📚 IGCSE Edexcel Physics: Alternating Current Core Concepts | IGCSE 爱德思物理：交流电考点精讲
Alternating current (AC) is the type of electricity delivered to homes and schools. In IGCSE Edexcel Physics, you must understand how AC differs from direct current (DC), how to interpret its waveform on an oscilloscope, and the basics of rectification and smoothing. This article covers all the key points for the exam, including mains electricity specifications, oscilloscope calculations, and diode circuits.
交流电(AC)是输送到家庭和学校的电力类型。在 IGCSE 爱德思物理中，你需要理解交流电与直流电(DC)的区别、如何在示波器上读懂它的波形，以及整流和平滑的基本知识。本文涵盖了考试涉及的所有重点，包括市电规格、示波器计算以及二极管电路。
1. What is Alternating Current? | 什么是交流电？
An alternating current (AC) is a flow of electric charge that periodically reverses direction. The voltage in an AC circuit also alternates, typically in the form of a sine wave. In contrast, direct current (DC) flows in one constant direction only. Batteries and solar cells provide DC, whereas mains electricity is AC.
交流电是一种电荷流动方向周期性反转的电流。交流电路中的电压也交替变化，通常呈正弦波形。相比之下，直流电(DC)仅沿一个恒定方向流动。电池和太阳能电池提供直流电，而市电则是交流电。
AC is preferred for power distribution because its voltage can be easily changed using transformers, reducing energy loss over long distances. The waveform of AC can be displayed on a cathode-ray oscilloscope (CRO) as a graph of voltage against time.
交流电更适合电力输送，因为它的电压可以通过变压器轻松改变，减少远距离输电的能量损耗。交流电的波形可以在阴极射线示波器(CRO)上显示为电压-时间图。
2. UK Mains Electricity Specifications | 英国市电规格
In the UK, the domestic mains supply provides an alternating current with a root mean square (rms) voltage of 230 V and a frequency of 50 Hz. The peak voltage, which is seen on an oscilloscope, is approximately 325 V because Vpeak = Vrms × √2. The mains supply uses three wires: live (brown), neutral (blue), and earth (green/yellow).
在英国，家庭市电提供均方根(rms)电压为 230 V、频率为 50 Hz 的交流电。在示波器上看到的峰值电压约为 325 V，因为 V峰值 = Vrms × √2。市电使用三根电线：火线(棕色)、零线(蓝色)和地线(绿/黄色)。
The potential difference between the live wire and neutral wire is 230 V rms. The earth wire is at 0 V and is a safety feature, providing a path for fault current to flow if an appliance’s metal casing becomes live. Fuses and circuit breakers are used to prevent overheating and electric shock.
火线与零线之间的电位差为 230 V rms。地线为 0 V，是一项安全装置，当电器金属外壳带电时可提供故障电流通路。保险丝和断路器用于防止过热和触电。
3. Displaying AC on an Oscilloscope | 在示波器上显示交流电
An oscilloscope plots voltage on the vertical axis (Y-gain) against time on the horizontal axis (time-base). For an AC signal, the trace is a regular sine wave that oscillates above and below the zero line. The height and width of the wave depend on the voltage and time settings.
示波器将电压标绘在纵轴(Y增益)上，时间标绘在横轴(时基)上。对于交流信号，轨迹是一条规则的正弦波，在零线上下振荡。波的高度和宽度取决于电压和时间的设置。
To measure the peak voltage, count the number of vertical divisions from the centre line to a crest, then multiply by the volts per division (V/div) setting. To find the time period T, count the horizontal divisions for one complete wave and multiply by the time per division (s/div or ms/div). Frequency f is then calculated as f = 1 / T.
要测量峰值电压，数出从中心线到波峰的垂直格数，再乘以每格电压(V/div)设定值。要得出周期 T，数出一个完整波的水平格数，再乘以每格时间(s/div 或 ms/div)。然后由 f = 1 / T 计算出频率。
4. Understanding Peak Voltage and RMS Voltage | 理解峰值电压与 RMS 电压
The peak voltage is the maximum voltage of an AC supply. For UK mains, the peak is about 325 V. However, the stated 230 V is the rms (root mean square) value, which is a sort of average that represents the equivalent DC voltage that would deliver the same power to a resistor. The relationship is:
Vpeak = Vrms × √2
峰值电压是交流电源的最大电压。对英国市电而言，峰值约为 325 V。然而，标称的 230 V 是 RMS(均方根)值，它代表能对电阻提供相同功率的等效直流电压。两者关系为：
V峰值 = Vrms × √2
In IGCSE, you may be asked to read the peak voltage directly from a CRO trace. You should be able to explain that the rms value is lower because the voltage varies, spending less time at the maximum.
在 IGCSE 中，你可能会被要求直接从 CRO 轨迹上读出峰值电压。你应能解释 RMS 值较低是因为电压在变化，处于最大值的时间较短。
5. Calculating Frequency from an Oscilloscope Trace | 从示波器轨迹计算频率
The frequency of an AC signal is the number of complete cycles per second. On an oscilloscope, first determine the period T – the time taken for one whole wave. Then apply:
f = 1 / T
交流信号的频率是每秒完整周期数。在示波器上，先确定周期 T——一个完整波所需的时间。然后应用：
f = 1 / T
Example: If the time-base is set to 2 ms/div and one complete wave spans 10 divisions, then T = 10 × 2 ms = 20 ms = 0.020 s. The frequency f = 1 / 0.020 = 50 Hz. This matches the UK mains frequency.
例如：若时基设为 2 ms/div，一个完整波跨越 10 格，则 T = 10 × 2 ms = 20 ms = 0.020 s。频率 f = 1 / 0.020 = 50 Hz，与英国市电频率相符。
Always check the time-base setting before calculating; the trace might be calibrated in seconds, milliseconds, or microseconds. The frequency obtained should be in hertz (Hz).
计算前务必检查时基设定；轨迹可能以秒、毫秒或微秒为单位。得到的频率应以赫兹(Hz)表示。
6. Rectification: Turning AC into DC | 整流：将交流变为直流
Many electronic devices require direct current, so AC must be rectified. The simplest method uses a single diode – this is half-wave rectification. The diode only allows current to pass during the positive half-cycles of the AC input, blocking the negative halves. The output is a pulsed DC waveform with gaps where the negative cycles were cut off.
许多电子设备需要直流电，因此交流电必须整流。最简单的方法是使用单个二极管——这就是半波整流。二极管只允许交流输入的正半周通过，阻断负半周。输出是一个脉动直流波形，负半周被切除掉了。
While half-wave rectification is easy to build, it is inefficient because half of the input power is wasted. A more efficient method is full-wave rectification using four diodes arranged in a bridge (a bridge rectifier). This circuit inverts the negative half-cycles so that both halves appear as positive pulses, using all of the AC input.
半波整流虽然容易搭建，但效率低，因为一半的输入功率被浪费了。更高效的方法是用四只二极管排列成桥式(桥式整流器)进行全波整流。该电路将负半周反转，使两个半周都呈现为正脉冲，从而利用全部交流输入。
7. Half-Wave Rectification in Detail | 详解半波整流
In a half-wave rectifier circuit, a single diode is placed in series with the load. When the AC input is positive (forward bias), the diode conducts and current flows. When the input goes negative (reverse bias), the diode blocks current. The resulting output voltage across the load is a series of positive bumps separated by flat gaps.
在半波整流电路中，单个二极管与负载串联。当交流输入为正(正向偏置)时，二极管导通，有电流流过。当输入变负(反向偏置)时，二极管阻断电流。负载上的输出电压是一串被平坦间隙隔开的正向凸起。
The peak output voltage is slightly less than the peak AC input because of the diode’s forward voltage drop (about 0.7 V for a silicon diode). This waveform is not suitable for sensitive electronics unless smoothed.
由于二极管存在正向压降(硅管约 0.7 V)，峰值输出电压略低于峰值交流输入。这种波形若未经平滑处理，不适合用于灵敏的电子设备。
8. Full-Wave Rectification Using a Diode Bridge | 使用二极管桥式电路的全波整流
A bridge rectifier consists of four diodes connected in a loop. During the positive half-cycle of the AC input, two diodes conduct and steer the current through the load in one direction. During the negative half-cycle, the other two diodes conduct, and the current through the load still flows in the same direction. The output is a waveform that never goes negative – both halves of the AC cycle are made positive.
桥式整流器由四只二极管环接而成。在交流输入的正半周，两只二极管导通，引导电流按一个方向流过负载。在负半周，另外两只二极管导通，通过负载的电流方向不变。输出的波形从不为负——交流电的两个半周都变成正向。
The frequency of the full-wave rectified output is double that of the input AC (for 50 Hz mains, the ripple frequency becomes 100 Hz). This reduces the gap between pulses, making smoothing more effective.
全波整流输出的频率是输入交流电的两倍(对 50 Hz 市电，纹波频率变为 100 Hz)。这减小了脉冲之间的间隙，使后续的平滑更加有效。
9. Smoothing with a Capacitor | 用电容器进行平滑
The rectified output is still bumpy DC. To produce a steady voltage, a large-value electrolytic capacitor is connected across the load. The capacitor charges up to the peak voltage during the rising part of each pulse and then discharges slowly through the load when the rectifier voltage drops. This fills in the gaps, reducing the ripple.
整流后的输出仍为起伏的直流电。要获得平稳电压，在负载两端并联一个大容量的电解电容。电容在每个脉冲上升段充电至峰值电压，然后在整流电压下降时通过负载缓慢放电，从而填补间隙，减小纹波。
With a larger capacitor, or a smaller load current (higher load resistance), the capacitor discharges more slowly, resulting in a smoother output with less ripple. In the exam, you might be asked to sketch smoothed waveforms and identify the effect of changing the capacitor value.
使用更大的电容，或负载电流更小(负载电阻更大)时，电容放电更慢，输出更平滑，纹波更小。考试中可能要求你画出平滑后的波形，并识别改变电容值产生的影响。
10. Why AC is Used with Transformers | 为什么交流电与变压器一起使用
An important application of AC is in transformers. A transformer works on the principle of electromagnetic induction, which requires a changing magnetic field. Only AC provides the continuously changing current needed to induce a voltage in the secondary coil. DC, being steady, cannot operate a standard transformer.
交流电的一个重要应用是在变压器中。变压器基于电磁感应原理工作，需要变化的磁场。只有交流电才能提供持续变化的电流，从而在次级线圈中感应出电压。直流电是恒定的，无法使标准变压器工作。
Transformers allow the voltage to be stepped up for efficient long-distance transmission and stepped down for safe domestic use. This is why the national grid uses alternating current.
变压器可以将电压升高以实现高效的远距离输电，再降低以供家庭安全使用。这就是国家电网采用交流电的原因。
11. Key Formulas and Units | 关键公式与单位
The essential formulas for AC calculations are summarised below. Consistently using the correct units is vital in the exam.
交流电计算的基本公式总结如下。在考试中始终使用正确的单位至关重要。
| Quantity | 物理量 | Formula | 公式 | Units | 单位 |
|---|---|---|
| Frequency f | f = 1 / T | hertz (Hz) |
| Period T | T = 1 / f | seconds (s) |
| Peak voltage Vpeak | Vpeak = Vrms × √2 | volts (V) |
| RMS voltage | Vrms = Vpeak / √2 | volts (V) |
When reading an oscilloscope, always multiply the number of divisions by the scale factor. For period, use time-base settings; for voltage, use Y-gain or volts/div.
读取示波器时，始终将格数乘以标度因子。对于周期，使用时间基准设定；对于电压，使用 Y 增益或 volts/div。
12. Exam Tips and Common Pitfalls | 考试技巧与常见误区
In IGCSE Edexcel questions, always note whether you are being asked for peak or rms voltage. If a CRO trace is given, the maximum height corresponds to Vpeak. Do not confuse period with half a cycle. When sketching rectification circuits, draw diodes with the correct symbol and orientation.
在 IGCSE 爱德思试题中，务必留意要求的是峰值电压还是 RMS 电压。若给出 CRO 轨迹，最大高度对应的是 V峰值。不要将周期与半个周期混淆。绘制整流电路时，要以正确的符号和方向画出二极管。
Remember that the smoothing capacitor must be connected in parallel with the load, not in series. The ripple frequency after full-wave rectification is double the input frequency. If asked to explain why AC is used in transformers, link it to the need for a changing magnetic flux.
记住平滑电容必须与负载并联，而不是串联。全波整流后的纹波频率是输入频率的两倍。若被问及为什么交流电用于变压器，要将其与需要变化的磁通量联系起来。
Practice interpreting oscilloscope traces with different time-base and voltage settings, and be comfortable calculating frequency from a diagram.
多练习解读不同时基和电压设定下的示波器轨迹，并熟练根据图解计算频率。
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