📚 B.5 Current and Circuits: SL Answers and Concept Explanations | IB 物理 B.5 电流与电路 SL 答案概念解析
IB Physics Standard Level Topic B.5: Current and Circuits builds the foundation for understanding how charge moves, how potential difference drives current, and how circuit components behave in series and parallel networks. This article explains the core concepts behind typical SL exam questions, clarifying the reasoning that leads to correct answers. Each section pairs English and Chinese explanations to help you master both the terminology and the underlying physics.
IB 物理标准水平主题 B.5：电流与电路为理解电荷如何运动、电势差如何驱动电流以及电路元件在串联和并联网络中如何表现奠定基础。本文解析典型 SL 考题背后的核心概念，阐明通向正确答案的推理过程。每个部分配对中英文解释，助你同时掌握术语和物理本质。
1. Charge Carriers and Electric Current | 电荷载流子与电流
Electric current is the net rate of flow of charge. In metallic conductors, free electrons drift under an electric field; in electrolytes, both positive and negative ions contribute. The conventional current direction is defined as the direction of positive charge flow, opposite to the electron drift velocity. Current I is measured in amperes (A), where 1 A = 1 C s⁻¹. The fundamental relation is:
I = Δq / Δt
电流是电荷的净流动速率。金属导体中自由电子在电场下漂移；电解液中正负离子都参与。习惯电流方向定义为正电荷流动方向，与电子漂移方向相反。电流 I 以安培 (A) 为单位，1 A = 1 C s⁻¹。基本关系式为 I = Δq / Δt。
For a given total charge, the number of elementary charges can be found using e = 1.60 × 10⁻¹⁹ C. If a wire carries a steady current of 2.0 A, the charge passing per minute is 120 C, corresponding to 7.5 × 10²⁰ electrons. This type of conversion appears frequently in SL multiple‑choice questions.
给定总电荷量，可用 e = 1.60 × 10⁻¹⁹ C 求元电荷数目。若导线载有 2.0 A 恒定电流，则每分钟通过电荷 120 C，相当于 7.5 × 10²⁰ 个电子。这类转换常出现在 SL 选择题中。
2. Potential Difference and Electromotive Force | 电势差与电动势
Potential difference (p.d.) between two points is the work done per unit charge to move charge from one point to the other. It is measured in volts (V), where 1 V = 1 J C⁻¹. For any component, V = W / q. The electromotive force (emf, symbol ε) of a source is the total energy converted into electrical energy per unit charge delivered to the complete circuit. Emf is not a force but a voltage; it represents the maximum potential difference a source can provide when no current flows.
电势差（电压）是将电荷从一点移动到另一点时每单位电荷所做的功，单位为伏特 (V)，1 V = 1 J C⁻¹。对任意元件有 V = W / q。电源的电动势 (emf, 符号 ε) 是每单位电荷传递到整个电路时转化为电能的总能量。Emf 不是力而是电压；它代表无电流时电源能提供的最大电势差。
When a cell delivers current, its terminal voltage drops due to internal resistance. The relationship Vterminal = ε − Ir is central to SL problems involving real cells. Understanding the difference between emf and terminal p.d. is essential for correct graph interpretation and circuit calculations.
当电池输出电流时，由于内阻，其端电压会下降。关系式 V端 = ε − Ir 是涉及真实电池的 SL 题目的核心。理解 emf 与端电压的区别对正确解读图像和电路计算至关重要。
3. Ohm’S Law and Resistance | 欧姆定律与电阻
For an ohmic conductor maintained at constant temperature, the current through it is directly proportional to the potential difference across it. The ratio V / I is constant and is defined as the resistance R. Ohm’s law can be written as:
V = I R
对于保持恒温的欧姆导体，通过它的电流与其两端电势差成正比。比值 V / I 恒定，定义为电阻 R。欧姆定律可写为 V = I R。
Resistance is measured in ohms (Ω). Not all components obey Ohm’s law; filament lamps and diodes are non‑ohmic because their resistance changes with temperature or applied voltage. In SL exams, you must be able to identify ohmic and non‑ohmic behaviour from I–V graphs.
电阻单位是欧姆 (Ω)。并非所有元件都遵从欧姆定律；灯丝灯泡和二极管的电阻随温度或外加电压而变，故为非欧姆元件。SL 考试中须能从 I–V 曲线图识别欧姆和非欧姆行为。
4. Resistivity and Conductivity | 电阻率与电导率
The resistance of a uniform wire is determined by its material and geometry:
R = ρ L / A
均匀导线的电阻由其材料和几何形状决定：R = ρ L / A。
Here ρ is the resistivity (unit Ω m), L the length and A the cross‑sectional area. Resistivity is an intrinsic material property; metals have low ρ (copper ~1.7×10⁻⁸ Ω m), while insulators have very high ρ. Conductivity σ is the reciprocal of resistivity, σ = 1/ρ. For metals, resistivity increases with temperature because lattice ions vibrate more, scattering electrons more frequently. In semiconductors, resistivity decreases with temperature as more charge carriers are liberated. This temperature dependence explains why a filament bulb’s resistance rises when it brightens.
其中 ρ 为电阻率（单位 Ω·m），L 为长度，A 为横截面积。电阻率是材料的本征属性；金属 ρ 较低（铜约 1.7×10⁻⁸ Ω·m），绝缘体 ρ 极高。电导率 σ 是电阻率的倒数，σ = 1/ρ。对金属，温度升高电阻率增大，因为晶格离子振动加剧，更频繁地散射电子。半导体中电阻率随温度升高而降低，因释放更多载流子。这一温度依赖关系解释了为何灯丝灯泡变亮时电阻升高。
5. Electrical Power and Energy Dissipation | 电功率与能量耗散
The power P converted in a circuit component is the product of the current through it and the potential difference across it: P = I V. For purely resistive loads, Ohm’s law allows alternative expressions:
P = I V = I² R = V² / R
电路元件中转换的功率 P 为通过电流与两端电势差的乘积：P = I V。对纯电阻负载，欧姆定律可导出另式：P = I V = I² R = V² / R。
Energy dissipated E = P t, where t is time. In SL problems, you may be asked to compare the brightness of bulbs in different configurations or to choose a suitable power rating for a resistor. For example, a 10 Ω resistor carrying 0.5 A dissipates 2.5 W, so a 5 W resistor would be safe. The kilowatt‑hour (kW h) is a commercial energy unit equal to 3.6 MJ; it often appears in questions linking power, time and energy cost.
耗散能量 E = P t，t 为时间。SL 题目可能要求比较不同配置下灯泡的亮度，或为电阻选择合适的额定功率。例如，10 Ω 电阻流过 0.5 A 时耗散 2.5 W，因此选用 5 W 电阻可安全工作。千瓦时 (kW h) 是商用能量单位，等于 3.6 MJ；常见于将功率、时间与电费联系起来的问题中。
6. Series Circuits: Current and Voltage Division | 串联电路：电流与电压分配
When components are connected in series, the current through each is identical. The total resistance is the sum:
Rtotal = R₁ + R₂ + R₃ + …
元件串联时，通过各元件的电流相同。总电阻为各电阻之和：R总 = R₁ + R₂ + R₃ + …
The supply voltage divides across the resistors in proportion to their resistance: V₁ = I R₁, V₂ = I R₂, and Vsupply = V₁ + V₂. This is the basis of potential divider circuits. If one component fails as an open circuit, the entire series loop is broken and all currents stop. In SL answers, always check that the voltages add up to the source emf or terminal voltage. A common pitfall is forgetting that the p.d. across a wire of negligible resistance is zero.
电源电压按电阻比例分配：V₁ = I R₁，V₂ = I R₂，且 V电源 = V₁ + V₂。这是分压电路的基础。若一个元件断路，整个串联回路断开，所有电流停止。SL 答卷中，务必核实各电压之和等于电源 emf 或端电压。常见误区是忽略电阻可忽略的导线两端电势差为零。
7. Parallel Circuits: Voltage and Current Division | 并联电路：电压与电流分配
In a parallel arrangement, each branch experiences the same potential difference V. The total current from the source is the sum of branch currents:
Itotal = I₁ + I₂ + I₃ + …
并联结构中，各支路两端电势差 V 相同。电源输出总电流为各支路电流之和：I总 = I₁ + I₂ + I₃ + …
The reciprocal total resistance is given by 1/Rtotal = 1/R₁ + 1/R₂ + … . For two resistors in parallel, a convenient form is Rtotal = (R₁R₂)/(R₁+R₂). Adding more branches reduces the total resistance and increases the total current drawn from the source. Branch currents divide inversely with resistance: I₁ / I₂ = R₂ / R₁. A useful answer check: the smaller resistor carries the larger current. In SL exam circuit analysis, correctly identifying parallel sections is key to finding equivalent resistances.
总电阻的倒数为 1/R总 = 1/R₁ + 1/R₂ + … 。两个电阻并联时，常用形式为 R总 = (R₁R₂)/(R₁+R₂)。增加支路使总电阻减小，从电源吸取的总电流增大。支路电流与电阻成反比：I₁ / I₂ = R₂ / R₁。一个有用的答案检验：较小电阻承载较大电流。在 SL 考试电路分析中，正确识别并联部分对求等效电阻至关重要。
8. Internal Resistance and Terminal Voltage | 内阻与端电压
A real cell has internal resistance r. When it supplies current I, the terminal potential difference V across the cell is less than its emf ε:
V = ε − I r
真实电池具有内阻 r。当它输出电流 I 时，电池两端的路端电压 V 小于其电动势 ε：V = ε − I r。
Typical SL questions ask you to calculate current using the total circuit resistance Rtotal = Rexternal + r, then find terminal p.d. as ε − I r or simply I Rexternal. For instance, a cell with ε = 1.50 V and r = 0.40 Ω connected to a 2.00 Ω external resistor gives I = 1.50 / (2.00 + 0.40) = 0.625 A, and V = 0.625 × 2.00 = 1.25 V. From a V–I graph, the y‑intercept gives ε and the gradient magnitude gives r. Emphasising that r is constant only for small current ranges is sometimes tested.
典型 SL 题目要求利用总电路电阻 R总 = R外 + r 计算电流，然后求端电压 V = ε − I r 或直接用 I R外。例如，ε = 1.50 V、r = 0.40 Ω 的电池接 2.00 Ω 外电阻，得 I = 1.50 / (2.00+0.40) = 0.625 A，V = 0.625 × 2.00 = 1.25 V。从 V–I 曲线，纵截距为 ε，斜率绝对值为 r。有时会考查 r 仅在小电流范围内可视为常数。
9. Kirchhoff’S Circuit Laws | 基尔霍夫电路定律
Kirchhoff’s current law (KCL) states that the sum of currents entering a junction equals the sum of currents
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