📚 Power and Alternating Current | 交流电与功率
Alternating current (a.c.) is central to electrical power systems because it can be generated, transformed and transmitted efficiently. In CIE A-Level Physics, the key ideas in this topic are sinusoidal waveforms, root-mean-square (r.m.s.) values, mean power in resistive circuits, transformer power relationships and power loss in transmission lines.
交流电在现代电力系统中至关重要，因为它便于发电、变压和高效传输。在 CIE A-Level 物理中，本专题的核心内容包含正弦波形、方均根值、纯电阻电路中的平均功率、变压器的功率关系以及输电线路的功率损耗。
1. Sinusoidal Alternating Current | 正弦交流电
An alternating current is one that reverses direction periodically. For a sinusoidal a.c. source, the current and voltage can be described by I = I₀ sin(ωt) and V = V₀ sin(ωt), where I₀ and V₀ are the peak values and ω is the angular frequency.
交流电是指方向周期性反转的电流。对于正弦交流电源，电流和电压可表示为 I = I₀ sin(ωt) 和 V = V₀ sin(ωt)，其中 I₀ 和 V₀ 是峰值，ω 是角频率。
The angular frequency is related to frequency by ω = 2πf. The period T is the time for one complete cycle, so T = 1/f. For a 50 Hz mains supply, T = 0.02 s. Because current and voltage change continuously, the instantaneous power p = vi also changes with time.
角频率与频率的关系为 ω = 2πf。周期 T 是完成一次完整循环所需的时间，因此 T = 1/f。对于 50 Hz 的市电，T = 0.02 s。由于电流和电压不断变化，瞬时功率 p = vi 也随时间变化。
2. Mean Power and RMS Values | 平均功率与方均根值
To find the average power dissipated in a resistor R, begin with P = I²R. For a sinusoidal current I = I₀ sin(ωt), the average of sin²(ωt) over one complete cycle is 1/2. Therefore the mean value of I² is I₀²/2.
要计算电阻 R 上消耗的平均功率，可从 P = I²R 出发。对于正弦电流 I = I₀ sin(ωt)，sin²(ωt) 在一个完整周期内的平均值是 1/2，因此 I² 的平均值为 I₀²/2。
This leads to the definitions of root-mean-square values: I_rms = I₀ / √2 and V_rms = V₀ / √2. The average power in a resistive load can then be written using exactly the same form as the d.c. power equations.
由此可定义方均根值：I_rms = I₀ / √2，V_rms = V₀ / √2。电阻负载上的平均功率就可用与直流功率公式完全相同的形式表示。
P_avg = I_rms V_rms = I_rms² R = V_rms² / R
3. Calculating RMS Values | 方均根值的计算
For a pure sinusoidal waveform, the r.m.s. value is the peak value divided by √2, which is approximately 0.707 times the peak value. For example, UK mains is quoted as 230 V r.m.s., so the peak voltage is about 325 V.
对于纯正弦波形，方均根值等于峰值除以 √2，即约为峰值的 0.707 倍。例如，英国家用电压标称值为 230 V r.m.s.，因此峰值电压约为 325 V。
| Quantity | Peak value | RMS value |
|---|---|---|
| Mains voltage | 325 V | 230 V |
| Current in 100 Ω heater | 3.25 A | 2.30 A |
Do not automatically multiply the peak value by 0.707 for non-sinusoidal waveforms, such as square waves or half-wave rectified signals. For those cases the r.m.s. value must be found from the definition: square each value, take the mean, then take the square root.
对于非正弦波形，如方波或半波整流信号，不能直接使用峰值乘以 0.707。此时必须依据定义求方均根值：先平方，再取平均，最后开平方。
4. Peak Power vs Average Power | 峰值功率与平均功率
The instantaneous power in a resistor is p = V₀I₀ sin²(ωt). This quantity pulses between zero and the peak power P_peak = V₀I₀, and it oscillates at twice the supply frequency because sin²(ωt) has double the frequency of sin(ωt).
电阻上的瞬时功率为 p = V₀I₀ sin²(ωt)。该值在零与峰值功率 P_peak = V₀I₀ 之间脉动，并且由于 sin²(ωt) 的频率是 sin(ωt) 的两倍，瞬时功率以两倍电源频率振荡。
The average power over one complete cycle is half the peak power: P_avg = ½V₀I₀ = V_rms I_rms. This average is the quantity that determines heating, fuse ratings and energy consumption in resistive appliances.
一个完整周期内的平均功率是峰值功率的一半：P_avg = ½V₀I₀ = V_rms I_rms。这一平均值决定了电阻性电器的发热、熔断器额定值和能量消耗。
5. Why RMS Values Are Used | 为何使用方均根值
The r.m.s. current is defined as the equivalent direct current that would dissipate the same average power in a given resistor. This equivalence allows technicians and engineers to apply familiar d.c. power formulas to a.c. circuits without needing to integrate over time.
方均根电流的定义是：在相同电阻上产生相同平均功率的等效直流电流。这种等效性使得技术人员和工程师能够将熟悉的直流功率公式直接用于交流电路，而无需对时间进行积分。
When a multimeter reads an a.c. value, it usually displays the r.m.s. value for a sinusoidal signal. Mains labels such as 230 V or 110 V refer to r.m.s. voltage, not peak voltage.
万用表测量交流信号时，通常显示正弦信号的方均根值。230 V 或 110 V 等市电标称值指的是方均根电压，而不是峰值电压。
6. Transformers and Power | 变压器与功率
An ideal transformer changes voltage and current while conserving power. The voltage ratio depends on the turns ratio: V_s / V_p = N_s / N_p, where the subscripts p and s refer to the primary and secondary coils. Since energy is conserved, V_p I_p = V_s I_s.
理想变压器在保持功率守恒的同时改变电压和电流。电压比取决于匝数比：V_s / V_p = N_s / N_p，其中下标 p 和 s 分别代表原线圈和副线圈。由于能量守恒，V_p I_p = V_s I_s。
V_s / V_p = N_s / N_p and V_p I_p = V_s I_s
A step-up transformer increases voltage and decreases current for the same power, while a step-down transformer decreases voltage and increases current. The current ratio is I_s / I_p = N_p / N_s.
升压变压器在功率相同时升高电压并降低电流，而降压变压器降低电压并增大电流。电流比为 I_s / I_p = N_p / N_s。
7. Power Loss in Transmission Lines | 输电线路的功率损耗
Transmission cables have resistance R, so the power lost as heat in the cables is P_loss = I² R. For a given transmitted power P = VI, increasing the transmission voltage reduces the current and therefore greatly reduces the line loss.
输电电缆具有电阻 R，因此电缆中以热量形式损失的功率为 P_loss = I² R。在传输功率 P = VI 不变的情况下，提高输电电压会降低电流，从而大幅减少线路损耗。
P_loss = I² R
For example, transmitting the same power at 400 kV instead of 25 kV reduces the current by a factor of 16, so the transmission line power loss falls by a factor of 16² = 256. This is why national grid systems use high voltages.
例如，若以 400 kV 代替 25 kV 传输相同功率，电流降低为原来的 1/16，因此线路功率损耗降低为原来的 1/16² = 1/256。这就是国家电网使用高电压的原因。
8. Rectification and Smoothing | 整流与滤波
Rectification converts alternating current into direct current by allowing current to pass in only one direction. Half-wave rectification uses one diode and passes only half of each cycle. Full-wave rectification uses a diode bridge and inverts the negative half-cycles, producing a more continuous output.
整流通过只允许电流单向流动将交流电转换为直流电。半波整流使用一个二极管，只通过每个周期的一半。全波整流使用二极管桥，将负半周反转，从而得到更连续的输出。
A smoothing capacitor reduces the ripple in the rectified output by charging and discharging across the load. The power delivered to the load depends on the averaged rectified waveform, so calculations must use the correct r.m.s. or mean value for that specific waveform.
滤波电容通过跨接在负载上充放电来减小整流输出中的纹波。输送到负载的功率取决于整流波形的平均值，因此计算时必须使用该特定波形的正确方均根值或平均值。
9. Worked Example | 典型例题
A 230 V r.m.s. mains supply is connected to a 100 Ω resistor. Calculate the peak voltage, r.m.s. current, peak current, average power and peak power.
将 230 V r.m.s. 的市电接入 100 Ω 电阻。计算峰值电压、方均根电流、峰值电流、平均功率和峰值功率。
V₀ = 230 × √2 ≈ 325 V
I_rms = V_rms / R = 230 / 100 = 2.30 A
I₀ = I_rms × √2 ≈ 3.25 A
P_avg = V_rms I_rms = 230 × 2.30 ≈ 529 W
P_peak = V₀ I₀ ≈ 325 × 3.25 ≈ 1056 W
Notice that the average power is exactly half the peak power for a sinusoidal supply and a purely resistive load.
注意，对于正弦电源和纯电阻负载，平均功率恰好是峰值功率的一半。
10. Exam Tips and Common Mistakes | 考试提示与常见错误
Always distinguish between peak and r.m.s. values. Use V_rms and I_rms when calculating average power, heater ratings, fuse currents and transformer power. Use peak values only when asked for peak power or peak voltage.
务必区分峰值和方均根值。在计算平均功率、加热器额定值、熔断电流和变压器功率时使用 V_rms 和 I_rms。只有在求峰值功率或峰值电压时才使用峰值。
A very common mistake is substituting peak current into P = I²R. This gives a value twice as large as the true average power. Another mistake is assuming the 0.707 relationship applies to all waveforms; it is valid only for pure sine waves.
一个常见错误是将峰值电流代入 P = I²R，这样得到的结果是真实平均功率的两倍。另一个错误是认为 0.707 关系适用于所有波形，实际上它只适用于纯正弦波。
In transformer calculations, the ideal power equation V_p I_p = V_s I_s assumes 100% efficiency. If the efficiency η is less than 1, use P_s = η P_p, where P_p is the primary input power and P_s is the secondary output power.
在变压器计算中，理想功率方程 V_p I_p = V_s I_s 假设效率为 100%。若效率 η 小于 1，则应使用 P_s = η P_p，其中 P_p 为原边输入功率，P_s 为副边输出功率。
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