📚 AQA A-level Physics June 2018 Insert 4: Capacitor Discharge and Time Constant | AQA物理A-level 2018年6月插入4：电容器放电与时间常数
The insert for the June 2018 AQA A-level Physics paper provided students with a graph and data related to the discharge of a capacitor through a fixed resistor. This article breaks down the essential physics of capacitor discharge, guides you through the interpretation of the insert, and prepares you for the types of questions that commonly appear in the exam. We will analyse the exponential decay curve, derive the time constant, and demonstrate how to handle uncertainties using the data from the insert.
2018年6月AQA物理A-level试卷所提供的插入题包含了一张电容器通过固定电阻放电的图表和数据。本文系统讲解电容器放电的核心物理知识，指导你解读插入内容，并为你备考中常见的题型做好准备。我们将分析指数衰减曲线、推导时间常数，并演示如何利用插入题中的数据进行不确定度分析。
1. The Discharge Circuit and Equation | 放电电路与方程
When a charged capacitor of capacitance C discharges through a resistor of resistance R, the charge Q on the capacitor decreases exponentially with time t. The instantaneous potential difference V across the capacitor follows the same exponential decay as the charge, given by the equation: V = V₀ e^(−t/RC), where V₀ is the initial potential difference at t = 0.
当一个电容为C的电容器通过阻值为R的电阻进行放电时，电容器上的电荷Q随时间t呈指数衰减。电容器两端的瞬时电势差V与电荷遵循同样的指数衰减规律，其方程为：V = V₀ e^(−t/RC)，其中V₀是t = 0时的初始电势差。
The product RC is known as the time constant, τ. It represents the time taken for the potential difference (or charge, or current) to fall to 1/e (approximately 36.8%) of its initial value. The time constant has units of seconds and is a key parameter that can be determined from the discharge graph.
乘积RC称为时间常数τ，它表示电势差（或电荷、电流）下降到初始值的1/e（约36.8%）所需的时间。时间常数的单位是秒，是可以通过放电图确定的关键参数。
The insert from June 2018 likely contained a voltmeter reading recorded at regular intervals as a capacitor discharged. By fitting an exponential curve to these points, you can extract τ and use it to find either R or C if the other is known.
2018年6月的插入题很可能包含一个电压表在电容器放电过程中按固定时间间隔记录的数据。通过对这些数据点拟合指数曲线，你可以提取出τ，并利用它求出R或C（已知其中一个量时）。
2. Experimental Setup | 实验装置
A typical capacitor discharge experiment involves a capacitor, a resistor, a battery or power supply, a switch, and a voltmeter or datalogger. The capacitor is first charged by connecting the power supply across it. Then the switch is thrown to disconnect the supply and connect the capacitor directly to the resistor, starting the discharge.
典型的电容器放电实验包括电容器、电阻、电池或电源、开关以及电压表或数据记录器。首先，通过将电源跨接在电容器上使其充电。然后，切换开关，断开电源并让电容器直接与电阻相连，开始放电。
In the AQA insert, data were likely collected using a data logger with a voltage sensor, sampling the potential difference every few seconds. The graph produced shows a smooth exponential decay, from which the time constant can be measured. Ensure you are familiar with the circuit diagram and the correct method of measuring V₀, because exam questions often test your practical understanding.
在AQA插入题中，数据很可能是通过带有电压传感器的数据记录器采集的，每隔几秒记录一次电势差。生成的图是平滑的指数衰减曲线，可从中测量时间常数。请务必熟悉电路图以及正确测量V₀的方法，因为考试题目常考查对实验的理解。
One important detail is the internal resistance of the voltmeter. If the voltmeter has a finite resistance, it will provide an additional discharge path, causing the observed decay to be faster than expected. In high-quality experiments a very high-impedance voltmeter (such as a digital multimeter) is used to minimise this effect.
一个重要的细节是电压表的内阻。如果电压表的电阻有限，它将提供额外的放电通路，导致观测到的衰减比预期更快。在高质量实验中，应使用极高阻抗的电压表（如数字万用表）以减小此影响。
3. Reading the Exponential Decay Graph | 读取指数衰减图
The insert graph typically plots potential difference V (in volts) on the vertical axis and time t (in seconds) on the horizontal axis. The curve starts at V₀ and asymptotically approaches zero. Because it is exponential, the initial gradient is steepest, and the curve flattens over time.
插入题图表通常以纵轴表示电势差V（单位伏特），横轴表示时间t（单位秒）。曲线从V₀开始，渐近地趋近于零。由于是指数衰减，初始斜率最陡，之后曲线逐渐趋于平缓。
To determine the time constant τ from the graph, you can find the time at which V equals 0.368V₀. Draw a horizontal line at 0.368V₀, intersect the curve, and read off the corresponding time. This time is τ. Alternatively, find the time at which the initial tangent meets the time axis; this intercept also gives τ, but the 0.368 method is usually more accurate with real data.
要从图中确定时间常数τ，可以找到V等于0.368V₀时对应的时间。画一条水平线在0.368V₀处，与曲线相交，读出对应的时间，这个时间就是τ。另一种方法是找出初始切线交时间轴的位置，这个截距也给出τ，但0.368方法通常对真实数据更准确。
When using the insert, always check the scales and units carefully. Look for the initial value V₀ from the y-intercept, not from a point at t = 0 recorded after discharge has already begun. If data logging started slightly after the switch was closed, you must extrapolate the curve back to t = 0.
使用插入题时，务必仔细检查刻度和单位。注意从y截距读取初始值V₀，而不是从放电已经开始后t = 0的点读取。如果数据记录在开关闭合后稍有延迟才开始，你需要将曲线外推回t = 0。
4. Determining the Time Constant from the Graph | 从图像确定时间常数
The most straightforward method is to use the known fraction 1/e ≈ 0.368. For any exponential decay, after one time constant the value is 36.8% of the original. On the insert graph, locate V₀, multiply by 0.368, and draw a horizontal line at that voltage. The intersection with the curve gives the time constant.
最直接的方法是利用分数1/e ≈ 0.368。对于任何指数衰减，经过一个时间常数后，数值为原始值的36.8%。在插入题的图中，找到V₀，乘以0.368，画一条水平线在该电压处。与曲线的交点即给出时间常数。
For example, if V₀ = 6.0 V, then 0.368V₀ = 2.21 V. If the curve reaches 2.21 V at t = 8.2 s, then τ = 8.2 s. If the graph has a grid, you should read the intersection to the nearest half division, and record the uncertainty from the half-width of the intersection line.
例如，若V₀ = 6.0 V，则0.368V₀ = 2.21 V。如果曲线在t = 8.2 s时降到2.21 V，则τ = 8.2 s。如果图中有网格，你应该将交点读到最接近的半格，并根据交点线的半宽记录不确定度。
Another method uses the initial slope. Draw a tangent to the curve at t = 0, and extend it to the time axis. The intercept is numerically equal to τ. This method is quick but often less reliable because drawing the tangent by hand introduces human error. In the exam, you may need to use both methods and compare, but the 0.368 method is generally preferred.
另一种方法是利用初始斜率。在t = 0处绘制曲线的切线，并延伸到时间轴。截距在数值上等于τ。这种方法快速，但手工绘制切线会引入人为误差，因此通常可靠性较低。在考试中可能要求你同时使用两种方法并比较，但0.368方法通常更受青睐。
5. Calculating Unknown Components | 计算未知元件参数
Once you have determined τ from the insert, you can calculate the capacitance or resistance using τ = RC. If the resistor value is printed on the insert, you can rearrange to C = τ/R. Conversely, if the capacitance is known, the resistance can be found.
一旦从插入题中确定了τ，你就可以利用τ = RC来计算电容或电阻。如果插入题中标注了电阻值，你可以通过C = τ/R来求电容。反之，如果已知电容，也可以求电阻。
Suppose the insert gives τ = 8.2 s and the resistor is 100 kΩ. Then C = τ/R = 8.2 / (100 × 10³) = 8.2 × 10⁻⁵ F = 82 μF. Check the exponent: 1 kΩ = 10³ Ω, so 100 kΩ = 10⁵ Ω. Division yields 8.2 × 10⁻⁵ F, which is 82 μF.
假设插入题给出τ = 8.2 s，电阻为100 kΩ。则C = τ/R = 8.2 / (100 × 10³) = 8.2 × 10⁻⁵ F = 82 μF。注意指数：1 kΩ = 10³ Ω，所以100 kΩ = 10⁵ Ω。除法结果得8.2 × 10⁻⁵ F，即82 μF。
Be mindful of units in this calculation. Time constants are in seconds, resistance in ohms, and capacitance in farads. If the resistor is given in MΩ, convert to ohms first. If the capacitor is in μF, convert to F before multiplying by R.
计算时务必注意单位。时间常数单位为秒，电阻单位为欧姆，电容单位为法拉。如果电阻以MΩ给出，先换算成欧姆。如果电容以μF给出，先换算成法拉再乘以R。
6. Logarithmic Analysis to Linearise the Data | 对数分析以线性化数据
The exponential equation V = V₀ e^(−t/RC) can be linearised by taking natural logarithms: ln V = ln V₀ − t/RC. This gives a straight line when ln V is plotted against time t. The gradient of this line is −1/RC, and the y-intercept is ln V₀. The insert may present log data or ask you to plot it yourself.
指数方程V = V₀ e^(−t/RC)可通过取自然对数线性化：ln V = ln V₀ − t/RC。当ln V对时间t作图时，得到一条直线。该直线的斜率为−1/RC，y截距为ln V₀。插入题可能直接给出对数数据，也可能要求你自行作图。
Using the linearised graph is often more accurate than reading directly from the exponential curve, because it uses all the data points fitted to a straight line rather than a single point at 0.368V₀. The gradient can be computed from any two well-separated points, and then −1/RC gives the time constant directly.
使用线性化图像通常比直接读指数曲线更准确，因为它利用所有数据点拟合直线，而不是仅使用0.368V₀这一单点。可以从相隔较远的两个点计算斜率，然后通过−1/RC直接得到时间常数。
When taking logs, remember that ln V is negative if V < 1 V. Use a calculator to obtain three significant figures for each ln V value. The graph paper in the insert may already have a natural log scale on one axis, or you may need to add a column to a data table.
取对数时，若V < 1 V，则ln V为负值。使用计算器计算每个ln V值时保留三位有效数字。插入题中的坐标纸可能已在某一坐标轴上采用自然对数刻度，或者你可能需要在数据表中添加一列。
7. Uncertainty and Error Analysis | 不确定度与误差分析
When reading τ from the exponential graph, uncertainty arises from the reading error in V₀, the thickness of the graph line, and the precision of the time axis. A typical approach is to estimate the maximum and minimum possible values of τ by using the error bars on the intersection, then quote τ ± Δτ.
从指数图读取τ时，不确定度来源于V₀的读数误差、图线粗细以及时间轴的分度值。典型的方法是使用交点处的误差棒估算τ的最大和最小值，然后给出τ ± Δτ。
If using the 0.368 method, the main uncertainty is in locating the horizontal line. Suppose V₀ = 6.0 ± 0.1 V. Then 0.368V₀ = 2.21 ± 0.04 V. The width of the graph line where it intersects this band gives a range of t values, from which Δτ is obtained.
如果使用0.368方法，主要不确定度在确定水平线的位置。假设V₀ = 6.0 ± 0.1 V，则0.368V₀ = 2.21 ± 0.04 V。图线与该带状区域相交的宽度给出了t的范围，由此可得到Δτ。
When calculating C or R from τ, the fractional uncertainties in τ and the known component add in quadrature if they are independent. For example, if τ = 8.2 ± 0.3 s and R = 100 kΩ ± 1%, then the fractional uncertainty in C is sqrt((0.3/8.2)² + (0.01)²) ≈ 0.038, so ΔC = 0.038 × 82 μF ≈ 3 μF.
当从τ计算C或R时，τ和已知元件的相对不确定度若相互独立，则以平方求和。例如，若τ = 8.2 ± 0.3 s，R = 100 kΩ ± 1%，则C的相对不确定度为sqrt((0.3/8.2)² + (0.01)²) ≈ 0.038，因此ΔC = 0.038 × 82 μF ≈ 3 μF。
8. Exam Tips and Common Mistakes | 考试技巧与常见错误
Common mistakes include: forgetting to convert units before using τ = RC, reading V₀ from a point after discharge has started, misidentifying the time constant as the time to reach half value (that is the half-life, T₁/₂ = τ ln 2), and confusing natural logarithms with base-10 logs.
常见错误包括：使用τ = RC前忘记换算单位，从放电开始后的某个点读取V₀，将时间常数误认为达到半值的时间（那是半衰期T₁/₂ = τ ln 2），以及对数的底混淆（自然对数与常用对数）。
When asked to “show that the time constant is approximately 8 seconds”, use a quick check: if V₀ = 6 V, then 0.368 × 6 = 2.21 V. Look for the time at which the graph crosses 2.2 V. You only need one significant figure in the comparison, not an exact calculation.
当要求“说明时间常数约为8秒”时，可进行快速检验：若V₀ = 6 V，则0.368 × 6 = 2.21 V。在图中找到电压降到约2.2 V的时刻。此类问题只需一位有效数字的比较，无需精确计算。
Another tip: if the insert provides a table of V versus t, you can rearrange the equation to V/V₀ = e^(−t/RC). Take logs of V/V₀ rather than V alone to get a y-intercept of zero, which simplifies the linear plot. This is often required in the first part of a question.
另一个技巧：如果插入题提供了V对t的数据表，你可以将方程改写为V/V₀ = e^(−t/RC)。取V/V₀的对数而不是V本身，可得到y截距为零，从而简化线性图。这常是题目第一部分的要求。
Finally, always quote your final time constant to the same number of significant figures as the data in the insert. If the data points are given to three significant figures, your τ should also be to three significant figures, and must include units.
最后，最终时间常数的有效数字位数应与插入题中的数据一致。如果数据点给出三位有效数字，你的τ也应保留三位有效数字，并包含单位。
9. Conclusion | 结论
The capacitor discharge insert from the June 2018 AQA paper is a rich source of exam-style questions. By mastering the exponential decay equation, the graphical methods for finding τ, the unit conversions, and the logarithmic linearisation technique, you will be well prepared for similar problems. Practise reading graphs carefully and always carry out an uncertainty check.
2018年6月AQA试卷中的电容器放电插入题是丰富的考试题型来源。通过掌握指数衰减方程、用图像求τ的方法、单位换算以及对数线性化技术，你将能从容应对类似问题。练习仔细读图，并始终进行不确定度检验。
The skills you develop here are not limited to capacitors—they apply to radioactive decay, Newton’s law of cooling, and any physical system that follows an exponential law. Treat the insert as a mini-case study and you will gain a deeper understanding of exponential processes.
这里所培养的技能不仅限于电容器——它们同样适用于放射性衰变、牛顿冷却定律以及任何遵循指数定律的物理系统。将插入题视为一个微型案例研究，你将更深入地理解指数过程。
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