Kirchhoff's Laws | 基尔霍夫定律

📚 Kirchhoff’s Laws | 基尔霍夫定律

Kirchhoff’s laws are two fundamental rules that govern the behaviour of electrical circuits. They extend Ohm’s law to complex networks and are essential tools for analysing circuits that cannot be simplified using series and parallel combinations alone.

基尔霍夫定律是支配电路行为的两条基本规则。它们将欧姆定律扩展到复杂网络，是分析无法仅通过串联和并联组合简化的电路所必不可少的工具。

1. Kirchhoff’s First Law (Current Law) | 基尔霍夫第一定律（电流定律）

Kirchhoff’s first law states that the total current entering a junction is equal to the total current leaving the junction. This is a direct consequence of the conservation of electric charge.

基尔霍夫第一定律指出：流入节点的总电流等于流出节点的总电流。这是电荷守恒定律的直接推论。

ΣI₍ᵢₙ₎ = ΣI₍ₒᵤₜ₎

At any junction in a circuit, charge cannot accumulate; whatever charge flows in must flow out. For a junction with three branches carrying currents I₁, I₂ and I₃, if I₁ and I₂ flow in and I₃ flows out, then I₁ + I₂ = I₃.

在电路中任何节点处，电荷不能累积；流入的电荷必然流出。对于一个有三条支路、电流分别为 I₁、I₂ 和 I₃ 的节点，若 I₁ 和 I₂ 流入而 I₃ 流出，则 I₁ + I₂ = I₃。

Physical principle	Conservation of charge
物理原理	电荷守恒
Application	Used to find unknown currents at junctions
应用	用于求解节点处的未知电流

2. Kirchhoff’s Second Law (Voltage Law) | 基尔霍夫第二定律（电压定律）

Kirchhoff’s second law states that the sum of the electromotive forces (e.m.f.) around any closed loop equals the sum of the potential differences (p.d.) around that loop. This follows from the conservation of energy.

基尔霍夫第二定律指出：沿任意闭合回路，电动势之和等于该回路中电势差之和。这是能量守恒定律的推论。

ΣE = ΣIR

When a charge q moves around a closed loop and returns to its starting point, its electric potential energy must be unchanged. Any energy gained from sources must be fully dissipated across resistors or other circuit elements.

当一个电荷 q 绕闭合回路运动并回到起点时，其电势能必须不变。从电源获得的任何能量都必须完全消耗在电阻或其他电路元件上。

Equivalently, the algebraic sum of all potential differences and e.m.f.s in a closed loop equals zero: ΣV = 0.

等价地，闭合回路中所有电势差和电动势的代数总和为零：ΣV = 0。

3. Sign Conventions | 符号约定

Applying Kirchhoff’s laws correctly requires a consistent sign convention. The choice of direction for loop traversal is arbitrary, but once chosen, it must be maintained throughout the calculation.

正确应用基尔霍夫定律需要一致的符号约定。回路绕行方向的选择是任意的，但一旦选定，必须在整个计算过程中保持。

If current flows from the positive terminal to the negative terminal of a source, the e.m.f. is taken as positive.

若电流从电源正极流向负极，则电动势取正值。
If current flows from the negative to the positive terminal, the e.m.f. is negative.

若电流从负极流向正极，则电动势取负值。
When traversing a resistor in the direction of current flow, the p.d. is negative (voltage drop).

当沿电流方向经过电阻时，电势差为负值（电压降）。
When traversing a resistor opposite to the current direction, the p.d. is positive (voltage rise).

当逆电流方向经过电阻时，电势差为正值（电压升）。

A common alternative approach: sum all e.m.f.s with their correct signs and set this equal to the sum of all IR products. This automatically accounts for direction.

一种常见的替代方法：将所有电动势按正确符号求和，并令其等于所有 IR 乘积之和。这自动考虑了方向。

4. Analysing a Simple Two-Loop Circuit | 分析简单双回路电路

Consider a circuit with two batteries E₁ = 6V and E₂ = 4V, and three resistors: R₁ = 2Ω on the top branch, R₂ = 3Ω in the middle branch, and R₃ = 5Ω on the bottom branch.

考虑一个包含两个电池 E₁ = 6V 和 E₂ = 4V、三个电阻的电路：顶部支路 R₁ = 2Ω，中间支路 R₂ = 3Ω，底部支路 R₃ = 5Ω。

Step 1: Label currents I₁, I₂ and I₃ in each branch with assumed directions.

步骤1：在各支路标上电流 I₁、I₂ 和 I₃，并假定方向。

Step 2: Apply Kirchhoff’s first law at a junction: I₁ = I₂ + I₃ (or a variant depending on assumed directions).

步骤2：在节点处应用基尔霍夫第一定律：I₁ = I₂ + I₃（或根据假定方向变化）。

Step 3: Apply Kirchhoff’s second law to each loop, producing two independent equations.

步骤3：对每个回路应用基尔霍夫第二定律，得到两个独立方程。

Step 4: Solve the simultaneous equations to find the unknown currents.

步骤4：解联立方程组，求出未知电流。

6 = 2I₁ + 3I₂ (loop 1) and 4 = 3I₂ − 5I₃ (loop 2)

Substituting I₁ = I₂ + I₃ gives three equations with three unknowns, solvable by substitution or elimination.

代入 I₁ = I₂ + I₃ 可得含三个未知数的三个方程，可通过代入法或消元法求解。

5. Worked Example: Finding Unknown Currents | 示例：求未知电流

Problem: In the circuit above, find the currents I₁, I₂ and I₃.

题目：在上面的电路中，求电流 I₁、I₂ 和 I₃。

Solution:

解答：

From Kirchhoff’s first law: I₁ = I₂ + I₃.

由基尔霍夫第一定律：I₁ = I₂ + I₃。

From loop 1 (containing E₁, R₁ and R₂):

由回路1（包含 E₁、R₁ 和 R₂）：

6 = 2I₁ + 3I₂

From loop 2 (containing E₂, R₂ and R₃):

由回路2（包含 E₂、R₂ 和 R₃）：

4 = 3I₂ − 5I₃

Substitute I₁ = I₂ + I₃ into the first loop equation:

将 I₁ = I₂ + I₃ 代入第一回路方程：

6 = 2(I₂ + I₃) + 3I₂ = 5I₂ + 2I₃

Rearranging: 5I₂ + 2I₃ = 6 and 3I₂ − 5I₃ = 4.

整理得：5I₂ + 2I₃ = 6 和 3I₂ − 5I₃ = 4。

Solving simultaneously: multiply the first by 5 and the second by 2:

联立求解：将第一个方程乘以5，第二个方程乘以2：

25I₂ + 10I₃ = 30 and 6I₂ − 10I₃ = 8

Adding: 31I₂ = 38, so I₂ = 1.23 A. Then I₃ = (4 − 3 × 1.23) ÷ (−5) = 3.69 ÷ 5 = −0.062 A. The negative sign shows I₃ flows in the opposite direction to the assumed one.

两式相加：31I₂ = 38，故 I₂ = 1.23 A。则 I₃ = (4 − 3 × 1.23) ÷ (−5) = 3.69 ÷ 5 = −0.062 A。负号表示 I₃ 的实际方向与假定方向相反。

Finally, I₁ = I₂ + I₃ = 1.23 − 0.062 = 1.17 A.

最后，I₁ = I₂ + I₃ = 1.23 − 0.062 = 1.17 A。

6. Internal Resistance and Kirchhoff’s Laws | 内阻与基尔霍夫定律

Real batteries have internal resistance r. In Kirchhoff’s second law, a battery of e.m.f. E and internal resistance r is treated as an ideal source E in series with a resistor r.

真实电池具有内阻 r。在基尔霍夫第二定律中，电动势为 E、内阻为 r 的电池被视作理想电源 E 与电阻 r 的串联组合。

When a current I flows through a battery, the terminal potential difference is V = E − Ir. The term Ir represents the “lost volts” due to internal resistance.

当电流 I 流过电池时，端电压为 V = E − Ir。其中 Ir 项表示由于内阻造成的”损失电压”。

V = E − Ir

In loop equations, always include the p.d. across the internal resistance as an additional IR term.

在回路方程中，务必把内阻两端的电势差作为额外的 IR 项计入。

7. Combining Kirchhoff’s Laws with Series and Parallel Rules | 基尔霍夫定律与串并联规则的结合

Series and parallel resistor combinations are special cases of Kirchhoff’s laws. Series circuits apply the second law; parallel circuits apply the first law at their junctions.

串联和并联电阻组合是基尔霍夫定律的特殊情况。串联电路应用第二定律；并联电路在节点处应用第一定律。

Configuration	Key relation	Kirchhoff law used
Series	R_total = R₁ + R₂ + R₃	Second law (voltage)
串联	R_total = R₁ + R₂ + R₃	第二定律（电压）
Parallel	1/R_total = 1/R₁ + 1/R₂	First law (current)
并联	1/R_total = 1/R₁ + 1/R₂	第一定律（电流）

When a circuit contains elements that are neither purely in series nor purely in parallel, Kirchhoff’s laws must be applied directly.

当电路包含既非纯串联也非纯并联的元件时，必须直接应用基尔霍夫定律。

8. Conservation Principles | 守恒原理

Understanding the physical basis of Kirchhoff’s laws helps in remembering them and applying them correctly.

理解基尔霍夫定律的物理基础有助于记忆和正确运用它们。

First law → conservation of charge: charge cannot be created or destroyed at a junction, so the algebraic sum of currents at a junction must be zero.

第一定律 → 电荷守恒：电荷在节点处不能创生或消灭，因此节点处电流的代数和必须为零。
Second law → conservation of energy: the net change in electric potential energy around a closed loop must be zero, since the charge returns to its starting point.

第二定律 → 能量守恒：电荷绕闭合回路回到起点，电势能的净变化必须为零。

These conservation principles are universal; Kirchhoff’s laws apply to any electrical circuit, regardless of its complexity.

这些守恒原理具有普遍性；基尔霍夫定律适用于任何电路，无论其多么复杂。

9. Worked Example: Battery with Internal Resistance | 示例：含内阻的电池

Problem: A battery of e.m.f. 9.0V and internal resistance 0.5Ω is connected to two parallel resistors of 4.0Ω and 6.0Ω. Find the current supplied by the battery and the p.d. across the terminals.

题目：一个电动势为 9.0V、内阻为 0.5Ω 的电池连接到两个并联电阻 4.0Ω 和 6.0Ω 上。求电池提供的电流和端电压。

Solution:

解答：

The combined parallel resistance is:

并联等效电阻为：

1/R_parallel = 1/4 + 1/6 = 5/12 → R_parallel = 12/5 = 2.4Ω

Total circuit resistance: R_total = r + 2.4 = 0.5 + 2.4 = 2.9Ω.

电路总电阻：R_total = r + 2.4 = 0.5 + 2.4 = 2.9Ω。

Using Kirchhoff’s second law around the single loop: E = I(R_total), so:

沿单回路应用基尔霍夫第二定律：E = I(R_total)，故：

I = 9.0 ÷ 2.9 = 3.10 A

Terminal p.d.: V = E − Ir = 9.0 − (3.10 × 0.5) = 9.0 − 1.55 = 7.45 V.

端电压：V = E − Ir = 9.0 − (3.10 × 0.5) = 9.0 − 1.55 = 7.45 V。

The current divides between the parallel branches: I₄ = V ÷ 4 = 7.45 ÷ 4 = 1.86 A and I₆ = V ÷ 6 = 7.45 ÷ 6 = 1.24 A. Note I₄ + I₆ = 3.10 A, verifying the first law.

电流在并联支路间分配：I₄ = V ÷ 4 = 7.45 ÷ 4 = 1.86 A，I₆ = V ÷ 6 = 7.45 ÷ 6 = 1.24 A。注意 I₄ + I₆ = 3.10 A，验证了第一定律。

10. Common Exam Pitfalls | 常见考试陷阱

Students frequently lose marks in Kirchhoff’s law questions due to a small number of recurring errors. Being aware of these can significantly improve your exam performance.

学生在基尔霍夫定律题目中常因少数反复出现的错误而失分。了解这些错误可以显著提高考试成绩。

Incorrect sign of e.m.f.: Always check whether you are traversing the battery from positive to negative (positive e.m.f.) or negative to positive (negative e.m.f.).

电动势符号错误：始终检查你是从正极到负极（电动势为正）还是从负极到正极（电动势为负）绕行电池。
Forgetting internal resistance: The internal resistance of a battery must be included as an extra IR term in the loop equation.

忘记内阻：电池的内阻必须作为额外的 IR 项计入回路方程。
Missing the junction equation: When there are three or more currents, remember to use Kirchhoff’s first law in addition to loop equations.

遗漏节点方程：当有三个或更多电流时，记得除了回路方程外还要使用基尔霍夫第一定律。
Assuming currents before solving: If a current comes out negative, the actual direction is opposite to your assumption. This does not invalidate the calculation.

在求解前进假设电流方向：如果电流计算结果为负，则实际方向与假设相反。这并不使计算失效。
Inconsistent units: Convert all quantities to SI units (A, V, Ω) before substituting into equations.

单位不一致：在代入方程前，将所有量转换为国际单位制单位（A、V、Ω）。

11. Strategy for Solving Kirchhoff Questions | 基尔霍夫题目的解题策略

Follow this systematic approach to tackle any Kirchhoff’s laws question in the exam:

按照以下系统方法来解决考试中的任何基尔霍夫定律题目：

Step	Action	步骤	操作
1	Draw the circuit and label all currents with assumed directions	1	画出电路并标注所有电流的假设方向
2	Apply the first law at junctions to relate currents	2	在节点处应用第一定律关联各电流
3	Choose loops and apply the second law to each	3	选择回路并对每个回路应用第二定律
4	Solve the simultaneous equations	4	解联立方程组
5	Interpret negative results as reversed direction	5	将负结果理解为方向反转

For CIE A-Level questions, state the law you are applying before writing the equation. Examiners reward clear methodical working.

对于 CIE A-Level 题目，在写出方程前先说明你应用的定律。考官对清晰有条理的解题过程给予分数。

12. Practice Questions | 练习题

Test your understanding with these exam-style questions:

用以下考试风格题目测试你的理解：

Q1: A circuit contains two batteries (E₁ = 3.0V, r₁ = 1.0Ω and E₂ = 6.0V, r₂ = 2.0Ω) connected in series opposing through a 5.0Ω resistor. Calculate the current in the circuit.

题1：一个电路包含两个电池（E₁ = 3.0V，r₁ = 1.0Ω；E₂ = 6.0V，r₂ = 2.0Ω），经一个 5.0Ω 电阻反向串联。求电路中的电流。
Q2: At a junction in a circuit, currents of 2.0A and 3.5A flow in, while a current of 1.5A flows out along one branch. Find the current in the remaining branch.

题2：在电路的某个节点，2.0A 和 3.5A 的电流流入，1.5A 的电流沿一条支路流出。求其余支路中的电流。
Q3: A 12V battery with internal resistance 0.8Ω supplies a network of three resistors: 4Ω and 6Ω in parallel, and this combination in series with 3Ω. Use Kirchhoff’s laws to find the total current.

题3：一个 12V、内阻 0.8Ω 的电池为三个电阻的网络供电：4Ω 和 6Ω 并联，再与 3Ω 串联。用基尔霍夫定律求总电流。

Answers: Q1: 0.375 A; Q2: 4.0 A; Q3: 1.48 A.

答案：题1：0.375 A；题2：4.0 A；题3：1.48 A。

Mastering Kirchhoff’s laws is essential for A-Level Physics success. Remember: the first law conserves charge at junctions, the second law conserves energy around loops, and careful attention to sign conventions will prevent most common errors.

掌握基尔霍夫定律是 A-Level 物理成功的关键。记住：第一定律在节点处守恒电荷，第二定律在回路中守恒能量，仔细注意符号约定将避免大多数常见错误。

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Kirchhoff’s Laws | 基尔霍夫定律