📚 Kirchhoff's Second Law: The Loop Rule Explained | 基尔霍夫第二定律：回路电压规则详解
Kirchhoff's second law is a fundamental circuit rule that follows directly from conservation of energy. It is sometimes called the loop rule or Kirchhoff's voltage law (KVL). For any closed loop in a circuit, the total energy gained per unit charge by the moving charges must equal the total energy lost per unit charge. This article explains the law, its sign conventions, worked examples, and common exam pitfalls for CIE A-Level Physics.
基尔霍夫第二定律是一条由能量守恒直接推出的基本电路定律，也称为回路规则或基尔霍夫电压定律(KVL)。对于电路中任意闭合回路，运动电荷每单位电荷获得的总能量必定等于每单位电荷失去的总能量。本文讲解该定律、符号约定、例题以及 CIE A-Level 物理考试中的常见失分点。
1. Statement of the Law | 定律表述
Around any closed loop in a circuit, the algebraic sum of the electromotive forces (e.m.f.s) is equal to the algebraic sum of the potential differences across the components.
在电路中，沿任意闭合回路，电动势(e.m.f.)的代数和等于各元件两端电势差的代数和。
Σ ε = Σ I R
An equivalent form is that the sum of all the potential changes (rises and drops) around a closed loop is zero, provided each change is given the correct sign.
与之等价的另一种表述是：沿闭合回路绕行一周，所有电势变化（升与降）的代数和为零，只要每个变化都赋予了正确的符号。
Σ V = 0
The word "algebraic" is essential. E.m.f.s and potential differences are directed quantities in this calculation: each one must be given a plus or minus sign according to the direction in which you choose to travel around the loop.
"代数"二字非常关键。在运算中，电动势和电势差都是有方向的量：必须根据你所选择的绕行方向，为每一个量标上正号或负号。
The law applies to any closed loop — a simple single circuit or one loop inside a complicated network — and it holds for direct current and for the instantaneous values of alternating current.
该定律适用于任意闭合回路——既可以是简单的单一电路，也可以是复杂网络中的某一个回路；它既适用于直流，也适用于交流电的瞬时值。
2. Where the Law Comes From: Conservation of Energy | 定律的来源：能量守恒
The law is not an independent rule invented to solve problems; it is a direct consequence of the conservation of energy. Charge carriers gain energy when they pass through a source of e.m.f. and lose energy when they pass through a resistor or any other component.
这条定律并不是为了解题而人为发明的独立规则，而是能量守恒的直接结果。载流子通过电源时获得能量，通过电阻或其他元件时失去能量。
Since a charge that travels once around a closed loop returns to the same point with the same potential energy it started with, the net energy transferred per unit charge must be zero. The whole loop is a closed energy circuit.
由于电荷绕闭合回路一周后回到起点，其电势能与出发时完全相同，因此每单位电荷的净能量转移必然为零。整个回路构成一个闭合的能量循环。
wo statements Σ ε = Σ I R and Σ V = 0 are the same law written in two ways. The first says "energy gained equals energy lost"; the second says "energy gained minus energy lost equals zero".
这正是为什么 Σ ε = Σ I R 与 Σ V = 0 是同一规律的两种写法。前者说"获得的能量等于失去的能量"，后者说"获得的能量减去失去的能量等于零"。
Kirchhoff's second law therefore applies to every loop of a network, no matter how complex, because energy conservation cannot be violated anywhere in the circuit.
因此，无论网络多么复杂，基尔霍夫第二定律对每一个回路都成立，因为电路的任何部分都不能违背能量守恒。
3. Two Equivalent Forms of the Law | 定律的两种等价形式
rm, Σ ε = Σ I R, is usually quickest for a single loop containing a battery and resistors.
考官接受两种形式中的任何一种，因此你应当能熟练地在两者之间转换。第一种形式 Σ ε = Σ I R 对于只含一个电源和若干电阻的单回路通常最快。
l and is the natural choice for networks with two or more loops, because you simply walk round a loop adding rises and subtracting drops until you return to the start.
第二种形式 Σ V = 0 更具一般性，是含两个或更多回路的网络的自然选择，因为你只需绕回路一周，把电势升相加、电势降相减，直到回到起点。
| Form | 形式 | Equation | 方程 | Best used for | 适用情形 |
| Energy gained = energy lost | 获得能量 = 失去能量 | Σ ε = Σ I R | Single loops, series circuits | 单回路、串联电路 |
| Net potential change = zero | 净电势变化 = 零 | Σ V = 0 | Multi-loop networks, simultaneous equations | 多回路网络、联立方程 |
Whichever form you use, the physical content is identical: the net work done on each unit of charge in one complete trip around the loop is zero.
无论使用哪种形式，其物理内涵完全相同：单位电荷绕回路一整圈所获得的净功为零。
4. Sign Conventions: The Rules You Must Apply | 符号约定：必须遵守的规则
Most marks lost on this topic come from sign errors, not from the physics. Choose a direction round the loop — clockwise is conventional — and then apply the rules consistently.
这一考点失分大多来自符号错误，而不是物理理解。先选定绕行方向（通常取顺时针），然后始终一致地应用规则。
- A source traversed from the negative terminal to the positive terminal gives a positive e.m.f. (+ε). | 由负极到正极穿过电源，电动势取正（+ε）。
- A source traversed from the positive terminal to the negative terminal gives a negative e.m.f. (−ε). | 由正极到负极穿过电源，电动势取负（−ε）。
- A resistor traversed in the direction of the current gives a potential drop (−I R). | 顺着电流方向穿过电阻，电势下降（−I R）。
- A resistor traversed against the direction of the current gives a potential rise (+I R). | 逆着电流方向穿过电阻，电势上升（+I R）。
| Element | 元件 | Traversal | 绕行方式 | Sign in Σ V = 0 | 在 Σ V = 0 中的符号 |
| Cell or battery | 电源 | − to + | 由 − 到 + | + ε |
| Cell or battery | 电源 | + to − | 由 + 到 − | − ε |
| Resistor | 电阻 | with the current | 顺电流 | − I R |
| Resistor | 电阻 | against the current | 逆电流 | + I R |
If you reverse your chosen direction of travel, every sign flips and the equation is simply multiplied by −1, so the physical prediction is unchanged. Consistency matters far more than the direction you pick.
如果你把选定的绕行方向反过来，所有符号都会变号，方程只不过整体乘以 −1，物理结论不变。符号的一致性远比绕行方向的选择重要。
5. E.M.F., Terminal P.D. and Lost Volts | 电动势、端电压与损失电压
A real cell has internal resistance r, so some energy per unit charge is dissipated inside the cell itself. This is why the terminal potential difference of a cell on load is always less than its e.m.f.
真实的电源具有内阻 r，因此有一部分单位电荷的能量在电源内部耗散。这就是接通负载时电源的端电压总是小于其电动势的原因。
ε = I (R + r) = I R + I r
Here ε is the e.m.f. of the cell, I R is the potential difference across the external resistance (the terminal p.d.), and I r is the "lost volts" dropped inside the cell.
其中 ε 是电源电动势，I R 是外电阻两端的电势差（即端电压），I r 是消耗在电源内部的"损失电压"。
terminal p.d. = ε − I r
hoff form. When the cell is on open circuit, I = 0 and the terminal p.d. equals the e.m.f.
对整个电路应用回路规则可得 ε = I R + I r，这正是同一结论的基尔霍夫表述形式。当电源处于开路状态时，I = 0，端电压等于电动势。
6. Applying KVL to Series and Parallel Circuits | 串联与并联电路中的应用
rrent is the same through every component, so the loop rule gives ε = I R₁ + I R₂ + I R₃ + … . Dividing by I shows immediately why resistances in series add.
在串联电路中，通过每个元件的电流相同，因此回路规则给出 ε = I R₁ + I R₂ + I R₃ + … 。两边同除以 I，就立刻可以看出串联电阻相加的原因。
Rtotal = R₁ + R₂ + R₃ (series)
In a parallel circuit the branches share the same potential difference. Applying the loop rule to each branch separately shows that the p.d. across each branch is identical, while the currents divide in inverse proportion to the branch resistances.
在并联电路中，各支路分享相同的电势差。对每条支路分别应用回路规则可以看出，各支路两端的电势差相同，而电流则按支路电阻的反比分配。
1 / Rtotal = 1 / R₁ + 1 / R₂ (parallel)
A useful check: add the potential drops across the components of one branch and the p.d. across the remaining part of the loop in series; the total must equal the net e.m.f. driving that loop.
一个实用的检验方法：把某条支路上各元件的电势降与该回路其余串联部分的电势差相加，其总和必须等于驱动该回路的净电动势。
7. Worked Example 1: Single Loop with Internal Resistance | 例题一：含内阻的单回路
A cell of e.m.f. 12 V and internal resistance 1.0 Ω is connected in series with resistors of 4.0 Ω and 7.0 Ω. Find the current, the p.d. across each resistor, and verify the loop rule.
一个电动势为 12 V、内阻为 1.0 Ω 的电源与 4.0 Ω 和 7.0 Ω 的电阻串联。求电流、各电阻两端的电势差，并验证回路规则。
Step 1: total resistance = 4.0 + 7.0 + 1.0 = 12 Ω. Step 2: current I = ε / Rtotal = 12 / 12 = 1.0 A.
第一步：总电阻 = 4.0 + 7.0 + 1.0 = 12 Ω。第二步：电流 I = ε / Rtotal = 12 / 12 = 1.0 A。
: I R = 1.0 × 4.0 = 4.0 V. Across 7.0 Ω: 1.0 × 7.0 = 7.0 V. Lost volts: 1.0 × 1.0 = 1.0 V.
第三步：各电势差。4.0 Ω 上：I R = 1.0 × 4.0 = 4.0 V；7.0 Ω 上：1.0 × 7.0 = 7.0 V；损失电压：1.0 × 1.0 = 1.0 V。
e terminal p.d. of the cell is 11 V, and the power wasted inside the cell is I² r = 1.0 W.
检验：4.0 V + 7.0 V + 1.0 V = 12 V = ε，满足 Σ ε = Σ I R。电源端电压为 11 V，电源内部损耗的功率为 I² r = 1.0 W。
8. Worked Example 2: Two Cells in One Loop | 例题二：同一回路中的两个电源
A 6.0 V cell (internal resistance 0.50 Ω) drives a circuit containing a 2.0 V cell connected so that it opposes the first, with internal resistance 0.50 Ω, in series with a 4.0 Ω resistor. Find the current.
一个 6.0 V 电源（内阻 0.50 Ω）驱动一个电路，电路中还接有一个反向连接的 2.0 V 电源（内阻 0.50 Ω），并与一个 4.0 Ω 电阻串联。求电流。
= 4.0 V. The total resistance is 4.0 + 0.50 + 0.50 = 5.0 Ω, giving I = 4.0 / 5.0 = 0.80 A.
两个电动势在回路中的方向相反，因此净电动势为 6.0 − 2.0 = 4.0 V。总电阻为 4.0 + 0.50 + 0.50 = 5.0 Ω，得 I = 4.0 / 5.0 = 0.80 A。
oop rule is satisfied. If the cells were instead connected to assist one another, the net e.m.f. would be 8.0 V and the current 1.6 A.
验证：6.0 − 2.0 − 0.80 × (4.0 + 0.50 + 0.50) = 6.0 − 2.0 − 4.0 = 0，回路规则成立。若两个电源改为互相增强，则净电动势为 8.0 V，电流为 1.6 A。
Note that the 2.0 V cell is being charged in this arrangement, so energy per unit charge is being supplied to it rather than delivered by it. The sign convention in the equation captures this automatically.
注意在此连接方式下，2.0 V 电源正在被充电，因此单位电荷的能量是输入给它，而不是由它输出。方程中的符号约定会自动体现这一点。
9. Worked Example 3: Two Loops and a Parallel Branch | 例题三：双回路与并联支路
A battery of e.m.f. 7.0 V and internal resistance 0.50 Ω is connected to a 4.0 Ω resistor in parallel with a 12 Ω resistor. Find the current from the battery, the terminal p.d. and the branch currents.
一个电动势 7.0 V、内阻 0.50 Ω 的电池接在并联的 4.0 Ω 与 12
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图