📚 Transformers: Principles, Equations and Power Transmission | 变压器：原理、方程与电力输送
A transformer is a device that changes an alternating voltage from one value to another by electromagnetic induction. It can step voltage up or down while transferring electrical energy between two circuits with no direct electrical connection. For CIE A-Level Physics, you need to understand the construction, the ideal transformer equations, energy losses, efficiency, and the role of transformers in the national grid.
变压器是一种利用电磁感应将一种交流电压变为另一种交流电压的装置。它可以在两个没有直接电气连接的电路之间传递电能，同时升高或降低电压。剑桥 CIE A-Level 物理要求你掌握变压器的结构、理想变压器方程、能量损耗、效率以及变压器在国家电网中的作用。
1. What is a Transformer? | 什么是变压器？
A transformer is a static electrical machine that operates only on alternating current. It changes the voltage and current levels but, in an ideal case, transfers the same electrical power from the primary circuit to the secondary circuit.
变压器是一种只能使用交流电工作的静态电气设备。它改变电压和电流的大小，但在理想情况下，从初级电路传递到次级电路的电功率保持不变。
Transformers do not generate energy. They convert electrical energy into a changing magnetic field and then back into electrical energy at a different voltage. This process relies on the principle of electromagnetic induction discovered by Michael Faraday.
变压器不产生能量。它先将电能转化为变化的磁场，再以不同的电压转化为电能。这一过程依赖于法拉第发现的电磁感应原理。
2. Structure and Core Design | 结构与铁芯设计
A basic transformer consists of two coils of insulated wire wound around a common soft iron core. The coil connected to the input voltage is called the primary coil, and the coil connected to the output circuit is called the secondary coil.
基本变压器由两个绕在同一软铁芯上的绝缘线圈组成。连接输入电压的线圈称为初级线圈，连接输出电路的线圈称为次级线圈。
The core is usually made of thin laminated sheets of soft iron. Lamination means the core is built from layers separated by insulating material. This design greatly reduces unwanted eddy currents in the core.
铁芯通常由薄的软铁叠片制成。叠片意味着铁芯由绝缘材料隔开的薄层构成。这种设计能大大减少铁芯中不必要的涡流。
Soft iron is used because it is easily magnetised and demagnetised, giving high magnetic permeability and low hysteresis loss. The core provides a low-reluctance path for the magnetic flux, so most of the flux produced by the primary coil links with the secondary coil.
使用软铁是因为它容易被磁化和去磁，磁导率高，磁滞损耗低。铁芯为磁通提供了低磁阻路径，因此初级线圈产生的大部分磁通都能与次级线圈交链。
3. Mutual Induction and Faraday’s Law | 互感与法拉第定律
When an alternating voltage is applied to the primary coil, an alternating current flows. This current produces an alternating magnetic flux in the core. Because the secondary coil is wound on the same core, this changing flux passes through the secondary coil.
当初级线圈施加交流电压时，会有交流电流流过。该电流在铁芯中产生交变磁通。由于次级线圈绕在同一铁芯上，这个变化的磁通会穿过次级线圈。
According to Faraday’s law of electromagnetic induction, a changing magnetic flux through a coil induces an e.m.f. in that coil. The induced e.m.f. is proportional to the number of turns and the rate of change of magnetic flux:
根据法拉第电磁感应定律，穿过线圈的变化磁通会在线圈中感应出电动势。感应电动势与线圈匝数和磁通变化率成正比：
ε = −N ΔΦ / Δt
Here ε is the induced e.m.f., N is the number of turns, and ΔΦ/Δt is the rate of change of magnetic flux. The negative sign shows that the induced e.m.f. opposes the change causing it, as stated by Lenz’s law.
其中 ε 是感应电动势，N 是线圈匝数，ΔΦ/Δt 是磁通变化率。负号表示感应电动势总是阻碍引起它的变化，这就是楞次定律。
Because the same alternating flux links both coils, the e.m.f. induced per turn is the same in the primary and secondary coils. This is the key idea behind the transformer equation.
由于相同的交变磁通同时穿过两个线圈，初级线圈和次级线圈每匝感应出的电动势相同。这是变压器方程背后的核心思想。
4. The Ideal Transformer Equations | 理想变压器方程
An ideal transformer is assumed to have no energy losses. The primary and secondary coils have zero resistance, all magnetic flux links both coils, and there are no eddy current or hysteresis losses in the core.
理想变压器假设没有能量损耗。初级和次级线圈电阻为零，所有磁通都同时穿过两个线圈，铁芯中没有涡流损耗或磁滞损耗。
For an ideal transformer, the ratio of the primary voltage Vₚ to the secondary voltage Vₛ is equal to the ratio of the number of turns in the primary coil Nₚ to the number of turns in the secondary coil Nₛ:
对于理想变压器，初级电压 Vₚ 与次级电压 Vₛ 之比等于初级线圈匝数 Nₚ 与次级线圈匝数 Nₛ 之比：
Vₚ / Vₛ = Nₚ / Nₛ
This is often called the turns ratio equation. If Nₛ is greater than Nₚ, the secondary voltage is greater than the primary voltage, giving a step-up transformer. If Nₛ is less than Nₚ, the transformer steps the voltage down.
这通常称为匝数比方程。如果 Nₛ 大于 Nₚ，则次级电压高于初级电压，变压器为升压变压器。如果 Nₛ 小于 Nₚ，则变压器为降压变压器。
Because an ideal transformer has no power loss, the input power equals the output power. Therefore the current ratio is the inverse of the turns ratio:
由于理想变压器没有功率损耗，输入功率等于输出功率。因此电流比是匝数比的倒数：
Iₚ / Iₛ = Nₛ / Nₚ
This means that if a transformer steps up voltage, it steps down current by the same factor, and vice versa. The relationship can also be written as VₚIₚ = VₛIₛ for an ideal transformer.
这意味着如果变压器升高电压，则电流按相同倍数减小，反之亦然。对于理想变压器，这一关系也可以写成 VₚIₚ = VₛIₛ。
5. Step-Up and Step-Down Transformers | 升压与降压变压器
A step-up transformer has more turns on the secondary coil than on the primary coil. It increases voltage and decreases current. Step-up transformers are used at power stations to raise the voltage before electricity is sent along transmission lines.
升压变压器的次级线圈匝数多于初级线圈。它升高电压并减小电流。升压变压器用于发电站，在电力送入输电线路之前提高电压。
A step-down transformer has fewer turns on the secondary coil than on the primary coil. It decreases voltage and increases current. Step-down transformers are used near homes and factories to reduce the high transmission voltage to a safe level for consumers.
降压变压器的次级线圈匝数少于初级线圈。它降低电压并增大电流。降压变压器用于家庭和工厂附近，将高输电电压降低到对用户安全的水平。
For example, a transformer with a primary coil of 200 turns connected to a 240 V supply and a secondary coil of 2000 turns gives a secondary voltage of 2400 V. This is a step-up transformer with a turns ratio of 1:10.
例如，一个初级线圈为 200 匝、接在 240 V 电源上、次级线圈为 2000 匝的变压器，其次级电压为 2400 V。这是一个匝数比为 1:10 的升压变压器。
6. Power and Current Relationships | 功率与电流关系
For an ideal transformer, the electrical power delivered to the primary coil equals the electrical power delivered by the secondary coil. This is a direct consequence of energy conservation.
对于理想变压器，输送给初级线圈的电功率等于次级线圈输出的电功率。这是能量守恒的直接结果。
Since electrical power is given by P = VI, the relationship between primary and secondary quantities is:
由于电功率由 P = VI 给出，初级量和次级量之间的关系为：
Pₚ = Pₛ ⇒ VₚIₚ = VₛIₛ
This equation is extremely useful for solving problems. If a transformer steps up voltage by a factor of 10, the current in the secondary circuit is one tenth of the primary current, assuming ideal operation.
这个方程在解题时非常有用。如果变压器将电压升高 10 倍，则在理想情况下，次级电路中的电流是初级电流的十分之一。
| Quantity | Symbol | SI unit |
| Primary voltage | Vₚ | volt (V) |
| Secondary voltage | Vₛ | volt (V) |
| Primary current | Iₚ | ampere (A) |
| Secondary current | Iₛ | ampere (A) |
| Turns on primary | Nₚ | dimensionless |
| Turns on secondary | Nₛ | dimensionless |
In real transformers, the output power is always slightly less than the input power because of energy losses. However, the ideal equations still give a very good first approximation for most exam problems.
在实际变压器中，由于存在能量损耗，输出功率总是略小于输入功率。然而，在大多数考试题目中，理想方程仍然是一个非常好的初步近似。
7. Energy Losses in Real Transformers | 实际变压器中的能量损耗
Real transformers are not 100% efficient. The main sources of energy loss are resistance heating in the coils, eddy currents in the core, hysteresis loss in the core, and sometimes flux leakage.
实际变压器的效率并非 100%。主要的能量损耗来源有线圈电阻发热、铁芯涡流、铁芯磁滞损耗，有时还有漏磁。
Copper loss occurs because the primary and secondary coils have some electrical resistance. When current flows through them, heat is produced according to P = I²R. Using thicker wire reduces this loss but increases cost and mass.
铜损的产生是因为初级和次级线圈有一定的电阻。当电流流过时，根据 P = I²R 产生热量。使用更粗的导线可以减少这种损耗，但会增加成本和重量。
Eddy current loss occurs because the changing magnetic flux also induces circulating currents in the iron core. These currents heat the core. Laminating the core into thin insulated sheets reduces eddy currents by increasing the resistance of the paths available to them.
涡流损耗的产生是因为变化的磁通也会在铁芯中感应出环形电流。这些电流使铁芯发热。将铁芯叠成薄的绝缘片可以增大涡流路径的电阻，从而减小涡流。
Hysteresis loss is caused by the continuous reversal of magnetisation of the core material as the alternating current changes direction. Soft iron has a narrow hysteresis loop, so it dissipates less energy per cycle. Using materials such as grain-oriented silicon steel or ferrites can further reduce hysteresis loss.
磁滞损耗是由于交流电改变方向时，铁芯材料不断反复磁化引起的。软铁的磁滞回线较窄，因此每个周期耗散的能量较少。使用晶粒取向硅钢或铁氧体等材料可以进一步减少磁滞损耗。
Flux leakage occurs when some magnetic flux produced by the primary coil does not pass through the secondary coil. Good core design and careful winding help minimise flux leakage, ensuring that as much flux as possible links both coils.
漏磁是指初级线圈产生的一部分磁通没有穿过次级线圈。良好的铁芯设计和仔细的绕线方式有助于减少漏磁，确保尽可能多的磁通同时穿过两个线圈。
8. Transformer Efficiency | 变压器效率
The efficiency of a transformer is defined as the ratio of useful output power to input power. It is usually expressed as a percentage:
变压器的效率定义为有用输出功率与输入功率之比。它通常用百分比表示：
η = (Pₛ / Pₚ) × 100%
For small transformers, efficiency may be around 80% to 95%. Large power transformers used in the national grid can have efficiencies above 99% because they are designed to minimise all major loss mechanisms.
小型变压器的效率可能在 80% 到 95% 左右。国家电网中使用的大型电力变压器效率可以超过 99%，因为它们的设计尽量减少所有主要损耗机制。
If a question gives input power and output power, efficiency can be calculated directly. If it gives voltage and current values, then P = VI can be used for the primary and secondary circuits. For an ideal transformer, efficiency is 100%.
如果题目给出输入功率和输出功率，可以直接计算效率。如果给出电压和电流值，则可以用 P = VI 分别计算初级和次级电路的功率。对于理想变压器，效率为 100%。
9. Transformers in National Grid Power Transmission | 变压器在电网输电中的应用
The national grid uses transformers to reduce energy loss during long-distance power transmission. At a power station, a step-up transformer raises the generated voltage to a very high value, often several hundred kilovolts.
国家电网利用变压器来减少远距离输电过程中的能量损耗。在发电站，升压变压器将发电电压升高到很高的值，通常为数百千伏。
The reason for using high voltage is clear from the power loss equation in transmission cables. The power lost as heat in a cable of resistance R carrying current I is given by P_loss = I²R. For a fixed transmitted power P = VI, increasing the voltage V reduces the current I, which greatly reduces the I²R loss.
使用高电压的原因可以从输电电缆的功率损耗方程中清楚地看出。电阻为 R 的电缆承载电流 I 时，以热量形式损失的功率为 P_loss = I²R。对于固定的传输功率 P = VI，提高电压 V 会减小电流 I，从而大大降低 I²R 损耗。
Near towns and cities, step-down transformers reduce the voltage in stages. A high-voltage transmission line may be reduced to around 11 kV for local distribution, and then a final step-down transformer reduces it to about 230 V for household use.
在城镇附近，降压变压器分阶段降低电压。高压输电线路可能先降到约 11 kV 用于本地配电，然后最终降压变压器将其降到约 230 V 供家庭使用。
High voltage transmission also allows thinner and lighter cables to be used for the same power transfer, reducing the cost of pylons and conductors. However, very high voltage requires greater insulation and safety clearances, so transmission voltage is chosen as a compromise.
高压输电还可以在传输相同功率的情况下使用更细更轻的电缆，从而降低铁塔和导线的成本。但是，非常高的电压需要更高的绝缘要求和安全间距，因此输电电压的选择是一个折中方案。
10. Exam-Style Worked Example | 考试型例题
A transformer has 500 turns on the primary coil and 50 turns on the secondary coil. The primary coil is connected to a 240 V alternating supply, and the secondary current is 2.0 A. Assuming the transformer is ideal, calculate the secondary voltage and the primary current.
一个变压器的初级线圈有 500 匝，次级线圈有 50 匝。初级线圈接在 240 V 交流电源上，次级电流为 2.0 A。假设变压器是理想的，计算次级电压和初级电流。
Step 1: Use the turns ratio equation to find the secondary voltage. Since Nₚ = 500 and Nₛ = 50, the turns ratio Nₛ/Nₚ = 50/500 = 0.1. Therefore Vₛ = Vₚ × Nₛ/Nₚ = 240 × 0.1 = 24 V.
第 1 步：使用匝数比方程求次级电压。由于 Nₚ = 500，Nₛ = 50，匝数比 Nₛ/Nₚ = 50/500 = 0.1。因此 Vₛ = Vₚ × Nₛ/Nₚ = 240 × 0.1 = 24 V。
Step 2: Use the current ratio equation. For an ideal transformer, Iₚ/Iₛ = Nₛ/Nₚ = 0.1. Therefore Iₚ = Iₛ × 0.1 = 2.0 × 0.1 = 0.20 A.
第 2 步：使用电流比方程。对于理想变压器，Iₚ/Iₛ = Nₛ/Nₚ = 0.1。因此 Iₚ = Iₛ × 0.1 = 2.0 × 0.1 = 0.20 A。
Check: input power = VₚIₚ = 240 × 0.20 = 48 W, and output power = VₛIₛ = 24 × 2.0 = 48 W. The powers are equal, which is consistent with an ideal transformer. This transformer is a step-down transformer because the secondary voltage is lower than the primary voltage.
检验：输入功率 = VₚIₚ = 240 × 0.20 = 48 W，输出功率 = VₛIₛ = 24 × 2.0 = 48 W。两者相等，符合理想变压器的条件。该变压器是降压变压器，因为次级电压低于初级电压。
11. Common Misconceptions | 常见误区
One common misconception is that a transformer can work with direct current. A steady direct current produces a constant magnetic field, so there is no changing flux to induce an e.m.f. in the secondary coil. A transformer requires an alternating or changing current.
一个常见的误区是认为变压器可以使用直流电。稳定的直流电产生恒定的磁场，因此没有变化的磁通来在次级线圈中感应电动势。变压器需要交流电或变化的电流。
Another misconception is that a step-up transformer creates energy. It does not. When voltage is increased, the maximum available current decreases, so power cannot be created from nothing. Energy is conserved, and in a real transformer some energy is lost as heat.
另一个误区是升压变压器能产生能量。它并不能。当电压升高时，可用的最大电流会减小，因此功率不可能凭空产生。能量是守恒的，在实际变压器中，部分能量会以热量形式损失。
Students sometimes think that the number of turns determines power. In fact, the turns ratio determines only the voltage and current ratio. The power transferred is determined by the load connected to the secondary circuit and is limited by the input power.
学生有时会认为匝数决定功率。实际上，匝数比只决定电压比和电流比。传输的功率由连接在次级电路的负载决定，并受输入功率的限制。
Finally, many learners confuse the current ratio with the voltage ratio. Remember that voltage is directly proportional to turns, while current is inversely proportional to turns for an ideal transformer.
最后，许多学习者会把电流比和电压比混淆。请记住，对于理想变压器，电压与匝数成正比，而电流与匝数成反比。
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