Transformers | 变压器

📚 Transformers | 变压器

Transformers are essential electromagnetic devices that change the voltage of an alternating current (AC) supply. They are a cornerstone of efficient electrical power transmission and feature prominently in the Edexcel IGCSE Science / Physics curriculum (topic 6.12). This article explains how transformers work, their key equations, energy considerations, and real‑world applications.

变压器是改变交流电（AC）电压的重要电磁装置。它们是高效电力输送的基石，并在 Edexcel IGCSE 科学 / 物理课程（主题 6.12）中占有重要地位。本文讲解变压器的工作原理、关键方程、能量考量以及实际应用。

1. What Is a Transformer? | 什么是变压器？

A transformer is a device that increases or decreases the voltage of an AC supply without changing its frequency. It only works with alternating current because a changing magnetic field is needed to induce a voltage in the second coil.

变压器是一种升高或降低交流电电压而不改变其频率的装置。它只能使用交流电，因为需要变化的磁场才能在第二个线圈中感应出电压。

Transformers are used everywhere – from the large grey boxes in electricity substations to the small chargers for mobile phones. They make long‑distance power transmission possible by stepping voltage up for the grid and stepping it down for safe domestic use.

变压器无处不在——从变电站的大型灰色设备到手机的小型充电器。它们通过为电网升压并为家庭安全使用降压，使远距离电力传输成为可能。

2. Operating Principle | 工作原理

A transformer operates on the principle of electromagnetic induction. When an alternating current flows through the primary coil, it produces a constantly changing magnetic field in the iron core. This changing field cuts the turns of the secondary coil and induces an alternating voltage across it.

变压器依据电磁感应原理工作。当交流电通过初级线圈时，会在铁芯中产生持续变化的磁场。这个变化的磁场切割次级线圈的匝数，并在其两端感应出交流电压。

The induced voltage appears only while the magnetic field is changing. If a steady direct current were applied, the magnetic field would be constant, and no voltage would be induced in the secondary coil after the initial switch‑on.

感应电压仅在磁场变化时出现。如果施加稳定的直流电，磁场将是恒定的，除初始接通瞬间外，次级线圈中不会感应出电压。

3. Structure of a Transformer | 变压器的结构

All transformers have three essential parts:

所有变压器都包含三个基本部分：

Primary coil – connected to the input AC supply.
初级线圈 – 连接到输入交流电源。
Secondary coil – delivers the output voltage.
次级线圈 – 提供输出电压。
Laminated soft‑iron core – links the two coils magnetically. The core is made of thin, insulated sheets (laminations) to reduce energy losses from eddy currents.
叠片软铁芯 – 将两个线圈磁耦合起来。铁芯由薄而相互绝缘的硅钢片叠成，以减少涡流造成的能量损失。

The primary and secondary coils are wound around the core but are electrically insulated from each other. Only the magnetic field transfers energy from one coil to the other.

初级线圈和次级线圈绕在铁芯上，但彼此电气绝缘。只有磁场将能量从一个线圈传递到另一个线圈。

4. Step‑Up and Step‑Down Transformers | 升压变压器与降压变压器

A transformer can either increase or decrease the voltage depending on the ratio of turns.

根据匝数比的不同，变压器可以升高或降低电压。

Type	Turns ratio	Effect on voltage	Example
Step‑up	N_s > N_p	V_s > V_p	Power station to grid
Step‑down	N_s < N_p	V_s < V_p	Grid to homes

The function of the transformer is determined entirely by the turn ratio; the device can work in reverse if the input and output connections are swapped.

变压器的功能完全由匝数比决定；如果交换输入和输出连接，该装置可以反向工作。

5. The Transformer Equation: Voltage and Turns | 变压器方程：电压与匝数

For an ideal transformer (100% efficient), the ratio of the primary voltage to the secondary voltage equals the ratio of the number of turns on the primary to the number of turns on the secondary.

对于理想变压器（效率 100%），初级电压与次级电压之比等于初级匝数与次级匝数之比。

V_p / V_s = N_p / N_s

Where V_p = primary voltage, V_s = secondary voltage, N_p = number of turns on the primary coil, N_s = number of turns on the secondary coil.

式中 Vₚ = 初级电压，Vₛ = 次级电压，Nₚ = 初级线圈匝数，Nₛ = 次级线圈匝数。

This equation can be rearranged to find any one quantity when the other three are known. For example, if a transformer has 200 primary turns, 1000 secondary turns and a 12 V input, the output voltage V_s = (N_s/N_p) × V_p = (1000/200)×12 = 60 V.

该公式可以变形，在其他三个量已知时求出任意一个量。例如，一台变压器初级 200 匝、次级 1000 匝，输入 12 V，输出电压 Vₛ = (Nₛ/Nₚ)×Vₚ = (1000/200)×12 = 60 V。

6. Current and Power in an Ideal Transformer | 理想变压器中的电流与功率

In an ideal transformer, the power input to the primary coil equals the power output from the secondary coil. Since power is the product of voltage and current, we can write:

在理想变压器中，初级线圈的输入功率等于次级线圈的输出功率。由于功率是电压与电流的乘积，可写成：

V_p I_p = V_s I_s

Substituting the voltage–turns relationship gives the current–turns relationship:

代入电压–匝数关系，可得电流–匝数关系：

I_p / I_s = N_s / N_p

This shows that if a transformer steps up the voltage, the current is stepped down by the same factor, and vice versa. A step‑up transformer has fewer turns on the primary but carries a larger primary current; the secondary delivers a higher voltage but a smaller current.

这表明，若变压器升高电压，电流则以相同的倍数降低，反之亦然。升压变压器初级匝数较少但承载较大的初级电流；次级输出较高的电压但较小的电流。

7. Efficiency and Energy Losses | 效率与能量损失

Real transformers are never 100% efficient. Energy is lost mainly through:

实际变压器的效率永远不会达到 100%。能量损失主要通过以下途径：

Resistive heating (copper losses): heat produced by current flowing through the resistance of the copper windings. Thicker wires reduce this loss.
电阻发热（铜损）：电流流过铜线圈的电阻产生的热量。使用更粗的导线可以减少这种损失。
Eddy currents: circulating currents induced in the iron core itself, which generate heat. Using a laminated core – thin iron sheets separated by insulation – greatly reduces eddy currents.
涡流：铁芯本身被感应出的环流，会产生热量。采用叠片铁芯——薄铁片间有绝缘层——可大幅减少涡流。
Hysteresis: energy used to continually reverse the magnetic domains in the core material. Soft iron is chosen because it magnetises and demagnetises easily, reducing this loss.
磁滞：用于反复翻转铁芯材料中磁畴所消耗的能量。选用软铁是因为它容易磁化和退磁，可减少这种损失。
Flux leakage: not all the magnetic field lines pass through the secondary coil. The core is designed to keep the flux concentrated, but some leakage always occurs.
磁漏：并非所有磁力线都穿过次级线圈。铁芯设计旨在集中磁通，但总有一些泄漏。

Efficiency can be expressed as: Efficiency = (useful power output / total power input) × 100%. Large grid transformers often achieve efficiencies above 99%.

效率可表示为：效率 =（有用输出功率 / 总输入功率）× 100%。大型电网变压器通常效率可达 99% 以上。

8. Transformers in the National Grid | 国家电网中的变压器

Electricity is generated in power stations at around 25 kV. To transmit it over hundreds of kilometres with minimal energy loss, step‑up transformers raise the voltage to 275 kV or 400 kV. The high voltage means a low current for the same power, and since the heat loss in cables is proportional to I²R, reducing the current dramatically cuts thermal losses.

发电厂生产的电力约为 25 kV。为将其远距离输送且能量损失最小，升压变压器将电压升至 275 kV 或 400 kV。高电压意味着相同功率下电流减小，而电缆中的热损耗正比于 I²R，降低电流可大幅减少热损失。

Near towns and cities, step‑down transformers at substations reduce the voltage first to 33 kV for local distribution, and then further to 230 V for safe domestic use. This sequence of stepping up and stepping down is fundamental to modern electrical infrastructure.

在城镇附近，变电站的降压变压器先将电压降至 33 kV 供本地分配，再进一步降至 230 V 供家庭安全使用。这一升压与降压的序列是现代电力基础设施的基础。

9. Practical Examples and Safety | 实际应用与安全

Common examples of transformers include: mains adapters for laptops (step‑down), ignition coils in cars (step‑up to produce a high‑voltage spark), and isolating transformers used in shaver sockets for safety.

变压器的常见例子包括：笔记本电脑的电源适配器（降压）、汽车点火线圈（升压产生高压火花）以及为安全而使用的剃须刀插座中的隔离变压器。

Transformers also provide electrical isolation between the primary and secondary circuits, which can protect users from electric shock. Always handle transformers with care, especially those operating at high voltages; water and damaged insulation can create serious hazards.

变压器还提供初级与次级回路之间的电气隔离，可保护用户免受电击。操作变压器时需始终小心，特别是高压运行的那些；水浸和绝缘损坏会带来严重的危险。

10. Solving Transformer Problems | 变压器问题求解

Let us work through a typical IGCSE question.

让我们通过一个典型的 IGCSE 题目来演练。

Example: A step‑down transformer has 5000 turns on its primary coil and 200 turns on its secondary coil. The primary voltage is 230 V. Calculate the secondary voltage and the secondary current if the primary current is 0.40 A and the transformer is 100% efficient.

例题：一台降压变压器初级线圈 5000 匝，次级线圈 200 匝。初级电压为 230 V。若初级电流为 0.40 A，且变压器效率为 100%，求次级电压与次级电流。

Use V_p/V_s = N_p/N_s: 230 / V_s = 5000 / 200 → V_s = 230 × (200/5000) = 9.2 V.

使用 Vₚ/Vₛ = Nₚ/Nₛ：230 / Vₛ = 5000 / 200 → Vₛ = 230 × (200/5000) = 9.2 V。

For an ideal transformer, power in = power out: V_pI_p = V_sI_s. 230 × 0.40 = 9.2 × I_s → I_s = (230 × 0.40) / 9.2 = 10 A.

对于理想变压器，输入功率 = 输出功率：VₚIₚ = VₛIₛ。230 × 0.40 = 9.2 × Iₛ → Iₛ = (230 × 0.40) / 9.2 = 10 A。

Notice that the voltage has been stepped down by a large factor, while the current has been stepped up significantly. This aligns with I_p/I_s = N_s/N_p.

注意电压大幅降低，而电流大幅升高。这符合 Iₚ/Iₛ = Nₛ/Nₚ。

11. Key Points for Revision | 复习要点

A transformer changes the voltage and current of an AC supply; it does not work with DC.
变压器改变交流电的电压和电流；不能用于直流电。
V_p / V_s = N_p / N_s (ideal transformer).
Vₚ / Vₛ = Nₚ / Nₛ（理想变压器）。
For an ideal transformer, power is conserved: V_pI_p = V_sI_s, leading to I_p / I_s = N_s / N_p.
对于理想变压器，功率守恒：VₚIₚ = VₛIₛ，从而 Iₚ / Iₛ = Nₛ / Nₚ。
Energy losses arise from heating, eddy currents, hysteresis and flux leakage; laminated soft‑iron cores greatly reduce eddy currents.
能量损失源自发热、涡流、磁滞和磁漏；叠片软铁芯可大幅降低涡流。
Step‑up transformers increase voltage and decrease current for efficient grid transmission; step‑down transformers do the opposite for safe use.
升压变压器升高电压、降低电流以实现电网高效传输；降压变压器则相反，以保障安全使用。

12. Summary | 总结

Transformers are elegant embodiments of electromagnetic induction that underpin modern electricity distribution. Mastering the relationships between voltage, turns and current, and understanding the practical design features that minimise energy losses, is essential for success in the Edexcel IGCSE Science examination.

变压器是电磁感应的精妙体现，支撑着现代电力分配。掌握电压、匝数和电流之间的关系，并理解减少能量损失的实际设计特点，是在 Edexcel IGCSE 科学考试中取得成功的关键。

Published by TutorHao | Physics Revision Series | aleveler.com

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Transformers | 变压器