📚 Transformers in Edexcel IGCSE Physics | Edexcel IGCSE 物理中的变压器
A transformer is a passive electrical device that transfers electrical energy from one alternating current circuit to another by electromagnetic induction, changing the voltage and current levels without any moving parts. Understanding transformers lies at the heart of many IGCSE Physics questions and real-world power distribution systems.
变压器是一种无源电气装置，它利用电磁感应将电能从一个交流电路传递到另一个交流电路，在没有任何运动部件的情况下改变电压和电流的大小。理解变压器是许多 IGCSE 物理考题以及现实世界电力分配系统的核心。
1. Electromagnetic Induction Recap | 电磁感应回顾
A transformer works on the principle of electromagnetic induction discovered by Michael Faraday: a changing magnetic field can induce an electromotive force (e.m.f.) in a nearby conductor. This induction only occurs when the magnetic field is varying, which is why transformers require an alternating current (a.c.) in the primary coil.
变压器依据法拉第发现的电磁感应原理工作：变化的磁场可以在邻近的导体中感生出电动势。只有当磁场发生变化时才会产生感应，因此变压器的初级线圈必须通入交流电。
2. Basic Construction of a Transformer | 变压器的基本结构
A simple transformer consists of two coils of insulated wire, known as the primary coil and the secondary coil, wound around a common soft iron core. The primary coil is connected to the alternating input voltage, while the secondary coil is connected to the output circuit. The soft iron core is laminated to reduce energy losses from eddy currents.
一个简单的变压器由两个绝缘导线线圈构成，分别称为初级线圈和次级线圈，它们缠绕在同一个软铁芯上。初级线圈连接到交流输入电压，次级线圈连接到输出电路。软铁芯采用叠片结构，以减小涡流造成的能量损失。
3. How a Transformer Works | 变压器的工作过程
When an alternating voltage Vₚ is applied to the primary coil, an alternating current Iₚ flows through it, producing an alternating magnetic field in the iron core. The core channels this changing magnetic field through the secondary coil. According to Faraday’s law, the changing magnetic flux linkage induces an alternating e.m.f. across the secondary coil, giving an output voltage Vₛ.
当初级线圈上施加交流电压 Vₚ 时，会有交流电流 Iₚ 流过它，在铁芯中产生交变磁场。铁芯将这一变化的磁场引导至次级线圈。根据法拉第定律，变化的磁通链会在次级线圈中感应出交变电动势，从而产生输出电压 Vₛ。
4. The Turns Ratio Equation | 匝数比公式
In an ideal transformer, the ratio of the secondary voltage to the primary voltage is directly proportional to the ratio of the number of turns on the secondary coil to the number of turns on the primary coil. This relationship is expressed by the transformer equation:
在理想变压器中，次级电压与初级电压之比正比于次级线圈匝数与初级线圈匝数之比。这一关系由变压器公式表达：
Vₛ / Vₚ = Nₛ / Nₚ
Where Vₛ is the secondary (output) voltage, Vₚ is the primary (input) voltage, Nₛ is the number of turns on the secondary coil, and Nₚ is the number of turns on the primary coil.
其中 Vₛ 为次级（输出）电压，Vₚ 为初级（输入）电压，Nₛ 为次级线圈匝数，Nₚ 为初级线圈匝数。
5. Power and Current in an Ideal Transformer | 理想变压器中的功率与电流
For an ideal transformer that is 100% efficient, the electrical power delivered to the primary coil equals the electrical power delivered by the secondary coil. Thus, the input power Pₚ = Vₚ × Iₚ equals the output power Pₛ = Vₛ × Iₛ. From this we can derive a second useful relationship.
对于效率为 100% 的理想变压器，输入初级线圈的电功率等于次级线圈输出的电功率。因此，输入功率 Pₚ = Vₚ × Iₚ 等于输出功率 Pₛ = Vₛ × Iₛ。由此可推导出第二个有用的关系式。
Vₚ Iₚ = Vₛ Iₛ
Combining this with the turns ratio equation gives a link between currents and turns: Nₛ / Nₚ = Iₚ / Iₛ. So if a transformer steps up the voltage, the current in the secondary coil steps down proportionally, and vice versa.
将其与匝数比公式结合，可得到电流与匝数之间的联系：Nₛ / Nₚ = Iₚ / Iₛ。因此，如果变压器将电压升高，次级线圈中的电流就会成比例地减小，反之亦然。
6. Step-Up and Step-Down Transformers | 升压变压器与降压变压器
Transformers are classified based on whether they increase or decrease the voltage. The table below summarises the key differences that IGCSE candidates must remember.
变压器根据电压的升降进行分类。下表总结了 IGCSE 考生必须记住的关键区别。
| Feature / 特征 | Step-Up Transformer / 升压变压器 | Step-Down Transformer / 降压变压器 |
|---|---|---|
| Turns ratio / 匝数比 | Nₛ > Nₚ (secondary has more turns) | Nₛ < Nₚ (secondary has fewer turns) |
| Voltage change / 电压变化 | Vₛ > Vₚ (output voltage is larger) | Vₛ < Vₚ (output voltage is smaller) |
| Current change / 电流变化 | Iₛ < Iₚ (current decreases) | Iₛ > Iₚ (current increases) |
| Typical application / 典型应用 | Power station to transmission lines | Transmission lines to homes, mobile phone chargers |
7. Energy Losses and Efficiency | 能量损耗与效率
Real transformers are not perfect and some energy is always lost, mainly in the form of heat. The main sources of loss include resistance heating in the coils (Joule heating), eddy currents in the iron core, magnetic flux leakage and magnetisation hysteresis. Laminating the core and using thick copper wire help reduce these losses, but the efficiency of large transformers can still exceed 99%.
实际变压器并非完美，总会有一些能量损耗，主要以热的形式散失。主要损耗来源包括线圈的电阻发热（焦耳热）、铁芯中的涡流、磁通泄漏以及磁化磁滞。采用叠片铁芯和粗铜导线有助于减少这些损耗，但大型变压器的效率仍可超过 99%。
8. Why a Transformer Needs Alternating Current | 为什么变压器需要交流电
A transformer cannot work with a steady direct current (d.c.). If a constant d.c. voltage is applied to the primary coil, the magnetic field produced is steady and unchanging, so there is no change in flux linkage and no induced e.m.f. in the secondary coil. An alternating current provides the necessary rate of change of magnetic flux.
变压器无法在稳定的直流电下工作。如果在初级线圈上施加恒定的直流电压，产生的磁场是稳定不变的，因此磁通链没有变化，次级线圈中不会感生出电动势。交流电则能提供所需的磁通变化率。
9. Using Transformers in the National Grid | 变压器在国家电网中的应用
Electricity is transmitted across long distances at very high voltages (e.g. 400 kV) to minimise current and hence heat losses in the cables (P = I² × R). Step-up transformers raise the voltage from power stations for transmission, and step-down transformers reduce it to safer levels (typically 230 V) for domestic and industrial use. This two-stage transformer arrangement is a core concept in IGCSE Physics.
电力在长距离输送时采用极高的电压（如 400 kV），以尽量减小电流，从而降低电缆中的热损耗（P = I² × R）。升压变压器将发电站的电压升高以便输电，降压变压器则将其降至安全水平（通常为 230 V），供家庭和工业使用。这种两阶段变压器布置是 IGCSE 物理的核心概念。
10. Worked Example Using the Transformer Equation | 使用变压器公式的实例计算
A charger contains a step-down transformer with a primary coil of 2000 turns connected to the 230 V mains supply. The secondary coil has 80 turns and is connected to a device requiring a current of 0.5 A. Calculate (a) the secondary voltage, (b) the current drawn from the mains, assuming the transformer is ideal.
某充电器内装有一降压变压器，其初级线圈为 2000 匝，连接到 230 V 市电。次级线圈为 80 匝，并连接到一个需要 0.5 A 电流的设备。假设为理想变压器，试计算：(a) 次级电压，(b) 从市电中吸取的电流。
Solution / 解答:
(a) Using Vₛ / Vₚ = Nₛ / Nₚ → Vₛ = Vₚ × Nₛ / Nₚ = 230 V × (80 / 2000) = 9.2 V
(b) Since this is an ideal transformer, Vₚ Iₚ = Vₛ Iₛ → Iₚ = Vₛ Iₛ / Vₚ = (9.2 V × 0.5 A) / 230 V = 0.02 A
The low current on the primary side once again illustrates why high-voltage transmission is so efficient.
初级侧很小的电流再次说明了为何高压输电如此高效。
11. Common Misconceptions in IGCSE Exams | IGCSE 考试中常见的误解
Many students believe a transformer can change the power output or that it works with d.c. if the voltage is high enough. Remember that power is conserved (in ideal cases), and the device strictly requires a changing magnetic field. Also, the turns ratio applies to voltages, not currents directly — the current ratio is the inverse of the turns ratio. Make sure to use the subscript ‘p’ for primary and ‘s’ for secondary consistently.
许多学生误以为变压器可以改变输出功率，或者只要电压足够高它就可以使用直流电工作。请记住，功率是守恒的（在理想情况下），并且该装置严格要求变化的磁场。此外，匝数比适用于电压而非直接适用于电流——电流比是匝数比的倒数。务必始终使用下标 ‘p’ 代表初级、’s’ 代表次级。
12. Summary of Key Formulas | 关键公式小结
The transformer topic in Edexcel IGCSE Physics can be mastered with just two fundamental equations, which appear in almost every examination session:
掌握 Edexcel IGCSE 物理中变压器这个主题只需两个基本方程式，它们几乎会在每次考试中出现：
- Vₛ / Vₚ = Nₛ / Nₚ – relates voltages and turns
- Vₚ Iₚ = Vₛ Iₛ – expresses conservation of power (assumed ideal)
- Vₛ / Vₚ = Nₛ / Nₚ – 电压与匝数的关系
- Vₚ Iₚ = Vₛ Iₛ – 表示功率守恒（假设理想变压器）
With these, you can solve numerical problems, design step-up or step-down arrangements, and explain energy-efficient power distribution across the country.
有了这些，你就能解决数值问题，设计升压或降压配置，并解释全国范围内节能的电力分配方式。
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