A-Level Physics: Calculating and Applying Electrical Power | A-Level 物理：电功率的计算与应用

📚 A-Level Physics: Calculating and Applying Electrical Power | A-Level 物理：电功率的计算与应用

Electrical power is a core concept in A-Level Physics, bridging the ideas of current, voltage, resistance, and energy transfer. It appears in almost every exam paper, whether in short calculations, circuit analysis, or practical applications such as household electricity and power transmission.

电功率是 A-Level 物理中的核心概念，它将电流、电压、电阻与能量转移紧密联系在一起。几乎每份试卷都会出现电功率相关题目，无论是简短计算、电路分析，还是家庭用电与电力传输等实际应用。

1. The Basic Definition of Power | 功率的基本定义

Power is defined as the rate at which energy is transferred or converted. In electrical circuits, this energy is carried by moving charges. The SI unit of power is the watt (W), where 1 W = 1 J s⁻¹.

功率定义为能量转移或转换的速率。在电路中，能量由移动的电荷携带。功率的国际单位是瓦特（W），1 W = 1 J s⁻¹。

For an electrical component, the power delivered to or dissipated by it is given by the product of the potential difference across it and the current through it:

对于某个电学元件，其获得或耗散的功率等于它两端的电势差与通过它的电流的乘积：

P = IV

This equation is fundamental and applies to any component, whether it is a resistor, a lamp, a motor, or a semiconductor diode. It is derived from the definition of potential difference: V = W/Q, where W is the energy transferred per unit charge Q. Since current I = Q/t, multiplying by V gives P = W/t = IV.

该方程是基本的，适用于任何元件，无论是电阻、灯泡、电动机还是半导体二极管。它由电势差的定义推导得出：V = W/Q，其中 W 是每单位电荷 Q 转移的能量。由于电流 I = Q/t，乘以 V 后得到 P = W/t = IV。

2. Combining P = IV with Ohm’s Law | 将 P = IV 与欧姆定律结合

For ohmic conductors that obey Ohm’s law (V = IR), the power equation can be rewritten in two alternative forms. Substituting V = IR into P = IV gives P = I²R.

对于服从欧姆定律（V = IR）的欧姆导体，功率方程可以改写为另外两种形式。将 V = IR 代入 P = IV，得到 P = I²R。

Substituting I = V/R into P = IV gives P = V²/R.

将 I = V/R 代入 P = IV，得到 P = V²/R。

These three expressions are all equivalent for resistors, but they are not interchangeable for non-ohmic components. For example, a diode does not have a constant resistance, so P = V²/R may not be reliable unless R is taken at the operating point.

对于电阻而言，这三个表达式是等价的，但对于非欧姆元件不能互相替换。例如，二极管没有恒定的电阻，因此除非在特定工作点取 R 值，否则 P = V²/R 可能不太可靠。

When solving exam problems, always check which quantities are known. If you know current and resistance, use P = I²R. If you know voltage and resistance, use P = V²/R. If you know voltage and current, use P = IV directly.

解题时应先检查已知量。若已知电流和电阻，用 P = I²R；若已知电压和电阻，用 P = V²/R；若已知电压和电流，直接用 P = IV。

3. Electrical Energy and Power Dissipation | 电能与功率耗散

The total electrical energy transferred by a device over a time t is:

设备在时间 t 内转移的总电能为：

W = IVt = Pt

This energy may appear as light (lamp), sound (speaker), mechanical work (motor), or thermal energy (heater). In a resistor, all electrical energy is converted to heat. This process is known as Joule heating.

这些能量可能表现为光（灯泡）、声音（扬声器）、机械功（电动机）或热能（加热器）。在电阻中，所有电能都转化为热量，这一过程称为焦耳热。

The equation W = VIt is used to calculate the energy transferred in a circuit. For instance, a 12 V battery supplying a current of 2 A for 5 minutes transfers an energy of:

方程 W = VIt 用于计算电路中转移的能量。例如，一个 12 V 电池以 2 A 电流供电 5 分钟，转移的能量为：

W = 12 V × 2 A × (5 × 60 s) = 7200 J

Note: Always convert time into seconds before using this equation.

注意：使用该方程前务必把时间换算为秒。

4. Power in Series and Parallel Circuits | 串并联电路中的功率

In a series circuit, the same current flows through each resistor, but the voltage is divided. Since P = I²R, the power dissipated in each resistor is proportional to its resistance. The largest resistor dissipates the most power.

在串联电路中，通过每个电阻的电流相同，但电压被分配。由 P = I²R 可知，每个电阻耗散的功率与其阻值成正比。阻值最大的电阻耗散功率最大。

In a parallel circuit, each resistor has the same voltage across it, but the current is divided. Since P = V²/R, the power dissipated in each resistor is inversely proportional to its resistance. The smallest resistor dissipates the most power.

在并联电路中，每个电阻两端电压相同，但电流被分配。由 P = V²/R 可知，每个电阻耗散的功率与其阻值成反比。阻值最小的电阻耗散功率最大。

These relationships are essential for analysing circuits that combine series and parallel groups. Identify the group’s total resistance first, then find the current or voltage, and finally distribute power according to the rules above.

这些关系对于分析串并联混合电路至关重要。先求总电阻，再求总电流或总电压，最后按上述规则分配功率。

5. Maximum Power Transfer Theorem | 最大功率传输定理

A common exam topic is the condition for maximum power transfer from a source with internal resistance r to an external load resistor R. The power delivered to the load is:

一个常见的考点是：内阻为 r 的电源向外部负载电阻 R 传输最大功率的条件。传递给负载的功率为：

P = I²R = (ε / (R + r))² R

where ε is the electromotive force (emf) of the source.

其中 ε 是电源的电动势。

Differentiating P with respect to R and setting dP/dR = 0, or by using the symmetry of the equation, we find that maximum power transfer occurs when the load resistance equals the internal resistance:

对 P 关于 R 求导并令 dP/dR = 0，或利用方程的对称性，可以发现当负载电阻等于内阻时，功率传输达到最大：

R = r

At this condition, the load receives half of the total power supplied by the source; the other half is dissipated inside the source. This is why in real power systems, engineers do not design for maximum power transfer, but instead aim for maximum efficiency by making R much larger than r.

此时负载仅获得电源提供的总功率的一半，另一半消耗在电源内部。这就是为什么实际电力系统中，工程师不会按最大功率传输来设计，而是通过让 R 远大于 r 来追求最大效率。

6. Power Ratings of Appliances | 电器的额定功率

Every electrical appliance carries a power rating, usually given in watts (W) or kilowatts (kW). This rating indicates the power consumption when the appliance is connected to the specified mains voltage.

每台电器都有一个额定功率，通常以瓦特（W）或千瓦（kW）为单位。该额定值表示电器在指定电源电压下工作时的功耗。

For example, a 2300 W electric kettle connected to a 230 V supply draws a current of:

例如，一个功率为 2300 W 的电水壶连接到 230 V 电源时，通过的电流为：

I = P / V = 2300 W / 230 V = 10 A

This calculation is important for selecting the correct fuse rating. A fuse is designed to melt and break the circuit if the current exceeds a safe level. The fuse rating should be slightly higher than the normal operating current, for example 13 A for a 10 A kettle.

该计算对于选择正确的保险丝额定值非常重要。保险丝的作用是当电流超过安全水平时熔断并切断电路。保险丝的额定值应略高于正常工作的电流，例如 10 A 的电水壶配 13 A 的保险丝。

Power ratings also appear in energy cost calculations. The energy used in kilowatt-hours (kWh) is:

额定功率也出现在能源费用计算中。以千瓦时（kWh）为单位的电能为：

Energy (kWh) = Power (kW) × Time (h)

A 2 kW heater running for 3 hours uses 6 kWh. If the cost per unit is 15 pence, the running cost is 90 pence.

一个 2 kW 的加热器运行 3 小时消耗 6 kWh。若每度电 15 便士，则运行成本为 90 便士。

7. Practical Worked Examples | 实用计算示例

Let’s work through a comprehensive example. A 6 Ω resistor and a 12 Ω resistor are connected in parallel across a 12 V battery. Find the total power dissipated.

下面看一个综合示例。一个 6 Ω 电阻和一个 12 Ω 电阻并联连接到 12 V 电池两端，求总耗散功率。

First find the equivalent resistance for parallel resistors:

先求并联等效电阻：

1/R_total = 1/6 + 1/12 = 1/4, so R_total = 4 Ω

Using P = V²/R for the whole circuit:

对整个电路用 P = V²/R：

P_total = 12² / 4 = 36 W

Now check each resistor individually. The 6 Ω resistor receives 12 V, so P₁ = 12² / 6 = 24 W. The 12 Ω resistor receives 12 V, so P₂ = 12² / 12 = 12 W. The total is 36 W, which matches.

再分别验证每个电阻。6 Ω 电阻两端为 12 V，故 P₁ = 12² / 6 = 24 W。12 Ω 电阻两端为 12 V，故 P₂ = 12² / 12 = 12 W。总功率为 36 W，结果一致。

This example illustrates two important points: total power can be found from total resistance, and individual power can be found from the common voltage in parallel branches.

这个例子说明了两个重要点：总功率可以通过总电阻求得，而各支路功率则基于并联时的共同电压进行计算。

8. Efficiency and Power Loss in Transmission | 输电效率与功率损耗

When electrical power is transmitted over long distances, some energy is lost as heat in the transmission cables. The power loss in a cable of resistance R carrying current I is:

当电能进行长距离传输时，部分能量会在传输电缆中以热能形式损耗。对于电阻为 R、电流为 I 的电缆，其损耗功率为：

P_loss = I²R

To reduce this loss, power stations use step-up transformers to increase the voltage and therefore reduce the current for a given power. Since loss depends on I², reducing the current by a factor of 10 reduces the power loss by a factor of 100.

为减少损耗，发电厂使用升压变压器提高电压，从而在给定功率下降低电流。由于损耗与 I² 成正比，将电流减小为原来的 1/10，功率损耗将减小为原来的 1/100。

The efficiency of a device or system is defined as:

设备或系统的效率定义为：

Efficiency = (useful output power / total input power) × 100%

For a transformer, efficiency = (I_s V_s) / (I_p V_p) × 100%, where subscripts s and p refer to secondary and primary coils respectively.

对于变压器，效率 = (I_s V_s) / (I_p V_p) × 100%，其中下标 s 和 p 分别代表副线圈和原线圈。

9. Common Pitfalls and Exam Tips | 常见错误与考试技巧

Many students lose marks due to careless mistakes. Here are some frequent pitfalls and how to avoid them.

许多学生因粗心而丢分。以下是一些常见陷阱及避免方法。

Using P = V²/R when the component is non-ohmic. Always check whether the resistance is constant before applying Ohm’s law variants.
在元件非欧姆时使用 P = V²/R。使用欧姆定律变形前，务必判断电阻是否恒定。
Forgetting to convert time to seconds in W = VIt. Time must be in seconds for energy in joules.
在 W = VIt 中忘记将时间换算为秒。要获得焦耳能量，时间必须以秒为单位。
Confusing total power with power per resistor in a mixed circuit. Redraw the circuit and label all currents and voltages before calculating.
在复杂电路中混淆总功率与单个电阻的功率。计算前先重绘电路，标出所有电流和电压。
Using the internal resistance of a battery as a load resistor in power calculations. The internal resistance dissipates energy but is not available to the external circuit.
在功率计算中将电池内阻当作负载电阻。内阻耗散能量，但不能为外部电路提供有用功率。
Neglecting the sign of work or power when a battery is being charged. During charging, electrical power is being supplied to the battery, so its internal chemical energy increases.
在电池充电时忽略功或功率的符号。充电时电能输入电池，因此其内部化学能增加。

To maximise marks, write the formula first, substitute numerical values with units, and state the final answer with the correct unit. Always show your working for multi-step calculations.

为取得最高分，先写公式，再代入带单位的数值，最终答案须带正确单位。多步计算务必展示过程。

10. Summary of Key Equations | 关键公式总结

The table

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A-Level Physics: Calculating and Applying Electrical Power | A-Level 物理：电功率的计算与应用