← All Units← 返回单元列表 ← Course Hub← 课程主页
I B  P H Y S I C S  H L
Unit B5 · SolutionsUnit B5 · 解析

Current and Circuits · Solutions电流与电路 · 解析

Companion to the IB-Style Practice SetIB 风格练习题的解析配套

MEDIUM HARD Paper 1 Paper 1B Paper 2 HL ONLY

Syllabus B5.1 to B5.6考纲 B5.1 至 B5.6PHYSICS HL

PART I  ·  PAPER 1 STYLE第一部分  ·  第一卷风格Short structured · 30 marks短结构题 · 30 分

Worked Solutions详细解析

Q1MEDIUMPaper 1charge flow + drift velocity电荷流动与漂移速度[4 marks]

Copper wire, $A = 1.0 \times 10^{-6}\ \mathrm{m^{2}}$, $n = 8.5 \times 10^{28}\ \mathrm{m^{-3}}$. (a) charge and electron count for $0.50\ \mathrm{A}$ over $3.0\ \mathrm{min}$; (b) drift speed at $8.5\ \mathrm{A}$.铜导线，$A = 1.0 \times 10^{-6}\ \mathrm{m^{2}}$，$n = 8.5 \times 10^{28}\ \mathrm{m^{-3}}$。(a) $0.50\ \mathrm{A}$ 持续 $3.0\ \mathrm{min}$ 的电荷量与电子数；(b) $8.5\ \mathrm{A}$ 时的漂移速率。

Answers:答案：  (a) $\Delta q = 90\ \mathrm{C}$, $N \approx 5.6 \times 10^{20}$  ·  (b) $v \approx 6.3 \times 10^{-4}\ \mathrm{m\,s^{-1}}$

(a) Charge and electron count M1·A1

Convert time: $\Delta t = 3.0\ \mathrm{min} = 180\ \mathrm{s}$. Use $I = \Delta q/\Delta t$: $\Delta q = I\,\Delta t = (0.50)(180) = 90\ \mathrm{C}$. (M1)

Number of electrons $N = \dfrac{\Delta q}{e} = \dfrac{90}{1.60 \times 10^{-19}} \approx 5.6 \times 10^{20}$. (A1)

(b) Drift speed M1·A1

Use $I = nAvq$ with $q = e$, rearranged for $v$: (M1)

$$ v = \frac{I}{nAe} = \frac{8.5}{(8.5 \times 10^{28})(1.0 \times 10^{-6})(1.60 \times 10^{-19})} \approx 6.3 \times 10^{-4}\ \mathrm{m\,s^{-1}}. $$

(A1)

Insight. The single most common loss here is leaving the time in minutes, which makes the charge $60$ times too small. The drift speed result, under a millimetre per second, is deliberately counter-intuitive: the lamp lights instantly not because electrons race down the wire but because the electric field is established along the whole circuit at near light speed, setting every electron drifting almost at once.

(a) 电荷量与电子数 M1·A1

换算时间：$\Delta t = 3.0\ \mathrm{min} = 180\ \mathrm{s}$。用 $I = \Delta q/\Delta t$：$\Delta q = I\,\Delta t = (0.50)(180) = 90\ \mathrm{C}$。(M1)

电子数 $N = \dfrac{\Delta q}{e} = \dfrac{90}{1.60 \times 10^{-19}} \approx 5.6 \times 10^{20}$。(A1)

(b) 漂移速率 M1·A1

用 $I = nAvq$，取 $q = e$，解出 $v$：(M1)

$$ v = \frac{I}{nAe} = \frac{8.5}{(8.5 \times 10^{28})(1.0 \times 10^{-6})(1.60 \times 10^{-19})} \approx 6.3 \times 10^{-4}\ \mathrm{m\,s^{-1}}. $$

(A1)

要点。这里最常见的失分是把时间留在分钟，使电荷小了 $60$ 倍。漂移速率不足每秒一毫米，是故意反直觉的：灯立即亮起并非因为电子在导线中飞奔，而是因为电场以接近光速沿整条电路建立，几乎同时让每个电子开始漂移。
Q2MEDIUMPaper 1power, energy and emf vs pd功率、能量与电动势-电势差[6 marks]

$8.0\ \Omega$ resistor carrying $1.5\ \mathrm{A}$. (a) power and energy in $1.0\ \mathrm{min}$; (b) energy of a $2.0\ \mathrm{kW}$ heater over $4.0\ \mathrm{h}$ in kW·h and J; (c) distinguish emf from pd.$8.0\ \Omega$ 电阻通过 $1.5\ \mathrm{A}$。(a) 功率与 $1.0\ \mathrm{min}$ 内能量；(b) $2.0\ \mathrm{kW}$ 电暖器 $4.0\ \mathrm{h}$ 的能量（kW·h 与 J）；(c) 区分电动势与电势差。

Answers:答案：  (a) $P = 18\ \mathrm{W}$, $W = 1.08 \times 10^{3}\ \mathrm{J}$  ·  (b) $8.0\ \mathrm{kW\,h} = 2.88 \times 10^{7}\ \mathrm{J}$  ·  (c) emf supplies energy to charge; pd removes energy from charge

(a) Power and energy M1·A1

Use $P = I^{2}R$ (the two known quantities): $P = (1.5)^{2}(8.0) = 18\ \mathrm{W}$. (M1)

Energy in $\Delta t = 60\ \mathrm{s}$: $W = P\,\Delta t = (18)(60) = 1.08 \times 10^{3}\ \mathrm{J}$. (A1)

(b) Heater energy M1·A1

Energy $=$ power $\times$ time $= (2.0\ \mathrm{kW})(4.0\ \mathrm{h}) = 8.0\ \mathrm{kW\,h}$. (M1)

In joules: $8.0\ \mathrm{kW\,h} \times 3.6 \times 10^{6}\ \mathrm{J\,(kW\,h)^{-1}} = 2.88 \times 10^{7}\ \mathrm{J}$. (A1)

(c) emf versus pd A1·A1

The emf of a source is the energy supplied per unit charge by the source (it gives energy to the charges). (A1)

The pd across a resistor is the energy transferred from electrical to other forms per unit charge by that component (it takes energy from the charges). Both are measured in volts. (A1)

Insight. All three power forms agree only because $V = IR$: here $P = VI = I^{2}R = V^{2}/R = 18\ \mathrm{W}$ with $V = 12\ \mathrm{V}$. The kilowatt-hour is an energy unit, not a power unit, and is the natural choice for billing; convert to joules only when SI is demanded. The emf-versus-pd distinction is a direction-of-energy-transfer statement, so naming "same units, opposite roles" is what earns the marks.

(a) 功率与能量 M1·A1

用 $P = I^{2}R$（两个已知量）：$P = (1.5)^{2}(8.0) = 18\ \mathrm{W}$。(M1)

$\Delta t = 60\ \mathrm{s}$ 内能量：$W = P\,\Delta t = (18)(60) = 1.08 \times 10^{3}\ \mathrm{J}$。(A1)

(b) 电暖器能量 M1·A1

能量 $=$ 功率 $\times$ 时间 $= (2.0\ \mathrm{kW})(4.0\ \mathrm{h}) = 8.0\ \mathrm{kW\,h}$。(M1)

换算为焦耳：$8.0\ \mathrm{kW\,h} \times 3.6 \times 10^{6}\ \mathrm{J\,(kW\,h)^{-1}} = 2.88 \times 10^{7}\ \mathrm{J}$。(A1)

(c) 电动势与电势差 A1·A1

电源的电动势是电源每单位电荷所供给的能量（把能量给予电荷）。(A1)

电阻两端的电势差是该元件每单位电荷把电能转化为其他形式的能量（从电荷取走能量）。两者都以伏特计。(A1)

要点。三种功率形式一致只因 $V = IR$：此处取 $V = 12\ \mathrm{V}$，则 $P = VI = I^{2}R = V^{2}/R = 18\ \mathrm{W}$。千瓦时是能量单位而非功率单位，是计费的自然选择；只有要求 SI 时才换算成焦耳。电动势与电势差的区别是能量传递方向的表述，说出"单位相同、角色相反"才得分。
Q3HARDPaper 1resistivity + stretched wire电阻率与拉伸导线[6 marks]

Nichrome wire: $L = 1.5\ \mathrm{m}$, $d = 0.40\ \mathrm{mm}$, $\rho = 1.1 \times 10^{-6}\ \mathrm{\Omega\,m}$. (a) resistance; (b) new resistance when stretched to $2L$ at constant volume.镍铬合金导线：$L = 1.5\ \mathrm{m}$，$d = 0.40\ \mathrm{mm}$，$\rho = 1.1 \times 10^{-6}\ \mathrm{\Omega\,m}$。(a) 电阻；(b) 体积不变拉伸到 $2L$ 后的新电阻。

Answers:答案：  (a) $R \approx 13\ \Omega$  ·  (b) $R' = 4R \approx 53\ \Omega$

(a) Resistance of the wire M1·M1·A1

Radius $r = d/2 = 0.20\ \mathrm{mm} = 2.0 \times 10^{-4}\ \mathrm{m}$. Cross-sectional area: (M1)

$$ A = \pi r^{2} = \pi (2.0 \times 10^{-4})^{2} \approx 1.26 \times 10^{-7}\ \mathrm{m^{2}}. $$

Apply $R = \dfrac{\rho L}{A}$: (M1)

$$ R = \frac{(1.1 \times 10^{-6})(1.5)}{1.26 \times 10^{-7}} \approx 13\ \Omega. $$

(A1)

(b) Stretched at constant volume M1·R1·A1

Volume $V = LA$ is fixed. If the length doubles to $2L$, the area must halve to $A/2$ to keep $LA$ constant. (R1)

Substitute both changes into $R' = \dfrac{\rho (2L)}{A/2} = 4\,\dfrac{\rho L}{A} = 4R$. (M1)

$$ R' = 4 \times 13 \approx 53\ \Omega. $$

(A1)

Insight. The trap is treating only the length change and quoting $R' = 2R$. At constant volume the two geometric factors reinforce each other: length up by a factor and area down by the same factor, so resistance scales by the factor squared. In general, stretching a wire by a factor $k$ at constant volume multiplies its resistance by $k^{2}$. Always halve the diameter to a radius before squaring it for the area.

(a) 导线电阻 M1·M1·A1

半径 $r = d/2 = 0.20\ \mathrm{mm} = 2.0 \times 10^{-4}\ \mathrm{m}$。横截面积：(M1)

$$ A = \pi r^{2} = \pi (2.0 \times 10^{-4})^{2} \approx 1.26 \times 10^{-7}\ \mathrm{m^{2}}. $$

套用 $R = \dfrac{\rho L}{A}$：(M1)

$$ R = \frac{(1.1 \times 10^{-6})(1.5)}{1.26 \times 10^{-7}} \approx 13\ \Omega. $$

(A1)

(b) 体积不变的拉伸 M1·R1·A1

体积 $V = LA$ 不变。若长度加倍到 $2L$，面积必减半到 $A/2$ 才能保持 $LA$ 不变。(R1)

把两处变化代入 $R' = \dfrac{\rho (2L)}{A/2} = 4\,\dfrac{\rho L}{A} = 4R$。(M1)

$$ R' = 4 \times 13 \approx 53\ \Omega. $$

(A1)

要点。陷阱在于只考虑长度变化而答 $R' = 2R$。体积不变时两个几何因子相互叠加：长度乘以某因子、面积除以同一因子，故电阻按该因子的平方放大。一般地，体积不变下把导线拉伸为 $k$ 倍长，电阻变为 $k^{2}$ 倍。求面积前务必先把直径折半为半径再平方。
Q4HARDPaper 1series + parallel reduction串并联化简[6 marks]

$2.0\ \Omega$ in series with ($6.0\ \Omega \parallel 3.0\ \Omega$), driven by an ideal $12\ \mathrm{V}$ supply. (a) equivalent resistance; (b) total current; (c) current in the $6.0\ \Omega$ and $3.0\ \Omega$.$2.0\ \Omega$ 与（$6.0\ \Omega \parallel 3.0\ \Omega$）串联，由理想 $12\ \mathrm{V}$ 电源驱动。(a) 等效电阻；(b) 总电流；(c) $6.0\ \Omega$ 与 $3.0\ \Omega$ 中的电流。

Answers:答案：  (a) $R_{\text{eq}} = 4.0\ \Omega$  ·  (b) $I = 3.0\ \mathrm{A}$  ·  (c) $I_{6} = 1.0\ \mathrm{A}$, $I_{3} = 2.0\ \mathrm{A}$

(a) Equivalent resistance M1·M1·A1

Reduce the parallel pair first: $R_{P} = \dfrac{(6.0)(3.0)}{6.0 + 3.0} = \dfrac{18}{9.0} = 2.0\ \Omega$. (M1)

Add the series resistor: $R_{\text{eq}} = 2.0 + R_{P} = 2.0 + 2.0$. (M1)

$$ R_{\text{eq}} = 4.0\ \Omega. $$

(A1)

(b) Total current A1

$I = \dfrac{V}{R_{\text{eq}}} = \dfrac{12}{4.0} = 3.0\ \mathrm{A}$ (this flows through the $2.0\ \Omega$ resistor). (A1)

(c) Branch currents M1·A1

Voltage across the parallel pair: $V_{P} = I R_{P} = (3.0)(2.0) = 6.0\ \mathrm{V}$. (M1)

Then $I_{6} = \dfrac{6.0}{6.0} = 1.0\ \mathrm{A}$ and $I_{3} = \dfrac{6.0}{3.0} = 2.0\ \mathrm{A}$. Check: $1.0 + 2.0 = 3.0\ \mathrm{A} = I$. (A1)

Insight. Reduce inward then expand outward: collapse the network to one resistor to get the supply current, then walk back out, using the shared voltage across a parallel block and the shared current through a series element. The current splits in inverse proportion to resistance, so the smaller $3.0\ \Omega$ branch carries twice the current of the $6.0\ \Omega$ branch. The junction check $I_{6} + I_{3} = I$ is free insurance against an arithmetic slip.

(a) 等效电阻 M1·M1·A1

先化简并联对：$R_{P} = \dfrac{(6.0)(3.0)}{6.0 + 3.0} = \dfrac{18}{9.0} = 2.0\ \Omega$。(M1)

加上串联电阻：$R_{\text{eq}} = 2.0 + R_{P} = 2.0 + 2.0$。(M1)

$$ R_{\text{eq}} = 4.0\ \Omega. $$

(A1)

(b) 总电流 A1

$I = \dfrac{V}{R_{\text{eq}}} = \dfrac{12}{4.0} = 3.0\ \mathrm{A}$（这股电流流过 $2.0\ \Omega$ 电阻）。(A1)

(c) 支路电流 M1·A1

并联对两端电压：$V_{P} = I R_{P} = (3.0)(2.0) = 6.0\ \mathrm{V}$。(M1)

则 $I_{6} = \dfrac{6.0}{6.0} = 1.0\ \mathrm{A}$，$I_{3} = \dfrac{6.0}{3.0} = 2.0\ \mathrm{A}$。核查：$1.0 + 2.0 = 3.0\ \mathrm{A} = I$。(A1)

要点。先向内化简、再向外展开：把网络收缩为单个电阻求出电源电流，再逐步还原，利用并联块两端共享的电压与串联元件共享的电流。电流按电阻成反比分配，故较小的 $3.0\ \Omega$ 支路电流是 $6.0\ \Omega$ 支路的两倍。节点核查 $I_{6} + I_{3} = I$ 是防止算错的免费保险。
Q5HARDPaper 1HL ONLYKirchhoff's laws, two cells基尔霍夫定律，双电源[8 marks]

Two source branches meet at a node and feed a shared $R_{3} = 2.0\ \Omega$. Left: $\varepsilon_{1} = 6.0\ \mathrm{V}$, $R_{1} = 2.0\ \Omega$. Middle: $\varepsilon_{2} = 4.0\ \mathrm{V}$, $R_{2} = 1.0\ \Omega$. $I_{3} = I_{1} + I_{2}$. (a) state both laws; (b) two loop equations, solve $I_{1}, I_{2}$; (c) $I_{3}$, pd across $R_{3}$, verify the left loop.两条含电源支路汇于一节点并共馈一个 $R_{3} = 2.0\ \Omega$。左：$\varepsilon_{1} = 6.0\ \mathrm{V}$，$R_{1} = 2.0\ \Omega$。中：$\varepsilon_{2} = 4.0\ \mathrm{V}$，$R_{2} = 1.0\ \Omega$。$I_{3} = I_{1} + I_{2}$。(a) 写出两定律；(b) 两回路方程，解 $I_{1}, I_{2}$；(c) $I_{3}$、$R_{3}$ 两端电压，验证左回路。

Answers:答案：  (a) junction $=$ charge conservation, loop $=$ energy conservation  ·  (b) $I_{1} = 1.25\ \mathrm{A}$, $I_{2} = 0.50\ \mathrm{A}$  ·  (c) $I_{3} = 1.75\ \mathrm{A}$, $V_{3} = 3.5\ \mathrm{V}$

(a) The two laws A1·A1

Junction law: the sum of currents into a node equals the sum out, $\sum I_{\text{in}} = \sum I_{\text{out}}$; this is conservation of charge. (A1)

Loop law: around any closed loop the sum of emfs equals the sum of $IR$ drops, $\sum \varepsilon = \sum IR$; this is conservation of energy. (A1)

(b) Loop equations and solution M1·M1·A1·A1

Left loop (cell 1 and $R_{3}$), with $I_{3} = I_{1} + I_{2}$: $\varepsilon_{1} = I_{1}R_{1} + I_{3}R_{3}$, i.e. $6.0 = 2.0\,I_{1} + 2.0(I_{1} + I_{2})$. (M1)

Middle loop (cell 2 and $R_{3}$): $\varepsilon_{2} = I_{2}R_{2} + I_{3}R_{3}$, i.e. $4.0 = 1.0\,I_{2} + 2.0(I_{1} + I_{2})$. (M1)

Simplify: $4I_{1} + 2I_{2} = 6.0$ and $2I_{1} + 3I_{2} = 4.0$. Solving the pair gives $I_{1} = 1.25\ \mathrm{A}$ (A1) and $I_{2} = 0.50\ \mathrm{A}$ (A1).

(c) $I_{3}$, pd, and loop check A1·A1

Junction law: $I_{3} = I_{1} + I_{2} = 1.25 + 0.50 = 1.75\ \mathrm{A}$, so $V_{3} = I_{3}R_{3} = (1.75)(2.0) = 3.5\ \mathrm{V}$. (A1)

Left-loop check: $I_{1}R_{1} + V_{3} = (1.25)(2.0) + 3.5 = 2.5 + 3.5 = 6.0\ \mathrm{V} = \varepsilon_{1}$. The loop law holds. (A1)

Insight. The reliable recipe for any multi-source circuit is: assign a current to each branch, impose the junction law so the unknowns reduce ($I_{3} = I_{1} + I_{2}$ here), then write one loop equation per independent loop and solve the linear system. A negative current would simply mean the real flow opposes your assumed direction, with the magnitude still correct. The final substitution back into the untouched loop is the examiner-pleasing self-check that turns a plausible answer into a verified one.

(a) 两条定律 A1·A1

节点定律：流入节点的电流之和等于流出之和，$\sum I_{\text{in}} = \sum I_{\text{out}}$；这是电荷守恒。(A1)

回路定律：沿任一闭合回路，电动势之和等于各 $IR$ 降之和，$\sum \varepsilon = \sum IR$；这是能量守恒。(A1)

(b) 回路方程与求解 M1·M1·A1·A1

左回路（电源 1 与 $R_{3}$），取 $I_{3} = I_{1} + I_{2}$：$\varepsilon_{1} = I_{1}R_{1} + I_{3}R_{3}$，即 $6.0 = 2.0\,I_{1} + 2.0(I_{1} + I_{2})$。(M1)

中回路（电源 2 与 $R_{3}$）：$\varepsilon_{2} = I_{2}R_{2} + I_{3}R_{3}$，即 $4.0 = 1.0\,I_{2} + 2.0(I_{1} + I_{2})$。(M1)

化简：$4I_{1} + 2I_{2} = 6.0$ 与 $2I_{1} + 3I_{2} = 4.0$。联立解得 $I_{1} = 1.25\ \mathrm{A}$ (A1)、$I_{2} = 0.50\ \mathrm{A}$ (A1)。

(c) $I_{3}$、电压与回路核查 A1·A1

节点定律：$I_{3} = I_{1} + I_{2} = 1.25 + 0.50 = 1.75\ \mathrm{A}$，故 $V_{3} = I_{3}R_{3} = (1.75)(2.0) = 3.5\ \mathrm{V}$。(A1)

左回路核查：$I_{1}R_{1} + V_{3} = (1.25)(2.0) + 3.5 = 2.5 + 3.5 = 6.0\ \mathrm{V} = \varepsilon_{1}$。回路定律成立。(A1)

要点。处理任意多电源电路的可靠流程是：给每条支路设一个电流，先用节点定律减少未知数（此处 $I_{3} = I_{1} + I_{2}$），再对每个独立回路写一条回路方程，解线性方程组。若电流为负，只表示真实方向与假设相反，大小仍正确。最后把结果代回未用过的那个回路自检，是把"看似合理"变成"已验证"的得分关键。
PART II  ·  PAPER 1B / DATA ANALYSIS第二部分  ·  第一卷 B / 数据分析Graphs · data · uncertainties · 22 marks图像 · 数据 · 不确定度 · 22 分

Worked Solutions详细解析

Q6HARDPaper 1BI-V characteristics: ohmic vs non-ohmicI-V 特性：欧姆与非欧姆[8 marks]

X (fixed resistor): $I = 0.50, 1.00, 1.50\ \mathrm{A}$ at $V = 2.0, 4.0, 6.0\ \mathrm{V}$. Y (lamp): $I = 0.40, 0.52, 0.60\ \mathrm{A}$. (a) $R$ of X at $2.0$ and $6.0\ \mathrm{V}$, ohmic? (b) $R$ of Y, explain trend; (c) % uncertainty in $I_{\mathrm{X}} = 1.50 \pm 0.05\ \mathrm{A}$; (d) diode characteristic vs X.X（定值电阻）：$V = 2.0, 4.0, 6.0\ \mathrm{V}$ 时 $I = 0.50, 1.00, 1.50\ \mathrm{A}$。Y（灯泡）：$I = 0.40, 0.52, 0.60\ \mathrm{A}$。(a) X 在 $2.0$、$6.0\ \mathrm{V}$ 的 $R$，是否欧姆？(b) Y 的 $R$ 与趋势解释；(c) $I_{\mathrm{X}} = 1.50 \pm 0.05\ \mathrm{A}$ 的百分比不确定度；(d) 二极管特性与 X 的差异。

Answers:答案：  (a) $R_{\mathrm{X}} = 4.0\ \Omega$ at both, ohmic  ·  (b) $R_{\mathrm{Y}} = 5.0\ \Omega \to 10\ \Omega$, rises with temperature  ·  (c) $\approx 3.3\%$  ·  (d) diode conducts one way past a threshold; X is a straight line through the origin

(a) Resistance of X M1·A1

Use $R = V/I$ at each point: $R = \dfrac{2.0}{0.50} = 4.0\ \Omega$ and $R = \dfrac{6.0}{1.50} = 4.0\ \Omega$. (M1)

The resistance is constant, so X is ohmic: $I \propto V$ gives a straight line through the origin. (A1)

(b) Resistance of Y and the trend M1·A1·A1

At $2.0\ \mathrm{V}$: $R = \dfrac{2.0}{0.40} = 5.0\ \Omega$; at $6.0\ \mathrm{V}$: $R = \dfrac{6.0}{0.60} = 10\ \Omega$. (M1·A1)

The resistance rises as $V$ increases. The larger current heats the filament; the hotter lattice ions vibrate more and scatter the electrons more, so the resistance increases. The lamp is non-ohmic. (A1)

(c) Percentage uncertainty A1

$\dfrac{0.05}{1.50}\times 100\% \approx 3.3\%$. (A1)

(d) Diode characteristic A1·A1

A diode conducts in only one direction: in forward bias almost no current flows until a threshold (about $0.6\ \mathrm{V}$ for silicon), after which the current rises steeply; in reverse bias it blocks the current. (A1)

Component X by contrast is a straight line through the origin in both directions, so it has a fixed resistance whereas the diode does not. (A1)

Insight. The defining test for ohmic behaviour is whether $R = V/I$ stays constant, equivalently whether the $I$-$V$ graph is a straight line through the origin. For a curved characteristic never use the chord-or-slope of the curve as "the resistance"; quote $R = V/I$ at the stated operating point. Filament lamps curve because heating raises $R$; diodes have a one-way threshold; both are textbook non-ohmic devices that still satisfy $V = IR$ as a point definition.

(a) X 的电阻 M1·A1

对每点用 $R = V/I$：$R = \dfrac{2.0}{0.50} = 4.0\ \Omega$，$R = \dfrac{6.0}{1.50} = 4.0\ \Omega$。(M1)

电阻恒定，故 X 为欧姆元件：$I \propto V$ 给出过原点的直线。(A1)

(b) Y 的电阻与趋势 M1·A1·A1

$2.0\ \mathrm{V}$ 处：$R = \dfrac{2.0}{0.40} = 5.0\ \Omega$；$6.0\ \mathrm{V}$ 处：$R = \dfrac{6.0}{0.60} = 10\ \Omega$。(M1·A1)

电阻随 $V$ 增大而升高。电流越大灯丝越热；更热的点阵离子振动加剧、对电子散射更多，故电阻增大。灯泡非欧姆。(A1)

(c) 百分比不确定度 A1

$\dfrac{0.05}{1.50}\times 100\% \approx 3.3\%$。(A1)

(d) 二极管特性 A1·A1

二极管只单向导通：正向偏置时在阈值（硅约 $0.6\ \mathrm{V}$）之前几乎无电流，超过阈值后电流陡升；反向偏置时阻断电流。(A1)

相比之下，元件 X 在正反两向都是过原点的直线，故其电阻恒定，而二极管的电阻不恒定。(A1)

要点。欧姆行为的判定标准是 $R = V/I$ 是否恒定，等价于 $I$-$V$ 图是否为过原点的直线。对弯曲特性绝不能用曲线的弦或斜率当作"电阻"；要在所述工作点用 $R = V/I$。灯丝灯泡因发热使 $R$ 升高而弯曲；二极管有单向阈值；二者都是教科书中的非欧姆器件，但 $V = IR$ 仍可作为点定义成立。
Q7HARDPaper 1BHL ONLYemf and internal resistance from a graph由图线测电动势与内阻[14 marks]

$V$-$I$ data for a cell: $V = 5.60, 5.20, 4.80, 4.40\ \mathrm{V}$ at $I = 0.50, 1.00, 1.50, 2.00\ \mathrm{A}$. (a) show $V = \varepsilon - rI$ is a straight line; (b) gradient and $r$; (c) intercept and $\varepsilon$; (d) internal power at $2.00\ \mathrm{A}$ and short-circuit current; (e) % uncertainty in $V = 4.40 \pm 0.10\ \mathrm{V}$.某电池的 $V$-$I$ 数据：$I = 0.50, 1.00, 1.50, 2.00\ \mathrm{A}$ 时 $V = 5.60, 5.20, 4.80, 4.40\ \mathrm{V}$。(a) 证明 $V = \varepsilon - rI$ 为直线；(b) 斜率与 $r$；(c) 截距与 $\varepsilon$；(d) $2.00\ \mathrm{A}$ 时内部功率与短路电流；(e) $V = 4.40 \pm 0.10\ \mathrm{V}$ 的百分比不确定度。

Answers:答案：  (b) gradient $= -0.80\ \mathrm{V\,A^{-1}}$, $r = 0.80\ \Omega$  ·  (c) $\varepsilon = 6.0\ \mathrm{V}$  ·  (d) $P_{r} = 3.2\ \mathrm{W}$, $I_{\max} = 7.5\ \mathrm{A}$  ·  (e) $\approx 2.3\%$

(a) Why $V$ against $I$ is linear M1·A1·A1

For a real cell, $\varepsilon = I(R + r) = IR + Ir$. The terminal pd is the voltage across the load, $V = IR$, so $\varepsilon = V + Ir$, giving (M1)

$$ V = \varepsilon - rI. $$

This has the form $y = c + mx$ with $y = V$ and $x = I$. (A1)

Both $\varepsilon$ and $r$ are constants for the cell, so a plot of $V$ against $I$ is a straight line. (A1)

(b) Gradient and internal resistance M1·A1·A1

Read the gradient from two well-separated points, $(0.50,\,5.60)$ and $(2.00,\,4.40)$: (M1)

$$ \text{gradient} = \frac{4.40 - 5.60}{2.00 - 0.50} = \frac{-1.20}{1.50} = -0.80\ \mathrm{V\,A^{-1}}. $$

(A1)

Since the gradient is $-r$, the internal resistance is $r = 0.80\ \Omega$. (A1)

(c) Intercept and emf M1·A1·R1

Extrapolate to $I = 0$ using $V = \varepsilon - rI$ with the point $(0.50,\,5.60)$: $\varepsilon = V + rI = 5.60 + (0.80)(0.50)$. (M1)

$$ \varepsilon = 5.60 + 0.40 = 6.0\ \mathrm{V}. $$

(A1)

At $I = 0$ there is no current, so no lost volts $Ir$ inside the cell, and the terminal pd equals the emf. The $V$-intercept is therefore $\varepsilon$. (R1)

(d) Internal power and short-circuit current M1·A1·A1

Power dissipated inside the cell at $I = 2.00\ \mathrm{A}$: $P_{r} = I^{2}r = (2.00)^{2}(0.80) = 3.2\ \mathrm{W}$. (M1·A1)

Short circuit means $R = 0$, so $I_{\max} = \dfrac{\varepsilon}{r} = \dfrac{6.0}{0.80} = 7.5\ \mathrm{A}$. (A1)

(e) Percentage uncertainty M1·A1

$\dfrac{0.10}{4.40}\times 100\% \approx 2.3\%$. (M1·A1)

Insight. The whole experiment is a deliberate linearisation: rearranging $\varepsilon = I(R + r)$ into $V = \varepsilon - rI$ puts the two unknowns into the intercept and the slope of a straight line, where a best-fit line averages out random scatter far better than any single reading. Read the gradient from widely spaced points, never from one $(I, V)$ pair. The physical meaning anchors the marks: the intercept is the emf because zero current means zero lost volts, and the magnitude of the slope is the internal resistance.

(a) 为何 $V$ 对 $I$ 为直线 M1·A1·A1

对真实电池，$\varepsilon = I(R + r) = IR + Ir$。端电压是负载两端电压 $V = IR$，故 $\varepsilon = V + Ir$，得 (M1)

$$ V = \varepsilon - rI. $$

此式形如 $y = c + mx$，其中 $y = V$、$x = I$。(A1)

电池的 $\varepsilon$ 与 $r$ 都是常量，故 $V$ 对 $I$ 作图为直线。(A1)

(b) 斜率与内阻 M1·A1·A1

用相距较远的两点 $(0.50,\,5.60)$ 与 $(2.00,\,4.40)$ 读斜率：(M1)

$$ \text{斜率} = \frac{4.40 - 5.60}{2.00 - 0.50} = \frac{-1.20}{1.50} = -0.80\ \mathrm{V\,A^{-1}}. $$

(A1)

因斜率为 $-r$，内阻 $r = 0.80\ \Omega$。(A1)

(c) 截距与电动势 M1·A1·R1

用 $V = \varepsilon - rI$ 与点 $(0.50,\,5.60)$ 外推到 $I = 0$：$\varepsilon = V + rI = 5.60 + (0.80)(0.50)$。(M1)

$$ \varepsilon = 5.60 + 0.40 = 6.0\ \mathrm{V}. $$

(A1)

$I = 0$ 时无电流，故电池内部无损失电压 $Ir$，端电压等于电动势。因此 $V$ 轴截距即 $\varepsilon$。(R1)

(d) 内部功率与短路电流 M1·A1·A1

$I = 2.00\ \mathrm{A}$ 时电池内部耗散功率：$P_{r} = I^{2}r = (2.00)^{2}(0.80) = 3.2\ \mathrm{W}$。(M1·A1)

短路即 $R = 0$，故 $I_{\max} = \dfrac{\varepsilon}{r} = \dfrac{6.0}{0.80} = 7.5\ \mathrm{A}$。(A1)

(e) 百分比不确定度 M1·A1

$\dfrac{0.10}{4.40}\times 100\% \approx 2.3\%$。(M1·A1)

要点。整个实验是有意的线性化：把 $\varepsilon = I(R + r)$ 重排为 $V = \varepsilon - rI$，使两个未知量分别落在直线的截距与斜率上，而最佳拟合直线比任何单点都更能平均掉随机散布。要用相距较远的点读斜率，绝不用单个 $(I, V)$ 点。物理意义锁定得分：截距是电动势，因为零电流意味零损失电压；斜率大小即内阻。
PART III  ·  PAPER 2 STYLE第三部分  ·  第二卷风格Extended structured · 30 marks长结构题 · 30 分

Worked Solutions详细解析

Q8HARDPaper 2full network with power含功率的完整网络[12 marks]

Ideal $24\ \mathrm{V}$ supply: $R_{1} = 4.0\ \Omega$ in series with ($R_{2} = 6.0\ \Omega \parallel R_{3} = 3.0\ \Omega$) in series with $R_{4} = 2.0\ \Omega$. (a) $R_{\text{eq}}$ and supply current; (b) pd across the parallel pair; (c) $I_{6}$, $I_{3}$, verify sum; (d) power in $R_{1}$ and total power; (e) which resistor dissipates most.理想 $24\ \mathrm{V}$ 电源：$R_{1} = 4.0\ \Omega$ 与（$R_{2} = 6.0\ \Omega \parallel R_{3} = 3.0\ \Omega$）串联，再与 $R_{4} = 2.0\ \Omega$ 串联。(a) $R_{\text{eq}}$ 与电源电流；(b) 并联对两端电压；(c) $I_{6}$、$I_{3}$ 并验证之和；(d) $R_{1}$ 功率与总功率；(e) 哪个电阻耗散最多。

Answers:答案：  (a) $R_{\text{eq}} = 8.0\ \Omega$, $I = 3.0\ \mathrm{A}$  ·  (b) $V_{P} = 6.0\ \mathrm{V}$  ·  (c) $I_{6} = 1.0\ \mathrm{A}$, $I_{3} = 2.0\ \mathrm{A}$  ·  (d) $P_{1} = 36\ \mathrm{W}$, $P_{\text{tot}} = 72\ \mathrm{W}$  ·  (e) $R_{1}$

(a) Equivalent resistance and current M1·A1·A1

Parallel pair: $R_{P} = \dfrac{(6.0)(3.0)}{6.0 + 3.0} = 2.0\ \Omega$. Series total: $R_{\text{eq}} = 4.0 + 2.0 + 2.0 = 8.0\ \Omega$. (M1·A1)

Supply current: $I = \dfrac{V}{R_{\text{eq}}} = \dfrac{24}{8.0} = 3.0\ \mathrm{A}$. (A1)

(b) Potential difference across the parallel pair M1·A1

The full current flows through $R_{P}$: $V_{P} = I R_{P} = (3.0)(2.0) = 6.0\ \mathrm{V}$. (M1·A1)

(c) Branch currents M1·A1·A1

$I_{6} = \dfrac{V_{P}}{R_{2}} = \dfrac{6.0}{6.0} = 1.0\ \mathrm{A}$ (M1·A1); $I_{3} = \dfrac{V_{P}}{R_{3}} = \dfrac{6.0}{3.0} = 2.0\ \mathrm{A}$.

Junction check: $I_{6} + I_{3} = 1.0 + 2.0 = 3.0\ \mathrm{A} = I$. (A1)

(d) Power in $R_{1}$ and total power M1·A1

$P_{1} = I^{2}R_{1} = (3.0)^{2}(4.0) = 36\ \mathrm{W}$. (M1)

Total power from the supply: $P_{\text{tot}} = VI = (24)(3.0) = 72\ \mathrm{W}$. (A1)

(e) Which resistor dissipates most A1·R1

$R_{1}$ dissipates the most ($36\ \mathrm{W}$). (A1)

It carries the full supply current and has the largest resistance among the series elements; since $P = I^{2}R$ for the same current, the largest series resistance dissipates the most. (R1)

Insight. Two power rules cut through these problems. For elements carrying the same current (series), $P = I^{2}R$ rises with $R$, so the biggest series resistor dominates. For elements sharing the same voltage (parallel), $P = V^{2}/R$ rises as $R$ falls, so the smallest parallel resistor dominates that block. A free consistency check is energy conservation: summing the power in every resistor ($36 + 12 + 18$ across the parallel pair where $P_{2} = V_{P}^{2}/R_{2} = 6\ \mathrm{W}$, $P_{3} = 12\ \mathrm{W}$, plus $P_{4} = I^{2}R_{4} = 18\ \mathrm{W}$) returns the supply's $72\ \mathrm{W}$.

(a) 等效电阻与电流 M1·A1·A1

并联对：$R_{P} = \dfrac{(6.0)(3.0)}{6.0 + 3.0} = 2.0\ \Omega$。串联总和：$R_{\text{eq}} = 4.0 + 2.0 + 2.0 = 8.0\ \Omega$。(M1·A1)

电源电流：$I = \dfrac{V}{R_{\text{eq}}} = \dfrac{24}{8.0} = 3.0\ \mathrm{A}$。(A1)

(b) 并联对两端电压 M1·A1

全电流流过 $R_{P}$：$V_{P} = I R_{P} = (3.0)(2.0) = 6.0\ \mathrm{V}$。(M1·A1)

(c) 支路电流 M1·A1·A1

$I_{6} = \dfrac{V_{P}}{R_{2}} = \dfrac{6.0}{6.0} = 1.0\ \mathrm{A}$ (M1·A1)；$I_{3} = \dfrac{V_{P}}{R_{3}} = \dfrac{6.0}{3.0} = 2.0\ \mathrm{A}$。

节点核查：$I_{6} + I_{3} = 1.0 + 2.0 = 3.0\ \mathrm{A} = I$。(A1)

(d) $R_{1}$ 功率与总功率 M1·A1

$P_{1} = I^{2}R_{1} = (3.0)^{2}(4.0) = 36\ \mathrm{W}$。(M1)

电源总功率：$P_{\text{tot}} = VI = (24)(3.0) = 72\ \mathrm{W}$。(A1)

(e) 哪个电阻耗散最多 A1·R1

$R_{1}$ 耗散最多（$36\ \mathrm{W}$）。(A1)

它承载全电源电流，且在串联元件中阻值最大；由于同一电流下 $P = I^{2}R$，串联中阻值最大者耗散最多。(R1)

要点。两条功率规则可贯通此类题。对电流相同的元件（串联），$P = I^{2}R$ 随 $R$ 增大，故最大的串联电阻占主导。对电压相同的元件（并联），$P = V^{2}/R$ 随 $R$ 减小而增大，故该块中最小的并联电阻占主导。一个免费的一致性核查是能量守恒：把每个电阻的功率相加（并联对中 $P_{2} = V_{P}^{2}/R_{2} = 6\ \mathrm{W}$、$P_{3} = 12\ \mathrm{W}$，再加 $P_{4} = I^{2}R_{4} = 18\ \mathrm{W}$，以及 $P_{1} = 36\ \mathrm{W}$）得回电源的 $72\ \mathrm{W}$。
Q9HARDPaper 2emf, internal resistance, power电动势、内阻与功率[10 marks]

Battery $\varepsilon = 12\ \mathrm{V}$, $r = 0.50\ \Omega$, load $R$. (a) at $R = 5.5\ \Omega$ find $I$ and terminal pd; (b) power in $R$, power in battery, efficiency; (c) at $R = 2.5\ \Omega$ find new $I$ and pd, comment; (d) maximum current and condition.电池 $\varepsilon = 12\ \mathrm{V}$，$r = 0.50\ \Omega$，负载 $R$。(a) $R = 5.5\ \Omega$ 时求 $I$ 与端电压；(b) $R$ 上功率、电池内功率、效率；(c) $R = 2.5\ \Omega$ 时新 $I$ 与端电压并评述；(d) 最大电流与条件。

Answers:答案：  (a) $I = 2.0\ \mathrm{A}$, $V = 11\ \mathrm{V}$  ·  (b) $P_{R} = 22\ \mathrm{W}$, $P_{r} = 2.0\ \mathrm{W}$, efficiency $\approx 92\%$  ·  (c) $I = 4.0\ \mathrm{A}$, $V = 10\ \mathrm{V}$ (lower)  ·  (d) $I_{\max} = 24\ \mathrm{A}$

(a) Current and terminal pd M1·A1·A1

Use $\varepsilon = I(R + r)$: $I = \dfrac{\varepsilon}{R + r} = \dfrac{12}{5.5 + 0.50} = \dfrac{12}{6.0} = 2.0\ \mathrm{A}$. (M1·A1)

Terminal pd: $V = \varepsilon - Ir = 12 - (2.0)(0.50) = 11\ \mathrm{V}$. (A1)

(b) Powers and efficiency M1·A1·A1

In the load: $P_{R} = I^{2}R = (2.0)^{2}(5.5) = 22\ \mathrm{W}$. Inside the battery: $P_{r} = I^{2}r = (2.0)^{2}(0.50) = 2.0\ \mathrm{W}$. (M1·A1)

Efficiency $= \dfrac{P_{R}}{P_{R} + P_{r}} = \dfrac{22}{24} \approx 0.92 = 92\%$. (A1)

(c) Reduced load M1·A1

$I = \dfrac{12}{2.5 + 0.50} = \dfrac{12}{3.0} = 4.0\ \mathrm{A}$; $V = \varepsilon - Ir = 12 - (4.0)(0.50) = 10\ \mathrm{V}$. (M1)

The terminal pd has dropped from $11\ \mathrm{V}$ to $10\ \mathrm{V}$: a smaller load draws more current, so the lost volts $Ir$ grow and less voltage reaches the terminals. (A1)

(d) Maximum current A1·R1

The current is largest when the external load is zero (a short circuit): $I_{\max} = \dfrac{\varepsilon}{r} = \dfrac{12}{0.50} = 24\ \mathrm{A}$. (A1)

This occurs when $R = 0$, so the only resistance limiting the current is the internal resistance. (R1)

Insight. Internal resistance is what makes terminal pd droop under load: $V = \varepsilon - Ir$ shows the cell only delivers its full emf at zero current. As the load falls the current rises, the lost volts $Ir$ grow, and both the terminal pd and the transfer efficiency drop. The short-circuit current $\varepsilon/r$ is the absolute ceiling, set entirely by $r$, which is why a low internal resistance is what lets a cell deliver large currents.

(a) 电流与端电压 M1·A1·A1

用 $\varepsilon = I(R + r)$：$I = \dfrac{\varepsilon}{R + r} = \dfrac{12}{5.5 + 0.50} = \dfrac{12}{6.0} = 2.0\ \mathrm{A}$。(M1·A1)

端电压：$V = \varepsilon - Ir = 12 - (2.0)(0.50) = 11\ \mathrm{V}$。(A1)

(b) 功率与效率 M1·A1·A1

负载上：$P_{R} = I^{2}R = (2.0)^{2}(5.5) = 22\ \mathrm{W}$。电池内部：$P_{r} = I^{2}r = (2.0)^{2}(0.50) = 2.0\ \mathrm{W}$。(M1·A1)

效率 $= \dfrac{P_{R}}{P_{R} + P_{r}} = \dfrac{22}{24} \approx 0.92 = 92\%$。(A1)

(c) 减小负载 M1·A1

$I = \dfrac{12}{2.5 + 0.50} = \dfrac{12}{3.0} = 4.0\ \mathrm{A}$；$V = \varepsilon - Ir = 12 - (4.0)(0.50) = 10\ \mathrm{V}$。(M1)

端电压从 $11\ \mathrm{V}$ 降到 $10\ \mathrm{V}$：负载越小电流越大，损失电压 $Ir$ 增大，到达端子的电压更少。(A1)

(d) 最大电流 A1·R1

当外负载为零（短路）时电流最大：$I_{\max} = \dfrac{\varepsilon}{r} = \dfrac{12}{0.50} = 24\ \mathrm{A}$。(A1)

此时 $R = 0$，唯一限制电流的只有内阻。(R1)

要点。内阻使端电压在带载时下垂：$V = \varepsilon - Ir$ 表明电池只在零电流时输出全部电动势。负载减小则电流增大，损失电压 $Ir$ 增大，端电压与传递效率都下降。短路电流 $\varepsilon/r$ 是绝对上限，完全由 $r$ 决定，这正是低内阻能让电池输出大电流的原因。
Q10HARDPaper 2HL ONLYpotential divider with thermistor含热敏电阻的分压器[8 marks]

$9.0\ \mathrm{V}$ supply across $R_{1} = 3.0\ \mathrm{k\Omega}$ in series with a thermistor (resistance falls as temperature rises); $V_{\text{out}}$ across the thermistor. (a) $V_{\text{out}}$ at thermistor $6.0\ \mathrm{k\Omega}$; (b) $V_{\text{out}}$ at $1.0\ \mathrm{k\Omega}$; (c) response to a temperature rise and use as a sensor.$9.0\ \mathrm{V}$ 电源接 $R_{1} = 3.0\ \mathrm{k\Omega}$ 与热敏电阻（电阻随温度升高而减小）串联；$V_{\text{out}}$ 取自热敏电阻两端。(a) 热敏电阻 $6.0\ \mathrm{k\Omega}$ 时 $V_{\text{out}}$；(b) $1.0\ \mathrm{k\Omega}$ 时 $V_{\text{out}}$；(c) 对温升的响应及作传感器之用。

Answers:答案：  (a) $V_{\text{out}} = 6.0\ \mathrm{V}$  ·  (b) $V_{\text{out}} = 2.25\ \mathrm{V}$  ·  (c) $V_{\text{out}}$ falls as temperature rises; drives a sensing/switching circuit

(a) Output at $6.0\ \mathrm{k\Omega}$ M1·M1·A1

The output is across the thermistor $R_{2}$, so use $V_{\text{out}} = V_{\text{in}}\dfrac{R_{2}}{R_{1} + R_{2}}$. (M1)

$$ V_{\text{out}} = 9.0 \times \frac{6.0}{3.0 + 6.0} = 9.0 \times \frac{6.0}{9.0}. $$

(M1)

$$ V_{\text{out}} = 9.0 \times \frac{2}{3} = 6.0\ \mathrm{V}. $$

(A1)

(b) Output at $1.0\ \mathrm{k\Omega}$ M1·A1

$V_{\text{out}} = 9.0 \times \dfrac{1.0}{3.0 + 1.0} = 9.0 \times \dfrac{1.0}{4.0} = 2.25\ \mathrm{V}$. (M1·A1)

(c) Response and sensor use A1·A1·R1

As the temperature rises the thermistor resistance falls. (A1)

It therefore takes a smaller share of the supply voltage, so $V_{\text{out}}$ decreases (from $6.0\ \mathrm{V}$ down to $2.25\ \mathrm{V}$ in this example). (A1)

This temperature-dependent output voltage can be fed to a comparator or a transistor so that the circuit switches (for example a fan or alarm) once the temperature crosses a chosen value, making it a temperature sensor. (R1)

Insight. A potential divider splits the supply in direct proportion to resistance, so the output across one element tracks that element's share. Replacing a fixed resistor with a thermistor (temperature-dependent) or an LDR (light-dependent) turns the divider into a sensor: the changing resistance steers $V_{\text{out}}$. Note the wiring decides the polarity of the response: taking $V_{\text{out}}$ across the thermistor gives a falling output with rising temperature, whereas taking it across the fixed resistor would give a rising output.

(a) $6.0\ \mathrm{k\Omega}$ 时的输出 M1·M1·A1

输出取自热敏电阻 $R_{2}$ 两端，故用 $V_{\text{out}} = V_{\text{in}}\dfrac{R_{2}}{R_{1} + R_{2}}$。(M1)

$$ V_{\text{out}} = 9.0 \times \frac{6.0}{3.0 + 6.0} = 9.0 \times \frac{6.0}{9.0}. $$

(M1)

$$ V_{\text{out}} = 9.0 \times \frac{2}{3} = 6.0\ \mathrm{V}. $$

(A1)

(b) $1.0\ \mathrm{k\Omega}$ 时的输出 M1·A1

$V_{\text{out}} = 9.0 \times \dfrac{1.0}{3.0 + 1.0} = 9.0 \times \dfrac{1.0}{4.0} = 2.25\ \mathrm{V}$。(M1·A1)

(c) 响应与传感器用途 A1·A1·R1

温度升高时热敏电阻阻值减小。(A1)

因此它分得的电源电压份额变小，$V_{\text{out}}$ 减小（本例从 $6.0\ \mathrm{V}$ 降到 $2.25\ \mathrm{V}$）。(A1)

这一随温度变化的输出电压可送入比较器或晶体管，使电路在温度越过设定值时切换（例如风扇或警报），从而构成温度传感器。(R1)

要点。分压器按电阻成正比分配电源电压，故某元件两端的输出跟随其所占份额。把固定电阻换成热敏电阻（随温度变化）或光敏电阻（随光照变化），分压器即成传感器：变化的电阻牵动 $V_{\text{out}}$。注意接线决定响应的极性：从热敏电阻取 $V_{\text{out}}$ 则温升时输出下降，若从固定电阻取则温升时输出上升。