AP Physics C: E&M · Unit 11 of 6

Unit 11: Electric Circuits

15-25% of the multiple-choice section8 topics

Topics in this unit

Electric Circuits is Unit 11 of AP Physics C: Electricity and Magnetism, worth 15 to 25 percent of the multiple-choice section over about 12 to 24 class periods. Eight topics, twelve learning objectives, and one calculus idea Physics 2 cannot reach: the RC circuit as a differential equation.

AP Physics: Unit 11 (topics 11.1 Electric Current, 11.2 Simple Circuits, 11.3 Resistance, Resistivity, and Ohm's Law, 11.4 Electric Power, 11.5 Compound Direct Current Circuits, 11.6 Kirchhoff's Loop Rule, 11.7 Kirchhoff's Junction Rule, 11.8 Resistor Capacitor (RC) Circuits). Unit 11 of the current AP Physics C: Electricity and Magnetism course and exam description, weighted 15 to 25% of the multiple-choice section at about 12 to 24 class periods, the top band alongside Unit 8. Eight topics carry twelve learning objectives, all using the task verb describe: 11.1.A, 11.2.A, 11.3.A, 11.3.B, 11.4.A, 11.5.A, 11.5.B, 11.5.C, 11.6.A, 11.7.A, 11.8.A and 11.8.B. Progress Check 11 lists about 24 multiple-choice questions, more than the roughly 18 listed for each of the other five units. Three boundary statements. Topic 11.2: unless otherwise specified, all circuit schematic diagrams will be drawn using conventional current. Topic 11.4: the course 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. Topic 11.5: unless otherwise stated, all batteries, wires, and meters are assumed to be ideal, and circuits with batteries of different potential differences connected in parallel will not be assessed. Topics 11.1, 11.3, 11.6, 11.7 and 11.8 print no boundary statement. Twenty-one equations appear in the required content and twelve are printed on the equation sheet; the RC differential equation, both of Kirchhoff's rules, the squared power forms and the terminal potential difference are not. Suggested skills by topic: 11.1 uses 1.A, 2.A, 2.D, 3.B; 11.2 uses 1.A, 2.C, 3.B, 3.C; 11.3 uses 1.B, 2.B, 2.D, 3.A, 3.B; 11.4 uses 1.A, 2.B, 2.C, 3.B; 11.5 uses 1.A, 2.A, 2.C, 3.C; 11.6 uses 1.C, 2.A, 2.C, 3.B; 11.7 uses 1.B, 2.B, 3.A, 3.B; 11.8 uses 1.B, 2.A, 2.D, 3.A, 3.C.

What the CED requires across Unit 11

Unit 11 of AP Physics C: Electricity and Magnetism is Electric Circuits. The course and exam description weights it at 15 to 25% of the multiple-choice section and suggests about 12 to 24 class periods. That is the top band, shared with Unit 8. Units 9, 12 and 13 each sit at 10 to 20%, Unit 10 at 10 to 15%.

Eight topics carry twelve learning objectives between them, and all twelve open with the task verb "describe".

Topic	Learning objectives	Suggested skills
11.1 Electric Current	11.1.A	1.A, 2.A, 2.D, 3.B
11.2 Simple Circuits	11.2.A	1.A, 2.C, 3.B, 3.C
11.3 Resistance, Resistivity, and Ohm's Law	11.3.A, 11.3.B	1.B, 2.B, 2.D, 3.A, 3.B
11.4 Electric Power	11.4.A	1.A, 2.B, 2.C, 3.B
11.5 Compound Direct Current Circuits	11.5.A, 11.5.B, 11.5.C	1.A, 2.A, 2.C, 3.C
11.6 Kirchhoff's Loop Rule	11.6.A	1.C, 2.A, 2.C, 3.B
11.7 Kirchhoff's Junction Rule	11.7.A	1.B, 2.B, 3.A, 3.B
11.8 Resistor Capacitor (RC) Circuits	11.8.A, 11.8.B	1.B, 2.A, 2.D, 3.A, 3.C

One number separates this unit from the rest of the course. Progress Check 11 lists about 24 multiple-choice questions; Progress Checks 8, 9, 10, 12 and 13 each list about 18. All six list 4 free-response questions, one of each type.

The CED says Unit 11 "serves to illuminate how, and why, simple electronic devices such as lightbulbs and household wiring function by exploring the nature and importance of electric currents, circuits, and resistance." Its essential questions are the household ones: how a house's wiring accounts for a flipped breaker cutting power to some rooms and not others, why warming bulbs take several minutes to shine brightly, and how touching a conductor to a capacitor before removing it from a circuit protects you.

Topic 11.5 carries three of the twelve objectives: equivalent resistance, resistive wires and battery internal resistance, and measurement with meters. Topic 11.7 gets two essential knowledge statements and no equation of its own on the sheet.

What calculus changes, and it is more than the RC circuit

This course is calculus-based and equivalent to the second course in a college sequence in calculus-based physics; its prerequisites say students should have taken or be concurrently taking calculus. In Unit 11 that shows up in four places.

Current is a derivative. Statement 11.1.A.1 gives $I = \dfrac{dq}{dt}$ as the definition, not a special case. A charge arriving at a non-constant rate has a current that is a function of time.

Current is a flux integral. Statement 11.1.A.2 defines current density and gives $I = \int \vec{J} \cdot d\vec{A}$ ; 11.1.A.3 says that if a function of current density is given, the total current is found by integrating it over the area, printing $I_{\text{tot}} = \int \vec{J}(r) \cdot d\vec{A}$ as a derived equation. Same machinery as electric flux and Gauss's law, reused on a wire. Statement 11.1.A.2.i ties current density to drift velocity, $\vec{J} = n q \vec{v}_d$ , and 11.1.A.2.iii ties it to the field inside the conductor, $\vec{E} = \rho \vec{J}$ . A current denser at the axis than at the surface is fair game here and unaskable in the algebra-based course.

Resistance can be an integral. Statement 11.3.A.2 gives $R = \rho \ell / A$ for uniform geometry. Statement 11.3.A.2.iii then says the total resistance of a resistor with uniform geometry, but made of a material whose resistivity varies along its length, is given by

R = \int \frac{\rho(\ell)\, d\ell}{A}

RC circuits are differential equations. Statement 11.8.B.1 reads: "The charge on a capacitor or the current in a resistor in an RC circuit can be described by a fundamental differential equation derived from Kirchhoff's loop rule." What follows carries the Derived equation label:

\mathcal{E} = \frac{dq}{dt} R + \frac{q}{C}

The CED's Required Equations page defines that label: not all framework equations appear on the equation sheet, many are provided for reference "or to demonstrate the final results of derivations expected of students on the exam", and those are the Derived Equations. So the loop equation is yours to produce, and it is not on the sheet.

Nowhere in Unit 11 does the framework print a solved exponential. What it prints is the behaviour: 11.8.B.2.i defines $\tau = R_{\text{eq}} C_{\text{eq}}$ , 11.8.B.2.ii says the time constant is the time for a charging capacitor's charge to rise from zero to approximately 63 percent of its final asymptotic value, 11.8.B.2.iii says approximately 37 percent for a discharge, and 11.8.B.3.iii says the potential difference, branch current and stored energy all asymptotically approach steady state. Four statements describe an exponential without writing one, which is enough to sketch every graph the unit can ask for.

Its sample Question 4 fixes the ceiling. Part B says "Derive, but do not solve, a differential equation" for the rate of change of the potential difference across a capacitor, and part C asks for a prediction justified by referring to that same equation. Set it up from a fundamental principle, then reason from it.

Unit 11's three boundary statements, quoted whole

Boundary statements are how the CED fences off a treatment, and they ship with every exception clause intact. Unit 11 prints three, under Topics 11.2, 11.4 and 11.5. Topics 11.1, 11.3, 11.6, 11.7 and 11.8 print none.

Topic 11.2, Simple Circuits: "Unless otherwise specified, all circuit schematic diagrams will be drawn using conventional current."

Topic 11.4, Electric Power: "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."

That second clause is the part that gets cut away, and losing it flips the meaning: the boundary limits what you are asked to analyze, not what you are expected to know. Statement 11.3.B.1.iii is the matching content, that resistors can also convert electrical energy to thermal energy.

Topic 11.5, Compound Direct Current Circuits: "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."

Both sentences count. The first sets a default that objective 11.5.B then switches off: 11.5.B.2 says the internal resistance of a nonideal battery may be treated as a resistor in series with an ideal battery, and 11.5.B.3 gives the derived $\Delta V_{\text{terminal}} = \mathcal{E} - Ir$ . The second removes one circuit family from the exam entirely.

The reference information adds two conventions that work like boundaries: resistors and lightbulbs are ohmic, and the direction of current is the direction in which positive charges would drift.

How the eight topics build

11.1 Electric Current defines the quantity. Current is the rate at which charge passes through a cross-sectional area of a wire; within a conductor it is charge carriers travelling with an average drift velocity, $I = n q v_d A$ ; and it is driven by a potential difference, sometimes called electromotive force, or emf. Statement 11.1.A.1.iii is the one worth reading twice: if the current is zero in a section of wire, the net motion of charge carriers is also zero, although individual carriers will not have zero speed.

11.2 Simple Circuits defines the object. A circuit is composed of electrical loops that may include wires, batteries, resistors, lightbulbs, capacitors, inductors, switches, ammeters and voltmeters. Closed, open and short circuits get a statement each, 11.2.A.3 notes that a single element may belong to more than one loop, and 11.2.A.4.ii prints the schematic symbols with a note that variable elements are marked by a diagonal strikethrough arrow across the standard symbol.

11.3 Resistance, Resistivity, and Ohm's Law splits into two objectives, which is the CED signalling that they are different ideas. 11.3.A is geometry and material: resistance measures the degree to which an object opposes the movement of electric charge, while resistivity is a fundamental property of a material that depends on its atomic and molecular structure. 11.3.B is circuit behaviour: Ohm's law, ohmic materials, and 11.3.B.1.iv, that the resistance of an ohmic element can be read off the slope of a graph of current against potential difference.

11.4 Electric Power is one objective and two equations, plus 11.4.A.2, that bulb brightness increases with power so power can be used to predict brightness qualitatively.

11.5 Compound Direct Current Circuits is the widest topic, with three objectives: equivalent resistance, nonideal batteries and resistive wires, and measurement. Statement 11.5.B.1.iii defines emf usefully, as the potential difference measured across the terminals when there is no current in the battery.

11.6 Kirchhoff's Loop Rule grounds the loop rule in conservation of energy and adds a representation that gets asked about: 11.6.A.2.ii says the electric potential at points in a circuit can be represented by a graph of potential against position within a loop.

11.7 Kirchhoff's Junction Rule does the same for charge, in two statements.

11.8 Resistor Capacitor (RC) Circuits replaces a resistor with a store. 11.8.A handles equivalent capacitance, and 11.8.B handles the circuit, with a long chain of essential knowledge statements describing its behaviour at both ends of time.

The Unit 11 equations, and which ones are printed

Twenty-one equations appear in Unit 11's required content. Twelve are printed on the AP Physics C: E&M formula sheet and nine are not.

Equation	Where the CED puts it	Printed
$I = \dfrac{dq}{dt}$	11.1.A.1, relevant	yes
$I = n q v_d A$	11.1.A.1.i, relevant	no
$I = \int \vec{J} \cdot d\vec{A}$	11.1.A.2, relevant	yes
$\vec{J} = n q \vec{v}_d$	11.1.A.2.i, relevant	no
$\vec{E} = \rho \vec{J}$	11.1.A.2.iii, relevant	yes
$I_{\text{tot}} = \int \vec{J}(r) \cdot d\vec{A}$	11.1.A.3, derived	no
$R = \dfrac{\rho \ell}{A}$	11.3.A.2, relevant	yes
$R = \int \dfrac{\rho(\ell)\, d\ell}{A}$	11.3.A.2.iii	no
$I = \dfrac{\Delta V}{R}$	11.3.B.1, relevant	yes
$P = I \Delta V$	11.4.A.1, relevant	yes
$P = I^2 R = \dfrac{\Delta V^2}{R}$	11.4.A.1, derived	no
$R_{\text{eq,s}} = \sum_i R_i$	11.5.A.2.i, relevant	yes
$\dfrac{1}{R_{\text{eq,p}}} = \sum_i \dfrac{1}{R_i}$	11.5.A.2.ii, relevant	yes
$\Delta V_{\text{terminal}} = \mathcal{E} - Ir$	11.5.B.3, derived	no
$\Delta U_E = q \Delta V$	11.6.A.1, relevant	yes
$\sum \Delta V = 0$	11.6.A.2.i, relevant	no
$\sum I_{\text{in}} = \sum I_{\text{out}}$	11.7.A.2, relevant	no
$\dfrac{1}{C_{\text{eq,s}}} = \sum_i \dfrac{1}{C_i}$	11.8.A.1.i, relevant	yes
$C_{\text{eq,p}} = \sum_i C_i$	11.8.A.1.iii, relevant	yes
$\mathcal{E} = \dfrac{dq}{dt} R + \dfrac{q}{C}$	11.8.B.1, derived	no
$\tau = R_{\text{eq}} C_{\text{eq}}$	11.8.B.2.i	yes

Three readings of that table are worth carrying into the exam.

Neither of [Kirchhoff's rules](/glossary/kirchhoffs-loop-rule) is printed. Two topics are named after them and neither $\sum \Delta V = 0$ nor $\sum I_{\text{in}} = \sum I_{\text{out}}$ appears on the sheet. They arrive as conservation laws rather than formulas: 11.6.A.2 derives the loop rule from conservation of energy, 11.7.A.1 derives the junction rule from conservation of electric charge. Write them from the principle.

Ohm's law is printed as $I = \dfrac{\Delta V}{R}$ , with the delta. Not $V = IR$ . The delta is the meaning: the numerator is the potential difference across that element, not a potential at a point, which is exactly what 11.3.B.1 says in words. The sheet also prints $P = I \Delta V$ and nothing else for power, so $P = I^2 R$ and $P = \Delta V^2 / R$ are substitutions you make yourself, and choosing the right one decides the answer. The Ohm's law guide and calculator cover the arithmetic.

Terminal potential difference is not printed. $\Delta V_{\text{terminal}} = \mathcal{E} - Ir$ carries the derived label, so an internal-resistance question expects you to rebuild it from 11.5.B.2.

Because the C: E&M sheet reprints the whole C: Mechanics table, $P_{\text{avg}} = W/\Delta t = \Delta E / \Delta t$ is on the same page. The Unit 10 capacitor block is printed too, and Topic 11.8 needs it: $C = Q/\Delta V$ and $U_C = \tfrac{1}{2} Q \Delta V$ .

Traps that span more than one topic

Current is a scalar with a direction, and that is not a contradiction. Statement 11.1.A.4 says that although current is a scalar quantity it does have a direction, and because that direction is relative to the current carrier and not to space, "current does not obey the laws of vector addition and has no vector components." So you never resolve a current into components.

The CED says two different things about resistivity and temperature, and both are true. Statement 11.3.A.2.ii says the resistivity of a conductor typically increases with temperature. Statement 11.3.B.1.ii says the resistivity of an ohmic material is constant regardless of temperature. The second defines the idealisation, the first describes real conductors, and the reference information resolves it by declaring resistors and lightbulbs ohmic unless stated otherwise. A question about a warming filament has switched that default off on purpose.

The series and parallel rules are swapped between resistors and capacitors. Rather than memorise which one flips, remember what is shared: 11.5.A.1.i says the current in each element in series must be the same, and 11.8.A.2 says each capacitor in series must have the same magnitude of charge on each plate. The reciprocal rules follow, and 11.8.A.1.ii prints the sanity check. The series and parallel guide owns the reduction routine.

More resistance does not mean more power. In series, $P = I^2 R$ and the largest resistor dissipates most. In parallel, $P = \Delta V^2 / R$ and the largest dissipates least. By 11.4.A.2 brightness follows power, so two bulbs swap ranking when you rewire them.

An uncharged capacitor acts like a wire; a charged one does not. Statement 11.8.B.3.i is careful: immediately after being placed in a circuit, an uncharged capacitor acts like a wire. Statement 11.8.B.3.iv says that after a long time a charging capacitor reaches a maximum potential difference at which there is zero current in its branch. Those two limits need no calculus and are asked constantly.

The time constant carries the equivalent values, not the nearest ones. The CED writes $\tau = R_{\text{eq}} C_{\text{eq}}$ with both subscripts in place. In a circuit with more than one resistor, the resistance the RC circuit actually sees is often neither printed value, and getting it wrong moves every point on the graph.

Ideal meters are opposites. An ammeter goes in series with zero resistance; a voltmeter goes in parallel with infinite resistance. Statement 11.5.C.3 says nonideal meters change the circuit being measured, which the CED's Topic 11.8 sample activity exploits to find a voltmeter's own resistance from a capacitor discharge.

How Unit 11 is assessed

The exam is 3 hours long: 42 multiple-choice questions in 85 minutes for half the score, then 4 free-response questions in 95 minutes for the other half, one of each type in a fixed order. A four-function, scientific, or graphing calculator is allowed on both sections.

Skill 3.B, applying an appropriate law, definition, theoretical relationship, or model to make a claim, is listed for six of the eight topics, more than any other skill. Skill 3.A, creating experimental procedures, appears on 11.3, 11.7 and 11.8, which fits the unit's exam-preparation note: the third free-response question is the Experimental Design and Analysis question, and "students will also be required to derive relevant equations, linearize and analyze data." The course requires that 25 percent of instructional time be spent in hands-on laboratory work.

Two separate parts of the CED show how much of this exam runs through circuits. First, the Instructional Approaches case study is built entirely on a Unit 11 circuit, two identical lightbulbs and a switch connected to a battery, with one multiple-choice question written against each of skills 2.A, 2.B, 2.C, 2.D, 3.B and 3.C off the same stimulus. The CED notes there that Science Practice 1 is free-response only, which matches the weighting table: 1.A, 1.B, 1.C and 3.A are all listed N/A on the multiple-choice section.

Second, the alignment table for the CED's own sample questions. Three of its fifteen sample multiple-choice questions align to Unit 11, aimed at 11.4.A, 11.7.A and 11.5.B, and two of its four sample free-response questions carry Unit 11 objectives: Question 3 aligns to 11.3.B, and Question 4 to 11.1.A, 11.3.B, 11.7.A and 11.8.B. No unit is guaranteed a free-response question in a given year, but Unit 11 appears in half of that sample set.

Where AP Physics 2 stops, and who each page is for

AP Physics 2 also has a Unit 11 called Electric Circuits, and its eight topic codes and titles line up one for one with these: Electric Current, Simple Circuits, Resistance Resistivity and Ohm's Law, Electric Power, Compound Direct Current Circuits, Kirchhoff's Loop Rule, Kirchhoff's Junction Rule, Resistor Capacitor (RC) Circuits. Two of the eight differ only by a hyphen or a parenthetical. Same architecture, different course.

The content is not the same. The gap is calculus in the four places above, plus one habit: on the C free-response section Science Practice 2 carries 40 to 45% of the weighting, the largest of the three practices, and skill 2.A alone carries 25 to 30% of the multiple-choice section. Symbolic answers are the norm.

The [AP Physics 2 Unit 11 hub](/ap-physics-2/unit-11-electric-circuits) is for AP Physics 2 students; this page is for AP Physics C students. In the algebra-based course you want the finite-difference treatment and can stop before the integrals. In Physics C you want $I = dq/dt$ , the current-density integral, the resistivity integral and the loop-rule differential equation, and you will be graded on the derivation rather than the number.

Unit 11 also feeds forward. Unit 13, Electromagnetic Induction reruns this structure with an inductor where the capacitor was: the loop rule produces a differential equation again, the time constant becomes $\tau = L / R_{\text{eq}}$ , and the two limiting instants return in a new disguise. The full course map is on the AP Physics C: Electricity and Magnetism hub.

A compound circuit with a nonideal battery

A battery of emf $\mathcal{E} = 12.0$ V and internal resistance $r = 1.0\ \Omega$ is in series with $R_1 = 2.0\ \Omega$ . That branch feeds a parallel combination of $R_2 = 6.0\ \Omega$ and $R_3 = 3.0\ \Omega$ . Find (a) the total resistance, (b) the battery current, (c) the terminal potential difference, (d) each parallel branch current, and (e) the power dissipated in $R_2$ and $R_3$ .

Declare the convention first: current means conventional current, per the Topic 11.2 boundary statement. Hold that to the end.
(a) Statement 11.5.A.2.ii gives $\dfrac{1}{R_{\text{eq,p}}} = \dfrac{1}{6.0} + \dfrac{1}{3.0} = \dfrac{3}{6.0}$ , so $R_{\text{eq,p}} = 2.0\ \Omega$ . Adding in series (11.5.A.2.i) the external resistance is $2.0 + 2.0 = 4.0\ \Omega$ , and 11.5.B.2 puts the internal resistance in series with it: $4.0 + 1.0 = 5.0\ \Omega$ .
(b) $I = \dfrac{\Delta V}{R} = \dfrac{12.0\ \mathrm{V}}{5.0\ \Omega} = 2.4$ A.
(c) The derived result 11.5.B.3 gives $\Delta V_{\text{terminal}} = \mathcal{E} - Ir = 12.0 - (2.4)(1.0) = 9.6$ V. Check across the external resistance instead: $(2.4)(4.0) = 9.6$ V, which is the loop rule closing.
(d) The parallel section carries $\Delta V_p = (2.4)(2.0) = 4.8$ V, shared by both paths per 11.5.A.1.ii. So $I_2 = 4.8/6.0 = 0.80$ A and $I_3 = 4.8/3.0 = 1.6$ A, and the junction rule checks the split: $0.80 + 1.6 = 2.4$ A.
(e) $P_2 = I_2^2 R_2 = (0.80)^2(6.0) = 3.84$ W and $P_3 = (1.6)^2(3.0) = 7.68$ W, and $\Delta V^2/R$ with the shared 4.8 V gives the same. Read the trap: in parallel the smaller resistance dissipates the more power, so a 3.0 ohm bulb here outshines a 6.0 ohm one, and putting the pair in series reverses the ordering.
Energy audit, since 11.4.A is about transfer. The battery delivers $\mathcal{E} I = 28.8$ W; dissipation is $5.76$ W internally, $11.52$ W in $R_1$ , plus 3.84 W and 7.68 W in the pair, totalling $28.8$ W. Nothing is missing.

(a) $5.0\ \Omega$ , of which $4.0\ \Omega$ is external. (b) 2.4 A. (c) 9.6 V. (d) $I_2 = 0.80$ A and $I_3 = 1.6$ A. (e) $P_2 = 3.84$ W and $P_3 = 7.68$ W. The 3.0 ohm resistor dissipates twice as much as the 6.0 ohm one because in parallel they share a potential difference, not a current.

Deriving an RC differential equation, then reading it

A battery of emf $\mathcal{E} = 12.0$ V and negligible internal resistance is in series with $R_1 = 2.0\ \mathrm{k}\Omega$ . That branch feeds a parallel combination of $R_2 = 6.0\ \mathrm{k}\Omega$ and an initially uncharged capacitor $C = 5.0\ \mu\mathrm{F}$ . The switch closes at $t = 0$ . (a) Derive, but do not solve, a differential equation for the charge $q$ on the capacitor. (b) Read the time constant off it. (c) Find the battery current immediately after closing and long after. (d) Find the final charge, the stored energy, and the charge at $t = \tau$ .

(a) Start from a fundamental principle, as the rubrics require. Junction rule where the parallel section begins: $I_1 = I_2 + \dfrac{dq}{dt}$ . The capacitor and $R_2$ share a potential difference, and $C = Q/\Delta V$ gives $\Delta V_C = q/C$ , so $I_2 = \dfrac{q}{R_2 C}$ .
Loop rule around the battery, $R_1$ and the capacitor: $\mathcal{E} - I_1 R_1 - \dfrac{q}{C} = 0$ . Substituting $I_1$ and collecting the $q$ terms gives $\mathcal{E} = R_1 \dfrac{dq}{dt} + q\,\dfrac{R_1 + R_2}{R_2 C}$ , the same shape as 11.8.B.1 with one resistor replaced by two. Stop here: the question said derive, not solve.
(b) Dividing by $R_1$ makes the coefficient of $q$ equal to $1/\tau$ , so $\tau = C\,\dfrac{R_1 R_2}{R_1 + R_2} = R_{\text{eq}} C_{\text{eq}}$ , exactly as 11.8.B.2.i writes it. Numerically $R_{\text{eq}} = \dfrac{(2.0)(6.0)}{8.0} = 1.5\ \mathrm{k}\Omega$ and $\tau = (1500)(5.0 \times 10^{-6}) = 7.5$ ms. Neither 2.0 nor 6.0 kilohms would have given this.
(c) Immediately after closing, 11.8.B.3.i has the uncharged capacitor acting like a wire, shorting $R_2$ : $I = 12.0/2000 = 6.0$ mA with $\Delta V_C = 0$ . Long after, 11.8.B.3.iv gives zero current in the capacitor branch, so the battery sees $R_1$ and $R_2$ in series: $I = 12.0/8000 = 1.5$ mA, four times smaller.
(d) The final capacitor potential difference is the one across $R_2$ , $(1.5 \times 10^{-3})(6000) = 9.0$ V. Confirm from the differential equation with $dq/dt = 0$ : $q_f = \dfrac{\mathcal{E} R_2 C}{R_1 + R_2} = 4.5 \times 10^{-5}$ C, and $q_f/C$ returns 9.0 V. So $q_f = 45\ \mu\mathrm{C}$ and $U_C = \tfrac{1}{2} Q \Delta V = 2.025 \times 10^{-4}$ J.
At $t = \tau$ , statement 11.8.B.2.ii puts the charge at approximately 63 percent of its final value, about $28\ \mu\mathrm{C}$ . That is a graph-reading answer, not a calculation, and it is all the framework asks for.

(a) $\mathcal{E} = R_1 \dfrac{dq}{dt} + q\,\dfrac{R_1 + R_2}{R_2 C}$ , from the junction rule plus the loop rule. (b) $\tau = R_{\text{eq}} C = 7.5$ ms with $R_{\text{eq}} = 1.5\ \mathrm{k}\Omega$ , the parallel pair. (c) 6.0 mA immediately after closing, 1.5 mA long after. (d) $q_f = 45\ \mu\mathrm{C}$ at 9.0 V, $U_C = 2.0 \times 10^{-4}$ J, and about $28\ \mu\mathrm{C}$ at $t = \tau$ .

Current density and drift speed in a wire

A cylindrical wire of radius $0.60$ mm carries a steady $2.4$ A. The metal has $n = 8.5 \times 10^{28}$ mobile carriers per cubic metre, each of charge magnitude $e = 1.60 \times 10^{-19}$ C, and resistivity $\rho = 1.7 \times 10^{-8}\ \Omega \cdot \mathrm{m}$ . Find (a) the current density, (b) the drift speed, and (c) the electric field inside the metal.

(a) The current is uniform across the wire, so the flux integral of 11.1.A.2 collapses to a product, $J = I/A$ . With $r = 6.0 \times 10^{-4}$ m, $A = \pi r^2 = 1.1310 \times 10^{-6}\ \mathrm{m^2}$ , so $J = \dfrac{2.4}{1.1310 \times 10^{-6}} = 2.1221 \times 10^{6}\ \mathrm{A/m^2}$ , or $2.1 \times 10^{6}\ \mathrm{A/m^2}$ to two figures.
(b) Statement 11.1.A.2.i gives $\vec{J} = n q \vec{v}_d$ , so $v_d = \dfrac{2.1221 \times 10^{6}}{(8.5 \times 10^{28})(1.60 \times 10^{-19})} = \dfrac{2.1221 \times 10^{6}}{1.36 \times 10^{10}} = 1.5603 \times 10^{-4}$ m/s, about $1.6 \times 10^{-4}$ m/s. That is roughly 0.16 millimetres per second, so a carrier takes hours to cross a metre of wire while the bulb lights at once.
(c) Statement 11.1.A.2.iii says a potential difference across a conductor creates a field inside it proportional to the resistivity and the current density, $\vec{E} = \rho \vec{J}$ . So $E = (1.7 \times 10^{-8})(2.1221 \times 10^{6}) = 3.6 \times 10^{-2}$ V/m. Units check: $\Omega \cdot \mathrm{m} \times \mathrm{A/m^2} = \mathrm{V/m}$ .

(a) $J = 2.1 \times 10^{6}\ \mathrm{A/m^2}$ . (b) $v_d = 1.6 \times 10^{-4}$ m/s. (c) $E = 3.6 \times 10^{-2}$ V/m from $\vec{E} = \rho \vec{J}$ . The drift speed is tiny even though the circuit responds immediately.

Frequently asked questions

How much of the AP Physics C E&M exam is Unit 11?

Unit 11, Electric Circuits, is weighted at 15 to 25% of the multiple-choice section of the AP Physics C: Electricity and Magnetism exam, over about 12 to 24 class periods. That is the top band in the course, shared with Unit 8. Units 9, 12 and 13 each sit at 10 to 20% and Unit 10 at 10 to 15%. The unit's AP Classroom Progress Check lists about 24 multiple-choice questions, more than the roughly 18 listed for each of the other five units.

How far does AP Physics C take RC circuits?

As far as writing the differential equation and reasoning from it. The course description prints the RC loop equation as a Derived Equation, which its front matter defines as a final result of a derivation expected of students on the exam, and that equation is not on the equation sheet. The framework never prints a solved exponential for charge or current anywhere in Unit 11. It prints the behaviour instead: essential knowledge 11.8.B.2.ii and 11.8.B.2.iii say the time constant is the time for a charging capacitor to reach approximately 63 percent of its final charge and for a discharging one to fall to approximately 37 percent of its initial charge. One of the course description's own sample free-response questions asks students to derive, but do not solve, a differential equation for the rate of change of the potential difference across a capacitor.

Is Ohm's law on the AP Physics C E&M equation sheet?

Yes, and it is printed as current equals the change in V divided by R, with the delta on the potential difference, rather than as V equals IR. The delta carries the meaning: the numerator is the potential difference across that particular element, not a potential at a point. Essential knowledge 11.3.B.1 says the same in words, that Ohm's law relates current, resistance, and potential difference across a conductive element of a circuit. The reference information also states that resistors and lightbulbs are ohmic unless otherwise stated, so a question about a filament whose resistance changes with temperature has switched that default off deliberately.

Are Kirchhoff's rules on the AP Physics C E&M equation sheet?

No. Two topics in Unit 11 are named after them, 11.6 Kirchhoff's Loop Rule and 11.7 Kirchhoff's Junction Rule, and neither rule appears anywhere on the AP Physics C: Electricity and Magnetism equation sheet. The course description introduces them as consequences of conservation laws rather than as formulas: 11.6.A.2 says the loop rule follows from conservation of energy and 11.7.A.1 says the junction rule follows from conservation of electric charge, so you write them from the principle. Three other Unit 11 results are also missing from the sheet: the power forms with current squared or potential difference squared, the terminal potential difference of a battery with internal resistance, and the RC differential equation.

What are the boundary statements for AP Physics C E&M Unit 11?

There are three, under Topics 11.2, 11.4 and 11.5. Topic 11.2 states that unless otherwise specified, all circuit schematic diagrams will be drawn using conventional current. Topic 11.4 states that the course 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. Topic 11.5 states that unless otherwise stated, all batteries, wires, and meters are assumed to be ideal, and adds that circuits with batteries of different potential differences connected in parallel will not be assessed. Topics 11.1, 11.3, 11.6, 11.7 and 11.8 print no boundary statement at all.

What is the difference between AP Physics C Unit 11 and AP Physics 2 Unit 11?

The architecture is identical and the treatment is not. Both courses have a Unit 11 called Electric Circuits with eight topics whose codes and titles line up one for one, two of them differing only by a hyphen or a parenthetical. AP Physics C adds calculus in four places: current as a derivative and as a current-density integral, current density as a vector tied to drift velocity and to the internal electric field, resistance as an integral when resistivity varies along the length, and the RC circuit as a differential equation from Kirchhoff's loop rule. It also weights symbolic derivation more heavily, with Science Practice 2 carrying 40 to 45% of its free-response section.