IB Physics HL · 鼎睿学苑

Unit B.5: Current and Circuits单元 B.5：电流与电路

The electrical-circuits unit of Theme B "The particulate nature of matter". Electric current as the rate of flow of charge and the microscopic drift-velocity model, potential difference and electromotive force (emf), electrical power and the kilowatt-hour, resistance and Ohm's law, resistivity, ohmic versus non-ohmic conductors, resistors in series and parallel, Kirchhoff's two laws, and real cells with internal resistance plus the potential divider. This is the foundation every later electricity-and-magnetism unit (electric and magnetic fields, electromagnetic induction) relies on.主题 B"物质的粒子本质"中的电路单元。电流作为电荷流动率与微观漂移速度模型、电势差与电动势（emf）、电功率与千瓦时、电阻与欧姆定律、电阻率、欧姆与非欧姆导体、电阻的串联与并联、基尔霍夫两条定律，以及含内阻的真实电池与分压器。后续所有电磁学单元（电场与磁场、电磁感应）都建立在本单元之上。

IB Physics · Theme B.5 · First Assessment 2025 Papers 1 · 2 6 Topics · SL + HL6 个核心专题 · SL + HL
Read me first先读这里

How to use this guide本指南使用说明

B.5 is the bookkeeping unit of electricity. Almost every problem reduces to three relationships you must wield fluently: charge and current ($I = \Delta q / \Delta t$), the defining equation of resistance ($V = IR$), and the power trio ($P = VI = I^{2}R = V^{2}/R$). The marks then come from choosing the right combination rule (series adds resistance, parallel adds conductance) and from being honest about internal resistance, which turns the "lost volts" inside a cell into examinable physics. Train the definitions alongside the circuit-reduction algebra.B.5 是电学的"记账"单元。几乎每道题都归结为三组你必须熟练运用的关系：电荷与电流（$I = \Delta q / \Delta t$）、电阻定义式（$V = IR$），以及功率三式（$P = VI = I^{2}R = V^{2}/R$）。分数随后来自选对组合规则（串联电阻相加、并联电导相加），以及如实处理内阻——它把电池内部"损失的电压"变成可考的物理。把定义与电路化简代数一起练。

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If you are cramming如果你在临阵磨枪

Memorise $I = \Delta q / \Delta t$, $V = IR$, and $P = VI = I^{2}R = V^{2}/R$. Series: $R_{\text{eq}} = R_{1} + R_{2} + \cdots$ (same current). Parallel: $1/R_{\text{eq}} = 1/R_{1} + 1/R_{2} + \cdots$ (same voltage). For a cell, emf $\varepsilon = I(R + r)$ where $r$ is internal resistance and terminal pd is $V = \varepsilon - Ir$.

背熟 $I = \Delta q / \Delta t$、$V = IR$ 与 $P = VI = I^{2}R = V^{2}/R$。串联：$R_{\text{eq}} = R_{1} + R_{2} + \cdots$（电流相同）。并联：$1/R_{\text{eq}} = 1/R_{1} + 1/R_{2} + \cdots$（电压相同）。对电池，电动势 $\varepsilon = I(R + r)$，其中 $r$ 为内阻，端电压 $V = \varepsilon - Ir$。

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If you are going for a 7如果你目标是 7 分

Derive the drift relation $I = nAvq$ and explain each symbol microscopically. Be fluent applying Kirchhoff's junction law (charge conservation) and loop law (energy conservation) to a two-loop circuit. Know the shape and physics of the $I$-$V$ characteristics of a filament lamp and a diode, and why neither is ohmic. Treat internal resistance and the potential divider as default, not as edge cases.

能推导漂移关系 $I = nAvq$ 并从微观解释每个符号。能熟练地把基尔霍夫节点定律（电荷守恒）与回路定律（能量守恒）用于双回路电路。掌握灯丝灯泡与二极管 $I$-$V$ 特性曲线的形状与物理，并说明二者为何都非欧姆。把内阻与分压器当作常态而非特例。

HL flagHL 标记说明 B.5 is a combined SL + HL super-topic; the bulk of it is common content. Sections carry an A.5-style reference chip showing the syllabus tier. Where a derivation or sub-result is HL-only depth, it is flagged inline with HL. SL students should still read every section but may treat the HL-flagged derivations as optional.B.5 是 SL + HL 合并的超级专题，大部分为共同内容。各节带有显示大纲层级的参考标签。凡推导或子结论属于 HL 深度，均以 HL 内联标记。SL 学生仍应通读每一节，但可把带 HL 标记的推导视为选学。
Topic B5.1 · Current and Charge FlowTopic B5.1 · 电流与电荷流动

Electric Current and the Drift-Velocity Model电流与漂移速度模型 B.5 SL+HL

Electric current. Current $I$ is the rate of flow of electric charge through a cross-section of a conductor. From the data booklet: $$ I = \frac{\Delta q}{\Delta t}. $$ Charge is measured in coulombs ($\mathrm{C}$), current in amperes ($\mathrm{A} = \mathrm{C\,s^{-1}}$). One coulomb is the charge of about $6.25 \times 10^{18}$ electrons; the elementary charge is $e = 1.60 \times 10^{-19}\ \mathrm{C}$.

Conventional current. Conventional current flows from $+$ to $-$ outside the cell, opposite to the actual drift of the (negative) electrons.

Drift-velocity model. For a conductor of cross-sectional area $A$ carrying charge carriers of number density $n$, each of charge $q$, drifting at mean speed $v$: $$ I = n A v q. $$ Drift speeds are tiny (mm per second), yet the current establishes almost instantly because the field propagates near light speed.
电流（current）。电流 $I$ 是电荷流过导体某横截面的速率。数据手册公式： $$ I = \frac{\Delta q}{\Delta t}. $$ 电荷以库仑（$\mathrm{C}$）计，电流以安培（$\mathrm{A} = \mathrm{C\,s^{-1}}$）计。一库仑约为 $6.25 \times 10^{18}$ 个电子的电荷；元电荷 $e = 1.60 \times 10^{-19}\ \mathrm{C}$。

常规电流方向。常规电流在电池外部由 $+$ 流向 $-$，与（带负电的）电子实际漂移方向相反。

漂移速度模型。对横截面积 $A$ 的导体，载流子数密度 $n$、每个电荷 $q$、平均漂移速率 $v$： $$ I = n A v q. $$ 漂移速率极小（每秒毫米量级），但电流几乎瞬间建立，因为电场以接近光速传播。
Worked Example B5.1a (charge from current)B5.1a 例题（由电流求电荷）

A torch bulb carries a steady current of $0.30\ \mathrm{A}$ for $5.0\ \mathrm{min}$. Find the charge that passes through the filament and the number of electrons this represents.手电筒灯泡中通过稳恒电流 $0.30\ \mathrm{A}$，持续 $5.0\ \mathrm{min}$。求通过灯丝的电荷量及对应的电子数目。

Identify. Known: $I = 0.30\ \mathrm{A}$, $\Delta t = 5.0\ \mathrm{min} = 300\ \mathrm{s}$. Want $\Delta q$, then $N$.

识别。已知：$I = 0.30\ \mathrm{A}$、$\Delta t = 5.0\ \mathrm{min} = 300\ \mathrm{s}$。求 $\Delta q$，再求 $N$。

Set up. Rearrange $I = \Delta q / \Delta t$ for the charge.

列式。由 $I = \Delta q / \Delta t$ 解出电荷。

$$ \Delta q = I\,\Delta t = (0.30)(300) = 90\ \mathrm{C}. $$

Electron count. Divide by the elementary charge $e = 1.60 \times 10^{-19}\ \mathrm{C}$:

电子数目。除以元电荷 $e = 1.60 \times 10^{-19}\ \mathrm{C}$：

$$ N = \frac{\Delta q}{e} = \frac{90}{1.60 \times 10^{-19}} \approx 5.6 \times 10^{20}. $$

Evaluate. Converting minutes to seconds first is essential; a common error is leaving $\Delta t$ in minutes and getting a charge $60\times$ too small.

评估。先把分钟换算成秒至关重要；常见错误是把 $\Delta t$ 留在分钟，导致电荷小了 $60$ 倍。

Worked Example B5.1b (drift velocity)B5.1b 例题（漂移速度）

A copper wire of cross-sectional area $1.0 \times 10^{-6}\ \mathrm{m^{2}}$ carries a current of $5.0\ \mathrm{A}$. Copper has $n = 8.5 \times 10^{28}$ free electrons per $\mathrm{m^{3}}$. Find the electron drift speed.横截面积 $1.0 \times 10^{-6}\ \mathrm{m^{2}}$ 的铜导线中通过 $5.0\ \mathrm{A}$ 电流。铜的自由电子数密度 $n = 8.5 \times 10^{28}\ \mathrm{m^{-3}}$。求电子漂移速率。

Set up. Rearrange $I = n A v q$ for the drift speed $v$, with $q = e$.

列式。由 $I = n A v q$ 解出漂移速率 $v$，取 $q = e$。

$$ v = \frac{I}{n A e} = \frac{5.0}{(8.5 \times 10^{28})(1.0 \times 10^{-6})(1.60 \times 10^{-19})}. $$

Compute. Denominator $= 8.5 \times 10^{28} \times 1.0 \times 10^{-6} \times 1.60 \times 10^{-19} \approx 1.36 \times 10^{4}$.

计算。分母 $= 8.5 \times 10^{28} \times 1.0 \times 10^{-6} \times 1.60 \times 10^{-19} \approx 1.36 \times 10^{4}$。

$$ v = \frac{5.0}{1.36 \times 10^{4}} \approx 3.7 \times 10^{-4}\ \mathrm{m\,s^{-1}}. $$

Evaluate. The drift speed is about $0.37\ \mathrm{mm\,s^{-1}}$ — far slower than most students expect. The lamp lights instantly because the electric field, not the electrons, travels at near light speed.

评估。漂移速率约 $0.37\ \mathrm{mm\,s^{-1}}$，远比多数同学想象的慢。灯立即亮起是因为电场（而非电子）以接近光速传播。

Going deeper: where $I = nAvq$ comes from深入：$I = nAvq$ 的由来

Consider a slice of conductor of area $A$ and length $\Delta x = v\,\Delta t$ (the distance carriers drift in time $\Delta t$). The volume of the slice is $A v\,\Delta t$, so the number of carriers inside is $n A v\,\Delta t$, each carrying charge $q$. The charge passing the cross-section in time $\Delta t$ is therefore

取一段导体，截面积 $A$，长度 $\Delta x = v\,\Delta t$（载流子在 $\Delta t$ 内漂移的距离）。该段体积为 $A v\,\Delta t$，内含载流子数 $n A v\,\Delta t$，每个带电荷 $q$。故在 $\Delta t$ 内通过截面的电荷为

$$ \Delta q = (n A v\,\Delta t)\, q. $$

Dividing by $\Delta t$ gives the current directly:

除以 $\Delta t$ 直接得电流：

$$ I = \frac{\Delta q}{\Delta t} = n A v q. $$

This links the macroscopic $I = \Delta q / \Delta t$ to the microscopic picture. For a fixed current, a thinner wire (smaller $A$) forces a larger drift speed $v$ — which is why thin filaments heat up.

这把宏观的 $I = \Delta q / \Delta t$ 与微观图像联系起来。在固定电流下，更细的导线（更小的 $A$）迫使更大的漂移速率 $v$——这正是细灯丝发热的原因。

A charge of $48\ \mathrm{C}$ flows past a point in a wire in $2.0\ \mathrm{min}$. The current is:$48\ \mathrm{C}$ 电荷在 $2.0\ \mathrm{min}$ 内流过导线某点。电流为：
B5.1 · Q1
$24\ \mathrm{A}$
$96\ \mathrm{A}$
$0.40\ \mathrm{A}$
$0.80\ \mathrm{A}$
$I = \Delta q / \Delta t$. Convert: $2.0\ \mathrm{min} = 120\ \mathrm{s}$. $I = 48 / 120 = 0.40\ \mathrm{A}$.$I = \Delta q / \Delta t$。换算：$2.0\ \mathrm{min} = 120\ \mathrm{s}$。$I = 48 / 120 = 0.40\ \mathrm{A}$。
Convert minutes to seconds before dividing. $I = \Delta q / \Delta t = 48 / 120$, not $48 / 2$.先把分钟换算成秒再相除。$I = \Delta q / \Delta t = 48 / 120$，不是 $48 / 2$。
For a fixed current $I$, the drift speed of charge carriers in a wire is doubled if you:在固定电流 $I$ 下，要使导线中载流子漂移速率加倍，应：
B5.1 · Q2
Double the cross-sectional area $A$把横截面积 $A$ 加倍
Halve the cross-sectional area $A$把横截面积 $A$ 减半
Double the carrier density $n$把载流子数密度 $n$ 加倍
Double the wire length把导线长度加倍
From $I = nAvq$, with $I$, $n$, $q$ fixed, $v = I/(nAq) \propto 1/A$. Halving $A$ doubles $v$. Length does not appear in this relation.由 $I = nAvq$，$I$、$n$、$q$ 固定时 $v = I/(nAq) \propto 1/A$。$A$ 减半则 $v$ 加倍。长度不出现在此关系中。
Rearrange $I = nAvq$ to $v = I/(nAq)$. With $I$ fixed, $v \propto 1/A$, so a smaller area means a larger drift speed.把 $I = nAvq$ 改写为 $v = I/(nAq)$。$I$ 固定时 $v \propto 1/A$，故面积越小漂移速率越大。
Topic B5.2 · Potential Difference, emf, PowerTopic B5.2 · 电势差、电动势、功率

Potential Difference, emf and Electrical Power电势差、电动势与电功率 B.5 SL+HL

Potential difference (pd). The potential difference $V$ across a component is the energy transferred from electrical to other forms per unit charge passing through it: $$ V = \frac{W}{q}. $$ Measured in volts ($\mathrm{V} = \mathrm{J\,C^{-1}}$). A voltmeter is connected in parallel across the component.

Electromotive force (emf). The emf $\varepsilon$ of a source is the energy supplied per unit charge by the source, converting chemical (or other) energy to electrical. Same units as pd. Crucially, emf is energy given to the charge; pd is energy taken from it by a component.

Electrical power. From the data booklet: $$ P = VI = I^{2}R = \frac{V^{2}}{R}. $$ The first form is general; the second and third assume the component obeys $V = IR$. Power in watts ($\mathrm{W} = \mathrm{J\,s^{-1}}$).

The kilowatt-hour. A unit of energy (not power) used for billing: $1\ \mathrm{kW\,h} = (1000\ \mathrm{W})(3600\ \mathrm{s}) = 3.6 \times 10^{6}\ \mathrm{J}$.
电势差（potential difference, pd）。元件两端电势差 $V$ 是每单位电荷通过该元件时由电能转化为其他形式能量的量： $$ V = \frac{W}{q}. $$ 以伏特（$\mathrm{V} = \mathrm{J\,C^{-1}}$）计。电压表与元件并联接入。

电动势（electromotive force, emf）。电源电动势 $\varepsilon$ 是电源每单位电荷所供给的能量，把化学（或其他）能转化为电能。单位同 pd。关键区别：emf 是给予电荷的能量，pd 是元件从电荷取走的能量。

电功率（power）。数据手册公式： $$ P = VI = I^{2}R = \frac{V^{2}}{R}. $$ 第一式普遍成立；第二、三式假设元件满足 $V = IR$。功率以瓦特（$\mathrm{W} = \mathrm{J\,s^{-1}}$）计。

千瓦时（kilowatt-hour）。计费用的能量单位（非功率）：$1\ \mathrm{kW\,h} = (1000\ \mathrm{W})(3600\ \mathrm{s}) = 3.6 \times 10^{6}\ \mathrm{J}$。
Worked Example B5.2a (power three ways)B5.2a 例题（功率三种算法）

A resistor of $12\ \Omega$ carries a current of $2.0\ \mathrm{A}$. Find the power it dissipates using all three forms of the power equation, and the energy converted in $30\ \mathrm{s}$.$12\ \Omega$ 电阻中通过 $2.0\ \mathrm{A}$ 电流。用功率方程的三种形式求其耗散功率，以及 $30\ \mathrm{s}$ 内转化的能量。

Identify. Known: $R = 12\ \Omega$, $I = 2.0\ \mathrm{A}$. First find $V = IR = (2.0)(12) = 24\ \mathrm{V}$.

识别。已知：$R = 12\ \Omega$、$I = 2.0\ \mathrm{A}$。先求 $V = IR = (2.0)(12) = 24\ \mathrm{V}$。

Three forms (must agree).

三种形式（须一致）。

$$ P = VI = (24)(2.0) = 48\ \mathrm{W}, \quad P = I^{2}R = (2.0)^{2}(12) = 48\ \mathrm{W}, \quad P = \frac{V^{2}}{R} = \frac{24^{2}}{12} = 48\ \mathrm{W}. $$

Energy. $W = P\,\Delta t = (48)(30) = 1440\ \mathrm{J} \approx 1.4\ \mathrm{kJ}$.

能量。$W = P\,\Delta t = (48)(30) = 1440\ \mathrm{J} \approx 1.4\ \mathrm{kJ}$。

Evaluate. All three forms give $48\ \mathrm{W}$, as they must. Pick whichever form uses the two quantities you already know to avoid an extra step.

评估。三式都给出 $48\ \mathrm{W}$，必然如此。选用你已知两个量的那一式，省去一步。

Worked Example B5.2b (kilowatt-hour billing)B5.2b 例题（千瓦时计费）

An electric heater rated $2.5\ \mathrm{kW}$ runs for $3.0\ \mathrm{hours}$. The electricity tariff is $0.20$ per $\mathrm{kW\,h}$. Find the energy used in $\mathrm{kW\,h}$ and in joules, and the cost.额定 $2.5\ \mathrm{kW}$ 的电暖器运行 $3.0\ \mathrm{h}$。电价为每 $\mathrm{kW\,h}$ $0.20$。求所用能量（以 $\mathrm{kW\,h}$ 和焦耳计）及费用。

Energy in kW·h. Energy $=$ power $\times$ time $= (2.5\ \mathrm{kW})(3.0\ \mathrm{h}) = 7.5\ \mathrm{kW\,h}$.

能量（kW·h）。能量 $=$ 功率 $\times$ 时间 $= (2.5\ \mathrm{kW})(3.0\ \mathrm{h}) = 7.5\ \mathrm{kW\,h}$。

Energy in joules. $7.5\ \mathrm{kW\,h} \times 3.6 \times 10^{6}\ \mathrm{J\,(kW\,h)^{-1}} = 2.7 \times 10^{7}\ \mathrm{J}$.

能量（焦耳）。$7.5\ \mathrm{kW\,h} \times 3.6 \times 10^{6}\ \mathrm{J\,(kW\,h)^{-1}} = 2.7 \times 10^{7}\ \mathrm{J}$。

Cost. $7.5\ \mathrm{kW\,h} \times 0.20 = 1.50$.

费用。$7.5\ \mathrm{kW\,h} \times 0.20 = 1.50$。

Evaluate. The kilowatt-hour keeps the arithmetic in convenient units; converting to joules is only needed when the question demands SI.

评估。千瓦时让算术保持在方便单位；只有题目要求 SI 时才需换算成焦耳。

Going deeper: why power has three equivalent forms深入：功率为何有三种等价形式

Power is energy per unit time. The energy delivered to a charge $q$ across a pd $V$ is $W = qV$, so the rate of energy transfer is

功率是单位时间的能量。电荷 $q$ 通过电势差 $V$ 所获能量为 $W = qV$，故能量传递率为

$$ P = \frac{W}{t} = \frac{qV}{t} = \left(\frac{q}{t}\right) V = IV. $$

This $P = VI$ is general — it holds for any component, ohmic or not. The other two forms come from substituting Ohm's law $V = IR$:

该 $P = VI$ 普遍成立——对任何元件（欧姆与否）都成立。另两式由代入欧姆定律 $V = IR$ 得到：

$$ P = VI = (IR)I = I^{2}R, \qquad P = VI = V\left(\frac{V}{R}\right) = \frac{V^{2}}{R}. $$

Exam trap. $P = I^{2}R$ and $P = V^{2}/R$ assume $V = IR$. For a non-ohmic component (lamp, diode) you must use $P = VI$ with the actual $V$ and $I$ read off the characteristic.

考试陷阱。$P = I^{2}R$ 与 $P = V^{2}/R$ 假设 $V = IR$。对非欧姆元件（灯泡、二极管）必须用 $P = VI$，取特性曲线上实际的 $V$ 与 $I$。

A $60\ \mathrm{W}$ lamp operates at $230\ \mathrm{V}$. The current it draws is approximately:$60\ \mathrm{W}$ 灯泡在 $230\ \mathrm{V}$ 下工作。其电流约为：
B5.2 · Q1
$0.26\ \mathrm{A}$
$3.8\ \mathrm{A}$
$13800\ \mathrm{A}$
$1.0\ \mathrm{A}$
From $P = VI$, $I = P/V = 60/230 \approx 0.26\ \mathrm{A}$.由 $P = VI$，$I = P/V = 60/230 \approx 0.26\ \mathrm{A}$。
Use $P = VI$ and solve for current: $I = P/V$. Divide power by voltage, do not multiply.用 $P = VI$ 求电流：$I = P/V$。功率除以电压，不是相乘。
Which statement best distinguishes emf from potential difference?下列哪项最能区分电动势与电势差？
B5.2 · Q2
They have different units两者单位不同
emf is always smaller than pd电动势总小于电势差
pd applies only to cells, emf only to resistors电势差只用于电池，电动势只用于电阻
emf is energy supplied per unit charge by a source; pd is energy transferred from charge by a component电动势是电源每单位电荷供给的能量；电势差是元件从电荷取走的每单位电荷能量
Both are measured in volts, so units are identical. The distinction is the direction of energy transfer: emf gives energy to the charges (source), pd takes energy from them (load).两者都以伏特计，单位相同。区别在于能量传递方向：电动势给予电荷能量（电源），电势差取走电荷能量（负载）。
Both are in volts. Think about energy transfer direction: emf supplies energy to charge; pd is the energy a component takes from the charge.两者均以伏特计。从能量传递方向考虑：电动势供给电荷能量；电势差是元件从电荷取走的能量。
Topic B5.3 · Resistance, Ohm's Law, ResistivityTopic B5.3 · 电阻、欧姆定律、电阻率

Resistance, Ohm's Law and the I-V Characteristic电阻、欧姆定律与 I-V 特性 B.5 SL+HL

Resistance. Resistance $R$ is the opposition to current, defined by $$ R = \frac{V}{I}, \qquad \text{equivalently} \qquad V = IR. $$ Measured in ohms ($\Omega = \mathrm{V\,A^{-1}}$). The defining equation $V = IR$ holds for any component as a definition of its resistance at that operating point.

Ohm's law. A component is ohmic if its resistance is constant — i.e. $I \propto V$ at constant temperature — giving a straight-line $I$-$V$ graph through the origin. Ohm's law is the statement that $R$ is constant, not the equation $V = IR$ itself.

Resistivity. From the data booklet, for a uniform conductor of length $L$ and cross-section $A$: $$ R = \frac{\rho L}{A}, $$ where $\rho$ is the resistivity ($\mathrm{\Omega\,m}$), a material property. Longer and thinner means more resistance.

Non-ohmic conductors.
  • Filament lamp: heats up as current rises, so $R$ increases; the $I$-$V$ curve flattens (slope falls) at higher $V$.
  • Diode: conducts only one way; negligible current until a threshold ($\approx 0.6\ \mathrm{V}$ for silicon), then current rises steeply. Blocks reverse current.
电阻（resistance）。电阻 $R$ 是对电流的阻碍，定义为 $$ R = \frac{V}{I}, \qquad \text{即} \qquad V = IR. $$ 以欧姆（$\Omega = \mathrm{V\,A^{-1}}$）计。定义式 $V = IR$ 对任何元件都成立，用于定义其在该工作点的电阻。

欧姆定律（Ohm's law）。若元件电阻恒定——即恒温下 $I \propto V$——则为欧姆元件，其 $I$-$V$ 图为过原点的直线。欧姆定律说的是 $R$ 恒定，而非 $V = IR$ 这个式子本身。

电阻率（resistivity）。数据手册中，对长度 $L$、截面 $A$ 的均匀导体： $$ R = \frac{\rho L}{A}, $$ 其中 $\rho$ 是电阻率（$\mathrm{\Omega\,m}$），为材料属性。越长越细，电阻越大。

非欧姆导体。
  • 灯丝灯泡：电流增大时升温，故 $R$ 增大；$I$-$V$ 曲线在高 $V$ 处变平（斜率减小）。
  • 二极管：只单向导通；在阈值（硅约 $0.6\ \mathrm{V}$）之前电流几乎为零，之后电流陡升。阻断反向电流。
Worked Example B5.3a (resistivity)B5.3a 例题（电阻率）

A nichrome wire has length $2.0\ \mathrm{m}$, diameter $0.50\ \mathrm{mm}$, and resistivity $\rho = 1.1 \times 10^{-6}\ \mathrm{\Omega\,m}$. Find its resistance.镍铬合金导线长 $2.0\ \mathrm{m}$，直径 $0.50\ \mathrm{mm}$，电阻率 $\rho = 1.1 \times 10^{-6}\ \mathrm{\Omega\,m}$。求其电阻。

Cross-sectional area. Radius $r = 0.25\ \mathrm{mm} = 2.5 \times 10^{-4}\ \mathrm{m}$.

横截面积。半径 $r = 0.25\ \mathrm{mm} = 2.5 \times 10^{-4}\ \mathrm{m}$。

$$ A = \pi r^{2} = \pi (2.5 \times 10^{-4})^{2} \approx 1.96 \times 10^{-7}\ \mathrm{m^{2}}. $$

Apply $R = \rho L / A$.

套用 $R = \rho L / A$。

$$ R = \frac{(1.1 \times 10^{-6})(2.0)}{1.96 \times 10^{-7}} \approx 11\ \Omega. $$

Evaluate. The diameter must be halved to a radius before squaring. Doubling the wire length would double $R$; doubling the diameter (four-fold area) would quarter it.

评估。必须先把直径折半为半径再平方。导线长度加倍则 $R$ 加倍；直径加倍（面积变四倍）则 $R$ 变为四分之一。

Worked Example B5.3b (resistance from an I-V graph)B5.3b 例题（由 I-V 图求电阻）

A filament lamp passes $0.20\ \mathrm{A}$ at $1.0\ \mathrm{V}$ and $0.50\ \mathrm{A}$ at $6.0\ \mathrm{V}$. Find its resistance at each operating point and state whether it is ohmic.灯丝灯泡在 $1.0\ \mathrm{V}$ 时通过 $0.20\ \mathrm{A}$，在 $6.0\ \mathrm{V}$ 时通过 $0.50\ \mathrm{A}$。求两个工作点的电阻并判断是否为欧姆元件。

Point 1. $R_{1} = V/I = 1.0 / 0.20 = 5.0\ \Omega$.

工作点 1。$R_{1} = V/I = 1.0 / 0.20 = 5.0\ \Omega$。

Point 2. $R_{2} = V/I = 6.0 / 0.50 = 12\ \Omega$.

工作点 2。$R_{2} = V/I = 6.0 / 0.50 = 12\ \Omega$。

Ohmic test. Resistance rose from $5.0\ \Omega$ to $12\ \Omega$ as voltage increased, so $R$ is not constant. The lamp is non-ohmic: heating the filament raises its resistance.

欧姆判定。电压升高时电阻从 $5.0\ \Omega$ 升到 $12\ \Omega$，故 $R$ 不恒定。灯泡非欧姆：灯丝升温使其电阻增大。

Evaluate. Always use $R = V/I$ at each individual point. Never use the slope $\Delta I / \Delta V$ as a resistance for a curved characteristic — that gives a different (dynamic) quantity.

评估。对每个工作点都用 $R = V/I$。对弯曲特性曲线绝不能用斜率 $\Delta I / \Delta V$ 当作电阻——那是另一个（动态）量。

Going deeper: why a filament's resistance rises with temperature深入：灯丝电阻为何随温度升高

In a metal, current is carried by free electrons that drift through a lattice of positive ions. As current rises, $I^{2}R$ heating raises the temperature, so the ions vibrate more vigorously. Electrons then collide with the lattice more frequently, scattering more, which reduces the drift speed achievable for a given field and so raises the resistance.

金属中电流由穿过正离子点阵的自由电子承载。电流增大时 $I^{2}R$ 发热使温度升高，离子振动加剧。电子与点阵碰撞更频繁、散射更多，使给定电场下可达到的漂移速率降低，从而电阻升高。

A diode is different: it is a semiconductor junction. Below the threshold voltage almost no current flows because the junction's potential barrier is not overcome; above it the barrier collapses and current rises sharply. In reverse bias the barrier grows and current is blocked. Neither device has a constant $R$, so neither obeys Ohm's law, even though $V = IR$ still defines their resistance at any chosen point.

二极管不同：它是半导体结。在阈值电压以下几乎无电流，因为未克服结的势垒；超过阈值后势垒崩溃，电流陡升。反向偏置时势垒增大、电流被阻断。两种器件 $R$ 都不恒定，故都不服从欧姆定律，尽管 $V = IR$ 仍可在任意所选点定义其电阻。

A wire is stretched to twice its original length, keeping its volume constant. Its resistance becomes:一根导线被拉长到原来两倍长，体积不变。其电阻变为：
B5.3 · Q1
Twice the original原来的 2 倍
Half the original原来的一半
Four times the original原来的 4 倍
Unchanged不变
Constant volume: if $L$ doubles, $A$ halves. Then $R = \rho L / A$ scales by $2 / (1/2) = 4$. Resistance quadruples.体积不变：$L$ 加倍则 $A$ 减半。由 $R = \rho L / A$，比例为 $2 / (1/2) = 4$。电阻变为 4 倍。
Volume $= LA$ is fixed, so doubling $L$ halves $A$. Both changes raise $R$: $R = \rho L / A$ scales by $2 \times 2 = 4$.体积 $= LA$ 不变，$L$ 加倍则 $A$ 减半。两者都使 $R$ 增大：$R = \rho L / A$ 比例为 $2 \times 2 = 4$。
The $I$-$V$ characteristic of a component is a straight line through the origin at constant temperature. The component is:某元件在恒温下的 $I$-$V$ 特性为过原点的直线。该元件为：
B5.3 · Q2
A diode二极管
An ohmic conductor (fixed resistor)欧姆导体（定值电阻）
A filament lamp灯丝灯泡
A thermistor with rising temperature温度升高的热敏电阻
A straight line through the origin means $I \propto V$, so $R = V/I$ is constant — the defining feature of an ohmic conductor. Lamps and diodes give curved characteristics.过原点的直线意味 $I \propto V$，故 $R = V/I$ 恒定——欧姆导体的标志。灯泡与二极管给出弯曲特性。
Constant resistance shows as a straight line through the origin. A lamp curves (R rises with heat); a diode has a threshold; both are non-ohmic.电阻恒定表现为过原点的直线。灯泡因发热而弯曲（R 升高）；二极管有阈值；二者均非欧姆。
Topic B5.4 · Series and Parallel ResistorsTopic B5.4 · 串联与并联电阻

Combining Resistors: Series and Parallel电阻组合：串联与并联 B.5 SL+HL

Series. Components share the same current; voltages add. From the data booklet: $$ R_{\text{series}} = R_{1} + R_{2} + \cdots + R_{n}. $$ The equivalent resistance is always larger than the biggest single resistor.

Parallel. Components share the same voltage; currents add. From the data booklet: $$ \frac{1}{R_{\text{parallel}}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} + \cdots + \frac{1}{R_{n}}. $$ The equivalent resistance is always smaller than the smallest single resistor. For two in parallel, the shortcut is $$ R_{\text{parallel}} = \frac{R_{1} R_{2}}{R_{1} + R_{2}}. $$ Sanity checks. $n$ equal resistors $R$ in series give $nR$; in parallel give $R/n$.
串联（series）。各元件电流相同；电压相加。数据手册公式： $$ R_{\text{series}} = R_{1} + R_{2} + \cdots + R_{n}. $$ 等效电阻总大于最大的单个电阻。

并联（parallel）。各元件电压相同；电流相加。数据手册公式： $$ \frac{1}{R_{\text{parallel}}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} + \cdots + \frac{1}{R_{n}}. $$ 等效电阻总小于最小的单个电阻。两电阻并联的捷径： $$ R_{\text{parallel}} = \frac{R_{1} R_{2}}{R_{1} + R_{2}}. $$ 复核。$n$ 个相等电阻 $R$ 串联得 $nR$；并联得 $R/n$。
Worked Example B5.4 (a mixed network)B5.4 例题（混联网络）

A $4.0\ \Omega$ resistor is in series with a parallel combination of a $6.0\ \Omega$ and a $3.0\ \Omega$ resistor. The network is connected to a $12\ \mathrm{V}$ ideal supply. Find the equivalent resistance, the total current, and the current through the $6.0\ \Omega$ resistor.一个 $4.0\ \Omega$ 电阻与 $6.0\ \Omega$、$3.0\ \Omega$ 并联组合串联。网络接到 $12\ \mathrm{V}$ 理想电源。求等效电阻、总电流，以及 $6.0\ \Omega$ 电阻中的电流。

Reduce the parallel pair.

先化简并联对。

$$ R_{P} = \frac{(6.0)(3.0)}{6.0 + 3.0} = \frac{18}{9.0} = 2.0\ \Omega. $$

Add in series. $R_{\text{eq}} = 4.0 + 2.0 = 6.0\ \Omega$.

串联相加。$R_{\text{eq}} = 4.0 + 2.0 = 6.0\ \Omega$。

Total current. $I = V / R_{\text{eq}} = 12 / 6.0 = 2.0\ \mathrm{A}$ (this flows through the $4.0\ \Omega$ resistor).

总电流。$I = V / R_{\text{eq}} = 12 / 6.0 = 2.0\ \mathrm{A}$（这股电流流过 $4.0\ \Omega$ 电阻）。

Voltage across the parallel pair. $V_{P} = I R_{P} = (2.0)(2.0) = 4.0\ \mathrm{V}$.

并联对两端电压。$V_{P} = I R_{P} = (2.0)(2.0) = 4.0\ \mathrm{V}$。

Current through the $6.0\ \Omega$. $I_{6} = V_{P} / 6.0 = 4.0 / 6.0 \approx 0.67\ \mathrm{A}$.

$6.0\ \Omega$ 中的电流。$I_{6} = V_{P} / 6.0 = 4.0 / 6.0 \approx 0.67\ \mathrm{A}$。

Evaluate. Check current conservation: $I_{3} = 4.0/3.0 \approx 1.33\ \mathrm{A}$, and $0.67 + 1.33 = 2.0\ \mathrm{A}$, matching the total. The larger resistor carries the smaller share of the current.

评估。核查电流守恒：$I_{3} = 4.0/3.0 \approx 1.33\ \mathrm{A}$，且 $0.67 + 1.33 = 2.0\ \mathrm{A}$，与总电流相符。电阻越大，分得的电流越小。

Going deeper: why series adds resistance but parallel adds conductance深入：为何串联相加电阻、并联相加电导

Series. The same current $I$ passes through each resistor, and the total pd is the sum: $V = V_{1} + V_{2} = IR_{1} + IR_{2} = I(R_{1} + R_{2})$. Dividing by $I$, $R_{\text{eq}} = R_{1} + R_{2}$. Adding length in series is like making one longer resistor.

串联。同一电流 $I$ 流过每个电阻，总电势差为各段之和：$V = V_{1} + V_{2} = IR_{1} + IR_{2} = I(R_{1} + R_{2})$。除以 $I$ 得 $R_{\text{eq}} = R_{1} + R_{2}$。串联相加如同把电阻接得更长。

Parallel. The same pd $V$ sits across each resistor, and the currents add: $I = I_{1} + I_{2} = V/R_{1} + V/R_{2} = V(1/R_{1} + 1/R_{2})$. Dividing by $V$, $1/R_{\text{eq}} = 1/R_{1} + 1/R_{2}$. Adding parallel paths is like widening the conductor — more area means more conductance, hence less resistance.

并联。同一电势差 $V$ 加在每个电阻上，电流相加：$I = I_{1} + I_{2} = V/R_{1} + V/R_{2} = V(1/R_{1} + 1/R_{2})$。除以 $V$ 得 $1/R_{\text{eq}} = 1/R_{1} + 1/R_{2}$。并联多条路径如同加宽导体——面积越大、电导越大，故电阻越小。

Two $10\ \Omega$ resistors are connected in parallel. Their equivalent resistance is:两个 $10\ \Omega$ 电阻并联。等效电阻为：
B5.4 · Q1
$5.0\ \Omega$
$20\ \Omega$
$10\ \Omega$
$0.20\ \Omega$
For $n$ equal resistors in parallel, $R_{\text{eq}} = R/n = 10/2 = 5.0\ \Omega$. The result is always less than the smallest single resistor.$n$ 个相等电阻并联：$R_{\text{eq}} = R/n = 10/2 = 5.0\ \Omega$。结果总小于最小的单个电阻。
Parallel resistance is always smaller than the smallest branch. For two equal $10\ \Omega$, it is $R/2 = 5.0\ \Omega$, not $20\ \Omega$ (that would be series).并联电阻总小于最小支路。两个相等的 $10\ \Omega$ 得 $R/2 = 5.0\ \Omega$，不是 $20\ \Omega$（那是串联）。
In a series circuit, which quantity is the same through every component?在串联电路中，哪个量在每个元件中都相同？
B5.4 · Q2
The potential difference电势差
The power dissipated耗散功率
The resistance电阻
The current电流
In series there is a single path, so the same current flows everywhere (charge conservation). Voltages divide in proportion to resistance; power and voltage differ between components.串联只有一条路径，故各处电流相同（电荷守恒）。电压按电阻成比例分配；功率与电压在各元件间不同。
Series = one path = same current everywhere. It is parallel branches that share the same voltage instead.串联 = 单一路径 = 各处电流相同。共享相同电压的是并联支路。
Topic B5.5 · Kirchhoff's LawsTopic B5.5 · 基尔霍夫定律

Kirchhoff's Laws and Circuit Analysis基尔霍夫定律与电路分析 B.5 SL+HL

Junction (current) law — charge conservation. The sum of currents into any junction equals the sum out: $$ \sum I_{\text{in}} = \sum I_{\text{out}}, \qquad \text{equivalently} \qquad \sum I = 0 \text{ at a node}. $$ Charge cannot accumulate at a point, so what flows in must flow out.

Loop (voltage) law — energy conservation. Around any closed loop, the sum of emfs equals the sum of pds (the algebraic sum of potential changes is zero): $$ \sum \varepsilon = \sum I R, \qquad \text{equivalently} \qquad \sum V = 0 \text{ around a loop}. $$ A charge returning to its start has zero net change in potential energy.

Sign rules for the loop law.
  • Pick a loop direction. Crossing a resistor with the current is a drop ($-IR$); against is a rise ($+IR$).
  • Crossing a cell from $-$ to $+$ is a rise ($+\varepsilon$); from $+$ to $-$ is a drop ($-\varepsilon$).
节点（电流）定律——电荷守恒。流入任一节点的电流之和等于流出之和： $$ \sum I_{\text{in}} = \sum I_{\text{out}}, \qquad \text{即} \qquad \text{节点处} \sum I = 0. $$ 电荷不能在一点积累，故流入必等于流出。

回路（电压）定律——能量守恒。沿任一闭合回路，电动势之和等于电势差之和（电势变化的代数和为零）： $$ \sum \varepsilon = \sum I R, \qquad \text{即} \qquad \text{沿回路} \sum V = 0. $$ 电荷绕回起点时电势能净变化为零。

回路定律的符号规则。
  • 选定绕行方向。顺电流过电阻为电势降（$-IR$）；逆电流为电势升（$+IR$）。
  • 过电池由 $-$ 到 $+$ 为电势升（$+\varepsilon$）；由 $+$ 到 $-$ 为电势降（$-\varepsilon$）。
Worked Example B5.5 (two-loop circuit)B5.5 例题（双回路电路）

A $12\ \mathrm{V}$ ideal cell drives a series resistor $R_{1} = 2.0\ \Omega$, after which the circuit splits into two parallel branches: $R_{2} = 6.0\ \Omega$ and $R_{3} = 12\ \Omega$. Use Kirchhoff's laws to find the current from the cell and the current in each branch.$12\ \mathrm{V}$ 理想电池驱动串联电阻 $R_{1} = 2.0\ \Omega$，之后电路分为两条并联支路：$R_{2} = 6.0\ \Omega$ 与 $R_{3} = 12\ \Omega$。用基尔霍夫定律求电池电流及各支路电流。

Junction law. The cell current $I$ splits at the node: $I = I_{2} + I_{3}$.

节点定律。电池电流 $I$ 在节点分流：$I = I_{2} + I_{3}$。

Reduce parallel pair (consistency). $R_{P} = (6.0 \times 12)/(6.0 + 12) = 72/18 = 4.0\ \Omega$, so $R_{\text{eq}} = 2.0 + 4.0 = 6.0\ \Omega$.

化简并联对（用于一致性核查）。$R_{P} = (6.0 \times 12)/(6.0 + 12) = 72/18 = 4.0\ \Omega$，故 $R_{\text{eq}} = 2.0 + 4.0 = 6.0\ \Omega$。

Cell current (loop law through the whole circuit). $12 = I R_{\text{eq}} = I (6.0) \Rightarrow I = 2.0\ \mathrm{A}$.

电池电流（对整条回路用回路定律）。$12 = I R_{\text{eq}} = I (6.0) \Rightarrow I = 2.0\ \mathrm{A}$。

Voltage across the parallel section. $V_{P} = 12 - I R_{1} = 12 - (2.0)(2.0) = 8.0\ \mathrm{V}$.

并联段两端电压。$V_{P} = 12 - I R_{1} = 12 - (2.0)(2.0) = 8.0\ \mathrm{V}$。

Branch currents. $I_{2} = 8.0/6.0 \approx 1.33\ \mathrm{A}$, $I_{3} = 8.0/12 \approx 0.67\ \mathrm{A}$.

支路电流。$I_{2} = 8.0/6.0 \approx 1.33\ \mathrm{A}$，$I_{3} = 8.0/12 \approx 0.67\ \mathrm{A}$。

Evaluate. Junction check: $I_{2} + I_{3} = 1.33 + 0.67 = 2.0\ \mathrm{A} = I$. Loop check: $I R_{1} + V_{P} = 4.0 + 8.0 = 12\ \mathrm{V} = \varepsilon$. Both laws are satisfied.

评估。节点核查：$I_{2} + I_{3} = 1.33 + 0.67 = 2.0\ \mathrm{A} = I$。回路核查：$I R_{1} + V_{P} = 4.0 + 8.0 = 12\ \mathrm{V} = \varepsilon$。两定律皆满足。

The two laws are conservation laws in disguise两定律实为守恒定律的化身 The junction law is conservation of charge (charge in $=$ charge out at a node). The loop law is conservation of energy (energy supplied by sources $=$ energy dissipated in resistors around a closed path). Naming the underlying conservation principle is a common short-answer mark.节点定律即电荷守恒（节点处流入电荷 $=$ 流出电荷）。回路定律即能量守恒（沿闭合路径，电源供给的能量 $=$ 电阻耗散的能量）。说出背后的守恒原理常是简答题的得分点。
Going deeper: applying the loop law to a circuit with two cells HL深入：把回路定律用于含两电池的电路 HL

Consider two cells of emf $\varepsilon_{1}$ and $\varepsilon_{2}$ with a single resistor $R$. If the cells are connected so their emfs oppose (driving current in opposite directions), the loop law gives

设两电池电动势 $\varepsilon_{1}$、$\varepsilon_{2}$，与单个电阻 $R$ 相连。若两电池连接使电动势相反（驱动电流方向相反），回路定律给出

$$ \varepsilon_{1} - \varepsilon_{2} = I R \quad\Rightarrow\quad I = \frac{\varepsilon_{1} - \varepsilon_{2}}{R}. $$

If instead they aid each other (same direction), the emfs add: $I = (\varepsilon_{1} + \varepsilon_{2})/R$. The sign bookkeeping is exactly the loop-law rule: traverse the loop, add $+\varepsilon$ for a rise and $-\varepsilon$ for a drop, subtract $IR$ for each resistor crossed with the current, and set the total to zero.

若二者相助（方向相同），电动势相加：$I = (\varepsilon_{1} + \varepsilon_{2})/R$。符号记账正是回路定律规则：沿回路绕行，电势升记 $+\varepsilon$、电势降记 $-\varepsilon$，每顺电流过一个电阻减 $IR$，令总和为零。

If a chosen current direction yields a negative $I$, the magnitude is still correct — the negative sign just means the real current flows opposite to your assumed direction. This is the standard, reliable way to handle multi-source circuits.

若所选电流方向算得 $I$ 为负，大小仍正确——负号只表示真实电流与假设方向相反。这是处理多电源电路标准而可靠的方法。

At a junction, currents of $3.0\ \mathrm{A}$ and $2.0\ \mathrm{A}$ flow in, and one current flows out along with a $1.5\ \mathrm{A}$ branch. The unknown outgoing current is:某节点流入电流 $3.0\ \mathrm{A}$ 与 $2.0\ \mathrm{A}$，流出有一条 $1.5\ \mathrm{A}$ 支路和一条未知支路。未知流出电流为：
B5.5 · Q1
$6.5\ \mathrm{A}$
$1.5\ \mathrm{A}$
$3.5\ \mathrm{A}$
$0.5\ \mathrm{A}$
Junction law: $\sum I_{\text{in}} = \sum I_{\text{out}}$. In: $3.0 + 2.0 = 5.0\ \mathrm{A}$. Out: $1.5 + I_{x} = 5.0 \Rightarrow I_{x} = 3.5\ \mathrm{A}$.节点定律：$\sum I_{\text{in}} = \sum I_{\text{out}}$。流入：$3.0 + 2.0 = 5.0\ \mathrm{A}$。流出：$1.5 + I_{x} = 5.0 \Rightarrow I_{x} = 3.5\ \mathrm{A}$。
Total in must equal total out. Sum the inflows ($5.0\ \mathrm{A}$), subtract the known outflow ($1.5\ \mathrm{A}$).流入总量必等于流出总量。把流入相加（$5.0\ \mathrm{A}$），减去已知流出（$1.5\ \mathrm{A}$）。
Kirchhoff's loop (voltage) law is a direct consequence of the conservation of:基尔霍夫回路（电压）定律直接源于以下哪个守恒：
B5.5 · Q2
Charge电荷
Energy能量
Momentum动量
Mass质量
The loop law states that energy per unit charge supplied by sources equals energy per unit charge dissipated around a loop — conservation of energy. The junction law is conservation of charge.回路定律表明电源每单位电荷供给的能量等于沿回路每单位电荷耗散的能量——即能量守恒。节点定律才是电荷守恒。
Charge conservation gives the junction law. The loop law (potential changes sum to zero around a loop) is energy conservation.电荷守恒给出节点定律。回路定律（沿回路电势变化之和为零）是能量守恒。
Topic B5.6 · Internal Resistance and Potential DividersTopic B5.6 · 内阻与分压器

Cells: emf, Internal Resistance and the Potential Divider电池：电动势、内阻与分压器 B.5 SL+HL

Real cells have internal resistance. A real cell of emf $\varepsilon$ behaves as an ideal source in series with an internal resistance $r$. Connected to an external load $R$, from the data booklet: $$ \varepsilon = I(R + r). $$ Terminal pd. The voltage actually available at the cell's terminals is $$ V = \varepsilon - I r. $$ The term $Ir$ is the "lost volts" dissipated inside the cell. At higher currents, more is lost, so the terminal pd drops.
  • Open circuit ($I = 0$): terminal pd $= \varepsilon$ (measure emf with a high-resistance voltmeter, no load).
  • Short circuit ($R = 0$): $I_{\max} = \varepsilon / r$.
Potential divider. Two resistors $R_{1}, R_{2}$ in series across a supply $V_{\text{in}}$ split the voltage. The output across $R_{2}$ is $$ V_{\text{out}} = V_{\text{in}}\,\frac{R_{2}}{R_{1} + R_{2}}. $$ A potentiometer (variable divider) gives a continuously adjustable output from $0$ to $V_{\text{in}}$.
真实电池有内阻。电动势 $\varepsilon$ 的真实电池等效为理想电源与内阻 $r$ 串联。接外负载 $R$ 时，数据手册公式： $$ \varepsilon = I(R + r). $$ 端电压（terminal pd）。电池端子实际可用电压为 $$ V = \varepsilon - I r. $$ 其中 $Ir$ 是电池内部耗散的"损失电压"。电流越大损失越多，故端电压下降。
  • 开路（$I = 0$）：端电压 $= \varepsilon$（用高阻电压表空载测电动势）。
  • 短路（$R = 0$）：$I_{\max} = \varepsilon / r$。
分压器（potential divider）。两电阻 $R_{1}, R_{2}$ 串联接在电源 $V_{\text{in}}$ 上分压。$R_{2}$ 两端输出为 $$ V_{\text{out}} = V_{\text{in}}\,\frac{R_{2}}{R_{1} + R_{2}}. $$ 电位器（可变分压器）可给出从 $0$ 到 $V_{\text{in}}$ 连续可调的输出。
Worked Example B5.6a (internal resistance)B5.6a 例题（内阻）

A cell of emf $1.5\ \mathrm{V}$ and internal resistance $0.50\ \Omega$ is connected to a $2.5\ \Omega$ resistor. Find the current, the terminal pd, and the power dissipated inside the cell.电动势 $1.5\ \mathrm{V}$、内阻 $0.50\ \Omega$ 的电池接到 $2.5\ \Omega$ 电阻。求电流、端电压及电池内部耗散功率。

Current. Use $\varepsilon = I(R + r)$:

电流。用 $\varepsilon = I(R + r)$：

$$ I = \frac{\varepsilon}{R + r} = \frac{1.5}{2.5 + 0.50} = \frac{1.5}{3.0} = 0.50\ \mathrm{A}. $$

Terminal pd. $V = \varepsilon - Ir = 1.5 - (0.50)(0.50) = 1.5 - 0.25 = 1.25\ \mathrm{V}$.

端电压。$V = \varepsilon - Ir = 1.5 - (0.50)(0.50) = 1.5 - 0.25 = 1.25\ \mathrm{V}$。

Power lost inside the cell. $P_{r} = I^{2} r = (0.50)^{2}(0.50) = 0.125\ \mathrm{W}$.

电池内部耗散功率。$P_{r} = I^{2} r = (0.50)^{2}(0.50) = 0.125\ \mathrm{W}$。

Evaluate. The terminal pd ($1.25\ \mathrm{V}$) is below the emf ($1.5\ \mathrm{V}$) by exactly the lost volts $Ir = 0.25\ \mathrm{V}$. Cross-check: $V = IR = (0.50)(2.5) = 1.25\ \mathrm{V}$, consistent.

评估。端电压（$1.25\ \mathrm{V}$）比电动势（$1.5\ \mathrm{V}$）低，恰好低出损失电压 $Ir = 0.25\ \mathrm{V}$。互校：$V = IR = (0.50)(2.5) = 1.25\ \mathrm{V}$，一致。

Worked Example B5.6b (potential divider)B5.6b 例题（分压器）

A $9.0\ \mathrm{V}$ supply is connected across a potential divider made of a $1.0\ \mathrm{k\Omega}$ resistor in series with a $2.0\ \mathrm{k\Omega}$ resistor. Find the output voltage taken across the $2.0\ \mathrm{k\Omega}$ resistor.$9.0\ \mathrm{V}$ 电源接在由 $1.0\ \mathrm{k\Omega}$ 与 $2.0\ \mathrm{k\Omega}$ 电阻串联组成的分压器上。求 $2.0\ \mathrm{k\Omega}$ 电阻两端的输出电压。

Apply the divider formula. Output across $R_{2} = 2.0\ \mathrm{k\Omega}$ with $R_{1} = 1.0\ \mathrm{k\Omega}$:

套用分压公式。取 $R_{2} = 2.0\ \mathrm{k\Omega}$ 两端输出，$R_{1} = 1.0\ \mathrm{k\Omega}$：

$$ V_{\text{out}} = V_{\text{in}}\,\frac{R_{2}}{R_{1} + R_{2}} = 9.0 \times \frac{2.0}{1.0 + 2.0} = 9.0 \times \frac{2}{3} = 6.0\ \mathrm{V}. $$

Evaluate. The voltage splits in proportion to resistance: the $2.0\ \mathrm{k\Omega}$ takes two-thirds of $9.0\ \mathrm{V}$. The k$\Omega$ units cancel in the ratio, so no unit conversion is needed.

评估。电压按电阻成比例分配：$2.0\ \mathrm{k\Omega}$ 分得 $9.0\ \mathrm{V}$ 的三分之二。比值中 k$\Omega$ 单位相消，无需换算。

Going deeper: measuring emf and internal resistance from a graph HL深入：由图线测定电动势与内阻 HL

Rearrange the terminal-pd equation into the form of a straight line. Starting from $V = \varepsilon - Ir$, a plot of terminal pd $V$ (vertical) against current $I$ (horizontal) gives

把端电压方程改写成直线形式。由 $V = \varepsilon - Ir$，以端电压 $V$（纵轴）对电流 $I$（横轴）作图，得

$$ V = -r\,I + \varepsilon. $$

This is $y = mx + c$ with

这是 $y = mx + c$，其中

  • $y$-intercept (where $I = 0$) $= \varepsilon$, the emf;
  • $y$ 轴截距（$I = 0$ 处）$= \varepsilon$，即电动势；
  • gradient $= -r$, so the magnitude of the slope gives the internal resistance.
  • 斜率 $= -r$，故斜率大小即内阻。

This is the standard data-response experiment: vary the load, record $V$ and $I$, plot the line, then read $\varepsilon$ from the intercept and $r$ from the slope. It is a frequent Paper 2 graph-and-extract question.

这是标准数据题实验：改变负载、记录 $V$ 与 $I$、作直线，再由截距读出 $\varepsilon$、由斜率读出 $r$。是 Paper 2 常见的作图与提取题。

A battery of emf $6.0\ \mathrm{V}$ delivers $2.0\ \mathrm{A}$ to a circuit, and its terminal pd reads $5.0\ \mathrm{V}$. Its internal resistance is:电动势 $6.0\ \mathrm{V}$ 的电池向电路供给 $2.0\ \mathrm{A}$，端电压读数 $5.0\ \mathrm{V}$。其内阻为：
B5.6 · Q1
$0.50\ \Omega$
$2.5\ \Omega$
$3.0\ \Omega$
$11\ \Omega$
Lost volts $= \varepsilon - V = 6.0 - 5.0 = 1.0\ \mathrm{V} = Ir$. So $r = 1.0 / 2.0 = 0.50\ \Omega$.损失电压 $= \varepsilon - V = 6.0 - 5.0 = 1.0\ \mathrm{V} = Ir$。故 $r = 1.0 / 2.0 = 0.50\ \Omega$。
The lost volts $Ir = \varepsilon - V = 1.0\ \mathrm{V}$. Divide by the current: $r = (\varepsilon - V)/I = 1.0/2.0$.损失电压 $Ir = \varepsilon - V = 1.0\ \mathrm{V}$。除以电流：$r = (\varepsilon - V)/I = 1.0/2.0$。
A $12\ \mathrm{V}$ supply feeds a potential divider of two equal resistors. The output across one resistor is:$12\ \mathrm{V}$ 电源接到由两个相等电阻组成的分压器。一个电阻两端的输出为：
B5.6 · Q2
$12\ \mathrm{V}$
$0\ \mathrm{V}$
$6.0\ \mathrm{V}$
$24\ \mathrm{V}$
$V_{\text{out}} = V_{\text{in}}\,R_{2}/(R_{1} + R_{2})$. With $R_{1} = R_{2}$, the ratio is $1/2$, so $V_{\text{out}} = 12 \times \tfrac{1}{2} = 6.0\ \mathrm{V}$.$V_{\text{out}} = V_{\text{in}}\,R_{2}/(R_{1} + R_{2})$。$R_{1} = R_{2}$ 时比值为 $1/2$，故 $V_{\text{out}} = 12 \times \tfrac{1}{2} = 6.0\ \mathrm{V}$。
Equal resistors split the supply equally. The divider ratio $R_{2}/(R_{1}+R_{2}) = 1/2$, giving half of $12\ \mathrm{V}$.相等电阻平分电源电压。分压比 $R_{2}/(R_{1}+R_{2}) = 1/2$，得 $12\ \mathrm{V}$ 的一半。
Exam Tactics考试技巧

Exam Strategy and Common Pitfalls考试策略与常见陷阱

Units and conversions (every paper)单位与换算（每张试卷）
  • Convert minutes and hours to seconds before using $I = \Delta q/\Delta t$ or $P = W/t$. A stray factor of $60$ or $3600$ is the most common B.5 arithmetic loss.
  • 使用 $I = \Delta q/\Delta t$ 或 $P = W/t$ 前先把分钟、小时换算成秒。多出或漏掉的 $60$、$3600$ 是 B.5 最常见的算术失分。
  • The kilowatt-hour is energy, not power. $1\ \mathrm{kW\,h} = 3.6 \times 10^{6}\ \mathrm{J}$. Use it for billing questions; convert to joules only when SI is required.
  • 千瓦时是能量而非功率。$1\ \mathrm{kW\,h} = 3.6 \times 10^{6}\ \mathrm{J}$。计费题用它；只有要求 SI 时才换算成焦耳。
Choosing the power form选对功率公式
  • $P = VI$ is always valid. Use $P = I^{2}R$ or $P = V^{2}/R$ only when the component obeys $V = IR$.
  • $P = VI$ 永远成立。仅当元件满足 $V = IR$ 时才用 $P = I^{2}R$ 或 $P = V^{2}/R$。
  • For a lamp or diode, read $V$ and $I$ off the characteristic and use $P = VI$. Do not assume a constant $R$.
  • 对灯泡或二极管，从特性曲线读出 $V$ 与 $I$ 后用 $P = VI$。不要假设 $R$ 恒定。
Circuit reduction (Paper 2 standard)电路化简（Paper 2 常考）
  • Reduce parallel blocks first, then add in series, then find the total current with $I = \varepsilon/R_{\text{eq}}$. Work back outward to get branch currents and voltages.
  • 先化简并联块，再串联相加，然后用 $I = \varepsilon/R_{\text{eq}}$ 求总电流。再由外向内回推各支路电流与电压。
  • Always finish with a junction check and a loop check. Branch currents must sum to the total; voltages around a loop must sum to the emf.
  • 最后务必做节点核查与回路核查。各支路电流之和应等于总电流；回路电压之和应等于电动势。
Internal resistance (the examiner's favourite trap)内阻（命题人最爱的陷阱）
  • Terminal pd is below emf by the lost volts $Ir$. Only an open-circuit ($I = 0$) voltmeter reads the true emf.
  • 端电压比电动势低出损失电压 $Ir$。只有开路（$I = 0$）的电压表才读出真实电动势。
  • To find $\varepsilon$ and $r$ from data, plot $V$ against $I$: intercept is $\varepsilon$, slope magnitude is $r$. State both clearly.
  • 由数据求 $\varepsilon$ 与 $r$ 时作 $V$-$I$ 图：截距为 $\varepsilon$，斜率大小为 $r$。两者都要写清。
Active Recall主动回忆

Flashcards闪卡

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Current as charge flow?电流（电荷流动）？
$$I = \frac{\Delta q}{\Delta t}$$
Drift-velocity current?漂移速度电流式？
$$I = n A v q$$
Potential difference definition?电势差定义？
$$V = \frac{W}{q}$$Energy per unit charge transferred.每单位电荷转化的能量。
Electrical power (three forms)?电功率（三种形式）？
$$P = VI = I^{2}R = \frac{V^{2}}{R}$$
One kilowatt-hour in joules?一千瓦时合多少焦耳？
$$1\ \mathrm{kW\,h} = 3.6 \times 10^{6}\ \mathrm{J}$$
Ohm's law / resistance definition?欧姆定律 / 电阻定义？
$$V = IR$$Ohmic if $R$ is constant.$R$ 恒定即为欧姆。
Resistivity formula?电阻率公式？
$$R = \frac{\rho L}{A}$$
Filament lamp characteristic?灯丝灯泡特性？
$R$ rises with heat; $I$-$V$ curve flattens. Non-ohmic.$R$ 随发热升高；$I$-$V$ 曲线变平。非欧姆。
Resistors in series?电阻串联？
$$R_{\text{s}} = R_{1} + R_{2} + \cdots$$
Resistors in parallel?电阻并联？
$$\frac{1}{R_{\text{p}}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} + \cdots$$
Kirchhoff's junction law?基尔霍夫节点定律？
$\sum I_{\text{in}} = \sum I_{\text{out}}$ — charge conservation.$\sum I_{\text{in}} = \sum I_{\text{out}}$——电荷守恒。
Kirchhoff's loop law?基尔霍夫回路定律？
$\sum \varepsilon = \sum IR$ — energy conservation.$\sum \varepsilon = \sum IR$——能量守恒。
emf with internal resistance?含内阻的电动势？
$$\varepsilon = I(R + r)$$Terminal pd $V = \varepsilon - Ir$.端电压 $V = \varepsilon - Ir$。
Potential divider output?分压器输出？
$$V_{\text{out}} = V_{\text{in}}\,\frac{R_{2}}{R_{1} + R_{2}}$$
Mixed Practice综合练习

Unit B.5 Practice Quiz单元 B.5 练习测验

A resistor dissipates $24\ \mathrm{W}$ when carrying $2.0\ \mathrm{A}$. Its resistance is:某电阻通过 $2.0\ \mathrm{A}$ 时耗散 $24\ \mathrm{W}$。其电阻为：
Q1
$48\ \Omega$
$12\ \Omega$
$6.0\ \Omega$
$2.0\ \Omega$
Use $P = I^{2}R$: $R = P/I^{2} = 24/(2.0)^{2} = 24/4 = 6.0\ \Omega$.用 $P = I^{2}R$：$R = P/I^{2} = 24/(2.0)^{2} = 24/4 = 6.0\ \Omega$。
With $P$ and $I$ known, use $P = I^{2}R$ and solve $R = P/I^{2}$. Remember to square the current.已知 $P$ 与 $I$，用 $P = I^{2}R$ 解 $R = P/I^{2}$。记得电流要平方。
A $6.0\ \Omega$ and a $3.0\ \Omega$ resistor are in parallel, and this combination is in series with a $4.0\ \Omega$ resistor. The total resistance is:$6.0\ \Omega$ 与 $3.0\ \Omega$ 电阻并联，该组合再与 $4.0\ \Omega$ 电阻串联。总电阻为：
Q2
$13\ \Omega$
$6.0\ \Omega$
$2.0\ \Omega$
$9.0\ \Omega$
Parallel pair: $R_{P} = (6 \times 3)/(6 + 3) = 18/9 = 2.0\ \Omega$. Add in series: $2.0 + 4.0 = 6.0\ \Omega$.并联对：$R_{P} = (6 \times 3)/(6 + 3) = 18/9 = 2.0\ \Omega$。串联相加：$2.0 + 4.0 = 6.0\ \Omega$。
Reduce the parallel block first ($R_{P} = R_{1}R_{2}/(R_{1}+R_{2}) = 2.0\ \Omega$), then add the series resistor.先化简并联块（$R_{P} = R_{1}R_{2}/(R_{1}+R_{2}) = 2.0\ \Omega$），再加串联电阻。
A cell of emf $1.5\ \mathrm{V}$ and internal resistance $0.30\ \Omega$ is short-circuited (load $R = 0$). The current is:电动势 $1.5\ \mathrm{V}$、内阻 $0.30\ \Omega$ 的电池被短路（负载 $R = 0$）。电流为：
Q3
$5.0\ \mathrm{A}$
$0.20\ \mathrm{A}$
$0.45\ \mathrm{A}$
$1.5\ \mathrm{A}$
With $R = 0$, $\varepsilon = I r$, so $I = \varepsilon / r = 1.5 / 0.30 = 5.0\ \mathrm{A}$. This is the maximum (short-circuit) current.$R = 0$ 时 $\varepsilon = I r$，故 $I = \varepsilon / r = 1.5 / 0.30 = 5.0\ \mathrm{A}$，即最大（短路）电流。
Short circuit means only the internal resistance limits the current: $I = \varepsilon / r$.短路时只有内阻限制电流：$I = \varepsilon / r$。
At a junction, $4.0\ \mathrm{A}$ enters and two branches carry $1.5\ \mathrm{A}$ and $2.5\ \mathrm{A}$ away. A third outgoing branch then carries:某节点流入 $4.0\ \mathrm{A}$，两条支路分别带走 $1.5\ \mathrm{A}$ 与 $2.5\ \mathrm{A}$。第三条流出支路带走：
Q4
$8.0\ \mathrm{A}$
$4.0\ \mathrm{A}$
$1.0\ \mathrm{A}$
$0\ \mathrm{A}$
Junction law: in $= 4.0\ \mathrm{A}$; out so far $= 1.5 + 2.5 = 4.0\ \mathrm{A}$. The inflow is already fully accounted for, so the third branch carries $0\ \mathrm{A}$.节点定律：流入 $= 4.0\ \mathrm{A}$；已流出 $= 1.5 + 2.5 = 4.0\ \mathrm{A}$。流入已被完全分配，故第三支路为 $0\ \mathrm{A}$。
Sum of outflows must equal inflow. $1.5 + 2.5 = 4.0\ \mathrm{A}$ already equals the $4.0\ \mathrm{A}$ in, leaving nothing for the third branch.流出之和须等于流入。$1.5 + 2.5 = 4.0\ \mathrm{A}$ 已等于流入的 $4.0\ \mathrm{A}$，第三支路无剩余。
HL A graph of a cell's terminal pd $V$ against current $I$ is a straight line. Which features give the emf and internal resistance?HL 某电池端电压 $V$ 对电流 $I$ 的图为直线。哪些特征给出电动势与内阻？
Q5
Slope $= \varepsilon$, intercept $= r$斜率 $= \varepsilon$，截距 $= r$
Slope $= r$, intercept $= 0$斜率 $= r$，截距 $= 0$
$y$-intercept $= \varepsilon$, slope magnitude $= r$$y$ 轴截距 $= \varepsilon$，斜率大小 $= r$
$x$-intercept $= \varepsilon$, slope $= 1/r$$x$ 轴截距 $= \varepsilon$，斜率 $= 1/r$
Rearranging $V = \varepsilon - Ir$ into $V = -rI + \varepsilon$ gives $y = mx + c$: the $y$-intercept (at $I = 0$) is $\varepsilon$, and the slope is $-r$, so the slope magnitude is $r$.把 $V = \varepsilon - Ir$ 改写为 $V = -rI + \varepsilon$，即 $y = mx + c$：$y$ 轴截距（$I = 0$ 处）为 $\varepsilon$，斜率为 $-r$，故斜率大小为 $r$。
Write $V = -rI + \varepsilon$. Compare with $y = mx + c$: intercept is the emf, slope is $-r$ (magnitude $r$).写成 $V = -rI + \varepsilon$。对照 $y = mx + c$：截距为电动势，斜率为 $-r$（大小为 $r$）。
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Readiness Checklist备考清单

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