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Solutions详解

Current Electricity and Circuits · Solutions电流与电路 · 详解

Companion to the Practice Set · Mark-by-mark walkthroughs · AP-Feeder / ON / BC styles练习题配套答案 · 逐分讲解 · AP 衔接 / 安 / 卑省考风格

EASY易 MEDIUM中 HARD难 🇺🇸 US美 🇨🇦 ON安 🇨🇦 BC卑 🇨🇦 AB阿 AP-style MCQAP 风格选择题 AP-feeder FRQAP 衔接简答题 ON Provincial-style安大略省考风格 BC Provincial-style卑诗省考风格 Universal Applied通用应用题 Honors荣誉级
PART I  ·  SHORT RESPONSE · SOLUTIONS第一部分  ·  短答题 · 详解AP MCQ + ON/BC short answer · 22 marksAP 选择题 + 安/卑省考短答 · 共 22 分

Section A · Worked SolutionsA 部分 · 详细解答

Q1EASY易 🇺🇸 US美 🇨🇦 AB阿 AP-style MCQAP 风格选择题 §1 Electric current电流 · 30-B2.7k [3 marks][3 分]

$6.0$ C of charge passes a point in $2.0$ s. What is the current?$6.0$ C 电荷在 $2.0$ s 内通过某一点。电流是多少？

Answer:答案：  (B)  $3.0\ \text{A}$

(a) Apply the definition of current套用电流定义 M1·A1·A1

Current is the rate of charge flow:电流是电荷流动的速率： $$ I \;=\; \frac{\Delta Q}{\Delta t} \;=\; \frac{6.0}{2.0} \;=\; 3.0 \;\text{A}. $$ Option (B). One ampere is one coulomb per second, so $6$ C in $2$ s is $3$ C/s $= 3$ A.选 (B)。一安培即每秒一库仑，故 $6$ C 历时 $2$ s 即 $3$ C/s $= 3$ A。
Why the distractors fail.干扰项分析。
(A) $0.33\ \text{A}$: inverts the ratio, computing $\Delta t / \Delta Q = 2.0/6.0$.把比值颠倒，算成 $\Delta t / \Delta Q = 2.0/6.0$。
(C) $12\ \text{A}$: multiplies $\Delta Q \times \Delta t$ instead of dividing.把 $\Delta Q \times \Delta t$ 相乘而非相除。
(D) $8.0\ \text{A}$: adds $6.0 + 2.0$ rather than forming the rate.把 $6.0 + 2.0$ 相加，而非求速率。
Current is a rate: coulombs per second, never coulombs alone.电流是速率：库仑每秒，绝非单看库仑。 The definition $I = \Delta Q / \Delta t$ makes current the flow rate of charge, exactly as velocity is the flow rate of position. The Alberta diploma outcome 30-B2.7k states this verbatim: "electric current as the amount of charge passing a reference point per unit of time." Keeping the units explicit ($\text{C} / \text{s} = \text{A}$) is the surest guard against inverting or multiplying the ratio. Conventional current points the way positive charge would move; in a metal the electrons actually drift the opposite way, but the numerical value of $I$ is unaffected.定义 $I = \Delta Q / \Delta t$ 使电流成为电荷的流动速率，正如速度是位置的变化速率。阿尔伯塔毕业考目标 30-B2.7k 原文如此表述："电流即单位时间内通过参考点的电荷量。"始终明写单位（$\text{C} / \text{s} = \text{A}$）是防止颠倒或相乘的最稳妥方法。惯用电流方向指向正电荷的运动方向；在金属中电子实际沿相反方向漂移，但 $I$ 的数值不受影响。
Q2EASY易 🇺🇸 US美 AP-style MCQAP 风格选择题 §3 Ohm's law欧姆定律 · BC Physics 11 [3 marks][3 分]

A current of $0.50$ A flows through a $24\ \Omega$ resistor. Voltage across it?$0.50$ A 电流流过 $24\ \Omega$ 电阻。其两端电压为多少？

Answer:答案：  (B)  $12\ \text{V}$

(a) Apply Ohm's law $V = IR$套用欧姆定律 $V = IR$ M1·A1·A1

$$ V \;=\; IR \;=\; (0.50)(24) \;=\; 12 \;\text{V}. $$ Option (B).选 (B)。
Why the distractors fail.干扰项分析。
(A) $48\ \text{V}$: divides $24 / 0.50$ instead of multiplying.把 $24 / 0.50$ 相除而非相乘。
(C) $0.021\ \text{V}$: computes $I / R = 0.50/24$, the wrong combination.算成 $I / R = 0.50/24$，组合错误。
(D) $24.5\ \text{V}$: adds $24 + 0.50$ rather than applying the law.把 $24 + 0.50$ 相加，而非套用定律。
Ohm's law is a multiplicative triangle: $V = IR$, $I = V/R$, $R = V/I$.欧姆定律是一个乘法三角：$V = IR$、$I = V/R$、$R = V/I$。 Picturing $V$ on top with $I$ and $R$ side by side underneath gives all three rearrangements at a glance. Here we are given current and resistance and asked for voltage, so cover $V$ and read off $I \times R$. Units confirm the form: $\text{A} \times \Omega = \text{A} \times (\text{V}/\text{A}) = \text{V}$. For an ohmic resistor $R$ is constant, so $V$ and $I$ are directly proportional; this is the slope $R$ of the $V$ versus $I$ graph.把 $V$ 画在顶部、$I$ 与 $R$ 并排在下方，三种变形一目了然。此题已知电流与电阻、求电压，故遮住 $V$ 读出 $I \times R$。单位可验证形式：$\text{A} \times \Omega = \text{A} \times (\text{V}/\text{A}) = \text{V}$。对欧姆式电阻 $R$ 恒定，故 $V$ 与 $I$ 成正比；这正是 $V$ 对 $I$ 图的斜率 $R$。
Q3EASY易 🇨🇦 ON安 ON Provincial-style安大略省考风格 §2 Voltage & electrical energy电压与电能 · SPH3U F2 [6 marks][6 分]

A device transfers $24$ J when $4.0$ C passes through it. (a) Voltage. (b) Meaning of 1 volt. (c) Why terminal voltage < EMF under load.某器件在 $4.0$ C 电荷通过时转移 $24$ J。(a) 电压。(b) 1 伏特的含义。(c) 为何负载下端电压小于电动势。

Answer:答案：  (a) $V = 6.0\ \text{V}$  ·  (b) 1 J of energy per 1 C of charge每 1 C 电荷 1 J 能量  ·  (c) internal-resistance drop $Ir$内阻压降 $Ir$

(a) Voltage from energy per unit charge由每单位电荷能量求电压 M1·A1

$$ V \;=\; \frac{W}{Q} \;=\; \frac{24}{4.0} \;=\; 6.0 \;\text{V}. $$

(b) Meaning of one volt一伏特的含义 A1·A1

A potential difference of one volt means one joule of energy is transferred for every one coulomb of charge that passes: $1\ \text{V} = 1\ \text{J/C}$. So $6.0$ V means each coulomb carries $6.0$ J between the device's terminals.一伏特的电位差表示每通过一库仑电荷转移一焦耳能量：$1\ \text{V} = 1\ \text{J/C}$。故 $6.0$ V 表示每库仑电荷在器件两端之间携带 $6.0$ J。

(c) Terminal voltage vs EMF端电压与电动势 A1·A1

A real battery has internal resistance $r$. When it delivers a current $I$, a voltage $Ir$ is dropped across that internal resistance inside the battery. The terminal voltage is therefore $V_T = \varepsilon - Ir$, which is less than the EMF $\varepsilon$ whenever current flows.实际电池有内阻 $r$。当它输出电流 $I$ 时，电池内部的内阻上产生 $Ir$ 的压降。因此端电压为 $V_T = \varepsilon - Ir$，只要有电流流过，它就小于电动势 $\varepsilon$。
Voltage is energy per charge; EMF is the energy per charge a source supplies before internal losses.电压是每电荷能量；电动势是电源在内部损耗前提供的每电荷能量。 The defining relation $V = W/Q$ ties the abstract idea of "potential difference" to a concrete energy bookkeeping: how many joules each coulomb gains or loses. EMF is the energy per coulomb the source generates from chemical or mechanical work; terminal voltage is what is actually available to the external circuit after the source spends some of that energy heating its own internal resistance. At zero current (open circuit) the two are equal; as load current rises, the gap $Ir$ widens. This is exactly why a nearly-dead battery still reads close to its rated voltage with no load but sags badly under a real load.定义式 $V = W/Q$ 把"电位差"这一抽象概念与具体的能量记账联系起来：每库仑获得或失去多少焦耳。电动势是电源由化学或机械做功为每库仑产生的能量；端电压则是电源把部分能量耗在自身内阻发热之后，外电路实际可用的部分。零电流（开路）时两者相等；负载电流增大时，差值 $Ir$ 随之增大。这正是为何快没电的电池空载时读数仍接近额定电压，但接上真实负载就明显下垂。
Q4MEDIUM中 🇺🇸 US美 AP-style MCQAP 风格选择题 §6 Electrical power电功率 · HS-PS3-5 [4 marks][4 分]

An $8.0\ \Omega$ resistor carries $3.0$ A. Power dissipated?$8.0\ \Omega$ 电阻通过 $3.0$ A。耗散功率为多少？

Answer:答案：  (C)  $72\ \text{W}$

(a) Use $P = I^2 R$ (known $I$ and $R$)用 $P = I^2 R$（已知 $I$ 与 $R$） M1·A1·A1

$$ P \;=\; I^2 R \;=\; (3.0)^2 (8.0) \;=\; 9 \times 8.0 \;=\; 72 \;\text{W}. $$ Option (C). Cross-check: $V = IR = 24$ V, so $P = IV = 3.0 \times 24 = 72$ W. $\checkmark$选 (C)。交叉验证：$V = IR = 24$ V，故 $P = IV = 3.0 \times 24 = 72$ W。$\checkmark$
Why the distractors fail.干扰项分析。
(A) $24\ \text{W}$: computes $I \times R = 3.0 \times 8.0$ (that is the voltage, not the power).算成 $I \times R = 3.0 \times 8.0$（那是电压，不是功率）。
(B) $2.7\ \text{W}$: uses $P = V^2/R$ but with $V = 3.0$ V (confusing the current with a voltage).用 $P = V^2/R$ 却取 $V = 3.0$ V（把电流误当电压）。
(D) $216\ \text{W}$: computes $I^2 R^2$ or $I \times R^2$ (squares the resistance too).算成 $I^2 R^2$ 或 $I \times R^2$（把电阻也平方了）。
Pick the power form that uses only what you are given: $P = IV = I^2R = V^2/R$.选用仅含已知量的功率形式：$P = IV = I^2R = V^2/R$。 All three forms are equivalent through Ohm's law, but choosing the right one avoids an extra step. Here $I$ and $R$ are given, so $P = I^2R$ is direct; using $P = IV$ would first require computing $V$. The most common slip is forgetting to square the current. Note that $P = I^2R$ shows power grows with the square of current, which is why doubling the current in a heating element quadruples the heat output, and why transmission lines minimise current (high voltage) to cut $I^2R$ losses.三种形式通过欧姆定律等价，但选对那一种可省去额外步骤。此题已知 $I$ 与 $R$，故 $P = I^2R$ 最直接；用 $P = IV$ 则需先算出 $V$。最常见的失误是忘记把电流平方。注意 $P = I^2R$ 表明功率随电流的平方增长，这就是为何加热元件电流加倍则发热量变为四倍，也是为何输电线尽量降低电流（采用高压）以减小 $I^2R$ 损耗。
Q5MEDIUM中 🇨🇦 BC卑 BC Provincial-style卑诗省考风格 §4 + §5 Series & parallel resistance串联与并联电阻 · Physics 11 [6 marks][6 分]

Resistors $5.0\ \Omega$, $10\ \Omega$, $15\ \Omega$. (a) Series equivalent. (b) $5.0\ \Omega \parallel 10\ \Omega$. (c) Why parallel < series.电阻 $5.0\ \Omega$、$10\ \Omega$、$15\ \Omega$。(a) 串联等效。(b) $5.0\ \Omega$ 与 $10\ \Omega$ 并联。(c) 为何并联 < 串联。

Answer:答案：  (a) $R_{\text{series}} = 30\ \Omega$  ·  (b) $R_{\text{eq}} = 3.3\ \Omega$  ·  (c) extra paths lower opposition额外通道降低阻碍

(a) Series resistances add directly串联电阻直接相加 M1·A1

$$ R_{\text{series}} \;=\; 5.0 + 10 + 15 \;=\; 30 \;\Omega. $$

(b) Parallel resistances combine by reciprocals并联电阻按倒数合成 M1·A1

$$ \frac{1}{R_{\text{eq}}} \;=\; \frac{1}{5.0} + \frac{1}{10} \;=\; \frac{2}{10} + \frac{1}{10} \;=\; \frac{3}{10} \;\Longrightarrow\; R_{\text{eq}} \;=\; \frac{10}{3} \;\approx\; 3.3 \;\Omega. $$

(c) Why parallel is always smaller为何并联总是更小 A1·A1

Adding resistors in series forces the current through more opposition in a single path, so resistances add and the total exceeds the largest single resistor. Adding resistors in parallel opens extra paths for the current, so the equivalent resistance is always less than the smallest branch. A parallel combination therefore must be smaller than any series combination of the same resistors.串联增加电阻迫使电流在单一路径上经历更多阻碍，故电阻叠加，总值超过最大的单个电阻。并联增加电阻则为电流开辟额外通道，故等效电阻始终小于最小的支路。因此同样几只电阻的并联组合一定小于串联组合。
Series adds opposition along one path; parallel adds paths, reducing opposition.串联沿一条路径叠加阻碍；并联增加路径，降低阻碍。 Two quick sanity bounds catch most arithmetic slips: $R_{\text{series}} >$ the largest single resistor, and $R_{\text{parallel}} <$ the smallest single resistor. In part (b), $3.3\ \Omega$ is indeed less than $5.0\ \Omega$, so the result passes the bound. The water analogy makes it intuitive: series resistors are like narrow pipes end to end (harder to push water through), while parallel resistors are like extra pipes side by side (easier overall flow). Note $10/3\ \Omega$ is an exact value; report $3.3\ \Omega$ to two significant figures only at the final step.两条快速边界核验能抓住大多数算术失误：$R_{\text{串联}} >$ 最大单个电阻，$R_{\text{并联}} <$ 最小单个电阻。(b) 中 $3.3\ \Omega$ 确实小于 $5.0\ \Omega$，故结果通过核验。水流类比让这一点直观：串联电阻像首尾相接的细管（水更难推过），并联电阻像并排的额外管道（整体流动更顺畅）。注意 $10/3\ \Omega$ 是精确值；仅在最后一步取两位有效数字报作 $3.3\ \Omega$。
PART II  ·  EXTENDED RESPONSE · SOLUTIONS第二部分  ·  简答题 · 详解AP-feeder FRQ + honors · 32 marksAP 衔接简答题 + 荣誉级 · 共 32 分

Section B · Worked SolutionsB 部分 · 详细解答

Q6MEDIUM中 🇺🇸 US美 AP-feeder FRQAP 衔接简答题 §4 Series circuit analysis串联电路分析 · HS-PS3-5 [8 marks][8 分]

$R_1 = 2.0\ \Omega$, $R_2 = 4.0\ \Omega$, $R_3 = 6.0\ \Omega$ in series across $12$ V (negligible $r$). (a) Total resistance. (b) Current. (c) Voltage across each, verify sum. (d) Total power.$R_1 = 2.0\ \Omega$、$R_2 = 4.0\ \Omega$、$R_3 = 6.0\ \Omega$ 串联接 $12$ V（内阻可忽略）。(a) 总电阻。(b) 电流。(c) 各电阻电压并核验之和。(d) 总功率。

Answer:答案：  (a) $R_{\text{total}} = 12\ \Omega$  ·  (b) $I = 1.0\ \text{A}$  ·  (c) $2.0,\ 4.0,\ 6.0\ \text{V}$  ·  (d) $P = 12\ \text{W}$

(a) Total resistance (series add)总电阻（串联相加） M1·A1

$$ R_{\text{total}} \;=\; R_1 + R_2 + R_3 \;=\; 2.0 + 4.0 + 6.0 \;=\; 12 \;\Omega. $$

(b) Current from Ohm's law由欧姆定律求电流 M1·A1

$$ I \;=\; \frac{V}{R_{\text{total}}} \;=\; \frac{12}{12} \;=\; 1.0 \;\text{A}. $$ In series the current is the same through every resistor.串联中流过每只电阻的电流相同。

(c) Voltage across each resistor每只电阻两端的电压 M1·A1·A1

$$ V_1 = IR_1 = (1.0)(2.0) = 2.0\ \text{V}, \quad V_2 = (1.0)(4.0) = 4.0\ \text{V}, \quad V_3 = (1.0)(6.0) = 6.0\ \text{V}. $$ Check: $V_1 + V_2 + V_3 = 2.0 + 4.0 + 6.0 = 12$ V $=$ supply voltage. $\checkmark$核验：$V_1 + V_2 + V_3 = 2.0 + 4.0 + 6.0 = 12$ V $=$ 电源电压。$\checkmark$

(d) Total power delivered输出的总功率 A1

$$ P \;=\; IV \;=\; (1.0)(12) \;=\; 12 \;\text{W}. $$
Series circuit: one current everywhere, voltage shared in proportion to resistance.串联电路：各处电流唯一，电压按电阻比例分配。 Because the same current flows through every series resistor, the voltage drop across each is $IR_i$, so the resistors act as a voltage divider: the largest resistor takes the largest share of the supply voltage. Kirchhoff's voltage law guarantees the drops sum to the source EMF, which is the built-in self-check in part (c). The total power can also be found component by component ($P_i = I^2 R_i$): $2 + 4 + 6 = 12$ W, matching $P = IV$. Energy conservation requires the power delivered by the battery to equal the sum dissipated in the resistors.由于相同电流流过每只串联电阻，每只电阻上的压降为 $IR_i$，故各电阻构成分压器：阻值最大者分得电源电压的最大份额。基尔霍夫电压定律保证各压降之和等于电源电动势，这正是 (c) 中内置的自我核验。总功率也可逐元件计算（$P_i = I^2 R_i$）：$2 + 4 + 6 = 12$ W，与 $P = IV$ 一致。能量守恒要求电池输出的功率等于各电阻耗散功率之和。
Q7MEDIUM中 🇨🇦 ON安 ON Provincial-style安大略省考风格 §5 Parallel circuit analysis并联电路分析 · SPH3U F3 [8 marks][8 分]

$R_1 = 8.0\ \Omega$, $R_2 = 12\ \Omega$, $R_3 = 24\ \Omega$ in parallel across $24$ V. (a) Equivalent resistance. (b) Branch currents. (c) Total current, verify it equals the sum.$R_1 = 8.0\ \Omega$、$R_2 = 12\ \Omega$、$R_3 = 24\ \Omega$ 并联接 $24$ V。(a) 等效电阻。(b) 各支路电流。(c) 总电流并核验其等于各支路之和。

Answer:答案：  (a) $R_{\text{eq}} = 4.0\ \Omega$  ·  (b) $3.0,\ 2.0,\ 1.0\ \text{A}$  ·  (c) $I_{\text{total}} = 6.0\ \text{A}$

(a) Equivalent resistance (reciprocals add)等效电阻（倒数相加） M1·A1·A1

$$ \frac{1}{R_{\text{eq}}} \;=\; \frac{1}{8.0} + \frac{1}{12} + \frac{1}{24} \;=\; \frac{3}{24} + \frac{2}{24} + \frac{1}{24} \;=\; \frac{6}{24} \;=\; \frac{1}{4} \;\Longrightarrow\; R_{\text{eq}} \;=\; 4.0 \;\Omega. $$ Note $4.0\ \Omega$ is less than the smallest branch ($8.0\ \Omega$), as expected for parallel.注意 $4.0\ \Omega$ 小于最小支路（$8.0\ \Omega$），符合并联预期。

(b) Branch currents (same $24$ V across each)各支路电流（各支路均为 $24$ V） M1·A1·A1

$$ I_1 = \frac{24}{8.0} = 3.0\ \text{A}, \quad I_2 = \frac{24}{12} = 2.0\ \text{A}, \quad I_3 = \frac{24}{24} = 1.0\ \text{A}. $$

(c) Total current总电流 M1·A1

$$ I_{\text{total}} \;=\; \frac{V}{R_{\text{eq}}} \;=\; \frac{24}{4.0} \;=\; 6.0 \;\text{A}. $$ Check: $I_1 + I_2 + I_3 = 3.0 + 2.0 + 1.0 = 6.0$ A $= I_{\text{total}}$. $\checkmark$核验：$I_1 + I_2 + I_3 = 3.0 + 2.0 + 1.0 = 6.0$ A $= I_{\text{total}}$。$\checkmark$
Parallel circuit: one voltage across all branches, current shared in inverse proportion to resistance.并联电路：各支路电压唯一，电流按电阻反比分配。 Each branch sees the full supply voltage, so the branch with the smallest resistance draws the most current ($I = V/R$). The junction rule (Kirchhoff's current law) guarantees the branch currents sum to the total supply current, the built-in self-check in part (c). The reciprocal sum is the most error-prone step: convert to a common denominator before adding, and remember to invert at the end to get $R_{\text{eq}}$, not $1/R_{\text{eq}}$. The sanity bound $R_{\text{eq}} < R_{\text{smallest}}$ catches the most common slip of forgetting the final inversion.每条支路都承受全部电源电压，故阻值最小的支路通过最大电流（$I = V/R$）。节点法则（基尔霍夫电流定律）保证各支路电流之和等于总电源电流，这正是 (c) 内置的自我核验。倒数求和是最易出错的步骤：相加前先通分，最后记得取倒数得到 $R_{\text{eq}}$ 而非 $1/R_{\text{eq}}$。核验边界 $R_{\text{eq}} < R_{\text{最小}}$ 可抓住忘记最后取倒数这一最常见失误。
Q8HARD难 🇨🇦 BC卑 BC Provincial-style卑诗省考风格 §2 + §6 EMF, internal resistance & power电动势、内阻与功率 · Physics 11 [8 marks][8 分]

$\varepsilon = 12$ V, internal $r = 0.40\ \Omega$, external $R = 5.6\ \Omega$. (a) Current. (b) Terminal voltage. (c) Internal & external power. (d) Effect of decreasing $R$.$\varepsilon = 12$ V，内阻 $r = 0.40\ \Omega$，外阻 $R = 5.6\ \Omega$。(a) 电流。(b) 端电压。(c) 内、外功率。(d) 减小 $R$ 的影响。

Answer:答案：  (a) $I = 2.0\ \text{A}$  ·  (b) $V_T = 11.2\ \text{V}$  ·  (c) $P_{\text{int}} = 1.6\ \text{W},\ P_{\text{ext}} = 22.4\ \text{W}$  ·  (d) terminal voltage falls端电压下降

(a) Current (total resistance $= R + r$)电流（总电阻 $= R + r$） M1·A1

$$ I \;=\; \frac{\varepsilon}{R + r} \;=\; \frac{12}{5.6 + 0.40} \;=\; \frac{12}{6.0} \;=\; 2.0 \;\text{A}. $$

(b) Terminal voltage端电压 M1·A1

$$ V_T \;=\; \varepsilon - Ir \;=\; 12 - (2.0)(0.40) \;=\; 12 - 0.80 \;=\; 11.2 \;\text{V}. $$

(c) Power dissipated internally and delivered externally内部耗散与外部输送的功率 M1·A1·A1

$$ P_{\text{int}} \;=\; I^2 r \;=\; (2.0)^2 (0.40) \;=\; 1.6 \;\text{W}, \qquad P_{\text{ext}} \;=\; I^2 R \;=\; (2.0)^2 (5.6) \;=\; 22.4 \;\text{W}. $$ Check: total power $= \varepsilon I = 12 \times 2.0 = 24$ W $= 1.6 + 22.4$. $\checkmark$核验：总功率 $= \varepsilon I = 12 \times 2.0 = 24$ W $= 1.6 + 22.4$。$\checkmark$

(d) Effect of decreasing the external resistance减小外电阻的影响 A1

Decreasing $R$ increases the current $I = \varepsilon/(R+r)$, which increases the internal drop $Ir$, so the terminal voltage $V_T = \varepsilon - Ir$ falls.减小 $R$ 使电流 $I = \varepsilon/(R+r)$ 增大，内阻压降 $Ir$ 随之增大，故端电压 $V_T = \varepsilon - Ir$ 下降。
A real source is an EMF in series with its internal resistance; that one model explains terminal-voltage sag.真实电源是电动势与其内阻的串联；这一模型即可解释端电压下垂。 Treating the battery as $\varepsilon$ in series with $r$ turns every internal-resistance problem into an ordinary series circuit: the same current flows through $r$ and $R$, and the EMF splits between them as $\varepsilon = Ir + IR$. The internal power $I^2 r$ is wasted as heat inside the battery (why batteries warm up under heavy load); the external power $I^2 R$ is the useful output. Maximum power transfer to the load occurs when $R = r$, a result worth remembering. The part (d) reasoning shows why a battery cannot deliver unlimited current: as $R \to 0$ (a short circuit), $I \to \varepsilon/r$ and the terminal voltage collapses toward zero, with almost all power dissipated internally.把电池视为 $\varepsilon$ 与 $r$ 串联，可将每个内阻问题化为普通串联电路：相同电流流过 $r$ 与 $R$，电动势按 $\varepsilon = Ir + IR$ 在二者间分配。内部功率 $I^2 r$ 作为热量浪费在电池内部（这是电池在重负载下发热的原因）；外部功率 $I^2 R$ 才是有用输出。当 $R = r$ 时负载获得最大功率传输，这一结论值得记住。(d) 的推理说明了电池为何不能输出无限电流：当 $R \to 0$（短路）时，$I \to \varepsilon/r$，端电压趋向零，几乎全部功率耗散在内部。
Q9HARD难Honors荣誉级 🇺🇸 US美 🇨🇦 BC卑 AP-feeder FRQAP 衔接简答题 §7 Kirchhoff's rules (two-loop)基尔霍夫定律（两回路） · BC Physics 11 [8 marks][8 分]

Two-loop network: $\varepsilon_1 = 10$ V (left), $\varepsilon_2 = 4.0$ V (right); $R_1 = R_2 = R_3 = 2.0\ \Omega$ ($R_2$ middle/shared). Currents $I_1$ (left, down), $I_2$ (middle, down), $I_3$ (right, down). (a) Junction equation. (b) Two loop equations. (c) Solve. (d) Verify.两回路网络：$\varepsilon_1 = 10$ V（左）、$\varepsilon_2 = 4.0$ V（右）；$R_1 = R_2 = R_3 = 2.0\ \Omega$（$R_2$ 居中共用）。电流 $I_1$（左，下）、$I_2$（中，下）、$I_3$（右，下）。(a) 节点方程。(b) 两个回路方程。(c) 求解。(d) 核验。

Answer:答案：  (a) $I_1 = I_2 + I_3$  ·  (c) $I_1 = 4.0\ \text{A},\ I_2 = 1.0\ \text{A},\ I_3 = 3.0\ \text{A}$

(a) Junction rule at the top node顶部节点的节点法则 A1

Charge is conserved at the node where $I_1$ splits:在 $I_1$ 分流的节点处电荷守恒： $$ I_1 \;=\; I_2 + I_3. $$

(b) Loop equations (clockwise; $+\varepsilon$ from $-$ to $+$, $-IR$ along the current)回路方程（顺时针；从负到正为 $+\varepsilon$，沿电流为 $-IR$） M1·A1

Left loop ($\varepsilon_1$, $R_1$, $R_2$):左回路（$\varepsilon_1$、$R_1$、$R_2$）： $$ \varepsilon_1 - I_1 R_1 - I_2 R_2 \;=\; 0 \;\Longrightarrow\; 10 - 2I_1 - 2I_2 \;=\; 0. $$ Right loop ($\varepsilon_2$, $R_3$, $R_2$):右回路（$\varepsilon_2$、$R_3$、$R_2$）： $$ \varepsilon_2 - I_3 R_3 + I_2 R_2 \;=\; 0 \;\Longrightarrow\; 4 - 2I_3 + 2I_2 \;=\; 0. $$

(c) Solve the simultaneous system联立求解 M1·A1·A1·A1

Substitute $I_1 = I_2 + I_3$ into the left loop:把 $I_1 = I_2 + I_3$ 代入左回路： $$ 10 - 2(I_2 + I_3) - 2I_2 \;=\; 10 - 4I_2 - 2I_3 \;=\; 0 \;\Longrightarrow\; 5 - 2I_2 - I_3 \;=\; 0. $$ From the right loop, $I_3 = 2 + I_2$. Substituting:由右回路 $I_3 = 2 + I_2$。代入： $$ 5 - 2I_2 - (2 + I_2) \;=\; 3 - 3I_2 \;=\; 0 \;\Longrightarrow\; I_2 \;=\; 1.0 \;\text{A}. $$ $$ I_3 \;=\; 2 + 1.0 \;=\; 3.0 \;\text{A}, \qquad I_1 \;=\; I_2 + I_3 \;=\; 1.0 + 3.0 \;=\; 4.0 \;\text{A}. $$

(d) Verify with the left-loop equation (used implicitly; re-check directly)用左回路方程核验 A1

Substitute into the left loop: $10 - 2(4.0) - 2(1.0) = 10 - 8 - 2 = 0$. $\checkmark$ The right loop also holds: $4 - 2(3.0) + 2(1.0) = 4 - 6 + 2 = 0$. $\checkmark$代入左回路：$10 - 2(4.0) - 2(1.0) = 10 - 8 - 2 = 0$。$\checkmark$ 右回路也成立：$4 - 2(3.0) + 2(1.0) = 4 - 6 + 2 = 0$。$\checkmark$
Kirchhoff's two rules are conservation of charge (junction) and energy (loop); together they solve any linear DC network.基尔霍夫两条定律即电荷守恒（节点）与能量守恒（回路）；二者合用可解任意线性直流网络。 For $n$ unknown branch currents you need $n$ independent equations: $(n-1)$ junction equations plus enough loop equations. Here $n = 3$: one junction plus two loops. The strict sign discipline is what trips students up. Fix it once: pick a loop direction, add $+\varepsilon$ when crossing a battery from $-$ to $+$, and $-IR$ when crossing a resistor along the assumed current (the $+I_2 R_2$ term in the right loop appears because that loop crosses $R_2$ against $I_2$). A negative current in the answer would simply mean the true direction is opposite to the assumed one; all three currents here came out positive, confirming the assumed downward directions were correct. Always close with the unused-equation check.对 $n$ 个未知支路电流需要 $n$ 个独立方程：$(n-1)$ 个节点方程加足够的回路方程。此处 $n = 3$：一个节点加两个回路。严格的符号纪律最易让学生出错。一次性约定好：选定回路方向，从负到正穿过电池加 $+\varepsilon$，沿假设电流穿过电阻加 $-IR$（右回路中出现 $+I_2 R_2$ 是因为该回路逆 $I_2$ 穿过 $R_2$）。结果中若出现负电流，仅表示真实方向与假设相反；此处三个电流均为正，确认假设的向下方向正确。最后务必用未使用的方程核验收尾。
PART III  ·  MODELING / APPLIED · SOLUTIONS第三部分  ·  建模与应用 · 详解Universal applied · 26 marks通用应用题 · 共 26 分

Section C · Modeling and ApplicationsC 部分 · 建模与应用

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]

$1200$ W microwave on a $120$ V supply. (a) Current. (b) Element resistance. (c) Energy in $5.0$ min (J). (d) Cost at $\$0.12$/kWh.$1200$ W 微波炉接 $120$ V 电源。(a) 电流。(b) 元件电阻。(c) $5.0$ min 内电能（J）。(d) 按 $\$0.12$/kWh 的费用。

Answer:答案：  (a) $I = 10\ \text{A}$  ·  (b) $R = 12\ \Omega$  ·  (c) $W = 3.6 \times 10^5\ \text{J}$  ·  (d) $\approx \$0.012$

(a) Current from $P = IV$由 $P = IV$ 求电流 M1·A1

$$ I \;=\; \frac{P}{V} \;=\; \frac{1200}{120} \;=\; 10 \;\text{A}. $$

(b) Resistance电阻 M1·A1

$$ R \;=\; \frac{V}{I} \;=\; \frac{120}{10} \;=\; 12 \;\Omega. \quad \left(\text{or } R = \frac{V^2}{P} = \frac{14400}{1200} = 12\ \Omega\right) $$

(c) Energy in $5.0$ min $= 300$ s$5.0$ min $= 300$ s 内的电能 M1·A1

$$ W \;=\; P\,\Delta t \;=\; 1200 \times 300 \;=\; 3.6 \times 10^5 \;\text{J}. $$

(d) Cost费用 A1

Energy in kWh: $W = 1.2\ \text{kW} \times (5.0/60)\ \text{h} = 0.10$ kWh. Cost $= 0.10 \times \$0.12 \approx \$0.012$ (just over one cent).以千瓦时计：$W = 1.2\ \text{kW} \times (5.0/60)\ \text{h} = 0.10$ kWh。费用 $= 0.10 \times \$0.12 \approx \$0.012$（略多于一美分）。
Power is the rate of energy use; energy is power times time, and the kWh is just a large joule.功率是用能的速率；电能是功率乘时间，而千瓦时只是一个大号的焦耳。 The rated wattage already bundles voltage and current ($P = IV$), so the current and resistance follow immediately from the supply voltage. The single most common error in part (c) is forgetting to convert minutes to seconds before using $W = P\,\Delta t$ in joules; $5.0$ min is $300$ s, not $5$. Utility bills use kWh because a joule is tiny: $1\ \text{kWh} = 3.6 \times 10^6\ \text{J}$, so the $3.6 \times 10^5$ J here is exactly $0.10$ kWh. Cross-checking the two energy units against each other is a quick way to catch a unit-conversion slip.额定功率已经把电压与电流打包（$P = IV$），故电流与电阻可由电源电压立即得出。(c) 中最常见的错误是用 $W = P\,\Delta t$（焦耳）前忘记把分钟换成秒；$5.0$ min 是 $300$ s，而非 $5$。电费账单用千瓦时，因为焦耳太小：$1\ \text{kWh} = 3.6 \times 10^6\ \text{J}$，故此处的 $3.6 \times 10^5$ J 恰为 $0.10$ kWh。把两种能量单位相互核对，是抓住单位换算失误的快捷方法。
Q11MEDIUM中 🇨🇦 ON安 Universal Applied通用应用题 §4 + §5 + §6 Mixed network混联网络 · SPH3U F3 [7 marks][7 分]

Two $6.0\ \Omega$ in parallel, in series with $9.0\ \Omega$, across $24$ V. (a) Total resistance. (b) Total current. (c) Voltage across $9.0\ \Omega$ and across the parallel pair. (d) Current in each $6.0\ \Omega$.两只 $6.0\ \Omega$ 并联，再与 $9.0\ \Omega$ 串联，接 $24$ V。(a) 总电阻。(b) 总电流。(c) $9.0\ \Omega$ 与并联对两端电压。(d) 每只 $6.0\ \Omega$ 的电流。

Answer:答案：  (a) $R_{\text{total}} = 12\ \Omega$  ·  (b) $I = 2.0\ \text{A}$  ·  (c) $18\ \text{V}$ & $6.0\ \text{V}$  ·  (d) $1.0\ \text{A}$ each各

(a) Total resistance (parallel pair, then series)总电阻（先并联对，再串联） M1·A1

Two equal resistors in parallel: $R_{\parallel} = 6.0/2 = 3.0\ \Omega$. In series with $9.0\ \Omega$:两只等值电阻并联：$R_{\parallel} = 6.0/2 = 3.0\ \Omega$。再与 $9.0\ \Omega$ 串联： $$ R_{\text{total}} \;=\; 3.0 + 9.0 \;=\; 12 \;\Omega. $$

(b) Total current总电流 A1

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

(c) Voltage across the series resistor and the parallel pair串联电阻与并联对两端电压 M1·A1

The full $2.0$ A flows through the $9.0\ \Omega$ resistor:全部 $2.0$ A 流过 $9.0\ \Omega$ 电阻： $$ V_{9} = IR = (2.0)(9.0) = 18\ \text{V}, \qquad V_{\parallel} = IR_{\parallel} = (2.0)(3.0) = 6.0\ \text{V}. $$ Check: $18 + 6.0 = 24$ V $=$ supply. $\checkmark$核验：$18 + 6.0 = 24$ V $=$ 电源。$\checkmark$

(d) Current through each $6.0\ \Omega$ branch每只 $6.0\ \Omega$ 支路的电流 M1·A1

Each branch sees the parallel-pair voltage $6.0$ V:每条支路承受并联对电压 $6.0$ V： $$ I_{\text{branch}} \;=\; \frac{6.0}{6.0} \;=\; 1.0 \;\text{A (each; } 1.0 + 1.0 = 2.0\ \text{A total).} $$
Reduce a mixed network in stages: collapse the parallel block first, then treat the rest as a series chain.分阶段化简混联网络：先合并并联块，再把其余部分当作串联链处理。 The strategy that never fails: replace the parallel pair with its single equivalent ($3.0\ \Omega$), solve the resulting simple series circuit for the total current, then "unfold" backward to find individual voltages and branch currents. The full current passes through the series resistor, so it gets the larger voltage share ($18$ V versus $6.0$ V) because it has the larger resistance. Inside the parallel block the $6.0$ V splits equally between the two identical branches, $1.0$ A each, and these sum back to the $2.0$ A total, the junction-rule check. Working stage by stage and checking voltage sums and current sums at each level catches almost every arithmetic error.永不失手的策略：先用单个等效电阻（$3.0\ \Omega$）替换并联对，求出所得简单串联电路的总电流，再"反向展开"求各电压与支路电流。全部电流流过串联电阻，故它因阻值较大而分得较大电压份额（$18$ V 对 $6.0$ V）。在并联块内，$6.0$ V 在两条相同支路间均分，各 $1.0$ A，二者相加回到 $2.0$ A 总电流，即节点法则核验。逐阶段计算并在每一层核对电压和与电流和，几乎能抓住所有算术错误。
Q12HARD难 🇨🇦 BC卑 Universal Applied通用应用题 §3 Resistivity电阻率 · Physics 11 [6 marks][6 分]

Nichrome wire: $L = 2.0$ m, $A = 2.2 \times 10^{-7}\ \text{m}^2$, $\rho = 1.1 \times 10^{-6}\ \Omega\cdot\text{m}$. (a) Resistance. (b) Current & power at $5.0$ V. (c) Resistance of same-material wire of double length.镍铬合金丝：$L = 2.0$ m，$A = 2.2 \times 10^{-7}\ \text{m}^2$，$\rho = 1.1 \times 10^{-6}\ \Omega\cdot\text{m}$。(a) 电阻。(b) $5.0$ V 下电流与功率。(c) 同材料、双倍长度丝的电阻。

Answer:答案：  (a) $R = 10\ \Omega$  ·  (b) $I = 0.50\ \text{A},\ P = 2.5\ \text{W}$  ·  (c) $20\ \Omega$

(a) Resistance from $R = \rho L / A$由 $R = \rho L / A$ 求电阻 M1·A1

$$ R \;=\; \frac{\rho L}{A} \;=\; \frac{(1.1 \times 10^{-6})(2.0)}{2.2 \times 10^{-7}} \;=\; \frac{2.2 \times 10^{-6}}{2.2 \times 10^{-7}} \;=\; 10 \;\Omega. $$

(b) Current and power at $5.0$ V$5.0$ V 下的电流与功率 M1·A1

$$ I \;=\; \frac{V}{R} \;=\; \frac{5.0}{10} \;=\; 0.50 \;\text{A}, \qquad P \;=\; \frac{V^2}{R} \;=\; \frac{(5.0)^2}{10} \;=\; 2.5 \;\text{W}. $$

(c) Resistance of the double-length wire双倍长度丝的电阻 A1·A1

Resistance is directly proportional to length ($R \propto L$) at fixed area and material. Doubling $L$ doubles $R$: $R' = 2 \times 10 = 20\ \Omega$. No full recalculation is needed.在材料与截面积固定时，电阻与长度成正比（$R \propto L$）。长度加倍则电阻加倍：$R' = 2 \times 10 = 20\ \Omega$。无需完整重算。
Resistance scales with the wire's geometry: $R = \rho L/A$, longer means more, thicker means less.电阻随导线几何尺寸变化：$R = \rho L/A$，越长越大，越粗越小。 Resistivity $\rho$ is an intrinsic material property (nichrome's high $\rho$ is exactly why it is used in heating elements), while resistance $R$ depends on the specific piece of wire. The two proportionalities to lock in: $R \propto L$ (a longer path opposes flow more) and $R \propto 1/A$ (a wider path lets more carriers through in parallel). This is why part (c) needs no arithmetic, only the proportionality. Watch the powers of ten: $2.2 \times 10^{-6}$ divided by $2.2 \times 10^{-7}$ is $10$, not $0.1$; subtract exponents carefully. The $2.5$ W dissipated is what makes a nichrome element glow and produce heat.电阻率 $\rho$ 是材料的内禀属性（镍铬合金的高 $\rho$ 正是其用作加热元件的原因），而电阻 $R$ 取决于具体的那段导线。需牢记两个正比关系：$R \propto L$（路径越长越阻碍流动）与 $R \propto 1/A$（路径越宽，越多载流子可并行通过）。这就是 (c) 无需运算、只需比例关系的原因。注意十的幂次：$2.2 \times 10^{-6}$ 除以 $2.2 \times 10^{-7}$ 等于 $10$ 而非 $0.1$；指数相减要仔细。耗散的 $2.5$ W 正是镍铬元件发红发热的原因。
Q13HARD难Honors荣誉级 🇺🇸 US美 🇨🇦 ON安 AP-feeder FRQAP 衔接简答题 §7 Kirchhoff's rules (applied)基尔霍夫定律（应用） · SPH3U F3 [6 marks][6 分]

Single loop: $9.0$ V battery (negligible $r$) in series with $3.0\ \Omega$ and $6.0\ \Omega$. Use Kirchhoff throughout. (a) Loop equation + sign convention. (b) Solve for current. (c) Voltage across each, confirm loop rule.单回路：$9.0$ V 电池（内阻可忽略）与 $3.0\ \Omega$、$6.0\ \Omega$ 串联。全程用基尔霍夫定律。(a) 回路方程与符号约定。(b) 求电流。(c) 各电阻电压并确认回路法则。

Answer:答案：  (a) $9.0 - 3I - 6I = 0$  ·  (b) $I = 1.0\ \text{A}$  ·  (c) $3.0\ \text{V}$ & $6.0\ \text{V}$, sum $= 9.0$ V

(a) Loop-rule equation (KVL)回路法则方程（KVL） M1·A1

Assume current $I$ flows clockwise, in the direction the battery drives it. Traverse the loop clockwise: crossing the battery from $-$ to $+$ gives $+9.0$; crossing each resistor along the current gives $-IR$. The algebraic sum of voltage changes around the closed loop is zero:假设电流 $I$ 沿顺时针流动，即电池驱动的方向。顺时针绕行回路：从负极到正极穿过电池为 $+9.0$；沿电流方向穿过每只电阻为 $-IR$。绕闭合回路一圈电压变化的代数和为零： $$ 9.0 - 3.0\,I - 6.0\,I \;=\; 0. $$

(b) Solve for the loop current求回路电流 M1·A1

$$ 9.0 \;=\; 9.0\,I \;\Longrightarrow\; I \;=\; 1.0 \;\text{A}. $$

(c) Voltage across each resistor; confirm the loop rule各电阻电压；确认回路法则 M1·A1

$$ V_{3} = IR = (1.0)(3.0) = 3.0\ \text{V}, \qquad V_{6} = (1.0)(6.0) = 6.0\ \text{V}. $$ Loop rule check: $+9.0 - 3.0 - 6.0 = 0$. The battery's energy gain per charge is exactly balanced by the drops across the two resistors. $\checkmark$回路法则核验：$+9.0 - 3.0 - 6.0 = 0$。电池为每单位电荷提供的能量恰好被两只电阻上的压降平衡。$\checkmark$
The loop rule is energy conservation per unit charge: every joule a battery gives is spent before the charge returns to its start.回路法则即每单位电荷的能量守恒：电池给出的每一焦耳都在电荷回到起点前用完。 Even for a simple series circuit where $I = V/R_{\text{total}}$ gives the answer in one line, writing the full KVL equation builds the habit needed for multi-loop networks. The sign convention is the whole game: a battery traversed $-$ to $+$ contributes $+\varepsilon$, a resistor traversed along the assumed current contributes $-IR$, and the sum around any closed loop is zero. Physically, a unit charge gains $9.0$ J crossing the battery and loses $3.0$ J then $6.0$ J crossing the resistors, returning to its starting potential with zero net change. The drops dividing as $3 : 6$ (one-third and two-thirds of the EMF) is the series voltage-divider rule, set by the resistance ratio.即便对于用 $I = V/R_{\text{总}}$ 一行即可求解的简单串联电路，写出完整的 KVL 方程也能培养处理多回路网络所需的习惯。符号约定是关键所在：从负到正穿过电池贡献 $+\varepsilon$，沿假设电流穿过电阻贡献 $-IR$，绕任意闭合回路之和为零。从物理上看，单位电荷穿过电池获得 $9.0$ J，穿过电阻先后失去 $3.0$ J 与 $6.0$ J，回到起点电位时净变化为零。压降按 $3 : 6$（电动势的三分之一与三分之二）分配，即由电阻比决定的串联分压规则。