单元4（WPH14/01）考试中，一道核心的实验探究题要求学生设计、实施并分析一个电容器通过已知电阻放电的实验。这一经典实验考查学生对指数衰减、时间常数、图像描绘和数据线性化的理解，这些都是电学和场论部分的核心内容。题目引导考生逐步选择合适的仪器，记录电压随时间的变化，绘制恰当的图像，并从中提取时间常数以确定电容或电阻。本文逐步拆解这一实验探究的全过程，提供详细说明、示范数据、分析方法和常见错误梳理——这对每一位AS物理考生来说都是不可多得的复习资源。
1. Background and Aims | 实验背景与目标
When a charged capacitor is connected across a resistor, the charge stored on its plates decays exponentially as current flows through the resistor. The potential difference V across the capacitor at any time t is given by the equation V = V₀ e−t/RC, where V₀ is the initial p.d., R is the total resistance in the discharge loop, and C is the capacitance. The product RC is known as the time constant τ, which represents the time taken for the p.d. to fall to about 37 % of its initial value. The aim of the practical investigation in the June 2019 paper was to verify this exponential relationship, measure the time constant accurately and use it to calculate the unknown capacitance of a capacitor, given a known resistor value. Students were also expected to appreciate that taking natural logarithms yields a linear equation: ln V = ln V₀ − t/RC, allowing τ to be found from the gradient of a straight-line graph.
当已充电的电容器与电阻连接时，极板上储存的电荷会随着电流流过电阻而呈指数形式衰减。任意时刻 t 电容器两端的电压 V 遵循方程 V = V₀ e−t/RC，其中 V₀ 为初始电压，R 为放电回路中的总电阻，C 为电容。乘积 RC 称为时间常数 τ，它表示电压下降至初始值约 37% 所需的时间。2019年6月真题中的实验探究旨在验证这一指数关系，准确测量时间常数，并利用已知电阻计算出电容器的未知电容值。同时，考生还需理解对等式两边取自然对数后可得到线性方程：ln V = ln V₀ − t/RC，通过直线图像的梯度可求得 τ。
2. Apparatus and Circuit | 实验器材与电路
The experimental setup described in the exam question is straightforward and safe. A typical list of apparatus includes: a large electrolytic capacitor (e.g., 1000 μF), a known resistor (e.g., 100 kΩ), a single-pole single-throw switch, a d.c. power supply or battery pack, a voltmeter with high internal resistance (preferably digital), a stopwatch, and connecting wires. The capacitor must be connected in series with the resistor and the switch, forming a simple RC loop. The voltmeter is placed across the capacitor to monitor its terminal p.d. In the charging phase, the power supply is briefly connected across the capacitor to charge it fully; for the discharge phase, the power supply is removed, and the switch is closed to allow the capacitor to discharge through the resistor alone. The question paper often provides a circuit diagram, and candidates must be able to identify where the voltmeter should be placed and why it must have a very high resistance – to minimise current drawn from the capacitor and avoid distorting the measured decay.
试卷中描述的实验装置简洁且安全。典型器材清单包括：一个大容量电解电容器（如 1000 μF）、一个已知电阻（如 100 kΩ）、一个单刀单掷开关、直流电源或电池组、一个高内阻电压表（最好为数字式）、秒表以及连接导线。电容器必须与电阻和开关串联，构成简单的 RC 回路。电压表跨接在电容器两端以监测其端电压。在充电阶段，电源短暂地与电容器并联使之完全充电；放电阶段则断开电源，闭合开关，让电容器仅通过电阻放电。试题通常会提供电路图，考生需要能够指出电压表的正确连接位置，并解释其内阻必须非常高的原因——尽可能减少从电容器分流出的电流，避免干扰测量的衰减曲线。
3. Step-by-Step Procedure | 实验步骤
First, connect the circuit as shown in the diagram, ensuring the capacitor is discharged safely by shorting its terminals momentarily with a wire (or using the switch to connect it across the resistor before beginning). Set the power supply to a safe d.c. voltage, such as 6.0 V, and charge the capacitor fully by closing the charging part of the circuit. Check that the voltmeter reading stabilises at the supply voltage. Then, simultaneously start the stopwatch and open the charging switch while closing the discharge switch. Record the voltmeter reading at regular time intervals, for example every 10 seconds for the first minute and then every 20 seconds after that, until the p.d. has fallen to less than 0.5 V. Repeat the experiment at least twice to obtain average values and reduce random errors. During the practical examination, candidates are often assessed on their ability to record readings with appropriate precision and to reset the capacitor fully before each new trial.
首先按图示连接电路，确保电容器已安全放电——可暂时用导线短路其两极（或在实验开始前用开关将其与电阻接通）。将电源设置为安全的直流电压，如 6.0 V，闭合充电回路给电容器充满电。确认电压表读数稳定在电源电压附近。然后同时启动秒表、断开充电开关并闭合放电开关。每隔固定时间记录电压表读数，例如前 1 分钟每 10 秒记录一次，之后每 20 秒记录一次，直到电压降至 0.5 V 以下。至少重复实验两次以获取平均值并减小随机误差。在实际考试中，考官通常会评估考生记录读数的恰当精度以及在每次新测量前确保电容器完全充电的能力。
4. Safety Precautions | 安全注意事项
Although the voltages are low, standard electrical safety rules still apply. Ensure the capacitor is fully discharged before handling any components to avoid electric shock. An electrolytic capacitor can retain charge for a long time even after the power supply is disconnected, so always discharge it through a resistor rather than shorting it directly with a wire, which could produce a large spark. Avoid touching bare connections while the circuit is live. If a variable power supply is used, set it correctly before connecting the capacitor to avoid over-voltage damage. In a school laboratory, a teacher or technician should check the circuit before power is applied. These precautions reflect typical requirements in both practical assessments and the June 2019 Unit 4 mark scheme.
尽管实验所用电压较低，但仍需遵守标准的电气安全规则。在处理任何元件前应确保电容器完全放电，以避免触电。电解电容器在断开电源后仍可长时间保留电荷，因此应始终经电阻放电，切勿直接用导线短路，否则可能产生强烈的火花。电路通电时不要触碰裸露的接线点。若使用可调电源，应在连接电容器前设置好正确电压，以免因过电压损坏器件。在学校实验室中，通电前应由教师或技术人员检查电路。这些预防措施体现了实验评估以及 2019 年 6 月单元 4 评分方案中的常见要求。
5. Data Collection Table | 数据记录表格
A well-structured table is crucial for obtaining full marks. The table should include columns for time t / s, potential difference V / V, and ideally ln(V) for later analysis. Headings must include the quantity and its unit. Below is a model set of data collected for a 1000 μF capacitor discharging through a 100 kΩ resistor (expected time constant τ ≈ 100 s). Note that voltage readings are given to two decimal places where the voltmeter resolution allows, and times are recorded to the nearest second.
一个结构清晰的数据表格对于获得满分至关重要。表格应包含时间 t / s、电压 V / V 等列，最好还预留 ln(V) 列供后续分析使用。表头必须标注物理量和单位。以下是一组放电数据范例，使用 1000 μF 电容和 100 kΩ 电阻（预计时间常数 τ ≈ 100 s）。注意电压读数根据电压表分辨率保留至两位小数，时间记录精确到秒。
| Time t / s | V / V | ln(V) |
|---|---|---|
| 0 | 6.00 | 1.79 |
| 10 | 5.43 | 1.69 |
| 20 | 4.92 | 1.59 |
| 30 | 4.46 | 1.50 |
| 40 | 4.05 | 1.40 |
| 50 | 3.68 | 1.30 |
| 60 | 3.34 | 1.21 |
| 80 | 2.72 | 1.00 |
| 100 | 2.21 | 0.79 |
| 120 | 1.80 | 0.59 |
| 150 | 1.33 | 0.28 |
6. Graphical Analysis of Raw Data | 原始数据的图形分析
The first graph candidates are usually required to plot is V against t. This reveals a smooth, decreasing exponential curve. From the raw graph, one can quickly estimate the time constant by noting the time at which V has fallen to about 2.22 V (i.e., 0.37 V₀ for V₀ = 6.00 V). In our model data, reading from the curve gives τ ≈ 100 s. The curve should show that the discharge is fastest near the beginning and gradually slows, consistent with the exponential model. It is important to use an appropriate scale, label axes correctly and draw a line of best fit through the points. In the June 2019 exam, marks were allocated specifically for the correct choice of scales and for plotting points accurately.
通常要求考生绘制 V 随 t 变化的图像第一张图。该图像显示为一条平滑的下降指数曲线。从原始图像出发，可以快速估算时间常数：找到电压下降至约 2.22 V（即当 V₀ = 6.00 V 时的 0.37 V₀）所对应的时间。在我们示例数据中，从曲线上读出 τ ≈ 100 s。曲线应展现出放电在最开始阶段最快，之后逐渐变慢的特征，这与指数模型一致。合理选取坐标轴尺度、正确标注轴并画出最佳拟合线十分关键。在 2019 年 6 月考试中，坐标轴尺度的正确选择和描点的准确性均有相应的分值分配。
7. Linearization and Determination of Time Constant | 线性化处理与时间常数测定
To obtain a more precise value for the time constant, the exponential equation is linearised by plotting ln V on the y-axis against time t on the x-axis. Taking natural logs of the discharge equation V = V₀ e−t/RC gives:
ln V = ln V₀ − (1/RC) t
This is of the form y = c + mx, where the gradient m = −1/RC and the intercept c = ln V₀. Therefore, a graph of ln V versus t should be a straight line with a negative gradient. The absolute value of the gradient equals 1/RC, so the time constant τ = RC = −1 / gradient (when gradient is negative). In the June 2019 question, candidates were expected to calculate the gradient using a large triangle, substitute correctly and then determine τ. Using our model data, a plot of ln V against t yields a straight line. Selecting two points from the line of best fit (not data points unless they lie exactly on the line), say (0, 1.79) and (150, 0.28), gives gradient = (0.28 − 1.79) / (150 − 0) = −0.0101 s⁻¹, hence τ = 1/0.0101 ≈ 99 s.
为了求得更为精确的时间常数值，可通过以 ln V 为纵轴、时间 t 为横轴作图来将指数方程线性化。对放电方程 V = V₀ e−t/RC 两边取自然对数可得：
ln V = ln V₀ − (1/RC) t
此式形如 y = c + mx，其中梯度 m = −1/RC，截距 c = ln V₀。因此，ln V 对 t 的图应为一条具有负梯度的直线。梯度的绝对值等于 1/RC，故时间常数 τ = RC = −1 / 梯度（当梯度为负时）。在 2019 年 6 月的考题中，要求考生利用大三角形计算梯度，正确代入并确定 τ。使用模型数据绘制 ln V – t 图可得到一条直线。从最佳拟合线（而非数据点，除非它们恰好在线上）上选取两点，例如 (0, 1.79) 和 (150, 0.28)，可得梯度 = (0.28 − 1.79) / (150 − 0) = −0.0101 s⁻¹，由此 τ = 1/0.0101 ≈ 99 s。
8. Calculation of Capacitance from Time Constant | 由时间常数计算电容
Once the time constant τ is known, the unknown capacitance can be calculated using τ = R × C. In the exam, the value of the resistor was given, e.g., R = 100 kΩ (100 × 10³ Ω). Rearranging gives C = τ / R. Using τ = 99 s gives C = 99 / (100 × 10³) = 9.9 × 10⁻⁴ F = 990 μF. Comparing with the nominal 1000 μF capacitor, this yields a percentage difference of about 1 %, which is well within typical tolerance values for electrolytic capacitors and experimental error. The mark scheme often awards marks for correct substitution, unit conversion and sensible final value with appropriate significant figures (e.g., 990 μF or 1.0 × 10⁻³ F). Students must show all working, including the conversion of kilohms to ohms, to gain full credit.
测出时间常数 τ 后，便可利用 τ = R × C 计算未知电容值。考试会给明电阻值，例如 R = 100 kΩ (100 × 10³ Ω)。将公式变形得 C = τ / R。代入 τ = 99 s 得到 C = 99 / (100 × 10³) = 9.9 × 10⁻⁴ F = 990 μF。与标称值 1000 μF 的电容器对比，百分偏差约为 1 %，完全在电解电容器的常规容差范围和实验误差之内。评分方案通常对正确代入、单位换算以及最终合理有效数字（如 990 μF 或 1.0 × 10⁻³ F）给予分数。考生需展示完整计算步骤，包括将千欧换算为欧姆，方能获得全部分数。
9. Uncertainty and Error Analysis | 不确定度与误差分析
Sources of uncertainty in this experiment include the reaction time when starting the stopwatch, the precision of the voltmeter and stopwatch, and any leakage current through the capacitor dielectric or the voltmeter itself. The exam question may ask candidates to estimate the percentage uncertainty in the time constant and hence in the calculated capacitance. For instance, if the voltmeter reads to ±0.01 V and the stopwatch to ±1 s, the largest uncertainty in the gradient arises from the first few points where the voltage changes most rapidly. A simple method is to draw lines of maximum and minimum gradient through the error bars on the ln V graph and determine the spread in gradient. Suppose gradient = −0.0101 s⁻¹ and the worst acceptable gradient is −0.0098 s⁻¹. Then uncertainty in gradient ≈ ±0.0003 s⁻¹, and percentage uncertainty in τ ≈ (0.0003/0.0101) × 100 % ≈ 3 %. This then propagates directly to the capacitance, giving C = 990 μF ± 30 μF. Answers should be expressed realistically; claiming uncertainty of 0.1 % would be unrealistic given the apparatus used.
本实验的不确定度来源包括启动秒表时的反应时间、电压表和秒表的精度，以及电容器介质或电压表本身可能存在的泄漏电流。考题可能要求考生估算时间常数以及由此计算出的电容的百分不确定度。例如，若电压表的最小分度为 ±0.01 V，秒表精度为 ±1 s，梯度的最大不确定度往往来自最初几个电压变化最快的点。一种简单的方法是在 ln V 图中过各误差棒描出最大和最小梯度的直线，并确定梯度的变化范围。假设梯度为 −0.0101 s⁻¹，最可接受的最差梯度为 −0.0098 s⁻¹，则梯度的不确定度约为 ±0.0003 s⁻¹，对应 τ 的百分不确定度约 (0.0003/0.0101) × 100 % ≈ 3 %。这一不确定度将直接传递至电容值，得到 C = 990 μF ± 30 μF。答案表述应当结合实际；鉴于所使用的仪器，声称不确定度仅 0.1% 是不切实际的。
10. Common Pitfalls and Improvements | 常见错误与改进方法
One frequent mistake is starting the discharge before the capacitor is fully charged, which leads to a lower V₀ and distorts the exponential fit. Another is using a voltmeter with insufficient internal resistance, causing the capacitor to discharge partially through the meter and making the measured p.d. drop faster than under an ideal RC discharge. Students sometimes forget to convert units or misplace the decimal point when calculating C from τ and R. To improve accuracy, the capacitor should be charged to the exact same voltage before each run, and a data-logger with a voltage sensor can replace manual stopwatch readings, eliminating human reaction time error. In the June 2019 context, candidates who suggested using a larger resistance value to make the discharge slower (giving more time to take readings) or who noted that the capacitor should be fitted without touching its leads to avoid heating effects were rewarded with improvement marks.
常见的错误之一是电容器尚未完全充满电便开始放电，这会导致 V₀ 偏低并扭曲指数拟合。另一错误是采用内阻过低的电压表，使电容器有一部分电流经电压表流失，导致测量的电压下降速度快于理想 RC 放电。学生在根据 τ 与 R 计算 C 时，有时会忘记换算单位或点错小数点。为提高精度，可在每次测量前将电容器充至完全相同的初始电压，并用带有电压传感器的数据记录仪代替手动秒表读数，以此消除人为反应时间误差。在 2019 年 6 月的试题背景下，考生若提出使用更大的电阻以减缓放电速度（从而有更长读数时间），或指出安装电容器时不要触碰引线以避免热效应的影响，这些改进建议均可获得加分。
11. Exam-style Questions and Answers | 真题讲解与解答
The June 2019 Unit 4 paper typically included follow-up questions. For example: “Explain why the graph of ln V against t should be a straight line.” The expected answer is that the relationship V = V₀ e−t/RC can be rearranged to ln V = ln V₀ − t/RC. Since ln V₀ and −1/RC are constants, the equation is of the form y = mx + c, which represents a straight line. Another common question: “State what the gradient of the ln V–t graph represents and how the time constant can be found.” Answer: gradient = −1/RC, so time constant = −1/gradient. Candidates could also be asked to plot a second graph or to estimate the half-life of the discharge. Since half-life t₁/₂ = RC ln 2 ≈ 0.693 RC, one can verify consistency. Providing clear, stepwise explanations is the key to high marks.
2019 年 6 月的单元 4 试卷中包含典型的后续问题。例如：“解释为何 ln V 对 t 的图像应为一条直线。” 期望的答案是：关系式 V = V₀ e−t/RC 可变形为 ln V = ln V₀ − t/RC，由于 ln V₀ 与 −1/RC 均为常数，方程呈 y = mx + c 的形式，故表示一条直线。另一个常见问题是：“说明 ln V–t 图的梯度代表什么，以及如何计算时间常数。” 答案：梯度 = −1/RC，所以时间常数 = −1/梯度。考生还可能被要求绘制第二张图或估算放电的半衰期。半衰期 t₁/₂ = RC ln 2 ≈ 0.693 RC，可用来检验一致性。提供清晰、逐步的解答是获取高分的关键。
12. Conclusion and Key Takeaways | 总结与要点
The experimental investigation embedded in the June 2019 AS Physics Unit 4 paper is a comprehensive test of practical physics skills and theoretical understanding. Students must demonstrate competence in circuit assembly, data logging, graph plotting, linearisation, gradient analysis and uncertainty calculations. The key formulas V = V₀ e−t/RC and τ = RC lie at the heart of the analysis. Remember always to use a high-resistance voltmeter, discharge the capacitor safely, take readings at regular intervals and, most importantly, transform the data to a straight-line form to extract the most reliable τ value. With the methodical approach outlined here, any similar experiment on capacitor discharge can be tackled with confidence in the exam.
2019 年 6 月 AS 物理单元 4 试卷中内嵌的实验探究，是对学生实践物理技能与理论理解的综合考查。考生必须展示出在电路组装、数据记录、图像绘制、线性化处理、梯度分析及不确定度计算等方面的能力。核心公式 V = V₀ e−t/RC 和 τ = RC 是整个分析的基石。务必记住：使用高内阻电压表、安全释放电容器电荷、每隔固定时间记录读数，以及最重要的——将数据转换为直线形式以提取最可靠的 τ 值。按照本文所述的系统方法，任何类似的电容器放电实验都能在考试中从容应对。
Published by TutorHao | Physics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导Cancel reply