📚 A-Level OCR Physics: Alternating Current Revision | 交流电考点精讲
Alternating current is central to OCR A Level Physics Module 6, appearing in electromagnetic induction, power transmission and rectification. Mastering peak, rms, transformer relationships and smoothing circuits is essential for exam success.
交流电是 OCR A Level 物理模块六的核心内容，涉及电磁感应、电力输送与整流。掌握峰值、有效值、变压器关系及滤波电路对考试至关重要。
1. Characteristics of AC | 交流电的特性
Alternating current (AC) periodically reverses direction. The voltage usually follows a sinusoidal form: V = V0 sin(ωt), where V0 is the peak voltage, ω = 2πf is the angular frequency, and f is the supply frequency in hertz (Hz). UK mains electricity has f = 50 Hz.
交流电周期性地反转方向。电压通常呈正弦波形：V = V0 sin(ωt)，其中 V0 为峰值电压，ω = 2πf 为角频率，f 为交流电频率（单位赫兹）。英国市电频率为 50 Hz。
The period T is the time for one complete oscillation. It is linked to frequency by T = 1/f. An AC waveform is fully described by its peak value, period and shape.
周期 T 是一次完整振荡所需的时间，它与频率的关系为 T = 1/f。交流波形完全由其峰值、周期和形状描述。
2. Peak, Peak-to-Peak and RMS Values | 峰值、峰峰值与有效值
The peak voltage V0 is the maximum displacement from zero. The peak-to-peak voltage Vpp is twice the peak value: Vpp = 2V0. On an oscilloscope trace, Vpp is the vertical height from the positive to the negative peak.
峰值电压 V0 是偏离零点的最大值。峰峰值 Vpp 是峰值的两倍：Vpp = 2V0。在示波器波形上，Vpp 即从正峰到负峰的垂直高度。
The root mean square (rms) value of a sinusoidal AC is the equivalent DC value that delivers the same heating power. For voltage and current:
Vrms = V0 / √2 and Irms = I0 / √2
The mains 230 V rating in the UK is an rms value, so its peak is approximately 325 V.
正弦交流电的均方根（rms）值是能产生相同热效应的等效直流值。电压和电流满足：
Vrms = V0 / √2  ； Irms = I0 / √2
英国市电标称的 230 V 是 rms 值，因此峰值约为 325 V。
3. Using an Oscilloscope | 使用示波器
An oscilloscope displays voltage on the vertical axis and time on the horizontal axis. From the screen you can read the peak voltage (multiplied by the Y-gain setting) and the time period (multiplied by the time-base setting). For example, if the time-base is 5 ms/div and one complete cycle spans 4 divisions, T = 20 ms and f = 1/0.020 = 50 Hz.
示波器的垂直轴表示电压，水平轴表示时间。从屏幕上可以读出峰值电压（乘以 Y 增益设定）和周期（乘以扫描时基设定）。例如，若时基为 5 ms/div，一个完整周期占据 4 格，则 T = 20 ms，f = 1/0.020 = 50 Hz。
Many exam questions ask you to determine the rms value from an oscilloscope picture: read Vpp, calculate V0 = Vpp/2, then apply Vrms = V0/√2.
许多考题要求学生根据示波器图像确定 rms 值：先读取 Vpp，求出 V0 = Vpp/2，再代入 Vrms = V0/√2。
4. Power in AC Circuits | 交流电路中的功率
For a purely resistive load, the average AC power is calculated using rms values. The instantaneous power is p(t) = v(t)i(t), but the useful quantity is average power: Pavg = IrmsVrms. This can also be written as Pavg = Vrms2 / R = Irms2 R.
对于纯电阻负载，平均交流功率用 rms 值计算。瞬时功率为 p(t) = v(t)i(t)，但有实际意义的是平均功率：Pavg = IrmsVrms，也可写成 Pavg = Vrms2 / R = Irms2 R。
This relationship explains why appliances rated at 230 V, 2.0 A dissipate approximately 460 W in a resistive heater. The same rms current in a DC circuit would deliver identical heating, demonstrating the equivalence of rms quantities.
这一关系解释了为何额定 230 V、2.0 A 的电阻性加热器耗散约 460 W。若直流电路通过相同的 rms 电流，产生的热效应完全一致，这体现了 rms 量的等效性。
5. The AC Generator | 交流发电机
A simple AC generator consists of a coil rotating in a uniform magnetic field. According to Faraday’s law, the induced emf is ε = BANω sin(ωt), where B is the magnetic flux density, A is the coil area, N is the number of turns, and ω is the angular speed. The peak emf is ε0 = BANω.
简单的交流发电机由在匀强磁场中转动的线圈构成。根据法拉第定律，感应电动势为 ε = BANω sin(ωt)，其中 B 为磁通密度，A 为线圈面积，N 为匝数，ω 为角速度。峰值电动势为 ε0 = BANω。
Slip rings and carbon brushes connect the rotating coil to the external circuit, maintaining the sinusoidal output. Increasing the rotation speed or the magnetic field strength raises both the peak emf and the frequency.
滑环和碳刷将旋转线圈与外部电路相连，保证了正弦输出。提高转速或增强磁场均可增大峰值电动势和频率。
6. Transformers | 变压器
An ideal transformer operates on the principle of electromagnetic induction, changing the voltage and current while conserving power. For two coils on a common iron core, the turns ratio equation is:
Vs / Vp = Ns / Np
It is assumed 100% efficiency, so primary power equals secondary power: VpIp = VsIs.
理想变压器基于电磁感应原理，可改变电压和电流并保持功率守恒。对于共用铁芯的两个线圈，匝数比方程为：
Vs / Vp = Ns / Np
假设效率为 100%，则初级功率等于次级功率：VpIp = VsIs。
A step-up transformer has Ns > Np, increasing voltage and reducing current. A step-down transformer does the reverse. Real transformers experience energy losses due to eddy currents, hysteresis in the core, and resistance in the windings.
升压变压器满足 Ns > Np，电压升高、电流减小；降压变压器则相反。实际变压器存在涡流、磁滞及绕组电阻带来的能量损耗。
7. Power Transmission | 电力输送
Electricity is transmitted over long distances at very high voltages (e.g. 400 kV) to minimise energy lost as heat in the cables. The power dissipated in a cable of resistance R is Ploss = I2R. By stepping up the voltage at the power station and stepping it down near consumers, the current in the transmission lines is greatly reduced, dramatically lowering Ploss.
电能通过高压（如 400 kV）远距离输送，以降低电缆热损耗。电阻为 R 的电缆耗散功率为 Ploss = I2R。在发电站升压、在用户附近降压，可大幅降低输电线中的电流，从而显著减少 Ploss。
For a given transmission power P = IV, doubling the voltage halves the current, and the I²R loss becomes one quarter. This is the fundamental reason for using AC grids with transformers.
对于给定的输送功率 P = IV，电压加倍则电流减半，I²R 损耗降为四分之一。这是交流电网结合变压器使用的根本原因。
8. Half-Wave Rectification | 半波整流
Half-wave rectification uses a single diode. During the positive half-cycle the diode is forward biased and conducts; during the negative half-cycle it is reverse biased and blocks current. The output voltage across the load is a series of positive half-sine pulses, with a mean DC value of Vdc = V0 / π.
半波整流使用单个二极管。正半周时，二极管正向偏置导通；负半周时，二极管反向偏置截止。负载两端的输出电压是一系列正半正弦脉冲，其平均直流值为 Vdc = V0 / π。
Because half-wave rectification discards the negative half-cycles, the output waveform has large gaps and a high ripple content, making it unsuitable for sensitive electronic devices without further smoothing.
由于半波整流舍弃了负半周，输出电压波形存在很大的间隙和较高的纹波成分，不经后续滤波难以用于敏感的电子设备。
9. Full-Wave Rectification | 全波整流
Full-wave rectification is typically achieved with a bridge rectifier consisting of four diodes. In both half-cycles, two diodes conduct so that the load current always flows in the same direction. The resulting output consists of overlapping positive humps, and the mean DC value is Vdc = 2V0 / π, exactly double that of half-wave rectification.
全波整流通常使用由四支二极管组成的桥式整流器实现。在两个半周内，始终有两支二极管导通，使负载电流方向始终一致。输出电压波形为彼此衔接的正向波峰，其平均直流值为 Vdc = 2V0 / π，正好是半波整流的两倍。
Importantly, the ripple frequency of the unfiltered full-wave output is twice the input AC frequency (e.g. 100 Hz for 50 Hz mains). This higher ripple frequency makes smoothing more effective.
重要的是，未经滤波的全波输出纹波频率是输入交流频率的两倍（如市电 50 Hz 对应 100 Hz）。较高的纹波频率使得后续滤波效果更佳。
10. Smoothing with Capacitors | 电容滤波
Smoothing reduces the ripple in a rectified DC output. A capacitor placed across the load charges up to the peak voltage and then slowly discharges through the load resistance, filling the gaps between pulses. The time constant τ = CR governs the discharge rate: a larger capacitance or load resistance reduces the ripple.
滤波可减小整流后直流输出中的纹波。并在负载两端的电容器会充电至峰值电压，然后通过负载电阻缓慢放电，填补脉冲之间的空隙。时间常数 τ = CR 决定了放电速率：增大电容或负载电阻可减小纹波。
In a full-wave rectified supply with smoothing, the output voltage has a small triangular ripple. The approximate ripple voltage is ΔV ≈ Iload / (frippleC), where Iload is the load current and fripple is twice the mains frequency. Choosing C sufficiently large maintains a nearly constant DC output.
在全波整流加滤波的电源中，
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