PART I · SHORT RESPONSE第一部分 · 短答题AP-style MCQ + ON/BC short answer · 22 marksAP 风格选择题 + 安/卑省考短答 · 共 22 分
Section A · Short ResponseA 部分 · 短答题
Mix of multiple-choice and short-answer items. For MCQs, circle the letter; show enough work in the margin that a marker could verify. For short-answer items, state units in every answer. Take the elementary charge as $e = 1.6 \times 10^{-19}\ \text{C}$ where needed. No calculator on Q1-Q2; calculator permitted on Q3-Q5.本节包含选择题与短答题。选择题请圈出字母答案，并在空白处写下足以让阅卷人核对的过程。短答题每道都要写出单位。需要时取基本电荷 $e = 1.6 \times 10^{-19}\ \text{C}$。Q1-Q2 不可使用计算器；Q3-Q5 可用计算器。
Three resistors have values $5.0\ \Omega$, $10\ \Omega$, and $15\ \Omega$.三只电阻的阻值分别为 $5.0\ \Omega$、$10\ \Omega$ 与 $15\ \Omega$。
(a)Find the equivalent resistance if the three are connected in series.求三者串联时的等效电阻。[2]
(b)The $5.0\ \Omega$ and $10\ \Omega$ resistors are now connected in parallel (the $15\ \Omega$ removed). Find their equivalent resistance.现将 $5.0\ \Omega$ 与 $10\ \Omega$ 电阻并联（移除 $15\ \Omega$）。求其等效电阻。[2]
(c)Without recalculating, explain how you know a parallel combination must have a smaller equivalent resistance than the series combination.不重新计算，说明你如何判断并联组合的等效电阻一定小于串联组合。[2]
PART II · EXTENDED RESPONSE第二部分 · 简答题AP-feeder FRQ + honors · 32 marksAP 衔接简答题 + 荣誉级 · 共 32 分
Section B · Extended ResponseB 部分 · 简答题
Show every step of reasoning. Identify the formula used before substituting values. State units in every final answer. Take $e = 1.6 \times 10^{-19}\ \text{C}$ where needed. For Kirchhoff problems, define your current directions before writing equations. Calculator permitted on Q6-Q9.每一步推理都要写出。在代入数值前先注明所用公式。每个最终答案都要写单位。需要时取 $e = 1.6 \times 10^{-19}\ \text{C}$。基尔霍夫题在写方程前先定义电流方向。Q6-Q9 可用计算器。
A two-loop network has two batteries (negligible internal resistance) and three resistors. Battery $\varepsilon_1 = 10\ \text{V}$ drives the left branch; battery $\varepsilon_2 = 4.0\ \text{V}$ drives the right branch. The resistors are $R_1 = 2.0\ \Omega$ (left branch), $R_2 = 2.0\ \Omega$ (middle shared branch), $R_3 = 2.0\ \Omega$ (right branch). Define branch currents $I_1$ (left, downward through $R_1$), $I_2$ (middle, downward through $R_2$), and $I_3$ (right, downward through $R_3$), with $I_1$ splitting into $I_2$ and $I_3$ at the top node.一个两回路网络含两只电池（内阻可忽略）与三只电阻。电池 $\varepsilon_1 = 10\ \text{V}$ 驱动左支路；电池 $\varepsilon_2 = 4.0\ \text{V}$ 驱动右支路。电阻为 $R_1 = 2.0\ \Omega$（左支路）、$R_2 = 2.0\ \Omega$（中间共用支路）、$R_3 = 2.0\ \Omega$（右支路）。定义支路电流 $I_1$（左，向下经 $R_1$）、$I_2$（中，向下经 $R_2$）、$I_3$（右，向下经 $R_3$），在顶部节点处 $I_1$ 分成 $I_2$ 与 $I_3$。
(a)Write the junction-rule equation at the top node.写出顶部节点的节点法则方程。[1]
(b)Write the loop-rule equation for the left loop ($\varepsilon_1$, $R_1$, $R_2$) and for the right loop ($\varepsilon_2$, $R_3$, $R_2$).写出左回路（$\varepsilon_1$、$R_1$、$R_2$）与右回路（$\varepsilon_2$、$R_3$、$R_2$）的回路法则方程。[2]
(c)Solve the system for $I_1$, $I_2$, and $I_3$.联立求解 $I_1$、$I_2$、$I_3$。[4]
(d)Verify your answer using the equation you did not use in the solve.用求解时未使用的方程核验你的答案。[1]
PART III · MODELING / APPLIED第三部分 · 建模与应用Universal applied · 26 marks通用应用题 · 共 26 分
Section C · Modeling and ApplicationsC 部分 · 建模与应用
Define symbols (with units) at the start of each question. State the formula used before substituting. Convert all quantities to SI (mA to A, k$\Omega$ to $\Omega$, minutes to seconds) before substituting. Conclude each question with a one-sentence answer in context. Calculator permitted throughout Part III.每题开始时定义符号（含单位）。代入数值前先写出所用公式。代入前把所有量换算为国际单位（mA 换 A、k$\Omega$ 换 $\Omega$、分钟换秒）。每题以一句结合情境的完整句子作答。第三部分全程可用计算器。
Q10MEDIUM中🇺🇸 US美Universal Applied通用应用题§6 Power & energy (household)功率与电能（家用） · HS-PS3-5[7 marks][7 分]
A microwave oven rated $1200\ \text{W}$ operates from a $120\ \text{V}$ household supply.一台额定功率 $1200\ \text{W}$ 的微波炉接在 $120\ \text{V}$ 家用电源上工作。
(a)Find the current the microwave draws.求微波炉的工作电流。[2]
(b)Find the resistance of its heating element.求其加热元件的电阻。[2]
(c)Find the electrical energy it consumes in $5.0$ minutes, in joules.求其在 $5.0$ 分钟内消耗的电能（以焦耳计）。[2]
(d)At a rate of $\$0.12$ per kWh, find the cost of running it for those $5.0$ minutes.按每千瓦时 $\$0.12$ 计费，求运行这 $5.0$ 分钟的费用。[1]
In a circuit, two $6.0\ \Omega$ resistors are connected in parallel with each other, and this parallel combination is in series with a single $9.0\ \Omega$ resistor. The whole network is connected to a $24\ \text{V}$ supply of negligible internal resistance.某电路中，两只 $6.0\ \Omega$ 电阻彼此并联，该并联组合再与一只 $9.0\ \Omega$ 电阻串联。整个网络接在一只内阻可忽略的 $24\ \text{V}$ 电源上。
(a)Find the total resistance of the network.求网络的总电阻。[2]
(b)Find the total current from the supply.求电源输出的总电流。[1]
(c)Find the voltage across the $9.0\ \Omega$ resistor and the voltage across the parallel combination.求 $9.0\ \Omega$ 电阻两端的电压与并联组合两端的电压。[2]
(d)Find the current through each of the two $6.0\ \Omega$ resistors.求两只 $6.0\ \Omega$ 电阻各自的电流。[2]
A heating element is made from a nichrome wire of length $L = 2.0$ m and cross-sectional area $A = 2.2 \times 10^{-7}\ \text{m}^2$. Nichrome has resistivity $\rho = 1.1 \times 10^{-6}\ \Omega \cdot \text{m}$.某加热元件由长 $L = 2.0$ m、截面积 $A = 2.2 \times 10^{-7}\ \text{m}^2$ 的镍铬合金丝制成。镍铬合金的电阻率 $\rho = 1.1 \times 10^{-6}\ \Omega \cdot \text{m}$。
(a)Find the resistance of the wire, using $R = \rho L / A$.用 $R = \rho L / A$ 求该丝的电阻。[2]
(b)The element is connected across a $5.0\ \text{V}$ supply. Find the current and the power dissipated.该元件接在 $5.0\ \text{V}$ 电源上。求电流与耗散功率。[2]
(c)A second wire of the same material has twice the length and the same cross-sectional area. State its resistance and explain your reasoning without a full recalculation.另一根同材料的丝长度为两倍、截面积相同。写出其电阻，并在不完整重算的情况下说明理由。[2]
A single loop contains a $9.0\ \text{V}$ battery (negligible internal resistance) in series with a $3.0\ \Omega$ resistor and a $6.0\ \Omega$ resistor. Apply Kirchhoff's rules throughout; do not simply quote the series shortcut.一个单回路含一只 $9.0\ \text{V}$ 电池（内阻可忽略）与一只 $3.0\ \Omega$ 电阻、一只 $6.0\ \Omega$ 电阻串联。全程应用基尔霍夫定律；不要直接套用串联捷径。
(a)Assign a current direction and write the loop-rule (KVL) equation, stating your sign convention.指定电流方向并写出回路法则（KVL）方程，说明你的符号约定。[2]
(b)Solve for the current in the loop.求回路中的电流。[2]
(c)Find the voltage across each resistor, and confirm that the loop rule (the algebraic sum of all voltage changes around the loop) is satisfied.求每只电阻两端的电压，并确认回路法则（绕回路一圈所有电压变化的代数和）成立。[2]
🇺🇸 US NGSS美国 NGSSHS-PS3-5(energy/field lens only)（仅能量/场视角）
Full Syllabus Map lives in ../Study Guides/Unit_9_Current_Electricity_and_Circuits.html. Note: NGSS treats circuits only through the energy/field lens (HS-PS3-5), with no quantitative circuit PE; Alberta Physics 20/30 has no dedicated DC-circuits unit (only the current definition). The Kirchhoff items (Q9, Q13) carry the Honors flag for every track except BC and ON, where Kirchhoff's laws are named content.完整大纲对照表见 ../Study Guides/Unit_9_Current_Electricity_and_Circuits.html。注：NGSS 仅以能量/场视角处理电路（HS-PS3-5），无定量电路评估目标；阿尔伯塔 Physics 20/30 没有专门的直流电路单元（仅含电流定义）。基尔霍夫题（Q9、Q13）对除 BC 与 ON（其将基尔霍夫定律列为指定内容）以外的所有轨道均标注荣誉级。