High School Physics

Current Electricity and Circuits电流与电路

Current electricity is the study of charges in motion and the circuits that direct them. This guide builds from the definition of electric current ($I = \Delta Q / \Delta t$) through voltage, electromotive force (EMF), Ohm's law ($V = IR$), and resistance, then develops the rules for series and parallel circuits, the formulas for electrical power ($P = IV$) and energy, and finally Kirchhoff's junction and loop rules for multi-branch networks. Ontario and BC treat this topic thoroughly at Grade 11; NGSS addresses it only through the energy/field lens (HS-PS3-5); Alberta Physics 20/30 has no dedicated DC-circuits unit. Worked examples use real numbers throughout.电流（electric current，电流）是对运动电荷及其所经电路的研究。本指南从电流的定义（$I = \Delta Q / \Delta t$）出发，经电压（voltage，电压）、电动势（EMF，电动势）、欧姆定律（Ohm's law，欧姆定律，$V = IR$）与电阻（resistance，电阻），进而建立串联（series，串联）与并联（parallel，并联）电路的法则，推导电功率（electrical power，电功率，$P = IV$）与电能公式，最后介绍适用于多支路网络的基尔霍夫定律（Kirchhoff's laws，基尔霍夫定律）。全部例题均用真实数字演算。

7 sections7 节内容 Strong in ON · BC  |  Conceptual/absent in NGSS · ABON · BC 强  |  NGSS · AB 概念/缺席 Honors block on Kirchhoff's Rules基尔霍夫定律为荣誉级
Read me first先读这里

How to use this guide如何使用本指南

Current electricity and DC circuits are a strong strand in Ontario (SPH3U Strand F) and BC (Physics 11), which name Ohm's law and Kirchhoff's rules explicitly. NGSS has no dedicated circuit-analysis PE; the closest anchor is HS-PS3-5 (field/energy framing). Alberta Physics 20/30 defines electric current in Physics 30 (outcome 30–B2.7k) but has no standalone circuits unit — Alberta students should treat this guide as enrichment or pre-AP preparation. The table below maps which sections are assessed core for each track.电流与直流电路是安大略（SPH3U F 单元）和 BC（Physics 11）的强项学习线索，两者均明确点名欧姆定律与基尔霍夫定律。NGSS 没有专门的电路分析 PE；最接近的锚点是 HS-PS3-5（场／能量框架）。阿尔伯塔 Physics 30 在成果 30–B2.7k 中定义了电流，但无独立电路单元——阿尔伯塔学生应将本指南视为拓展或 AP 前准备。下表说明每条学习轨道中哪些节属于被评估的核心内容。

If you are in…如果你在… Focus on these sections重点学习 Defer / lighter可推迟 / 减负 Source依据
🇺🇸 US NGSS HS Physical Science美国 NGSS 物理科学 §1 (current) and §6 (power/energy) connect to HS-PS3-5 field/energy framing; use conceptually§1（电流）与 §6（电功率/能量）通过 HS-PS3-5 场／能量框架相连；以概念理解为主 §4–§5 (series/parallel rules) and §7 (Kirchhoff): valuable but not directly assessed at the NGSS PE level§4–§5（串并联规则）与 §7（基尔霍夫）：有价值但不在 NGSS PE 层面直接评估 ngss_hs_ps_extract.md — HS-PS3-5 (field/energy); no dedicated DC-circuits PE— HS-PS3-5（场/能量）；无专门的直流电路 PE
🇨🇦 ON Grade 11 — SPH3U安大略 11 年级 — SPH3U §1 through §7 in full. SPH3U Strand F covers current, circuits, power, and electromagnetism as core Grade 11 content§1 至 §7 完整学习。SPH3U F 单元把电流、电路、电功率与电磁学列为 11 年级核心内容 Nothing — Kirchhoff's rules (§7) are within SPH3U scope (F2/F3 investigation expectations)无 — 基尔霍夫定律（§7）在 SPH3U F2/F3 探究期望范围内 science_11-12_physics_extract.md — SPH3U Strand F Overall Expectations F1–F3— SPH3U F 单元总体期望 F1–F3
🇨🇦 BC Grade 11 — Physics 11BC 11 年级 — Physics 11 §1 through §7. BC Physics 11 Content explicitly names "electric circuits (DC), Ohm's law, and Kirchhoff's laws," so all sections are core§1 至 §7。BC Physics 11 内容明确点名"直流电路、欧姆定律与基尔霍夫定律"，故全部节为核心 Nothing — BC also includes terminal voltage vs EMF in the elaboration, so lean on §2 and §7无 — BC 在细化说明中还含端电压与 EMF，故加重 §2 与 §7 physics_11-12_extract.md — Physics 11 Content: electric circuits (DC), Ohm's law, Kirchhoff's laws, power & efficiency in circuits— Physics 11 内容：直流电路、欧姆定律、基尔霍夫定律、电路中的功率与效率
🇨🇦 AB — Physics 20/30阿尔伯塔 — Physics 20/30 §1 (current definition, 30–B2.7k) is directly assessed. §3 (Ohm's law) and §6 (power) are useful context for 30–B2 field/energy outcomes§1（电流定义，30–B2.7k）直接被评估。§3（欧姆定律）与 §6（电功率）是 30–B2 场/能量成果的有用背景 §4–§5 (series/parallel) and §7 (Kirchhoff): enrichment — no dedicated DC-circuits unit in Alberta Physics 20 or 30§4–§5（串并联）与 §7（基尔霍夫）：拓展内容 — 阿尔伯塔 Physics 20/30 无专门直流电路单元 physics_20-30_extract.md — Physics 30 Unit B GO2, knowledge outcome 30–B2.7k (electric current defined); no circuit-analysis outcomes— Physics 30 B 单元 GO2，知识成果 30–B2.7k（电流定义）；无电路分析成果
🇺🇸 AP / IB feeder trackAP / IB 衔接轨道 All seven sections plus going-deeper derivations. AP Physics 1 Unit 5 (Electric Circuits) and IB Physics HL Topic B.5 both build on Ohm's law, Kirchhoff, and power全部 7 节，并完成深入推导。AP Physics 1 第 5 单元（电路）与 IB Physics HL B.5 主题均建立在欧姆定律、基尔霍夫定律与功率之上 Nothing — this unit provides the circuit-analysis foundation AP and IB build on无 — 本单元提供 AP 与 IB 所依赖的电路分析基础 ngss_hs_ps_extract.md — see also Unit 8 (Electrostatics) as the field foundation; Unit 10 (Magnetism) as the next step— 另见 Unit 8（静电学）作为场的基础；Unit 10（磁学）为后续单元

Once you have located your row, use the two cards below for the speed at which you should work through the recommended sections.找到所在行后，用下面两张卡片决定推进速度。

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If you are cramming the night before如果你在临阵磨枪

Memorise five things: the current definition $I = \Delta Q / \Delta t$; Ohm's law $V = IR$; power $P = IV = \frac{V^2}{R} = I^2 R$; series rules ($R_{\text{total}} = R_1 + R_2 + \cdots$, same current everywhere); and parallel rules ($\frac{1}{R_{\text{total}}} = \frac{1}{R_1} + \frac{1}{R_2} + \cdots$, same voltage across each branch). Read every cram-cheat box. Skip the Kirchhoff derivations if pressed for time.背熟五件事：电流定义 $I = \Delta Q / \Delta t$；欧姆定律 $V = IR$；电功率 $P = IV = \frac{V^2}{R} = I^2 R$；串联规则（$R_{\text{total}} = R_1 + R_2 + \cdots$，各处电流相同）；并联规则（$\frac{1}{R_{\text{total}}} = \frac{1}{R_1} + \frac{1}{R_2} + \cdots$，各支路电压相同）。读每个速记框。时间紧时可跳过基尔霍夫推导。

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If you are going for the top mark如果你目标顶分

Understand why resistors add differently in series versus parallel (current path vs voltage constraint). Practise applying Kirchhoff's rules systematically: label all branch currents with assumed directions before writing equations. Know all three forms of the power formula and choose the one that uses given quantities. Be clear on the distinction between EMF (energy per charge supplied by a source) and terminal voltage (EMF minus the internal-resistance drop).理解为何串联与并联中电阻的叠加方式不同（电流路径 vs 电压约束）。练习系统地应用基尔霍夫定律：写方程前先标出所有支路电流的假设方向。熟悉电功率公式的三种形式，选取利用已知量的那种。明确区分 EMF（电源提供的每单位电荷能量）与端电压（EMF 减去内阻压降）。

Honors flag.荣誉级标记。 Section 7 (Kirchhoff's Rules) carries the Honors chip because multi-loop network analysis goes beyond the single-loop scope of most Grade 11 courses. Ontario SPH3U and BC Physics 11 introduce Kirchhoff's rules at the conceptual level; quantitative multi-loop problems are an AP Physics 1 / IB Physics HL expectation. If your row above sends you to §7, read it as enrichment or as AP/IB preparation.§7（基尔霍夫定律）标 Honors，因为多回路网络分析超出大多数 11 年级课程的单回路范围。安大略 SPH3U 与 BC Physics 11 在概念层面引入基尔霍夫定律；定量多回路问题是 AP Physics 1 / IB Physics HL 的期望。如果你的行指向 §7，请将其视为拓展或 AP/IB 准备。
Section 1 · Electric current & charge flow第 1 节 · 电流与电荷流动

Electric Current and Charge Flow电流（电流）与电荷流动

Current is the rate of charge flow — the fundamental quantity of circuits.电流是电荷流动的速率 — 电路最根本的量。 $$ I = \frac{\Delta Q}{\Delta t} \qquad \text{(electric current, SI unit: ampere A)} $$
  • Conventional current惯用电流方向 — defined as the direction positive charges would flow. In metal wires, actual carriers are electrons flowing in the opposite direction, but we always draw and label current in the conventional direction.— 定义为正电荷流动的方向。在金属导线中，实际载流子是向相反方向流动的电子，但我们始终按惯用方向画出并标注电流。
  • Charge $Q$电荷量 $Q$ — SI unit coulomb (C). The elementary charge is $e = 1.6 \times 10^{-19}$ C. One ampere means one coulomb per second: $1\ \mathrm{A} = 1\ \mathrm{C/s}$.— 国际单位库仑（C）。基本电荷 $e = 1.6 \times 10^{-19}$ C。一安培即每秒一库仑：$1\ \mathrm{A} = 1\ \mathrm{C/s}$。
  • Direct current (DC)直流电（DC） — current that flows in one direction only (batteries, this entire unit). Alternating current (AC) reverses periodically — that is a separate topic.— 仅沿一个方向流动的电流（电池、本整个单元）。交流电（AC）周期性反向 — 那是另一个话题。
Alberta 30–B2.7k defines "electric current as the amount of charge passing a reference point per unit of time" — exactly the definition above. Ontario SPH3U Strand F and BC Physics 11 both list current and electron flow as core content.阿尔伯塔 30–B2.7k 将"电流"定义为"单位时间内通过参考点的电荷量"——正是上面的定义。安大略 SPH3U F 单元与 BC Physics 11 都把电流与电子流列为核心内容。
Worked Example 1 · Charge and current例题 1 · 电荷与电流

A charge of $3.6$ C passes through a wire in $1.5$ s. (a) What is the current? (b) How many electrons pass in this time? (Use $e = 1.6 \times 10^{-19}$ C.)$3.6$ C 的电荷在 $1.5$ s 内通过一根导线。(a) 电流是多少？(b) 这段时间内有多少个电子通过？（取 $e = 1.6 \times 10^{-19}$ C。）

(a) Apply the definition.(a) 套用定义。

$$ I = \frac{\Delta Q}{\Delta t} = \frac{3.6}{1.5} = 2.4\ \mathrm{A}. $$

(b) Number of electrons.(b) 电子数。

$$ n = \frac{\Delta Q}{e} = \frac{3.6}{1.6 \times 10^{-19}} = 2.25 \times 10^{19} \text{ electrons.} $$

Sanity-check.合理性核验。 A $2.4$ A current is typical for a small appliance (a phone charger draws roughly $1$–$2$ A). The electron count is enormous precisely because the elementary charge is so tiny.$2.4$ A 的电流是小型电器的典型值（手机充电器约 $1$–$2$ A）。电子数之所以庞大，正因基本电荷极小。

A current of $0.50$ A flows for $4.0$ s. How much charge passes?$0.50$ A 的电流流动 $4.0$ s。通过了多少电荷？
§1 · Q1
$0.125$ C
$2.0$ C
$4.5$ C
$8.0$ C
$\Delta Q = I \cdot \Delta t = 0.50 \times 4.0 = 2.0$ C.$\Delta Q = I \cdot \Delta t = 0.50 \times 4.0 = 2.0$ C。
Rearrange $I = \Delta Q / \Delta t$ to get $\Delta Q = I \Delta t$.将 $I = \Delta Q / \Delta t$ 变形得 $\Delta Q = I \Delta t$。
In a metal wire, what carries the current?在金属导线中，电流由什么来承载？
§1 · Q2
Protons moving in the conventional current direction沿惯用电流方向运动的质子
Positive ions moving toward the negative terminal向负极移动的正离子
Free electrons moving opposite to the conventional current direction沿惯用电流反方向运动的自由电子
Protons and electrons moving in the same direction沿相同方向运动的质子与电子
In metals, only electrons are free to move. They flow toward the positive terminal, which is opposite to the conventional current direction (positive to negative outside the source).在金属中，只有电子可以自由移动。它们流向正极，与惯用电流方向（在电源外从正极到负极）相反。
Metal lattice ions cannot move. Conventional current is defined in the direction positive charges would flow; electrons (negative) flow the opposite way.金属晶格中的离子不能移动。惯用电流方向定义为正电荷的流动方向；电子（负电荷）则向相反方向流动。
Going deeper — drift velocity and why current is slow but signals are fast深入 — 漂移速度：为何电流很慢但信号很快

In a copper wire carrying $1$ A with cross-section $1\ \mathrm{mm}^2$, the average drift velocity of electrons is only about $0.07\ \mathrm{mm/s}$ — far slower than a snail. Yet a light turns on almost instantly. The reason: the entire electron sea throughout the wire shifts together. The electromagnetic field that "pushes" the sea propagates at nearly the speed of light, so the signal (energy) travels fast even though individual electrons barely move. This distinction — between the speed of charge carriers and the speed of the electrical signal — is a favourite AP/IB conceptual question.在截面积为 $1\ \mathrm{mm}^2$、电流为 $1$ A 的铜导线中，电子的平均漂移速度只有约 $0.07\ \mathrm{mm/s}$ —— 比蜗牛还慢。但电灯几乎瞬间就亮了。原因：整个导线中的电子海整体同步位移。"推动"电子海的电磁场以接近光速传播，因此信号（能量）传播极快，即使单个电子几乎不移动。载流子速度与电信号速度之间的这一区别，是 AP/IB 最爱考的概念题。

Section 2 · Voltage, EMF & electrical energy第 2 节 · 电压、电动势与电能

Voltage, EMF and Electrical Energy电压（电压）、电动势（电动势）与电能

Voltage is the energy per unit charge — the "pressure" that drives current.电压是每单位电荷的能量 — 驱动电流的"压力"。 $$ V = \frac{W}{Q} \qquad \text{(potential difference, SI unit: volt V)} $$
  • Potential difference (voltage) $V$电位差（电压）$V$ — the work $W$ done per unit charge $Q$ moved between two points. Unit: $1\ \mathrm{V} = 1\ \mathrm{J/C}$.— 在两点之间移动单位电荷 $Q$ 所做的功 $W$。单位：$1\ \mathrm{V} = 1\ \mathrm{J/C}$。
  • EMF (electromotive force) $\varepsilon$电动势 $\varepsilon$ — the energy per unit charge supplied by a source (battery, generator). EMF is not a force; it is an energy per charge (units: volts). A $9\ \mathrm{V}$ battery supplies $9\ \mathrm{J}$ per coulomb of charge it drives around the circuit.— 电源（电池、发电机）提供的每单位电荷能量。电动势不是力；它是每单位电荷的能量（单位：伏特）。$9\ \mathrm{V}$ 电池对其驱动绕回路的每库仑电荷提供 $9\ \mathrm{J}$。
  • Terminal voltage vs EMF.端电压与电动势。 A real battery has internal resistance $r$. When delivering current $I$, the terminal voltage is $V_T = \varepsilon - Ir$ (less than the EMF). BC Physics 11 elaborations explicitly name "terminal voltage versus electromotive force (EMF)."实际电池有内阻 $r$。输出电流 $I$ 时，端电压为 $V_T = \varepsilon - Ir$（小于电动势）。BC Physics 11 细化说明明确点名"端电压与电动势（EMF）"。
$$ V_T = \varepsilon - Ir \qquad \text{(terminal voltage with internal resistance)} $$
Worked Example 2 · Internal resistance and terminal voltage例题 2 · 内阻与端电压

A battery has EMF $\varepsilon = 12.0\ \mathrm{V}$ and internal resistance $r = 0.50\ \Omega$. When it delivers a current of $4.0\ \mathrm{A}$, find the terminal voltage and the power lost internally.一只电池的电动势 $\varepsilon = 12.0\ \mathrm{V}$，内阻 $r = 0.50\ \Omega$。当它输出 $4.0\ \mathrm{A}$ 电流时，求端电压与内部损耗功率。

Terminal voltage.端电压。

$$ V_T = \varepsilon - Ir = 12.0 - (4.0)(0.50) = 12.0 - 2.0 = 10.0\ \mathrm{V}. $$

Power lost internally.内部损耗功率。 The $2.0$ V drop across internal resistance dissipates power:内阻上的 $2.0$ V 压降消耗功率：

$$ P_{\text{internal}} = I^2 r = (4.0)^2 (0.50) = 8.0\ \mathrm{W}. $$

Interpretation.解读。 The battery generates $\varepsilon I = 48\ \mathrm{W}$; $8\ \mathrm{W}$ heats the battery itself; $40\ \mathrm{W}$ is delivered to the external circuit. Terminal voltage drops as current increases — this is why a "flat" battery lights a bulb dimly.电池产生 $\varepsilon I = 48\ \mathrm{W}$；$8\ \mathrm{W}$ 使电池自身发热；$40\ \mathrm{W}$ 输出到外电路。电流增大时端电压下降 — 这就是"快没电"的电池使灯泡变暗的原因。

Moving $2.0$ C of charge through a device requires $14$ J of work. What is the voltage across the device?将 $2.0$ C 的电荷通过某器件需要做 $14$ J 的功。该器件两端电压是多少？
§2 · Q1
$28$ V
$0.14$ V
$12$ V
$7.0$ V
$V = W/Q = 14/2.0 = 7.0$ V.$V = W/Q = 14/2.0 = 7.0$ V。
Voltage is energy per unit charge: $V = W/Q$.电压是每单位电荷的能量：$V = W/Q$。
A battery has EMF $6.0$ V and internal resistance $0.20\ \Omega$. It delivers $3.0$ A. What is the terminal voltage?一只电池电动势 $6.0$ V，内阻 $0.20\ \Omega$，输出 $3.0$ A 电流。端电压是多少？
§2 · Q2
$6.0$ V
$5.4$ V
$6.6$ V
$3.0$ V
$V_T = \varepsilon - Ir = 6.0 - (3.0)(0.20) = 6.0 - 0.60 = 5.4$ V.$V_T = \varepsilon - Ir = 6.0 - (3.0)(0.20) = 6.0 - 0.60 = 5.4$ V。
Terminal voltage $= \varepsilon - Ir$. The internal resistance drop reduces the voltage available at the terminals.端电压 $= \varepsilon - Ir$。内阻压降使端子处可用电压降低。
Section 3 · Resistance & Ohm's Law第 3 节 · 电阻（电阻）与欧姆定律（欧姆定律）

Resistance and Ohm's Law电阻与欧姆定律

Ohm's law: voltage, current, and resistance are linked by $V = IR$.欧姆定律：电压、电流与电阻通过 $V = IR$ 相连。 $$ V = IR \qquad \text{(Ohm's law, SI: V in volts, I in amperes, R in ohms }\Omega\text{)} $$
  • Resistance $R$电阻 $R$ — the opposition a material offers to current flow. Unit: ohm ($\Omega$). $1\ \Omega = 1\ \mathrm{V/A}$.— 材料对电流流动的阻碍。单位：欧姆（$\Omega$）。$1\ \Omega = 1\ \mathrm{V/A}$。
  • Ohmic vs non-ohmic.欧姆式与非欧姆式。 An ohmic resistor has constant $R$ regardless of voltage; $V$ vs $I$ is a straight line through the origin. LEDs and diodes are non-ohmic ($R$ varies with $V$).欧姆式电阻器的 $R$ 不随电压变化；$V$ 对 $I$ 图是过原点的直线。LED 和二极管是非欧姆式的（$R$ 随 $V$ 变化）。
  • Factors affecting resistance.影响电阻的因素。 For a uniform wire: $R = \rho L / A$, where $\rho$ is resistivity (material property), $L$ is length, $A$ is cross-sectional area. Longer $\Rightarrow$ more $R$; thicker $\Rightarrow$ less $R$.均匀导线：$R = \rho L / A$，其中 $\rho$ 是电阻率（材料属性），$L$ 是长度，$A$ 是截面积。越长 $\Rightarrow$ $R$ 越大；越粗 $\Rightarrow$ $R$ 越小。
BC Physics 11 and Ontario SPH3U both name Ohm's law as core circuit content. Alberta 30–B2.7k defines current; Ohm's law itself is enrichment for the AB track.BC Physics 11 与安大略 SPH3U 都把欧姆定律列为核心电路内容。阿尔伯塔 30–B2.7k 定义了电流；欧姆定律本身对 AB 轨道属于拓展内容。
Worked Example 3 · Finding resistance from Ohm's law例题 3 · 用欧姆定律求电阻

A toaster draws $5.0$ A when connected to a $120\ \mathrm{V}$ supply. (a) What is its resistance? (b) If the supply voltage drops to $110\ \mathrm{V}$, what current will flow (assuming the resistance stays constant)?一只烤面包机接在 $120\ \mathrm{V}$ 电源上时电流为 $5.0$ A。(a) 其电阻是多少？(b) 若电源电压降到 $110\ \mathrm{V}$（假设电阻不变），电流为多少？

(a) Resistance.(a) 电阻。

$$ R = \frac{V}{I} = \frac{120}{5.0} = 24\ \Omega. $$

(b) New current at lower voltage.(b) 较低电压下的新电流。

$$ I = \frac{V}{R} = \frac{110}{24} \approx 4.6\ \mathrm{A}. $$

Sanity-check.合理性核验。 Lower voltage $\Rightarrow$ lower current (both drop proportionally for a fixed $R$). The toaster will heat more slowly at $110\ \mathrm{V}$.较低电压 $\Rightarrow$ 较小电流（对固定 $R$，两者成比例下降）。$110\ \mathrm{V}$ 时烤面包机加热较慢。

A resistor has $R = 15\ \Omega$ and carries $0.40$ A. What is the voltage across it?一只电阻 $R = 15\ \Omega$，通过 $0.40$ A 电流。其两端电压是多少？
§3 · Q1
$37.5$ V
$0.027$ V
$6.0$ V
$15$ V
$V = IR = (0.40)(15) = 6.0$ V.$V = IR = (0.40)(15) = 6.0$ V。
Ohm's law: $V = IR$. Multiply current by resistance.欧姆定律：$V = IR$。电流乘以电阻。
On a $V$–$I$ graph for an ohmic resistor, what does the slope represent?欧姆式电阻器的 $V$–$I$ 图上，斜率表示什么？
§3 · Q2
Resistance $R$电阻 $R$
Current $I$电流 $I$
Power $P$电功率 $P$
Voltage $V$电压 $V$
From $V = IR$, the slope of $V$ vs $I$ is $V/I = R$. For an ohmic resistor this is constant, giving a straight line with slope $R$.由 $V = IR$，$V$ 对 $I$ 图的斜率为 $V/I = R$。欧姆式电阻器的斜率恒定，得到斜率为 $R$ 的直线。
Slope of $V$ vs $I$ is $\Delta V / \Delta I = R$ by Ohm's law.$V$ 对 $I$ 图的斜率为 $\Delta V / \Delta I = R$（欧姆定律）。
Section 4 · Series circuits第 4 节 · 串联（串联）电路

Series Circuits串联电路

In a series circuit, the same current flows through every component.串联电路中，同一电流流过每个元件。 $$ R_{\text{series}} = R_1 + R_2 + R_3 + \cdots \qquad \text{(resistances add directly)} $$
  • Current rule.电流法则。 Same current $I$ everywhere in the loop: $I = I_1 = I_2 = \cdots$.回路各处电流 $I$ 相同：$I = I_1 = I_2 = \cdots$。
  • Voltage rule.电压法则。 Voltage divides: $V_{\text{total}} = V_1 + V_2 + \cdots$. Each resistor gets a share proportional to its resistance (voltage divider).电压分配：$V_{\text{total}} = V_1 + V_2 + \cdots$。每只电阻分得与其阻值成比例的份额（分压器）。
  • Why total resistance increases.为何总电阻增大。 Each added resistor forces current through more opposition; the single-path constraint means every ohm adds directly.每增加一只电阻都使电流多经历一段阻碍；单路径约束使每欧姆直接叠加。
Worked Example 4 · Three resistors in series例题 4 · 三只电阻串联

Three resistors $R_1 = 4.0\ \Omega$, $R_2 = 6.0\ \Omega$, $R_3 = 2.0\ \Omega$ are connected in series to a $24\ \mathrm{V}$ battery (assume negligible internal resistance). Find (a) the total resistance, (b) the current, and (c) the voltage across $R_2$.三只电阻 $R_1 = 4.0\ \Omega$、$R_2 = 6.0\ \Omega$、$R_3 = 2.0\ \Omega$ 串联接在 $24\ \mathrm{V}$ 电池上（内阻忽略不计）。求 (a) 总电阻，(b) 电流，(c) $R_2$ 两端电压。

(a) Total resistance.(a) 总电阻。

$$ R_{\text{total}} = 4.0 + 6.0 + 2.0 = 12\ \Omega. $$

(b) Current (same throughout).(b) 电流（各处相同）。

$$ I = \frac{V}{R_{\text{total}}} = \frac{24}{12} = 2.0\ \mathrm{A}. $$

(c) Voltage across $R_2$.(c) $R_2$ 两端电压。

$$ V_2 = I R_2 = (2.0)(6.0) = 12\ \mathrm{V}. $$

Check: voltages sum to supply.核验：各电压之和等于电源电压。 $V_1 + V_2 + V_3 = 8 + 12 + 4 = 24\ \mathrm{V}$. ✓

Two resistors $5.0\ \Omega$ and $10\ \Omega$ are in series across a $30\ \mathrm{V}$ supply. What is the current?$5.0\ \Omega$ 与 $10\ \Omega$ 两只电阻串联接在 $30\ \mathrm{V}$ 电源上。电流是多少？
§4 · Q1
$6.0$ A
$3.0$ A
$1.0$ A
$2.0$ A
$R_{\text{total}} = 5.0 + 10 = 15\ \Omega$; $I = V/R = 30/15 = 2.0$ A.$R_{\text{total}} = 5.0 + 10 = 15\ \Omega$；$I = V/R = 30/15 = 2.0$ A。
In series, add resistances first, then use Ohm's law: $I = V / (R_1 + R_2)$.串联先加电阻，再用欧姆定律：$I = V / (R_1 + R_2)$。
Adding a third resistor in series with two existing ones will:在两只现有电阻的串联电路中再串联一只电阻，将会：
§4 · Q2
Decrease the total resistance and increase the current减小总电阻并增大电流
Increase the total resistance and decrease the current增大总电阻并减小电流
Not change the total resistance不改变总电阻
Increase the voltage across each original resistor增大每只原有电阻两端的电压
Series resistances add: $R_{\text{total}}$ increases. By Ohm's law $I = V/R_{\text{total}}$, a larger $R$ gives a smaller $I$ for the same supply voltage.串联电阻叠加：$R_{\text{total}}$ 增大。由欧姆定律 $I = V/R_{\text{total}}$，相同电源电压下 $R$ 较大则 $I$ 较小。
In series, each new resistor adds to $R_{\text{total}}$, which then reduces the current for a fixed voltage.串联中每只新增电阻都加入 $R_{\text{total}}$，固定电压下电流因此减小。
Section 5 · Parallel circuits第 5 节 · 并联（并联）电路

Parallel Circuits并联电路

In a parallel circuit, the same voltage appears across every branch.并联电路中，每条支路两端电压相同。 $$ \frac{1}{R_{\text{parallel}}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \cdots \qquad \text{(reciprocals add)} $$
  • Voltage rule.电压法则。 Same voltage $V$ across every branch: $V = V_1 = V_2 = \cdots$.每条支路电压 $V$ 相同：$V = V_1 = V_2 = \cdots$。
  • Current rule.电流法则。 Current divides: $I_{\text{total}} = I_1 + I_2 + \cdots$. More current flows through lower-resistance branches.电流分配：$I_{\text{total}} = I_1 + I_2 + \cdots$。阻值较小的支路通过更多电流。
  • Why total resistance decreases.为何总电阻减小。 Each new parallel branch opens an additional path for current; more paths $\Rightarrow$ less total opposition. $R_{\text{parallel}}$ is always less than the smallest individual $R$.每条新并联支路为电流开辟额外通道；通道越多 $\Rightarrow$ 总阻碍越小。$R_{\text{parallel}}$ 始终小于最小的单个 $R$。
  • Two equal resistors in parallel.两只等值电阻并联。 $R_{\text{parallel}} = R/2$. (Quick sanity-check for any two-resistor parallel calculation.)$R_{\text{parallel}} = R/2$。（任意两电阻并联计算的快速核验。）
Worked Example 5 · Two resistors in parallel例题 5 · 两只电阻并联

Two resistors $R_1 = 6.0\ \Omega$ and $R_2 = 12\ \Omega$ are connected in parallel to a $12\ \mathrm{V}$ supply. Find (a) the equivalent resistance, (b) the total current from the supply, and (c) the current through each branch.$R_1 = 6.0\ \Omega$ 与 $R_2 = 12\ \Omega$ 两只电阻并联接在 $12\ \mathrm{V}$ 电源上。求 (a) 等效电阻，(b) 电源总电流，(c) 各支路电流。

(a) Equivalent resistance.(a) 等效电阻。

$$ \frac{1}{R_{\text{eq}}} = \frac{1}{6.0} + \frac{1}{12} = \frac{2}{12} + \frac{1}{12} = \frac{3}{12} \;\Longrightarrow\; R_{\text{eq}} = 4.0\ \Omega. $$

Note: $4.0\ \Omega$ is less than both $6.0\ \Omega$ and $12\ \Omega$ — as expected for parallel.注意：$4.0\ \Omega$ 小于 $6.0\ \Omega$ 与 $12\ \Omega$ — 并联时的预期结果。

(b) Total current.(b) 总电流。

$$ I_{\text{total}} = \frac{V}{R_{\text{eq}}} = \frac{12}{4.0} = 3.0\ \mathrm{A}. $$

(c) Branch currents (same voltage $12$ V across each).(c) 支路电流（各支路电压均为 $12$ V）。

$$ I_1 = \frac{V}{R_1} = \frac{12}{6.0} = 2.0\ \mathrm{A}, \qquad I_2 = \frac{V}{R_2} = \frac{12}{12} = 1.0\ \mathrm{A}. $$

Check: $I_1 + I_2 = 2.0 + 1.0 = 3.0$ A $= I_{\text{total}}$. ✓核验：$I_1 + I_2 = 2.0 + 1.0 = 3.0$ A $= I_{\text{total}}$。✓

Two resistors $4.0\ \Omega$ and $4.0\ \Omega$ are connected in parallel. What is the equivalent resistance?$4.0\ \Omega$ 与 $4.0\ \Omega$ 两只电阻并联。等效电阻是多少？
§5 · Q1
$8.0\ \Omega$
$4.0\ \Omega$
$2.0\ \Omega$
$1.0\ \Omega$
Two equal resistors in parallel: $R_{\text{eq}} = R/2 = 4.0/2 = 2.0\ \Omega$. (Check: $1/R_{\text{eq}} = 1/4 + 1/4 = 1/2$, so $R_{\text{eq}} = 2$.)两只等值电阻并联：$R_{\text{eq}} = R/2 = 4.0/2 = 2.0\ \Omega$。（核验：$1/R_{\text{eq}} = 1/4 + 1/4 = 1/2$，故 $R_{\text{eq}} = 2$。）
Parallel resistors combine as $1/R_{\text{eq}} = 1/R_1 + 1/R_2$. For two equal resistors, $R_{\text{eq}} = R/2$.并联电阻：$1/R_{\text{eq}} = 1/R_1 + 1/R_2$。两只等值电阻时，$R_{\text{eq}} = R/2$。
In a parallel circuit with branches of $3\ \Omega$ and $6\ \Omega$ across a $12\ \mathrm{V}$ source, which branch carries more current?$12\ \mathrm{V}$ 电源并联电路中，支路分别为 $3\ \Omega$ 与 $6\ \Omega$，哪条支路电流更大？
§5 · Q2
The $3\ \Omega$ branch (lower resistance, more current)$3\ \Omega$ 支路（阻值更小，电流更大）
The $6\ \Omega$ branch (higher resistance, more current)$6\ \Omega$ 支路（阻值更大，电流更大）
Both branches carry the same current两条支路电流相同
Cannot be determined without the total current不知道总电流无法判断
Both branches have the same voltage ($12$ V). $I = V/R$: the $3\ \Omega$ branch carries $12/3 = 4$ A; the $6\ \Omega$ branch carries $12/6 = 2$ A. Lower resistance $\Rightarrow$ more current.两条支路电压相同（$12$ V）。$I = V/R$：$3\ \Omega$ 支路 $12/3 = 4$ A；$6\ \Omega$ 支路 $12/6 = 2$ A。阻值越小 $\Rightarrow$ 电流越大。
In parallel, voltage is the same across each branch. Use $I = V/R$: lower $R$ means larger $I$ for the same $V$.并联中各支路电压相同。用 $I = V/R$：相同 $V$ 下 $R$ 越小则 $I$ 越大。
Section 6 · Electrical power & energy第 6 节 · 电功率（电功率）与电能

Electrical Power and Energy电功率与电能

Power is the rate at which electrical energy is delivered or dissipated.电功率是电能被输送或耗散的速率。 $$ P = IV = I^2 R = \frac{V^2}{R} \qquad \text{(SI unit: watt W)} $$
  • Three forms, one choose.三种形式，按需选择。 All three are equivalent via Ohm's law. Use $P = IV$ when you know $I$ and $V$; use $P = V^2/R$ when you know $V$ and $R$; use $P = I^2 R$ when you know $I$ and $R$.三种形式通过欧姆定律等价。已知 $I$ 与 $V$ 用 $P = IV$；已知 $V$ 与 $R$ 用 $P = V^2/R$；已知 $I$ 与 $R$ 用 $P = I^2 R$。
  • Electrical energy.电能。 $W = P \Delta t$ (joules). Utility companies use kWh: $1\ \mathrm{kWh} = 3.6 \times 10^6\ \mathrm{J}$.$W = P \Delta t$（焦耳）。电力公司用千瓦时：$1\ \mathrm{kWh} = 3.6 \times 10^6\ \mathrm{J}$。
  • Joule heating.焦耳热效应。 In a resistor all the electrical power is converted to thermal energy ($P = I^2 R$). This is the basis of heaters, incandescent bulbs, and fuse ratings.在电阻中，所有电功率都转化为热能（$P = I^2 R$）。这是电暖器、白炽灯和保险丝额定值的物理基础。
BC Physics 11 explicitly links "power and efficiency: ... numerical examples (e.g., resistance, power, and efficiency in circuits)" to the circuits content. Ontario SPH3U Strand D and F cover energy transformations and electrical power.BC Physics 11 明确将"功率与效率：……数值例题（例如电路中的电阻、功率与效率）"与电路内容相连。安大略 SPH3U D 单元与 F 单元涵盖能量转换与电功率。
Worked Example 6 · Power and energy in a household appliance例题 6 · 家用电器的功率与电能

A $1500\ \mathrm{W}$ electric kettle operates at $120\ \mathrm{V}$. Find (a) the current it draws, (b) its resistance, and (c) the electrical energy consumed in $3.0$ min.一只 $1500\ \mathrm{W}$ 的电热水壶在 $120\ \mathrm{V}$ 下工作。求 (a) 其电流，(b) 其电阻，(c) $3.0$ min 内消耗的电能。

(a) Current.(a) 电流。

$$ I = \frac{P}{V} = \frac{1500}{120} = 12.5\ \mathrm{A}. $$

(b) Resistance.(b) 电阻。

$$ R = \frac{V}{I} = \frac{120}{12.5} = 9.6\ \Omega. \quad \text{(or } R = V^2/P = 14400/1500 = 9.6\ \Omega\text{)} $$

(c) Energy consumed in $3.0$ min $= 180$ s.(c) $3.0$ min $= 180$ s 内消耗的电能。

$$ W = P\,\Delta t = 1500 \times 180 = 2.7 \times 10^5\ \mathrm{J} = 270\ \mathrm{kJ}. $$

(In kWh: $W = 1.5 \times (3/60) = 0.075\ \mathrm{kWh}$, at roughly \$0.15/kWh this costs about \$0.01.)（以千瓦时计：$W = 1.5 \times (3/60) = 0.075\ \mathrm{kWh}$，按约 $\$0.15/\mathrm{kWh}$ 计费约 $\$0.01$。）

A $60\ \Omega$ resistor carries $2.0$ A. What power is dissipated?一只 $60\ \Omega$ 电阻通过 $2.0$ A 电流。耗散功率是多少？
§6 · Q1
$30$ W
$120$ W
$60$ W
$240$ W
$P = I^2 R = (2.0)^2 (60) = 4 \times 60 = 240$ W.$P = I^2 R = (2.0)^2 (60) = 4 \times 60 = 240$ W。
When you know $I$ and $R$, use $P = I^2 R$. Remember to square the current first.已知 $I$ 与 $R$ 时，用 $P = I^2 R$。记得先把电流平方。
A $800\ \mathrm{W}$ appliance runs for $0.50$ h. How much energy does it consume? 🇨🇦 BC Physics 11 power & efficiency一只 $800\ \mathrm{W}$ 电器运行 $0.50$ h。消耗了多少电能？🇨🇦 BC Physics 11 功率与效率
§6 · Q2
$400$ J
$1.44 \times 10^6$ J
$800$ J
$1600$ J
$W = P\,\Delta t = 800 \times (0.50 \times 3600) = 800 \times 1800 = 1.44 \times 10^6$ J (or $0.40$ kWh).$W = P\,\Delta t = 800 \times (0.50 \times 3600) = 800 \times 1800 = 1.44 \times 10^6$ J（或 $0.40$ kWh）。
Convert hours to seconds first: $0.50\ \mathrm{h} = 1800\ \mathrm{s}$. Then $W = P\,\Delta t$.先把小时换算成秒：$0.50\ \mathrm{h} = 1800\ \mathrm{s}$。再用 $W = P\,\Delta t$。
Section 7 · Kirchhoff's Rules Honors第 7 节 · 基尔霍夫定律（基尔霍夫定律） Honors

Kirchhoff's Rules Honors基尔霍夫定律 荣誉

Curriculum note.课纲提示。 BC Physics 11 explicitly names "Kirchhoff's laws" as Content. Ontario SPH3U Strand F includes Kirchhoff's rules within the circuit investigation expectations (F2/F3). NGSS has no dedicated PE for this material. Alberta Physics 20/30 does not include Kirchhoff's rules as a learning outcome. Quantitative multi-loop analysis is an AP Physics 1 / IB Physics HL expectation. This section carries the Honors chip for all tracks except where Kirchhoff's rules are explicitly named in the curriculum.BC Physics 11 在内容中明确点名"基尔霍夫定律"。安大略 SPH3U F 单元在电路探究期望（F2/F3）中包含基尔霍夫定律。NGSS 对此内容无专门 PE。阿尔伯塔 Physics 20/30 不将基尔霍夫定律列为学习目标。定量多回路分析是 AP Physics 1 / IB Physics HL 的期望。除课纲明确点名基尔霍夫定律的轨道外，本节均标 Honors。
Two rules that handle any circuit, no matter how complex.两条定律可处理任意复杂电路。
  • Junction rule (KCL — Kirchhoff's Current Law):节点法则（KCL — 基尔霍夫电流定律）： The sum of currents entering a junction equals the sum leaving it. Conservation of charge: $\sum I_{\text{in}} = \sum I_{\text{out}}$.流入节点的电流之和等于流出节点的电流之和。电荷守恒：$\sum I_{\text{in}} = \sum I_{\text{out}}$。
  • Loop rule (KVL — Kirchhoff's Voltage Law):回路法则（KVL — 基尔霍夫电压定律）： The algebraic sum of all voltage changes around any closed loop is zero. Energy conservation: $\sum_{\text{loop}} \Delta V = 0$.沿任何闭合回路一圈，所有电压变化的代数和为零。能量守恒：$\sum_{\text{loop}} \Delta V = 0$。
$$ \sum I_{\text{in}} = \sum I_{\text{out}} \qquad \text{(junction)} $$ $$ \sum_{\text{loop}} \Delta V = 0 \qquad \text{(loop)} $$ Sign convention for the loop rule: going through a resistor in the direction of assumed current, the voltage drop is $-IR$; going against the current, it is $+IR$. Going through a battery from $-$ to $+$ is $+\varepsilon$; from $+$ to $-$ is $-\varepsilon$.回路法则的符号约定：沿假设电流方向穿过电阻，电压降为 $-IR$；逆电流方向穿过，为 $+IR$。从电池负极到正极为 $+\varepsilon$；从正极到负极为 $-\varepsilon$。
Worked Example 7 · Two-loop network例题 7 · 两回路网络

A circuit has two batteries and three resistors: battery $\varepsilon_1 = 12\ \mathrm{V}$ (internal resistance negligible) and $\varepsilon_2 = 6.0\ \mathrm{V}$ (internal resistance negligible). Resistors $R_1 = 4.0\ \Omega$ (left branch), $R_2 = 2.0\ \Omega$ (middle shared branch), $R_3 = 4.0\ \Omega$ (right branch). Label branch currents $I_1$ (left, downward), $I_2$ (middle, downward), $I_3$ (right, downward). Apply Kirchhoff's rules to find all currents.一个电路有两只电池和三只电阻：电池 $\varepsilon_1 = 12\ \mathrm{V}$（内阻忽略不计）和 $\varepsilon_2 = 6.0\ \mathrm{V}$（内阻忽略不计）。电阻 $R_1 = 4.0\ \Omega$（左支路）、$R_2 = 2.0\ \Omega$（中间共用支路）、$R_3 = 4.0\ \Omega$（右支路）。标注支路电流 $I_1$（左，向下）、$I_2$（中，向下）、$I_3$（右，向下）。用基尔霍夫定律求所有电流。

Junction rule at the top node:顶部节点的节点法则： $I_1 = I_2 + I_3$.

Loop 1 (left loop, clockwise):回路 1（左回路，顺时针）：

$$ \varepsilon_1 - I_1 R_1 - I_2 R_2 = 0 \;\Longrightarrow\; 12 - 4I_1 - 2I_2 = 0. $$

Loop 2 (right loop, clockwise):回路 2（右回路，顺时针）：

$$ \varepsilon_2 - I_3 R_3 + I_2 R_2 = 0 \;\Longrightarrow\; 6 - 4I_3 + 2I_2 = 0. $$

Solve simultaneously.联立求解。 Substitute $I_1 = I_2 + I_3$ into Loop 1: $12 - 4(I_2 + I_3) - 2I_2 = 12 - 6I_2 - 4I_3 = 0$. From Loop 2: $I_3 = (6 + 2I_2)/4$. Substituting: $12 - 6I_2 - (6 + 2I_2) = 0 \Rightarrow 6 - 8I_2 = 0 \Rightarrow I_2 = 0.75\ \mathrm{A}$. Then $I_3 = (6 + 1.5)/4 = 1.875\ \mathrm{A}$ and $I_1 = 0.75 + 1.875 = 2.625\ \mathrm{A}$.将 $I_1 = I_2 + I_3$ 代入回路 1：$12 - 4(I_2 + I_3) - 2I_2 = 12 - 6I_2 - 4I_3 = 0$。由回路 2：$I_3 = (6 + 2I_2)/4$。代入：$12 - 6I_2 - (6 + 2I_2) = 0 \Rightarrow 6 - 8I_2 = 0 \Rightarrow I_2 = 0.75\ \mathrm{A}$。则 $I_3 = (6 + 1.5)/4 = 1.875\ \mathrm{A}$，$I_1 = 0.75 + 1.875 = 2.625\ \mathrm{A}$。

Check with junction rule:用节点法则核验： $I_2 + I_3 = 0.75 + 1.875 = 2.625\ \mathrm{A} = I_1$. ✓

At a junction, currents of $3.0$ A and $1.5$ A flow in; $2.0$ A flows out in one branch. What current flows out in the second branch?在一个节点，$3.0$ A 与 $1.5$ A 流入；$2.0$ A 从一条支路流出。第二条支路流出多少电流？
§7 · Q1
$2.5$ A
$0.5$ A
$4.5$ A
$1.5$ A
Junction rule: $\sum I_{\text{in}} = \sum I_{\text{out}}$. Total in $= 3.0 + 1.5 = 4.5$ A. One branch out $= 2.0$ A. Second branch $= 4.5 - 2.0 = 2.5$ A.节点法则：$\sum I_{\text{in}} = \sum I_{\text{out}}$。总流入 $= 3.0 + 1.5 = 4.5$ A。一支路流出 $= 2.0$ A。第二支路 $= 4.5 - 2.0 = 2.5$ A。
By KCL, all currents in must equal all currents out. Sum the inputs and subtract the known output.由 KCL，所有流入电流之和等于所有流出电流之和。将输入电流求和再减去已知的流出电流。
A simple loop has a $9.0\ \mathrm{V}$ battery, a $3.0\ \Omega$ resistor, and a $6.0\ \Omega$ resistor in series. Applying KVL (going around the loop in the direction of current): which equation is correct?一个简单回路串联一只 $9.0\ \mathrm{V}$ 电池、一只 $3.0\ \Omega$ 电阻与一只 $6.0\ \Omega$ 电阻。沿电流方向绕回路一圈应用 KVL，哪个方程正确？
§7 · Q2
$9.0 + 3I + 6I = 0$
$9.0 = 3.0 + 6.0$
$9.0 - 3I - 6I = 0$
$-9.0 - 3I - 6I = 0$
KVL: going from $-$ to $+$ through the battery gives $+9.0$; going through each resistor in the direction of current gives $-IR$. Sum $= 0$: $9.0 - 3I - 6I = 0 \Rightarrow I = 1.0\ \mathrm{A}$.KVL：从电池负极到正极为 $+9.0$；沿电流方向经过每只电阻为 $-IR$。总和 $= 0$：$9.0 - 3I - 6I = 0 \Rightarrow I = 1.0\ \mathrm{A}$。
Battery traversed $- \to +$ contributes $+\varepsilon$. Each resistor traversed in the current direction contributes $-IR$. The loop sum must equal zero.从负极到正极穿过电池贡献 $+\varepsilon$。沿电流方向穿过每只电阻贡献 $-IR$。回路总和必须为零。
Going deeper — why Kirchhoff's rules are conservation laws in disguise深入 — 为何基尔霍夫定律是披着外衣的守恒定律

The junction rule is simply conservation of electric charge applied at a node: charge cannot accumulate at a junction in steady state, so every coulomb that arrives must leave. It is a direct application of the principle that $\Delta Q / \Delta t = 0$ at each node.节点法则不过是在节点处应用电荷守恒：稳态下电荷不能在节点积累，因此每个流入的库仑都必须流出。这是在每个节点直接应用 $\Delta Q / \Delta t = 0$ 的原理。

The loop rule is conservation of energy per unit charge (conservation of electric potential). Going around any closed path and returning to the starting point, the net energy gained per unit charge must be zero — otherwise you would create or destroy energy with each trip around the loop. Batteries add energy per charge ($+\varepsilon$); resistors dissipate it ($-IR$). The algebraic sum must vanish.回路法则是每单位电荷的能量守恒（电位守恒）。沿任意闭合路径绕行一圈回到起点，每单位电荷获得的净能量必须为零 — 否则每次绕行都会创造或消灭能量。电池增加每单位电荷的能量（$+\varepsilon$）；电阻耗散能量（$-IR$）。代数和必须消失。

Together, these two conservation principles allow you to solve any linear DC circuit: write one junction equation per independent node and one loop equation per independent loop, then solve the simultaneous system. The number of independent equations needed equals the number of unknown branch currents.这两条守恒原理合在一起，使你能够求解任意线性直流电路：对每个独立节点写一个节点方程，对每个独立回路写一个回路方程，再联立求解。所需独立方程数等于未知支路电流数。

Exam Tactics考试策略

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

Setup discipline解题前的纪律
  • Identify circuit type first.先判断电路类型。 Is it pure series? Pure parallel? Mixed? Draw or re-draw the circuit. Label all currents and voltages before writing any equation. The most common error is applying series formulas to a parallel branch.是纯串联？纯并联？还是混联？画出或重新画出电路图。在写任何方程前，标注所有电流与电压。最常见的错误是把串联公式用于并联支路。
  • Choose the power formula that uses knowns.选用已知量对应的功率公式。 All three forms $P = IV = I^2R = V^2/R$ are equivalent. Pick the one that avoids introducing an extra unknown. For a fixed-voltage supply ($V$ known, $R$ known), use $P = V^2/R$.三种形式 $P = IV = I^2R = V^2/R$ 等价。选择避免引入额外未知量的那种。对固定电压电源（已知 $V$、已知 $R$），用 $P = V^2/R$。
  • Carry units; use SI.带单位；用国际单位。 Convert mA to A, k$\Omega$ to $\Omega$, minutes to seconds before substituting. A mismatch of units is a frequent dropped mark.代入前把 mA 换成 A、k$\Omega$ 换成 $\Omega$、分钟换成秒。单位不匹配是常见失分点。
Circuits (§1–§6)电路（§1–§6）
  • Series: same $I$, voltages add.串联：$I$ 相同，电压叠加。 Resistances add directly: $R_{\text{total}} = \sum R_i$. Adding more resistors reduces the current.电阻直接叠加：$R_{\text{total}} = \sum R_i$。增加更多电阻会减小电流。
  • Parallel: same $V$, currents add.并联：$V$ 相同，电流叠加。 Reciprocals add: $1/R_{\text{eq}} = \sum 1/R_i$. Adding more branches reduces $R_{\text{eq}}$ and increases total current.倒数叠加：$1/R_{\text{eq}} = \sum 1/R_i$。增加更多支路减小 $R_{\text{eq}}$，增大总电流。
  • Terminal voltage drops under load.端电压在负载下下降。 A real battery's terminal voltage is $V_T = \varepsilon - Ir$. Higher current $\Rightarrow$ bigger internal-resistance drop $\Rightarrow$ lower terminal voltage. Measure EMF with no current; measure terminal voltage under load.实际电池的端电压为 $V_T = \varepsilon - Ir$。电流越大 $\Rightarrow$ 内阻压降越大 $\Rightarrow$ 端电压越低。无电流时测 EMF；有负载时测端电压。
Kirchhoff's rules (§7) Honors基尔霍夫定律（§7）荣誉
  • Assign current directions before writing equations.写方程前先分配电流方向。 Your assumed direction can be wrong; a negative answer just means the actual current is opposite. Never change directions mid-problem.你的假设方向可能是错的；负值结果只表示实际电流方向相反。绝不在解题中途更改方向。
  • Count equations needed.计算所需方程数。 For $n$ unknown branch currents, you need $n$ independent equations: $(n - 1)$ junction equations plus enough loop equations to make $n$ total.对 $n$ 个未知支路电流，需要 $n$ 个独立方程：$(n-1)$ 个节点方程加上足够多的回路方程，共 $n$ 个。
  • Sign rule for the loop equation.回路方程的符号法则。 Battery $- \to +$: add $+\varepsilon$. Resistor in current direction: add $-IR$. Resistor against current direction: add $+IR$. Sum of all terms $= 0$.电池从负到正：加 $+\varepsilon$。沿电流方向经过电阻：加 $-IR$。逆电流方向经过电阻：加 $+IR$。所有项之和 $= 0$。
Answer hygiene作答规范
  • Round at the very end.最后一步再四舍五入。 Carry extra digits through intermediate steps; round only the final number to the precision the question asks.中间步骤多留几位；仅在最终答案处按题目要求精度四舍五入。
  • Sanity-check direction of inequality.核验不等式方向。 $R_{\text{parallel}} <$ smallest branch $R$. $R_{\text{series}} >$ largest single $R$. If your answer violates this, recheck.$R_{\text{parallel}} <$ 最小支路 $R$。$R_{\text{series}} >$ 最大单个 $R$。若答案违反此规则，重新检查。
  • Check Kirchhoff answers with the unused equation.用未使用的方程核验基尔霍夫结果。 After solving, substitute back into a loop or junction equation you did not use in the solve — it should be satisfied exactly.求解后，代入求解时未使用的某条回路或节点方程 — 它应精确成立。
Active Recall主动复习

Flashcards闪卡

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Electric current definition?电流（电流）定义？
$$I = \frac{\Delta Q}{\Delta t}$$ SI unit: ampere (A). $1\ \mathrm{A} = 1\ \mathrm{C/s}$国际单位：安培（A）。$1\ \mathrm{A} = 1\ \mathrm{C/s}$
Conventional current direction?惯用电流方向？
Direction positive charges would flow. In metals, electrons flow opposite to conventional current.正电荷流动的方向。在金属中，电子流动方向与惯用电流方向相反。
Voltage (potential difference)?电压（电位差）？
$$V = \frac{W}{Q}$$ energy per unit charge; SI unit: volt V ($1\ \mathrm{V} = 1\ \mathrm{J/C}$)每单位电荷能量；国际单位：伏特 V（$1\ \mathrm{V} = 1\ \mathrm{J/C}$）
EMF vs terminal voltage?电动势（电动势）与端电压？
$$V_T = \varepsilon - Ir$$ EMF $\varepsilon$ = energy/charge from source; terminal voltage $V_T$ is less due to internal resistance drop $Ir$电动势 $\varepsilon$ = 电源提供的每单位电荷能量；端电压 $V_T$ 因内阻压降 $Ir$ 而减小
Ohm's Law?欧姆定律（欧姆定律）？
$$V = IR$$ $R$ in ohms $\Omega$; valid for ohmic (linear) resistors$R$ 单位欧姆 $\Omega$；适用于欧姆式（线性）电阻
Series resistance rule?串联（串联）电阻规则？
$$R_{\text{series}} = R_1 + R_2 + \cdots$$ same $I$ everywhere; voltages add各处 $I$ 相同；电压叠加
Parallel resistance rule?并联（并联）电阻规则？
$$\frac{1}{R_{\text{par}}} = \frac{1}{R_1} + \frac{1}{R_2} + \cdots$$ same $V$ each branch; currents add各支路 $V$ 相同；电流叠加
Two equal resistors in parallel?两只等值电阻并联？
$$R_{\text{eq}} = \frac{R}{2}$$ Always less than either individual $R$始终小于任一单个 $R$
Electrical power — three forms?电功率（电功率）——三种形式？
$$P = IV = I^2R = \frac{V^2}{R}$$ SI unit: watt (W)国际单位：瓦特（W）
Electrical energy?电能？
$$W = P\,\Delta t$$ in joules (J); utility unit: kWh ($1\ \mathrm{kWh} = 3.6 \times 10^6\ \mathrm{J}$)单位焦耳（J）；电力单位：千瓦时（$1\ \mathrm{kWh} = 3.6 \times 10^6\ \mathrm{J}$）
Kirchhoff's junction rule (KCL)?基尔霍夫节点法则（KCL，基尔霍夫定律）？
$$\sum I_{\text{in}} = \sum I_{\text{out}}$$ conservation of charge at a node节点处电荷守恒
Kirchhoff's loop rule (KVL)?基尔霍夫回路法则（KVL，基尔霍夫定律）？
$$\sum_{\text{loop}} \Delta V = 0$$ conservation of energy per charge around any closed loop沿任意闭合回路，每单位电荷能量守恒
Sign convention in KVL loop?KVL 回路中的符号约定？
Battery $- \to +$: $+\varepsilon$. Resistor with current: $-IR$. Resistor against current: $+IR$. Sum $= 0$.电池从负到正：$+\varepsilon$。沿电流方向过电阻：$-IR$。逆电流方向：$+IR$。总和 $= 0$。
Joule heating in a resistor?电阻中的焦耳热效应？
$$P = I^2 R$$ all electrical power converts to thermal energy; basis of heaters and fuse ratings所有电功率转化为热能；电暖器和保险丝额定值的物理基础
Mixed Practice综合练习

Practice Quiz综合测验

A charge of $5.0$ C passes through a wire in $2.0$ s. What is the current?$5.0$ C 的电荷在 $2.0$ s 内通过一根导线。电流是多少？
Q1
$10$ A
$0.40$ A
$2.5$ A
$7.0$ A
$I = \Delta Q / \Delta t = 5.0 / 2.0 = 2.5$ A.$I = \Delta Q / \Delta t = 5.0 / 2.0 = 2.5$ A。
Current is charge divided by time: $I = Q/t$.电流是电荷除以时间：$I = Q/t$。
A battery has EMF $9.0$ V and internal resistance $0.30\ \Omega$. It delivers $5.0$ A. What is the terminal voltage? 🇨🇦 BC Physics 11 / ON SPH3U一只电池 EMF $9.0$ V，内阻 $0.30\ \Omega$，输出 $5.0$ A。端电压是多少？🇨🇦 BC Physics 11 / ON SPH3U
Q2
$9.0$ V
$7.5$ V
$10.5$ V
$1.5$ V
$V_T = \varepsilon - Ir = 9.0 - (5.0)(0.30) = 9.0 - 1.5 = 7.5$ V.$V_T = \varepsilon - Ir = 9.0 - (5.0)(0.30) = 9.0 - 1.5 = 7.5$ V。
Terminal voltage $= \varepsilon - Ir$. Internal resistance always reduces the available voltage.端电压 $= \varepsilon - Ir$。内阻总是使可用电压减小。
A $20\ \Omega$ resistor is connected to $60$ V. What current flows through it?一只 $20\ \Omega$ 电阻接在 $60$ V 两端。通过的电流是多少？
Q3
$3.0$ A
$1200$ A
$0.33$ A
$80$ A
$I = V/R = 60/20 = 3.0$ A.$I = V/R = 60/20 = 3.0$ A。
Ohm's law rearranged: $I = V/R$.欧姆定律变形：$I = V/R$。
Three $9.0\ \Omega$ resistors are in series across an $18$ V supply. What is the current? 🇨🇦 ON SPH3U F2三只 $9.0\ \Omega$ 电阻串联接在 $18$ V 电源上。电流是多少？🇨🇦 ON SPH3U F2
Q4
$6.0$ A
$3.0$ A
$18$ A
$\tfrac{2}{3}$ A
$R_{\text{total}} = 3 \times 9.0 = 27\ \Omega$; $I = V/R = 18/27 = 2/3 \approx 0.67$ A.$R_{\text{total}} = 3 \times 9.0 = 27\ \Omega$；$I = V/R = 18/27 = 2/3 \approx 0.67$ A。
Series: resistances add. $R_{\text{total}} = 3 \times 9 = 27\ \Omega$; then $I = V/R_{\text{total}}$.串联：电阻叠加。$R_{\text{total}} = 3 \times 9 = 27\ \Omega$；再用 $I = V/R_{\text{total}}$。
A $6.0\ \Omega$ and a $3.0\ \Omega$ resistor are in parallel across $12$ V. What is the total current drawn from the supply? 🇨🇦 BC Physics 11$6.0\ \Omega$ 与 $3.0\ \Omega$ 两只电阻并联接在 $12$ V 上。从电源流出的总电流是多少？🇨🇦 BC Physics 11
Q5
$1.3$ A
$4.0$ A
$6.0$ A
$2.0$ A
$I_1 = 12/6 = 2$ A; $I_2 = 12/3 = 4$ A; $I_{\text{total}} = 2 + 4 = 6$ A. (Or: $R_{\text{eq}} = 2\ \Omega$, $I = 12/2 = 6$ A.)$I_1 = 12/6 = 2$ A；$I_2 = 12/3 = 4$ A；$I_{\text{total}} = 2 + 4 = 6$ A。（或：$R_{\text{eq}} = 2\ \Omega$，$I = 12/2 = 6$ A。）
In parallel, same voltage across each branch. Find each branch current with $I = V/R$, then sum them.并联中各支路电压相同。用 $I = V/R$ 求每条支路电流，再求和。
A $100\ \Omega$ resistor is connected to a $12$ V supply. What power is dissipated?一只 $100\ \Omega$ 电阻接在 $12$ V 电源上。耗散功率是多少？
Q6
$0.12$ W
$1.44$ W
$144$ W
$12$ W
$P = V^2/R = 12^2 / 100 = 144/100 = 1.44$ W. (Use $V^2/R$ when $V$ and $R$ are given.)$P = V^2/R = 12^2 / 100 = 144/100 = 1.44$ W。（已知 $V$ 与 $R$ 时用 $V^2/R$。）
Choose the power formula matching your knowns. Here $V$ and $R$ are given, so use $P = V^2/R$.选择与已知量匹配的功率公式。此处已知 $V$ 与 $R$，故用 $P = V^2/R$。
A $2000\ \mathrm{W}$ heater runs for $30$ min. How much energy does it consume (in joules)?一只 $2000\ \mathrm{W}$ 的电暖器运行 $30$ min。消耗了多少能量（焦耳）？
Q7
$3.6 \times 10^6$ J
$60000$ J
$1000$ J
$3600$ J
$W = P\,\Delta t = 2000 \times (30 \times 60) = 2000 \times 1800 = 3.6 \times 10^6$ J (= 1.0 kWh).$W = P\,\Delta t = 2000 \times (30 \times 60) = 2000 \times 1800 = 3.6 \times 10^6$ J（= 1.0 kWh）。
Convert minutes to seconds first: $30\ \mathrm{min} = 1800\ \mathrm{s}$. Then $W = P\Delta t$.先把分钟换成秒：$30\ \mathrm{min} = 1800\ \mathrm{s}$。再用 $W = P\Delta t$。
At a junction, $4.0$ A enters from one branch and $1.5$ A enters from another. What is the current leaving the junction (one exit branch)? 🇨🇦 BC Physics 11 / ON SPH3U Kirchhoff在一个节点，一条支路流入 $4.0$ A，另一条流入 $1.5$ A。离开节点的电流是多少（单一出口支路）？🇨🇦 BC Physics 11 / ON SPH3U 基尔霍夫
Q8
$2.5$ A
$4.0$ A
$1.5$ A
$5.5$ A
KCL: $\sum I_{\text{in}} = \sum I_{\text{out}}$. In: $4.0 + 1.5 = 5.5$ A; so out $= 5.5$ A.KCL：$\sum I_{\text{in}} = \sum I_{\text{out}}$。流入：$4.0 + 1.5 = 5.5$ A；故流出 $= 5.5$ A。
At a junction, current in equals current out (KCL = charge conservation). Sum all inflows.在节点处，流入电流等于流出电流（KCL = 电荷守恒）。将所有流入电流求和。
A loop contains a $6.0$ V battery and two resistors ($R_1 = 1.0\ \Omega$, $R_2 = 2.0\ \Omega$) in series (negligible internal resistance). What current flows? 🇨🇦 BC Physics 11 Kirchhoff一个回路中串联一只 $6.0$ V 电池和两只电阻（$R_1 = 1.0\ \Omega$，$R_2 = 2.0\ \Omega$），内阻忽略不计。电流是多少？🇨🇦 BC Physics 11 基尔霍夫
Q9
$6.0$ A
$3.0$ A
$2.0$ A
$1.0$ A
KVL: $6.0 - I(1.0) - I(2.0) = 0 \Rightarrow 6.0 = 3I \Rightarrow I = 2.0$ A. (Same as Ohm's law: $I = V/R_{\text{total}} = 6/3 = 2$ A.)KVL：$6.0 - I(1.0) - I(2.0) = 0 \Rightarrow 6.0 = 3I \Rightarrow I = 2.0$ A。（等同于欧姆定律：$I = V/R_{\text{total}} = 6/3 = 2$ A。）
KVL: battery voltage $=$ sum of resistor voltage drops. $6 = I(1+2) = 3I$.KVL：电池电压 $=$ 各电阻电压降之和。$6 = I(1+2) = 3I$。
A household circuit fuse is rated $15$ A at $120$ V. What is the maximum power the circuit can safely deliver?一个家用电路保险丝额定 $120$ V、$15$ A。该电路能安全输出的最大功率是多少？
Q10
$135$ W
$1800$ W
$8.0$ W
$3600$ W
$P = IV = 15 \times 120 = 1800$ W. This is the maximum safe power at the rated voltage and current.$P = IV = 15 \times 120 = 1800$ W。这是额定电压与电流下的最大安全功率。
Power $= IV$. The fuse blows above $15$ A, so maximum safe power is $P = 15 \times 120$.功率 $= IV$。超过 $15$ A 时保险丝熔断，故最大安全功率为 $P = 15 \times 120$。
Self-check自查

Readiness Checklist准备就绪清单

Tick each item when you can do it cold, without notes, on a first attempt.能在无笔记、首次尝试下完成，再勾选每一项。

0 / 11 mastered已掌握 0 / 11
Next Steps后续单元

What This Feeds Into本单元的去向

Current electricity and circuits is the quantitative bridge between the electric-field ideas of Unit 8 (Electrostatics) and the magnetic phenomena of Unit 10 (Magnetism). Ohm's law and Kirchhoff's rules established here are the working toolkit for every later circuit problem. The cross-references below point at adjacent units in this repo that build directly on this unit's content.电流与电路是第 8 单元（静电学）的电场概念与第 10 单元（磁学）的磁现象之间的定量桥梁。这里建立的欧姆定律（欧姆定律）与基尔霍夫定律（基尔霍夫定律）是后续所有电路问题的工作工具箱。下方链接指向本仓库中直接建立在本单元内容之上的相邻单元。

Within High School Physics.在 HS Physics 内部。

Unit 8 (Electrostatics and Electric Fields) introduces the electric field and Coulomb's law that create the potential difference driving current here. Unit 10 (Magnetism and Electromagnetic Induction) extends circuits to include inductors and transformers (AC circuits); the DC circuit analysis of this unit is the prerequisite. The concept of electrical power ($P = IV$) connects back to Unit 3 (Work, Energy and Power) and forward to any discussion of energy efficiency.第 8 单元（静电学与电场）引入产生驱动本单元电流的电位差的电场与库仑定律。第 10 单元（磁学与电磁感应）将电路扩展到包含电感和变压器（交流电路）；本单元的直流电路分析是其先决条件。电功率（$P = IV$）的概念与第 3 单元（功、能与功率）相连，并向前延伸到任何关于能量效率的讨论。

Feeder units from Unit 8 and forward to Unit 10.来自第 8 单元、通向第 10 单元的衔接链接。

Unit 8 · Electrostatics and Electric Fields (Coulomb's law, electric potential — the field that drives current in circuits)第 8 单元 · 静电学与电场（库仑定律、电位 — 驱动电路中电流的场） Unit 10 · Magnetism and Electromagnetic Induction (magnetic fields produced by current; Faraday's law and the generator that supplies EMF)第 10 单元 · 磁学与电磁感应（电流产生磁场；法拉第定律与提供 EMF 的发电机）

If you are aiming for AP Physics 1, Unit 5 of that course (Electric Circuits) maps directly to this guide; AP adds more complex multi-loop analysis, RC circuits, and the energy-storage framing. For IB Physics HL, Topic B.5 (Current and Circuits) covers Ohm's law, Kirchhoff's rules, internal resistance, and power at a level that matches and extends this guide. For IB Physics SL, the scope is similar to §1–§6 here, with Kirchhoff's rules treated more qualitatively.备考 AP Physics 1：该课程第 5 单元（电路）直接对应本指南；AP 增加了更复杂的多回路分析、RC 电路和能量存储框架。备考 IB Physics HL：主题 B.5（电流与电路）在与本指南匹配并超越本指南的深度上涵盖欧姆定律、基尔霍夫定律、内阻和功率。备考 IB Physics SL：范围与本指南 §1–§6 类似，基尔霍夫定律处理得更偏概念。

DINGRUI SCHOLARS

High School Physics · Current Electricity and Circuits · v1高中物理 · 电流与电路 · v1