📚 A-Level Physics: How Transformers Work and the Turns Ratio | A-Level 物理：变压器的工作原理与变压比
A transformer is a device that transfers electrical energy between two circuits through electromagnetic induction. It is a cornerstone of modern power distribution and a frequent examination topic in CIE A-Level Physics. To master this topic, you need to understand not only the structure and working principle but also the derivation of the turns ratio equation and the sources of energy loss.
变压器是一种通过电磁感应在两个电路之间传递电能的装置。它是现代电力输送的基石，也是 CIE A-Level 物理考试中的高频考点。要掌握这一主题，你不仅需要理解其结构与工作原理，还需要熟练掌握变压比公式的推导以及能量损失的各种来源。
1. Basic Structure of a Transformer | 变压器的基本结构
A simple transformer consists of two coils of insulated wire wound around a common soft iron core. The coil connected to the alternating current (a.c.) supply is called the primary coil, and the coil connected to the load is called the secondary coil. The soft iron core is laminated, meaning it is made of thin sheets insulated from each other, to reduce energy losses due to eddy currents.
一个简单的变压器由两个绕在公共软铁芯上的绝缘线圈组成。与交流电源相连的线圈称为初级线圈，与负载相连的线圈称为次级线圈。软铁芯采用叠片结构，即由相互绝缘的薄片叠加而成，目的是减少涡流造成的能量损失。
- Primary coil: receives energy from the a.c. source.
- 初级线圈：从交流电源接收能量。
- Secondary coil: delivers energy to the external circuit.
- 次级线圈：向外部电路输送能量。
- Soft iron core: provides a low-reluctance path for the magnetic flux.
- 软铁芯：为磁通量提供低磁阻路径。
2. The Principle of Operation | 工作原理
When an alternating current flows through the primary coil, it produces a continuously changing magnetic flux in the soft iron core. This changing flux links with the secondary coil and, according to Faraday’s law of electromagnetic induction, induces an e.m.f. in the secondary coil. If the secondary circuit is closed, an induced current flows through the load.
当初级线圈中通入交变电流时，会在软铁芯中产生持续变化的磁通量。这个变化的磁通量穿过次级线圈，根据法拉第电磁感应定律，在次级线圈中感应出电动势。如果次级电路闭合，感应电流就会流过负载。
The key point is that a transformer only works with alternating current, not direct current. With d.c., the current is constant, so there is no changing magnetic flux and hence no induced e.m.f. in the secondary coil.
关键在于，变压器只能使用交流电工作，不能使用直流电。对于直流电，电流恒定不变，因此没有变化的磁通量，次级线圈中也就不会产生感应电动势。
3. Faraday’s Law and Mutual Induction | 法拉第定律与互感
Faraday’s law states that the induced e.m.f. is proportional to the rate of change of magnetic flux linkage. For the secondary coil with Nₛ turns:
法拉第定律指出，感应电动势与磁链的变化率成正比。对于匝数为 Nₛ 的次级线圈：
Eₛ = −Nₛ × (ΔΦ/Δt)
where Eₛ is the induced e.m.f. in the secondary coil, Nₛ is the number of turns on the secondary, and ΔΦ/Δt is the rate of change of magnetic flux through one turn. The negative sign indicates Lenz’s law, meaning the induced e.m.f. opposes the change producing it.
其中 Eₛ 是次级线圈中的感应电动势，Nₛ 是次级线圈的匝数，ΔΦ/Δt 是通过每一匝线圈的磁通量变化率。负号表示楞次定律，即感应电动势总是阻碍引起它的磁通量变化。
Because the same magnetic flux links both coils (assuming no leakage), the e.m.f. induced per turn is the same for both primary and secondary. This leads directly to the transformer equation.
由于相同的磁通量同时穿过两个线圈（假设无漏磁），初级和次级线圈每匝感应的电动势相同。这直接引出变压器公式。
4. The Ideal Transformer Equation | 理想变压器方程
For an ideal transformer, there is no energy loss, and the primary and secondary e.m.f.s are related to their turn numbers by:
对于理想变压器，没有能量损失，初级和次级电动势与匝数的关系为：
Eₚ / Eₛ = Nₚ / Nₛ
where Eₚ and Eₛ are the e.m.f.s of the primary and secondary coils, and Nₚ and Nₛ are their respective numbers of turns. This is the turns ratio equation, also written as:
其中 Eₚ 和 Eₛ 分别是初级和次级线圈的电动势，Nₚ 和 Nₛ 分别是它们的匝数。这就是变压比公式，也可写作：
Eₛ / Eₚ = Nₛ / Nₚ
In practice, the terminal voltage Vₛ across the secondary is approximately equal to Eₛ, and the applied primary voltage Vₚ is approximately equal to Eₚ, so the equation is often written as Vₚ / Vₛ = Nₚ / Nₛ.
在实际应用中，次级线圈两端的电压 Vₛ 近似等于 Eₛ，初级线圈两端的外加电压 Vₚ 近似等于 Eₚ，因此该公式常写作 Vₚ / Vₛ = Nₚ / Nₛ。
5. Step-Up and Step-Down Transformers | 升压变压器与降压变压器
If Nₛ > Nₚ, the transformer is a step-up transformer: the secondary voltage is higher than the primary voltage. If Nₛ < Nₚ, it is a step-down transformer: the secondary voltage is lower than the primary voltage.
如果 Nₛ > Nₚ，则为升压变压器：次级电压高于初级电压。如果 Nₛ < Nₚ，则为降压变压器：次级电压低于初级电压。
| Type | 类型 | Turns | 匝数 | Voltage | 电压 | Typical Use | 典型用途 |
| Step-up | 升压 | Nₛ > Nₚ | Vₛ > Vₚ | Power transmission | 电力输送 |
| Step-down | 降压 | Nₛ < Nₚ | Vₛ < Vₚ | Domestic supply | 家庭供电 |
6. Current Relationship in an Ideal Transformer | 理想变压器中的电流关系
An ideal transformer has no power loss, so the power input to the primary equals the power output from the secondary:
理想变压器没有功率损失，因此初级输入功率等于次级输出功率：
Vₚ × Iₚ = Vₛ × Iₛ
Rearranging gives:
整理可得：
Iₛ / Iₚ = Vₚ / Vₛ = Nₚ / Nₛ
This shows that a step-up transformer increases voltage but decreases current, and vice versa. This is why power companies use step-up transformers for transmission: the higher voltage means a lower current, which significantly reduces the power lost as heat in the transmission lines (P_loss = I²R).
这表明升压变压器提高电压的同时会降低电流，反之亦然。这就是电力公司使用升压变压器输电的原因：更高的电压意味着更低的电流，从而大幅减少输电线路上因发热而损耗的功率（P_loss = I²R）。
7. Energy Losses in Real Transformers | 实际变压器中的能量损失
A real transformer is not perfectly efficient. Energy is lost through several mechanisms, and knowing these enables us to explain design features.
实际变压器并非完全高效。能量通过多种机制损失，了解这些机制有助于解释变压器的设计特征。
| Loss Mechanism | 损失机制 | Cause | 原因 | Reduction Method | 减小方法 |
| Copper loss | 铜损 | Resistance of the coils | 线圈电阻 | Use thick, low-resistivity wire | 使用粗而电阻率低的导线 |
| Eddy current loss | 涡流损失 | Induced currents in the iron core | 铁芯中感应出的电流 | Laminated core | 使用叠片铁芯 |
| Hysteresis loss | 磁滞损失 | Repeated magnetisation of the core | 铁芯反复磁化 | Use soft magnetic material | 使用软磁性材料 |
| Flux leakage | 漏磁 | Some flux does not link both coils | 部分磁通未穿过两个线圈 | Wind coils on a closed iron core | 将线圈绕在闭合铁芯上 |
8. Efficiency of a Transformer | 变压器的效率
The efficiency of a transformer is defined as:
变压器的效率定义为：
Efficiency = (Output Power / Input Power) × 100%
For a real transformer, the output power is always slightly less than the input power because of the losses described above. In CIE examinations, you may be asked to calculate efficiency using:
对于实际变压器，由于上述损失，输出功率总是略小于输入功率。在 CIE 考试中，你可能会被要求用以下公式计算效率：
Efficiency = (Vₛ × Iₛ) / (Vₚ × Iₚ) × 100%
Modern power transformers achieve efficiencies of over 99%, but no transformer is perfectly efficient.
现代电力变压器的效率可超过 99%，但没有任何变压器是完全高效的。
9. Worked Example | 计算示例
A step-down transformer has 1200 turns on its primary coil and 60 turns on its secondary coil. The primary is connected to a 240 V a.c. supply. Calculate: (a) the secondary voltage, (b) the secondary current when the primary current is 0.5 A (assuming ideal conditions).
一个降压变压器的初级线圈有 1200 匝，次级线圈有 60 匝。初级线圈接在 240 V 交流电源上。计算：(a) 次级电压；(b) 当初级电流为 0.5 A 时的次级电流（假设为理想条件）。
(a) Using the turns ratio equation:
(a) 使用变压比公式：
Vₛ = Vₚ × (Nₛ / Nₚ) = 240 × (60 / 1200) = 240 × 0.05 = 12 V
(b) Using the power conservation equation:
(b) 使用功率守恒方程：
Iₛ = Iₚ × (Vₚ / Vₛ) = 0.5 × (240 / 12) = 0.5 × 20 = 10 A
Notice that the voltage has been reduced by a factor of 20, while the current has been increased by the same factor, keeping the power constant.
注意，电压降低了 20 倍，而电流增加了同样的倍数，从而保持功率恒定。
10. Power Transmission and the Role of Transformers | 电力输送与变压器的角色
In the national grid, electricity is generated at power stations at a relatively low voltage (about 25 kV). Step-up transformers raise this to 400 kV or even higher for transmission over long distances. This reduces the current and therefore minimises the I²R power loss in the transmission cables. At the consumer end, step-down transformers reduce the voltage to 230 V for domestic use.
在国家电网中，发电站以相对较低的电压（约 25 kV）发电。升压变压器将其升高到 400 kV 甚至更高，以进行长距离输电。这降低了电流，从而最大限度地减少了输电线缆中的 I²R 功率损耗。在用户端，降压变压器将电压降至 230 V 供家庭使用。
A common exam question asks why high voltage is used for transmission. The answer is always: for a given power, higher voltage means lower current, and since power loss in cables is proportional to I², reducing the current drastically cuts the heating loss.
一个常见的考试问题是：为什么输电要使用高电压？答案总是：对于给定的功率，电压越高意味着电流越小，而电缆中的功率损耗与 I² 成正比，因此降低电流可以大幅减少发热损耗。
11. Common Misconceptions and Exam Tips | 常见误区与考试提示
Students often make the mistake of applying VₚIₚ = VₛIₛ to non-ideal transformers or to cases where the secondary circuit is open. Remember that this equation assumes 100% efficiency and must be used with care.
学生常常错误地将 VₚIₚ = VₛIₛ 用于非理想变压器或次级电路断开的情况。请记住，该方程假设效率为 100%，使用时必须谨慎。
- Always check whether the transformer is step-up or step-down before substituting numbers.
- 在代入数值之前，务必先判断变压器是升压还是降压。
- When asked about eddy currents, mention lamination explicitly.
- 当被问及涡流时，要明确提到叠片结构。
- Do not forget that transformers require a.c., not d.c.
- 不要忘记变压器需要交流电，而不是直流电。
- If the secondary circuit is open, Iₛ = 0, but Vₛ is still given by the turns ratio.
- 如果次级电路断开，Iₛ = 0，但 Vₛ 仍由变压比公式给出。
12. Summary | 总结
A transformer works on the principle of mutual induction, transferring electrical energy between two coils via a changing magnetic flux in a soft iron core. The turns ratio equation Vₚ/Vₛ = Nₚ/Nₛ is derived from Faraday’s law and is the most important equation to remember. In an ideal transformer, power is conserved, so VₚIₚ = VₛIₛ. Real transformers suffer from copper, eddy current, hysteresis, and flux leakage losses, all of which can be reduced through careful design.
变压器基于互感原理工作，通过软铁芯中变化的磁通量在两个线圈之间传递电能。变压比公式 Vₚ/Vₛ = Nₚ/Nₛ 由法拉第定律推导得出，是必须牢记的最重要公式。在理想变压器中，功率守恒，因此 VₚIₚ = VₛIₛ。实际变压器存在铜损、涡流损失、磁滞损失和漏磁损失，均可通过精心设计来减小。
Mastering these concepts and practising calculation questions will ensure you are well-prepared for any transformer question in your CIE A-Level Physics examination.
掌握这些概念并练习计算题，将确保你在 CIE A-Level 物理考试中从容应对任何变压器相关题目。
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