| Learning Objective | Essential Knowledge |
|---|---|
11.1.A |
|
Electric Circuits
AP Physics C: Electricity and Magnetism · Topic 11
11.1
Electric Current
Syllabus
Source: College Board AP Course and Exam Description
Electric current 电流 is the rate at which charge passes a cross-section of wire, $I=\dfrac{dq}{dt}$, measured in amperes 安培. Conventional current points the way positive charge would move. Microscopically, a current is a slow drift of many carriers:
with $n$ the number of carriers per volume, $q$ the charge each carries, $v_d$ the drift velocity 漂移速度, and $A$ the cross-sectional area 横截面积.
Charge carriers drift slowly through a conductor to make a current
Worked example. A copper wire with $A=1.0\times10^{-6}\ \text{m}^2$ and $n=8.5\times10^{28}\ \text{m}^{-3}$ carries $1.7\ \text{A}$. Then $v_d=\dfrac{I}{nqA}=\dfrac{1.7}{(8.5\times10^{28})(1.6\times10^{-19})(1.0\times10^{-6})}\approx1.3\times10^{-4}\ \text{m/s}$ – the carriers drift slower than a snail, even though the signal travels near light speed.
Current density 电流密度 is charge flow per unit area, $\vec{J}=nq\vec{v}_d$, linked to the field driving it by $\vec{E}=\rho\vec{J}$. In general $I=\int\vec{J}\cdot d\vec{A}$; if $J(r)$ varies across the wire, integrate it over the cross-section to get the total current. One care point: current has a direction along its wire, but it is a scalar 标量 – currents do not add as vectors, and there are no "components of current".
| English | Chinese | Pinyin |
|---|---|---|
| Electric current | 电流 | diàn liú |
| amperes | 安培 | ān péi |
| drift velocity | 漂移速度 | piāo yí sù dù |
| cross-sectional area | 横截面积 | héng jié miàn jī |
| Current density | 电流密度 | diàn liú mì dù |
| scalar | 标量 | biāo liàng |
11.2
Electric Circuits
Syllabus
| Learning Objective | Essential Knowledge |
|---|---|
11.2.A |
Boundary statement: Unless otherwise specified, all circuit schematic diagrams will be drawn using conventional current. |
Source: College Board AP Course and Exam Description
A circuit is a set of closed loops built from wires, batteries, resistors, lightbulbs, capacitors, inductors, switches, and meters; charge can flow only around a closed path. One element can belong to several loops at once – that is what makes multi-loop problems interesting. Every analysis starts by reading the circuit diagram 电路图: trace each loop and identify which elements share the same current (series 串联) and which share the same potential difference (parallel 并联).
Build series and parallel circuits
In series the same current flows through every bulb and voltage divides; in parallel each branch gets the full voltage. Switch mode to see the bulbs' brightness change.
| English | Chinese | Pinyin |
|---|---|---|
| circuit diagram | 电路图 | diàn lù tú |
| series | 串联 | chuàn lián |
| parallel | 并联 | bìng lián |
11.3
Resistance, Resistivity, and Ohm's Law
Syllabus
| Learning Objective | Essential Knowledge |
|---|---|
11.3.A |
|
11.3.B |
|
Source: College Board AP Course and Exam Description
Resistance 电阻 measures how strongly an object opposes charge flow. It grows with the material's resistivity 电阻率 and the conductor's length, and shrinks with its area:
Ohm's law 欧姆定律 relates the current through an element to the potential difference across it:
A longer conductor has more resistance; a wider one has less
Worked example. Stretch a wire to double its length: the volume is fixed, so the area halves, and $R=\rho\ell/A$ becomes $\rho(2\ell)/(A/2)=4R$ – four times the resistance. An element is ohmic 欧姆性 if $R$ stays constant (a straight line through the origin on an $I$–$\Delta V$ graph); a lightbulb filament, which heats up, is not.
Apply Ohm's law
Ohm's law $V=IR$: for a fixed resistance, current is proportional to voltage. Raise the resistance and the same voltage pushes less current.
| English | Chinese | Pinyin |
|---|---|---|
| Resistance | 电阻 | diàn zǔ |
| resistivity | 电阻率 | diàn zǔ lǜ |
| Ohm's law | 欧姆定律 | ōu mǔ dìng lǜ |
| ohmic | 欧姆性 | ōu mǔ xìng |
11.4
Electric Power
Syllabus
| Learning Objective | Essential Knowledge |
|---|---|
11.4.A |
Boundary statement: AP Physics C: Electricity & Magnetism only expects students to analyze the transfer of mechanical and electrical energy, although students should be aware that electrical energy can also be dissipated in the form of thermal energy. |
Source: College Board AP Course and Exam Description
A charge $q$ falling through a potential difference $\Delta V$ gives up energy $q\Delta V$, so the rate of energy transfer in an element is
In a resistor all of it becomes heat. Use the form whose quantities you actually know – and use power to rank lightbulb brightness: brighter = more power, not necessarily more resistance. In series, the larger resistance is brighter ($P=I^2R$, same $I$); in parallel, the smaller one is ($P=\Delta V^2/R$, same $\Delta V$).
Read an I-V characteristic
Power is $P=IV$. A resistor's I-V line is straight, but a lamp curves as it heats and its resistance rises. The area under I-V relates to the energy delivered.
11.5
Compound Direct Current Circuits
Syllabus
| Learning Objective | Essential Knowledge |
|---|---|
11.5.A |
|
11.5.B |
|
11.5.C |
Boundary statement: Unless otherwise stated, all batteries, wires, and meters are assumed to be ideal. Circuits with batteries of different potential differences connected in parallel will not be assessed. |
Source: College Board AP Course and Exam Description
Reduce resistor networks to an equivalent resistance 等效电阻: series resistances add ($R_{\text{eq}}=R_1+R_2+\cdots$), while parallel resistances add as reciprocals ($\tfrac{1}{R_{\text{eq}}}=\tfrac{1}{R_1}+\tfrac{1}{R_2}+\cdots$ – always less than the smallest branch). Collapse the network step by step to find the battery current, then expand back out to find each element's current and voltage.
Worked example. A $12\ \text{V}$ battery drives a $4.0\ \Omega$ and a $2.0\ \Omega$ resistor in series: $R_{\text{eq}}=6.0\ \Omega$, $I=2.0\ \text{A}$, the voltages split $8.0\ \text{V}$ and $4.0\ \text{V}$, and the $4.0\ \Omega$ resistor dissipates $P=I^2R=16\ \text{W}$.
Resistors in parallel combine to a smaller equivalent resistance
Real batteries are not ideal. Model a battery as an ideal battery 理想电池 of emf $\varepsilon$ in series with its own internal resistance 内阻 $r$. When current flows, some emf is used up inside, so the terminal voltage 端电压 – what a voltmeter across the battery actually reads – drops:
Worked example. A battery with $\varepsilon=12\ \text{V}$ and $r=0.50\ \Omega$ supplies $2.0\ \text{A}$: the terminals sit at $\Delta V=12-2.0(0.50)=11\ \text{V}$. With no current, a voltmeter reads the full $12\ \text{V}$.
Meters: an ammeter 电流表 goes in series at the point whose current you want (ideal ammeter: zero resistance); a voltmeter 电压表 goes in parallel across the element (ideal voltmeter: infinite resistance). Non-ideal meters disturb the circuit they measure – a real ammeter adds series resistance, a real voltmeter steals current.
| English | Chinese | Pinyin |
|---|---|---|
| equivalent resistance | 等效电阻 | děng xiào diàn zǔ |
| ideal battery | 理想电池 | lǐ xiǎng diàn chí |
| internal resistance | 内阻 | nèi zǔ |
| terminal voltage | 端电压 | duān diàn yā |
| ammeter | 电流表 | diàn liú biǎo |
| voltmeter | 电压表 | diàn yā biǎo |
11.6
Kirchhoff's Loop Rule
Syllabus
| Learning Objective | Essential Knowledge |
|---|---|
11.6.A |
|
Source: College Board AP Course and Exam Description
Charges moving through potential differences exchange energy ($\Delta U_E=q\Delta V$), and energy must balance around any closed path. That is Kirchhoff's loop rule 基尔霍夫回路定则:
Sign discipline wins these problems: crossing a battery from $-$ to $+$ is $+\varepsilon$; crossing a resistor with the assumed current is $-IR$ (against it, $+IR$). Write one equation per independent loop.
| English | Chinese | Pinyin |
|---|---|---|
| Kirchhoff's loop rule | 基尔霍夫回路定则 | jī ěr huò fū huí lù dìng zé |
11.7
Kirchhoff's Junction Rule
Syllabus
| Learning Objective | Essential Knowledge |
|---|---|
11.7.A |
|
Source: College Board AP Course and Exam Description
Kirchhoff's junction rule 基尔霍夫节点定则 is conservation of charge at a junction 节点:
Current divides at a junction: what flows in equals what flows out
Together the two rules solve any multi-loop circuit: assign a current to each branch, write junction equations, then loop equations, and solve. A negative answer just means that current flows opposite to your assumed direction.
Two loop equations and one junction equation solve this two-battery circuit
Worked example. In the circuit above, $\varepsilon_1=12\ \text{V}$ with $R_1=1.0\ \Omega$ on the left, $\varepsilon_2=9.0\ \text{V}$ with $R_2=1.0\ \Omega$ on the right, and a shared middle resistor $R_3=2.0\ \Omega$ carrying $I_3=I_1+I_2$ (junction rule). The two loop equations are
Solving: $I_1=3.6\ \text{A}$, $I_2=0.60\ \text{A}$, so $I_3=4.2\ \text{A}$ through the middle. Check with the second loop: $2(3.6)+3(0.60)=9.0$ ✓.
| English | Chinese | Pinyin |
|---|---|---|
| Kirchhoff's junction rule | 基尔霍夫节点定则 | jī ěr huò fū jié diǎn dìng zé |
| junction | 节点 | jié diǎn |
11.8
Resistor-Capacitor (RC) Circuits
Syllabus
| Learning Objective | Essential Knowledge |
|---|---|
11.8.A |
|
11.8.B |
|
Source: College Board AP Course and Exam Description
Capacitor networks reduce like resistors but with the rules swapped: parallel capacitances add ($C_{\text{eq}}=C_1+C_2$), series add as reciprocals – and capacitors in series must carry the same charge on each plate, by conservation of charge. Use the equivalent capacitance 等效电容 to analyse the network, then expand back.
In an RC circuit RC电路, Kirchhoff's loop rule gives the differential equation
whose solutions are exponentials with time constant 时间常数 $\tau=RC$:
The current is largest at the first instant and decays – it never "waits" for the capacitor. Learn the two limits: at $t=0$ an uncharged capacitor acts like a plain wire (maximum current); after a long time it is fully charged, no current flows in its branch, and it acts like a break. In any steady state 稳态, cover the capacitor branch with your finger, solve the resistor circuit, then read the capacitor's voltage from the element it sits across.
The charge on a capacitor decays exponentially as it discharges
Worked example. With $R=5.0\ \text{k}\Omega$, $C=200\ \mu\text{F}$, and a $10\ \text{V}$ battery: $\tau=RC=1.0\ \text{s}$; initial current $\dfrac{\varepsilon}{R}=2.0\ \text{mA}$; after one time constant the charge is $q=CV(1-e^{-1})\approx1.3\times10^{-3}\ \text{C}$, about $63\%$ of full charge, and the current has fallen to $37\%$ of its initial value.
A real circuit on a breadboard: a resistor and capacitor together set how fast the voltage rises and falls
| English | Chinese | Pinyin |
|---|---|---|
| equivalent capacitance | 等效电容 | děng xiào diàn róng |
| RC circuit | RC电路 | RC diàn lù |
| time constant | 时间常数 | shí jiān cháng shù |
| steady state | 稳态 | wěn tài |
11.8
Exam tips
- Relate current to charge flow $I=\tfrac{dQ}{dt}$ and use $J=\sigma E$, $R=\tfrac{\rho L}{A}$ for resistance.
- Apply Kirchhoff's laws (junction: charge; loop: energy) with consistent sign conventions.
- Analyse RC circuits with calculus: charging/discharging give exponentials with time constant $\tau=RC$.
- Combine resistors (series add, parallel reciprocal) and track power $P=IV=I^2R$.
- At $t=0$ a capacitor acts like a wire; after a long time ($t\to\infty$) it acts like an open branch.