📚 Kirchhoff’s Laws | 基尔霍夫定律
Kirchhoff’s laws are two fundamental rules that describe how current and voltage behave in electrical circuits. They are named after the German physicist Gustav Kirchhoff and are essential for analysing more complex circuits than simple series or parallel ones. These laws are based on the principles of conservation of charge and conservation of energy, and they give us a systematic way to find unknown currents and potential differences.
基尔霍夫定律是描述电路中电流和电压行为的两条基本规则。它们以德国物理学家古斯塔夫·基尔霍夫的名字命名,对于分析比简单串联或并联更复杂的电路至关重要。这些定律基于电荷守恒和能量守恒的原理,为我们提供了一种系统的方法来求解未知的电流和电位差。
1. Charge Conservation and Current | 电荷守恒与电流
Electric current is the flow of electric charge. In a metallic conductor, this charge is carried by electrons. The principle of conservation of charge states that charge cannot be created or destroyed; it can only move from one place to another. This is the foundation of Kirchhoff’s first law.
电流是电荷的流动。在金属导体中,电荷由电子携带。电荷守恒原理指出,电荷不能被创造或消灭,只能从一个地方移动到另一个地方。这是基尔霍夫第一定律的基础。
2. Kirchhoff’s First Law (Junction Rule) | 基尔霍夫第一定律(节点定律)
Kirchhoff’s first law states that at any junction (node) in a circuit, the total current entering the junction must equal the total current leaving the junction. Mathematically, this can be written as:
基尔霍夫第一定律指出,在电路中的任一节点处,流入节点的总电流必定等于流出节点的总电流。用数学形式可表达为:
ΣIin = ΣIout
This law is a direct consequence of charge conservation. If more charge were to flow into a point than out of it, charge would accumulate there, which does not happen in a steady circuit.
该定律是电荷守恒的直接结果。如果流入某点的电荷比流出的多,电荷就会在那里积累,这在稳定电路中是不会发生的。
3. Applying the First Law in Circuits | 第一定律在电路中的应用
In a simple series circuit, there is only one path for current, so the current is the same everywhere. The first law is automatically satisfied. In parallel circuits, the current splits at junctions. For example, if a current of 3 A enters a junction and splits into two branches, the sum of the branch currents must equal 3 A (e.g., 1 A + 2 A).
在简单的串联电路中,电流只有一条路径,所以各处电流相同,第一定律自然满足。在并联电路中,电流在节点处分流。例如,如果 3 A 的电流流进一个节点并分成两条支路,那么支路电流之和必定等于 3 A(例如 1 A + 2 A)。
4. Energy Conservation and Potential Difference | 能量守恒与电位差
Potential difference (voltage) is a measure of the energy transferred per unit charge between two points in a circuit. The law of conservation of energy tells us that the total energy gained by charges as they pass through a source (like a battery) must equal the total energy lost by the charges as they pass through components in the external circuit.
电位差(电压)是电路中两点间单位电荷转移能量的量度。能量守恒定律告诉我们,电荷通过电源(如电池)时获得的总能量必须等于电荷在外电路元件中损失的总能量。
5. Kirchhoff’s Second Law (Loop Rule) | 基尔霍夫第二定律(回路定律)
Kirchhoff’s second law states that around any closed loop in a circuit, the sum of the electromotive forces (e.m.f.s) is equal to the sum of the potential differences across the components. This can be expressed as:
基尔霍夫第二定律指出,沿电路中任一闭合回路,电动势之和等于各元件两端电位差之和。可表达为:
Σε = Σ(V1 + V2 + …)
where ε represents the e.m.f. of a source. This law follows from energy conservation: the energy supplied by the battery is totally converted into other forms in the circuit components.
其中 ε 表示电源的电动势。该定律来自能量守恒:电池提供的能量在电路元件中全部转化为其他形式的能量。
6. Applying the Second Law in Circuits | 第二定律在电路中的应用
In a series circuit with a single battery and two resistors, the sum of the voltages across the resistors equals the battery voltage. For instance, if a 12 V battery is connected to two resistors and one has a voltage drop of 4 V, the other must have 8 V. In a parallel circuit, each branch is connected directly to the battery, so the voltage across each branch equals the battery e.m.f.
在由一个电池和两个电阻组成的串联电路中,各电阻两端电压之和等于电池电压。例如,一个 12 V 电池连接两个电阻,一个电阻的电压降为 4 V,另一个必定为 8 V。在并联电路中,每条支路直接与电池相连,因此每条支路的电压都等于电池电动势。
7. Series Circuits with Kirchhoff’s Laws | 串联电路与基尔霍夫定律
In a series circuit, the current is constant throughout by the first law. By the second law, the total supplied voltage Vtotal is divided among the components:
在串联电路中,根据第一定律,电流处处相同。根据第二定律,总供电电压 Vtotal 分配在各元件上:
Vtotal = V1 + V2 + V3 + …
The resistance of the whole circuit is the sum of individual resistances: Rtotal = R1 + R2 + R3 + … . This can be derived from the second law and Ohm’s law (V = IR).
整个电路的电阻等于各个电阻之和:Rtotal = R1 + R2 + R3 + …。这可以通过第二定律和欧姆定律 (V = IR) 推导出来。
8. Parallel Circuits with Kirchhoff’s Laws | 并联电路与基尔霍夫定律
In a parallel circuit, the first law tells us that the total current from the source equals the sum of the currents in the branches. The second law tells us that the voltage across each branch is the same and equals the source voltage. Therefore:
在并联电路中,第一定律告诉我们,电源提供的总电流等于各支路电流之和。第二定律告诉我们,各支路两端的电压相同,等于电源电压。因此:
Itotal = I1 + I2 + I3 + …
Vtotal = V1 = V2 = V3 = …
For resistance, the reciprocal relationship 1/Rtotal = 1/R1 + 1/R2 + … is the consequence of applying both laws and Ohm’s law.
对于电阻,倒数关系 1/Rtotal = 1/R1 + 1/R2 + … 是应用这两条定律和欧姆定律的结果。
9. Worked Example: Series Circuit | 例题:串联电路
A 9 V battery is connected in series with a 10 Ω resistor and a 20 Ω resistor. Apply Kirchhoff’s laws to find the current and the voltage across each resistor.
一个 9 V 电池与一个 10 Ω 电阻和一个 20 Ω 电阻串联。应用基尔霍夫定律求电流和每个电阻两端的电压。
Solution:
Total resistance Rtotal = 10 Ω + 20 Ω = 30 Ω.
By Ohm’s law, current I = V / Rtotal = 9 V / 30 Ω = 0.3 A. (The current is the same everywhere by 1st law.)
Voltage across 10 Ω: V1 = I × R1 = 0.3 A × 10 Ω = 3 V.
Voltage across 20 Ω: V2 = I × R2 = 0.3 A × 20 Ω = 6 V.
Check by 2nd law: V1 + V2 = 3 V + 6 V = 9 V, which equals the battery voltage.
解:总电阻 Rtotal = 10 Ω + 20 Ω = 30 Ω。
由欧姆定律,电流 I = V / Rtotal = 9 V / 30 Ω = 0.3 A。(根据第一定律,电流处处相同。)
10 Ω 电阻两端电压:V1 = I × R1 = 0.3 A × 10 Ω = 3 V。
20 Ω 电阻两端电压:V2 = I × R2 = 0.3 A × 20 Ω = 6 V。
用第二定律检验:V1 + V2 = 3 V + 6 V = 9 V,等于电池电压。
10. Worked Example: Parallel Circuit | 例题:并联电路
Two resistors of 6 Ω and 12 Ω are connected in parallel across a 12 V battery. Use Kirchhoff’s laws to determine the current through each resistor and the total current from the battery.
两个分别为 6 Ω 和 12 Ω 的电阻并联在一个 12 V 电池上。利用基尔霍夫定律求通过每个电阻的电流和电池提供的总电流。
Solution:
The voltage across each branch equals the battery voltage (2nd law): V = 12 V.
Current through 6 Ω resistor: I1 = V / R1 = 12 V / 6 Ω = 2 A.
Current through 12 Ω resistor: I2 = V / R2 = 12 V / 12 Ω = 1 A.
Total current by 1st law: Itotal = I1 + I2 = 2 A + 1 A = 3 A.
(Alternatively, equivalent resistance Req = (1/6 + 1/12)⁻¹ = 4 Ω; Itotal = 12 V / 4 Ω = 3 A, confirming the result.)
解:各支路电压等于电池电压(第二定律):V = 12 V。
通过 6 Ω 电阻的电流:I1 = V / R1 = 12 V / 6 Ω = 2 A。
通过 12 Ω 电阻的电流:I2 = V / R2 = 12 V / 12 Ω = 1 A。
由第一定律得总电流:Itotal = I1 + I2 = 2 A + 1 A = 3 A。
(亦可求等效电阻 Req = (1/6 + 1/12)⁻¹ = 4 Ω;Itotal = 12 V / 4 Ω = 3 A,验证了结果。)
11. Common Misconceptions and Pitfalls | 常见误解与易错点
A common mistake is to think that current is ‘used up’ as it passes through circuit elements. Kirchhoff’s first law reinforces that current is conserved at junctions and does not decrease along a series circuit. Another error is forgetting that voltage in a parallel circuit is the same across all branches, even if the branch resistances are different. Always check that the sum of voltages around a closed loop matches the e.m.f., and that the sum of currents at a junction is zero when direction signs are taken into account.
一个常见误解是认为电流在通过电路元件时会被“用掉”。基尔霍夫第一定律强调电流在节点处是守恒的,不会在串联电路中沿路减小。另一个错误是忘记并联电路中各支路两端电压相同,即使支路电阻不同。始终记得检查闭合回路中电压之和是否等于电动势,以及在节点处考虑方向符号后电流之和为零。
12. Summary and Exam Tips | 总结与考试技巧
Kirchhoff’s laws are powerful tools for circuit analysis. In the exam, you may be asked to state the laws, apply them to find missing values in a circuit diagram, or use them to justify familiar rules for series and parallel circuits. Remember the two laws: current in = current out at a junction; and sum of e.m.f.s = sum of p.d.s around a loop. Practise by drawing circuits and labelling currents and voltages before applying the equations. Always show your working clearly, especially when using Ohm’s law alongside Kirchhoff’s laws.
基尔霍夫定律是电路分析的有力工具。考试中,你可能需要陈述这些定律,应用它们找出电路图中缺失的数值,或用它们证明串并联电路中的熟规律。请记住这两条定律:节点处流入电流等于流出电流;在回路中,电动势之和等于各段电位差之和。练习绘制电路并标出电流和电压,然后应用方程。务必清晰展示解题步骤,特别是将欧姆定律与基尔霍夫定律结合使用时。
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