Kirchhoff’s Laws: Key Concepts for IB and CCEA Physics | IB CCEA 物理：基尔霍夫定律考点精讲

📚 Kirchhoff’s Laws: Key Concepts for IB and CCEA Physics | IB CCEA 物理：基尔霍夫定律考点精讲

Kirchhoff’s laws are fundamental tools for analysing electrical circuits, essential for both IB and CCEA physics syllabi. They describe how current and voltage behave at junctions and around closed loops, enabling us to solve complex circuits beyond simple series and parallel combinations.

基尔霍夫定律是分析电路的基本工具，对 IB 和 CCEA 物理课程都至关重要。它们描述了电流在节点处和电压沿闭合回路的变化规律，使我们能够求解超出简单串并联的复杂电路。

1. Introduction to Kirchhoff’s Laws | 基尔霍夫定律简介

Gustav Kirchhoff formulated two rules in 1845, based on the conservation of charge and conservation of energy. These rules are known as Kirchhoff’s Current Law (KCL) and Kirchhoff’s Voltage Law (KVL).

古斯塔夫·基尔霍夫于1845年根据电荷守恒和能量守恒提出了两条规则，分别称为基尔霍夫电流定律（KCL）和基尔霍夫电压定律（KVL）。

---

2. Kirchhoff’s Current Law (KCL) – Junction Rule | 基尔霍夫电流定律（节点定则）

KCL states that the total current entering a junction equals the total current leaving that junction. Mathematically, ΣI_in = ΣI_out.

KCL 指出，流入某一节点的总电流等于流出该节点的总电流。数学表达式为 ΣI_in = ΣI_out。

This law arises from the conservation of electric charge; charge cannot accumulate or disappear at a node.

该定律源于电荷守恒；电荷不能在节点处积聚或消失。

---

3. Applying KCL: Conservation of Charge | 应用 KCL：电荷守恒

Consider a junction where three currents meet: I₁ = 2 A entering, I₂ = 3 A entering, and I₃ = 5 A leaving. Verify KCL: 2 + 3 = 5, so it holds.

考虑一个节点，有三股电流汇合：I₁ = 2 A 流入，I₂ = 3 A 流入，I₃ = 5 A 流出。验证 KCL：2 + 3 = 5，符合定律。

If the currents were not balanced, that would imply creation or destruction of charge, which is impossible.

如果电流不平衡，就意味着电荷的产生或消失，这是不可能的。

---

4. Kirchhoff’s Voltage Law (KVL) – Loop Rule | 基尔霍夫电压定律（回路定则）

KVL states that the sum of the electromotive forces (emfs) and potential differences (p.d.) around any closed loop in a circuit is zero. Written as ΣV = 0.

KVL 指出，沿电路中任一闭合回路，电动势（emf）和电势差（p.d.）的代数和为零。写作 ΣV = 0。

This is a consequence of energy conservation: the net work done on a unit charge moving around a loop is zero.

这是能量守恒的结果：单位电荷环绕回路一周所做的净功为零。

---

5. Applying KVL: Conservation of Energy | 应用 KVL：能量守恒

When traversing a loop, increases in potential (e.g., crossing a battery from – to +) are taken as positive, and drops (e.g., across a resistor in the direction of current) as negative, or vice versa, as long as consistency is maintained.

在沿回路行进时，电势升高（例如从电池负极到正极）记为正值，电势下降（例如沿电流方向经过电阻）记为负值，也可反过来规定，但需保持一致。

Summing all signed voltages around the loop yields zero, ensuring the energy supplied equals energy dissipated.

将回路中所有带符号的电压相加结果为零，保证提供的能量等于消耗的能量。

---

6. Sign Conventions for KVL | KVL 的正负号规则

Consistent sign conventions are essential. The table below summarises a common convention using the loop direction method.

一致的符号规则至关重要。下表总结了使用回路方向法的常用规则。

Element	Condition	Voltage Value (Convention)
Resistor	Loop direction same as current	–IR
Resistor	Loop direction opposite to current	+IR
Battery / EMF	Loop direction from – to +	+ε
Battery / EMF	Loop direction from + to –	–ε

For example, if you traverse a resistor with current, the potential drops, hence –IR. If against current, the potential rises, +IR.

例如，沿电流方向经过电阻，电势降低，故为 –IR；若逆电流方向，电势升高，为 +IR。

---

7. Solving Circuit Problems Using KCL and KVL | 使用 KCL 和 KVL 求解电路问题

To solve for unknown currents or voltages, label all currents in the circuit and choose loop directions. Apply KCL at junctions to reduce unknowns. Then apply KVL to independent loops, obtaining a system of equations. Solve simultaneously.

求解未知电流或电压时，标出电路中所有电流并选择回路方向。在节点处应用 KCL 以减少未知量。然后对独立回路应用 KVL，得到方程组，联立求解。

Always check that the number of equations matches the number of unknowns. Use Ohm’s law to relate voltage and current across resistors.

始终确保方程数量与未知数数量匹配。利用欧姆定律联系电阻上的电压和电流。

---

8. Worked Example: Single-Loop Circuit | 例题：单回路电路

A single-loop circuit contains a 12 V battery and two resistors in series: R₁ = 4 Ω and R₂ = 8 Ω. Determine the current.

一个单回路电路包含一个 12 V 电池和两个串联电阻：R₁ = 4 Ω，R₂ = 8 Ω。求电流。

Using KVL around the loop clockwise: +12 V – I×4 Ω – I×8 Ω = 0 → 12 = 12I → I = 1 A.

顺时针绕行回路应用 KVL：+12 V – I×4 Ω – I×8 Ω = 0 → 12 = 12I → I = 1 A。

---

9. Worked Example: Multi-Loop Circuit (Two Loops) | 例题：多回路电路（双回路）

Consider a circuit with two loops sharing a middle branch. Battery 1: ε₁ = 10 V, Battery 2: ε₂ = 5 V, resistors R₁ = 2 Ω, R₂ = 4 Ω, R₃ = 3 Ω. Use KCL and KVL to find currents I₁, I₂, I₃.

考虑一个双回路电路，中间支路共用。电池1：ε₁ = 10 V，电池2：ε₂ = 5 V，电阻 R₁ = 2 Ω，R₂ = 4 Ω，R₃ = 3 Ω。利用 KCL 和 KVL 求电流 I₁、I₂、I₃。

Assign currents: I₁ through R₁ from ε₁, I₂ through R₂ from ε₂, and I₃ through R₃ upwards at the junction. KCL at top junction: I₁ + I₂ = I₃.

设电流：I₁ 从 ε₁ 流经 R₁，I₂ 从 ε₂ 流经 R₂，I₃ 在节点处向上流经 R₃。顶部节点 KCL：I₁ + I₂ = I₃。

KVL left loop (clockwise): +10 – 2I₁ – 3I₃ = 0 → 10 – 2I₁ – 3(I₁+I₂) = 0 → 10 – 5I₁ – 3I₂ = 0

左回路 KVL（顺时针）：+10 – 2I₁ – 3I₃ = 0 → 10 – 2I₁ – 3(I₁+I₂) = 0 → 10 – 5I₁ – 3I₂ = 0

KVL right loop (clockwise): +5 – 4I₂ – 3I₃ = 0 → 5 – 4I₂ – 3(I₁+I₂) = 0 → 5 – 3I₁ – 7I₂ = 0

右回路 KVL（顺时针）：+5 – 4I₂ – 3I₃ = 0 → 5 – 4I₂ – 3(I₁+I₂) = 0 → 5 – 3I₁ – 7I₂ = 0

Solve the two equations: from first, 5I₁ = 10 – 3I₂ → I₁ = 2 – 0.6I₂. Substitute into second: 5 – 3(2 – 0.6I₂) – 7I₂ = 0 → 5 – 6 + 1.8I₂ – 7I₂ = 0 → –1 – 5.2I₂ = 0 → I₂ = –0.192 A. Then I₁ = 2 – 0.6(–0.192) = 2.115 A, I₃ = I₁ + I₂ = 1.923 A. The negative sign for I₂ means actual direction is opposite to assumed.

解方程组：由第一个方程得 5I₁ = 10 – 3I₂ → I₁ = 2 – 0.6I₂。代入第二个：5 – 3(2 – 0.6I₂) – 7I₂ = 0 → 5 – 6 + 1.8I₂ – 7I₂ =

Published by TutorHao | IB Physics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）