A-Level Physics: Calculating AC Power | A-Level 物理：交流电功率的计算方法

📚 A-Level Physics: Calculating AC Power | A-Level 物理：交流电功率的计算方法

Alternating current is used throughout the world because it can be transformed easily to high voltages for transmission. However, calculating the power delivered by an alternating current is trickier than using the DC equation P = IV, because the voltage and current are changing continuously. This article explains the correct method, introduces root-mean-square values, and applies them to resistive, inductive, capacitive and general AC circuits.

交流电之所以被全世界广泛使用，是因为它能够方便地通过变压器升高电压进行远距离传输。然而，计算交流电所传输的功率远比直流电的 P = IV 复杂，因为电压和电流都在不断变化。本文将解释正确的计算方法，介绍有效值（均方根值），并把这些方法应用于纯电阻、纯电感、纯电容及一般交流电路。

1. RMS Values and Why They Matter | 有效值及其重要性

For a sine wave, the average value of the current over a complete cycle is zero. If we used the arithmetic mean of current in P = I²R, we would wrongly conclude that an AC circuit produces no heat at all. The solution is to use root-mean-square values, written as Iᵣₘₛ and Vᵣₘₛ.

对正弦波而言，电流在一个完整周期内的算术平均值是零。如果我们把电流的算术平均值直接代入 P = I²R，就会错误地认为交流电路完全不产生热量。解决办法是使用有效值，记作 Iᵣₘₛ 和 Vᵣₘₛ。

For a sinusoidal supply with peak values V₀ and I₀, the RMS values are:

Vᵣₘₛ = V₀ / √2

Iᵣₘₛ = I₀ / √2

The RMS value is the steady DC value that would produce the same heating effect in a resistor as the alternating value. This is why all household AC voltages are stated as RMS values.

有效值是指：在同一个电阻上，若直流电和交流电产生相同的热效应，则这个直流电的大小就等于该交流电的有效值。因此，家用交流电压给出的数值通常都是有效值。

2. Instantaneous Power in a Resistive Circuit | 纯电阻电路中的瞬时功率

Consider a resistor R connected to an AC supply. If the instantaneous voltage is v(t) = V₀ sin(ωt), then by Ohm’s law the instantaneous current is i(t) = I₀ sin(ωt), where I₀ = V₀ / R. The instantaneous power is the product of instantaneous voltage and instantaneous current:

考虑一个连接到交流电源的电阻 R。若瞬时电压为 v(t) = V₀ sin(ωt)，根据欧姆定律，瞬时电流为 i(t) = I₀ sin(ωt)，其中 I₀ = V₀ / R。瞬时功率等于瞬时电压与瞬时电流的乘积：

p(t) = v(t)i(t) = V₀I₀ sin²(ωt)

Because sin²(ωt) is always positive, the power is always positive. Energy is therefore converted into heat continuously, even though the current changes direction every half-cycle.

由于 sin²(ωt) 总是非负的，所以瞬时功率始终为正。因此，即使电流每半个周期改变一次方向，电阻仍会持续地将电能转化为热能。

3. Average Power and the Half-Factor | 平均功率与二分之一因子

Over one full cycle, the average of sin²(ωt) is exactly ½. Therefore the average power dissipated in a pure resistor is:

在一个完整周期内，sin²(ωt) 的平均值正好是 ½。因此，纯电阻消耗的平均功率为：

Pₐᵥ = ½ V₀I₀

Substituting V₀ = √2 Vᵣₘₛ and I₀ = √2 Iᵣₘₛ:

将 V₀ = √2 Vᵣₘₛ 和 I₀ = √2 Iᵣₘₛ 代入：

Pₐᵥ = ½ × (√2 Vᵣₘₛ) × (√2 Iᵣₘₛ) = VᵣₘₛIᵣₘₛ

Notice the factor of ½ cancels with the two √2 factors. This is why RMS values are defined as peak divided by √2: the resulting power equation has the same form as the DC equation.

注意，½ 与两个 √2 相乘后相互抵消。这正是把有效值定义为峰值除以 √2 的原因：这样得到的功率方程与直流电方程形式完全相同。

4. The General RMS Power Equation | 通用有效值功率方程

For any sinusoidal AC supply, the average power delivered to a load can be written using RMS values:

对于任意正弦交流电源，传递给负载的平均功率都可以用有效值写成：

Pₐᵥ = VᵣₘₛIᵣₘₛ

For a pure resistor, this can also be written in three equivalent forms:

对于纯电阻，该式还有三种等价形式：

Pₐᵥ = VᵣₘₛIᵣₘₛ = Iᵣₘₛ²R = Vᵣₘₛ² / R

These equations are identical in structure to DC power equations. The only change is that peak values must be replaced by RMS values.

这些方程在结构上与直流电功率方程完全相同。唯一的变化是，必须把峰值替换为有效值。

5. Why RMS Is the “Heating Equivalent” | 为什么有效值是”加热等效值”

The heating effect of a current depends on the square of the current. In AC circuits, we must average I²R over one cycle, not simply average I. The RMS current is defined as the square root of the mean of i²(t):

电流的热效应取决于电流的平方。在交流电路中，我们必须对一个周期内的 I²R 取平均，而不是对 I 直接取平均。有效值电流定义为 i²(t) 在一个周期内平均值的平方根：

Iᵣₘₛ = √( ⟨i²⟩ )

For a sine wave, the mean of i² is I₀²/2, so Iᵣₘₛ = I₀/√2. This explains exactly why the factor √2 appears.

对正弦波而言，i² 的平均值是 I₀²/2，所以 Iᵣₘₛ = I₀/√2。这正解释了 √2 这个因子的来源。

Examiners often ask why RMS values are used. The best answer is that RMS current produces the same average heating in a resistor as the same numerical DC current.

考官常问为什么使用有效值。最好的回答是：同样数值的有效值电流和一个直流电流在同一個电阻上产生的平均热效应相同。

6. Pure Inductors and Capacitors: Zero Average Power | 纯电感与纯电容：平均功率为零

For a pure inductor, the current lags the voltage by 90°. For a pure capacitor, the current leads the voltage by 90°. In both cases, the instantaneous power alternates between positive and negative values.

对于纯电感，电流相位落后电压 90°。对于纯电容，电流相位超前电压 90°。在这两种情况下，瞬时功率都在正值和负值之间交替变化。

When the instantaneous power is positive, the component is absorbing energy from the supply. When it is negative, the component is returning energy back to the supply. Over a full cycle, the positive and negative contributions cancel exactly.

当瞬时功率为正时，元件从电源吸收能量；当瞬时功率为负时，元件把能量返还给电源。在一个完整周期内，正负能量刚好完全抵消。

Pₐᵥ = 0 for a pure inductor or pure capacitor

纯电感或纯电容的平均功率 Pₐᵥ = 0

This means ideal inductors and capacitors store energy temporarily but do not dissipate it as heat. Only real resistance converts electrical energy into thermal energy.

这意味着理想电感和电容只会暂时储存能量，而不会把能量转化为热。只有真实电阻才会把电能转化为热能。

7. Power in a General AC Circuit: The Phase Angle | 一般交流电路的功率：相位角

In a circuit containing both resistance and reactance, the voltage and current are not in phase. If the instantaneous voltage is v(t) = V₀ sin(ωt) and the current is i(t) = I₀ sin(ωt – φ), then φ is called the phase angle.

在同时含有电阻和电抗的电路中，电压和电流相位并不相同。若瞬时电压为 v(t) = V₀ sin(ωt)，瞬时电流为 i(t) = I₀ sin(ωt – φ)，则 φ 称为相位角。

Expanding the product and averaging over one cycle gives the general AC power equation:

展开这个乘积并取一个周期的平均值，可以得到一般的交流电功率方程：

Pₐᵥ = VᵣₘₛIᵣₘₛ cos φ

Here cos φ is called the power factor. For a pure resistor, φ = 0° and cos φ = 1, so Pₐᵥ = VᵣₘₛIᵣₘₛ. For a pure inductor or capacitor, φ = 90° and cos φ = 0, so Pₐᵥ = 0.

这里的 cos φ 称为功率因数。对于纯电阻，φ = 0°，cos φ = 1，所以 Pₐᵥ = VᵣₘₛIᵣₘₛ。对于纯电感或纯电容，φ = 90°，cos φ = 0，所以 Pₐᵥ = 0。

8. Apparent Power, Real Power and Power Factor | 视在功率、有功功率与功率因数

The product VᵣₘₛIᵣₘₛ is called the apparent power S. Its unit is the volt-ampere (VA). The actual average power P is called the real or true power, measured in watts (W).

乘积 VᵣₘₛIᵣₘₛ 称为视在功率 S，单位是伏安（VA）。实际的平均功率 P 称为有功功率或真实功率，单位是瓦特（W）。

S = VᵣₘₛIᵣₘₛ

P = S cos φ

power factor = cos φ = P / S

In an AC circuit containing reactive components, S can be much larger than P. A high apparent power means the supply must carry a larger current than would be needed for the same amount of useful power. This is why improving the power factor is important in industry.

在含有电抗元件的交流电路中，视在功率 S 可能远大于有功功率 P。高视在功率意味着电源需要提供比产生同样有用功率时更大的电流。这就是为什么提高功率因数在工业中非常重要。

9. Worked Example 1: Heater on Mains Supply | 例题1：市电加热器

A 230 V, 50 Hz mains supply is connected to an electric heater of resistance 50 Ω. Calculate the RMS current and the average power dissipated.

一个 230 V、50 Hz 的市电电源连接到一个电阻为 50 Ω 的电加热器。试计算有效值电流和平均功率。

Using Ohm’s law with RMS values:

在有效值下使用欧姆定律：

Iᵣₘₛ = Vᵣₘₛ / R = 230 / 50 = 4.6 A

Then the average power is:

平均功率为：

Pₐᵥ = Iᵣₘₛ²R = 4.6² × 50 = 1058 W

Equivalently, Pₐᵥ = Vᵣₘₛ² / R = 230² / 50 = 1058 W. Since the heater is purely resistive, no power-factor correction is needed.

等价地，Pₐᵥ = Vᵣₘₛ² / R = 230² / 50 = 1058 W。由于加热器是纯电阻负载，不需要修正功率因数。

10. Worked Example 2: Motor with Power Factor | 例题2：功率因数电动机

A motor is connected to a 240 V RMS supply and draws a current of 8.0 A RMS. The power factor of the motor is 0.80. Calculate the apparent power, the real power, and the reactive power.

一台电动机连接到 240 V 有效值电源，电流为 8.0 A 有效值。电动机的功率因数为 0.80。试计算视在功率、有功功率和无功功率。

The apparent power is:

视在功率为：

S = VᵣₘₛIᵣₘₛ = 240 × 8.0 = 1920 VA

The real power is:

有功功率为：

P = S cos φ = 1920 × 0.80 = 1536 W

Since S² = P² + Q² for sinusoidal AC, the reactive power Q is:

对于正弦交流电，S² = P² + Q²，因此无功功率 Q 为：

Q = √(S² – P²) = √(1920² – 1536²) = √(3686400 – 2359296) = √1151104 ≈ 1073 var

The motor therefore consumes 1536 W of useful power but requires the supply to handle 1920 VA because of the lagging current.

因此，电动机实际消耗 1536 W 的有用功率，但由于电流滞后，电源需要承担 1920 VA 的视在功率。

11. Common Exam Pitfalls | 常见考试误区

Using peak values instead of RMS values in power equations.
将峰值直接代入功率方程，而没有先转换为有效值。
Forgetting that the average of sin²(ωt) over a full cycle is ½.
忘记 sin²(ωt) 在一个完整周期内的平均值是 ½。
Assuming P = VᵣₘₛIᵣₘₛ works for every AC circuit; it only works for a purely resistive circuit.
认为 P = VᵣₘₛIᵣₘₛ 对所有交流电路都成立；实际上只有在纯电阻电路中才成立。
Ignoring the power factor when the circuit contains inductors or capacitors.
当电路含有电感或电容时忽略功率因数。
Confusing leading and lagging phase angles; always state which quantity is leading.
混淆超前和滞后的相位角；应该明确指出哪个量超前。

12. Quick Revision Table | 快速复习表

Quantity	Symbol	Equation
Peak voltage	V₀	V₀ = √2 Vᵣₘₛ
RMS voltage	Vᵣₘₛ	Vᵣₘₛ = V₀ / √2
RMS current	Iᵣₘₛ	Iᵣₘₛ = I₀ / √2
Average power, resistive	Pₐᵥ	Pₐᵥ = VᵣₘₛIᵣₘₛ = Iᵣₘₛ²R = Vᵣₘₛ² / R
Average power, general	Pₐᵥ	Pₐᵥ = VᵣₘₛIᵣₘₛ cos φ
Apparent power	S	S = VᵣₘₛIᵣₘₛ
Reactive power	Q	Q = VᵣₘₛIᵣₘₛ sin φ
Power factor	cos φ	cos φ = P / S

The key idea is always to use RMS values for AC power calculations and to include cos φ whenever the load is not purely resistive. Once you remember these two rules, most AC power problems become straightforward.

关键思路是：在交流电功率计算中始终使用有效值，并且只要负载不是纯电阻，就要乘上 cos φ。一旦记住这两条规则，大多数交流电功率问题就会变得非常简单。

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A-Level Physics: Calculating AC Power | A-Level 物理：交流电功率的计算方法