📚 AQA A-Level Physics: Alternating Current – Key Points | AQA A-Level 物理：交流电 考点精讲
Alternating current (AC) is a crucial part of the AQA A-Level Physics specification, covering everything from sinusoidal waveforms and root-mean-square values to transformers and rectification. This article breaks down the must-know exam points with bilingual explanations, circuit descriptions, and worked formula interpretations to help you feel confident with any AC question.
交流电是 AQA A-Level 物理的核心章节之一，涉及正弦波形、有效值、变压器和整流等关键概念。本文以中英双语解析考点，包括波形分析、公式推导和电路原理，帮助你全面掌握交流电相关的解题方法。
1. AC vs DC | 交流电与直流电
Direct current (DC) flows steadily in a single direction, so the voltage and current remain constant over time. In AQA circuits, DC is typically supplied by a battery or a stabilized power supply.
直流电沿单一方向稳定流动，电压和电流不随时间变化。在 AQA 考题中，直流电通常来自电池或稳压电源。
Alternating current (AC) periodically changes direction, producing a voltage that varies with time, most commonly in a sinusoidal pattern. Mains electricity in the UK is AC with a frequency of 50 Hz and a nominal rms voltage of 230 V.
交流电的方向周期性改变，产生的电压随时间变化，通常呈正弦波形。英国市电就是 50 Hz、有效值 230 V 的交流电。
2. Sinusoidal Waveform and Frequency | 正弦波形与频率
An AC voltage can be described by V = V₀ sin(ωt) or V = V₀ sin(2πft), where V₀ is the peak voltage, ω is the angular frequency (in rad s⁻¹), f is the frequency (in Hz), and t is time. The period T of the wave is related to frequency by T = 1 / f.
交流电压可以表示为 V = V₀ sin(ωt) 或 V = V₀ sin(2πft)，其中 V₀ 是峰值电压，ω 是角频率 (rad s⁻¹)，f 是频率 (Hz)，t 为时间。周期 T 与频率的关系是 T = 1 / f。
The angular frequency ω is simply ω = 2πf. For UK mains f = 50 Hz, ω ≈ 314 rad s⁻¹ and T = 0.02 s. Exam questions often ask you to read f from an oscilloscope trace by measuring T and using f = 1/T.
角频率 ω 即 ω = 2πf。以英国 50 Hz 市电为例，ω ≈ 314 rad s⁻¹，周期 T = 0.02 s。试题常要求你从示波器屏上测量周期 T，再通过 f = 1/T 计算频率。
3. Peak, Peak‑to‑peak and RMS Values | 峰值、峰峰值与有效值
The peak voltage V₀ is the maximum value of the waveform. The peak‑to‑peak voltage is the difference between the maximum positive and maximum negative values, equal to 2V₀. These are clearly visible on an oscilloscope display.
峰值电压 V₀ 是波形的最大值。峰峰值电压 是正最大值与负最大值之间的差值，等于 2V₀。这两者都可以在示波器屏幕上直接读出。
The root‑mean‑square (rms) value is the equivalent DC value that would deliver the same average power to a resistor. For a sinusoidal waveform:
有效值 (rms) 是指与交流电在电阻上产生相同平均功率的等效直流值。对于正弦波：
Vrms = V₀ / √2 and Irms = I₀ / √2
The rms is always smaller than the peak value, and UK mains rated at 230 Vrms has a peak voltage of about 325 V.
有效值始终小于峰值，英国额定 230 Vrms 的市电对应的峰值电压约为 325 V。
4. Calculating RMS and Power | 有效值计算与功率
For a pure sinusoidal AC, rms values can be found directly from the peak. The average power dissipated in a resistor R is given by P = IrmsVrms = I²rmsR = V²rms / R. These equations mirror the familiar DC power formulas but use rms quantities.
对于纯粹的正弦交流电，有效值可以直接从峰值求得。电阻 R 上的平均功率为 P = IrmsVrms = I²rmsR = V²rms / R。这些公式与直流功率公式形式相同，只是使用有效值。
If a current or voltage is not sinusoidal, the simple V₀/√2 conversion does not apply, and rms must be determined by taking the square root of the mean of the squared values. AQA questions focus on sinusoidal AC, but you may be asked to compare rms and peak values for a given waveform.
如果电流或电压不是正弦波，就不能直接用 V₀/√2 换算，而必须通过平方取平均再开方求有效值。AQA 考题以正弦交流电为主，但也可能要求你比较给定波形的有效值与峰值。
5. Using an Oscilloscope | 示波器的使用
An oscilloscope plots voltage on the vertical axis against time on the horizontal axis. The timebase setting (e.g. 5 ms/div) allows you to determine the period T by counting horizontal divisions. The y‑gain or voltage sensitivity (e.g. 2 V/div) lets you find the peak voltage.
示波器在垂直轴显示电压，水平轴显示时间。时基设置（如 5 ms/div）可通过计算水平格数确定周期 T。垂直增益或电压灵敏度（如 2 V/div）可用于读取峰值电压。
To obtain the frequency, measure T and use f = 1/T. To find the rms voltage, first determine V₀ from the trace and then divide by √2. Pay close attention to whether the trace shows a full cycle or only half a cycle; always measure one complete wave.
通过测量周期 T 用 f = 1/T 可得频率。要得到有效值，先从波形上读取 V₀，再除以 √2。注意观察屏上波形是否显示完整一个周期，一定要测量一个完整波形。
6. Transformer Principles | 变压器原理
A transformer consists of two coils wound on a laminated soft iron core. An alternating current in the primary coil produces a changing magnetic flux, which links the secondary coil and induces an alternating voltage by Faraday’s law.
变压器由绕在叠片软铁芯上的两个线圈组成。初级线圈中的交流电产生变化的磁通量，该磁通量穿过次级线圈，根据法拉第定律感应出交流电压。
Step‑up transformers increase voltage (more secondary turns), while step‑down transformers decrease voltage (fewer secondary turns). Transformers are essential in the national grid for reducing power losses by transmitting electricity at very high voltages.
升压变压器增加电压（次级匝数更多），降压变压器降低电压（次级匝数更少）。在国家电网中，变压器用于将电升至高电压传输，以减少功率损耗。
7. Ideal Transformer Equation | 理想变压器方程
For an ideal transformer with 100% efficiency, the ratio of secondary voltage Vs to primary voltage Vp equals the ratio of secondary turns Ns to primary turns Np:
对于效率 100% 的理想变压器，次级电压 Vs 与初级电压 Vp 之比等于匝数比：
Vs / Vp = Ns / Np
Because power in equals power out, VpIp = VsIs, so current is inversely proportional to the turns ratio: Is / Ip = Np / Ns.
由于输入功率等于输出功率，VpIp = VsIs，因此电流与匝数比成反比：Is / Ip = Np / Ns。
These equations are frequently tested. If a transformer has 2000 primary turns and 200 secondary turns, and the input is 230 Vrms, the output voltage is 23 Vrms. The output current will be proportionally larger, assuming an ideal transformer.
这些公式经常在考题中出现。若一变压器初级 2000 匝，次级 200 匝，输入 230 Vrms，则输出电压为 23 Vrms。假设是理想变压器，输出电流将按比例增大。
8. Transformer Efficiency and Losses | 变压器效率与损耗
Real transformers are not 100% efficient. Energy losses occur due to copper losses (resistive heating in the coils), eddy currents induced in the core, and hysteresis losses from the continuous magnetisation reversal of the core material.
实际变压器效率达不到 100%。能量损耗包括线圈电阻发热引起的铜损、铁芯中感生的涡流以及铁芯材料反复磁化造成的磁滞损耗。
To minimise eddy currents, the iron core is laminated with insulating layers. Soft iron is chosen because it easily magnetises and demagnetises, reducing hysteresis losses. Efficiency can be expressed as:
为减少涡流，铁芯采用互相绝缘的叠片结构。选用软铁是因为它易磁化易退磁，可降低磁滞损耗。效率可表示为：
Efficiency = (Pout / Pin) × 100% = (VsIs / VpIp) × 100%
AQA may ask you to suggest why a transformer operates less efficiently at lower frequencies or with a non‑laminated core.
AQA 可能会问为什么变压器在较低频率下或使用非叠片铁芯时效率更低。
9. Rectification: Half‑wave and Full‑wave | 整流：半波与全波
Rectification converts AC into DC. Half‑wave rectification uses a single diode to pass current only during one half of the AC cycle, resulting in a waveform with gaps where there is no output voltage.
整流将交流电转换为直流电。半波整流使用一只二极管，仅在交流电的半个周期内允许电流通过，输出的波形有一半时间电压为零。
Full‑wave rectification, typically achieved with a bridge rectifier (four diodes), inverts the negative half‑cycle so that both halves contribute to a unidirectional output. This doubles the frequency of the output pulses compared with the input AC.
全波整流通常由桥式整流器（四只二极管）实现，它将负半周翻转，使得两个半周都产生单向输出。与输入交流电相比，输出脉动频率加倍。
Exam diagrams often require you to identify the correct arrangement of diodes in a bridge rectifier and to sketch the resulting voltage waveform for a resistive load.
考题常要求你识别桥式整流器中二极管的正确连接方式，并绘制电阻性负载上的电压波形。
10. Smoothing Circuits | 平滑滤波电路
The pulsating DC from a rectifier is not suitable for most electronic devices. A smoothing capacitor connected across the load charges during the voltage peaks and discharges through the load when the rectifier output falls, reducing ripple.
整流得到的脉动直流电不适合大多数电子设备。在负载两端并联平滑电容，当电压处于峰值时电容充电，在整流输出下降时电容通过负载放电，从而减小波纹。
A larger capacitor stores more charge and discharges more slowly, resulting in a smoother output voltage with smaller ripple. The time constant τ = RLC must be much larger than the time between peaks for effective smoothing.
电容越大，储存电荷越多，放电越慢，输出电压越平滑，波纹越小。时间常数 τ = RLC 必须远大于两个峰值之间的时间，才能有效滤波。
In AQA questions you might be asked to sketch the smoothed waveform and explain how increasing capacitance or load resistance affects the ripple.
AQA 试题中可能会要求你画出平滑后的波形，并解释增大电容或负载电阻对波纹的影响。
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