AP Physics 2 · Unit 11 of 7

Unit 11: Electric Circuits

15-18% of the multiple-choice section8 topics

Topics in this unit

  1. 11.1Electric Current
  2. 11.2Simple Circuits
  3. 11.3Resistance, Resistivity, and Ohm's Law
  4. 11.4Electric Power
  5. 11.5Compound Direct Current (DC) Circuits
  6. 11.6Kirchhoff's Loop Rule
  7. 11.7Kirchhoff's Junction Rule
  8. 11.8Resistor-Capacitor (RC) Circuits

Electric Circuits is Unit 11 of AP Physics 2, worth 15 to 18 percent of the multiple-choice section and about 12 to 20 class periods. All eight topics rest on two conservation laws: Kirchhoff's junction rule is conservation of charge, and Kirchhoff's loop rule is conservation of energy.

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 (DC) Circuits, 11.6 Kirchhoff's Loop Rule, 11.7 Kirchhoff's Junction Rule, 11.8 Resistor-Capacitor (RC) Circuits). Unit 11 Electric Circuits carries 15-18% of the AP Physics 2 multiple-choice section, the top weighting band in the course, and the CED estimates about 12 to 20 class periods. Its eight topics all follow from two conservation laws: Kirchhoff's junction rule for charge and Kirchhoff's loop rule for energy.

What Unit 11 covers, and what the CED asks for

Unit 11 is Electric Circuits, the third of the seven units in AP Physics 2. The course and exam description effective fall 2024 weights it at 15 to 18 percent of the multiple-choice section and budgets roughly 12 to 20 class periods. That is the top weighting band in the course, shared with Unit 9 Thermodynamics and Unit 10 Electric Force, Field, and Potential.

The unit opener frames this as a return trip rather than a new subject. The CED says Unit 11 revisits the behavior of charged particles in order to deepen conservation of energy and apply it to circuits, and it states directly that the unit takes more than calculating currents, resistances and potential differences in a simple circuit. The three examples it gives of what "more" means read like a syllabus in miniature: articulate what happens when a bulb is removed from a circuit, design an experiment to test whether a bulb is ohmic, and justify how and why series and parallel arrangements change what a circuit does.

The questions the unit opens with have the same shape. Why do the lights on an unplugged device dim slowly before going out? Why do several bulbs on a string go out when one is unplugged? How would you make a 120 W bulb brighter than a 40 W one? Not one of those is a plug-in question. Each is a comparison or a justification, which is what the science practices attached to this unit ask for.

The whole unit is two conservation laws

If you take one thing from this page, take this. Every technique in Unit 11 is one of two conservation laws applied to a circuit, and the CED says so in two essential knowledge statements rather than leaving you to notice it.

  • Kirchhoff's junction rule is conservation of electric charge (EK 11.7.A.1). Charge does not pile up at a junction, so the charge arriving per unit time equals the charge leaving per unit time: Iin=Iout\sum I_{\text{in}} = \sum I_{\text{out}}.
  • Kirchhoff's loop rule is conservation of energy (EK 11.6.A.2). Carry a charge around a closed loop and it returns to the potential it started at, so the potential differences around that loop sum to zero: ΔV=0\sum \Delta V = 0. EK 11.6.A.1 supplies the energy link explicitly, ΔUE=qΔV\Delta U_E = q \Delta V.

Everything else in the unit is a consequence of those two, a definition, or a statement about a material:

  • Series resistances add because the junction rule leaves the current nowhere else to go, so one current II crosses every element and the loop rule adds their individual IRIR drops. Req,s=iRiR_{\text{eq,s}} = \sum_i R_i is the loop rule with a single current in it.
  • Parallel resistances combine by reciprocals because the loop rule forces the same ΔV\Delta V across every branch and the junction rule adds the branch currents back together. 1/Req,p=i1/Ri1/R_{\text{eq,p}} = \sum_i 1/R_i is the junction rule with a single voltage in it.
  • Capacitors in series carry equal magnitudes of charge, and EK 11.8.A.2 names conservation of charge as the reason.
  • A battery's terminal voltage sags under load, ΔVterminal=EIr\Delta V_{\text{terminal}} = \mathcal{E} - I r, because the loop rule counts the drop across the battery's own internal resistance alongside every other drop (EK 11.5.B.3).

Ohm's law is the one relationship here that is not a conservation law. I=ΔV/RI = \Delta V / R describes how a particular object responds to a potential difference, and EK 11.3.B.1.i limits the constant-resistance version to ohmic materials. A circuit of non-ohmic elements still obeys both Kirchhoff rules exactly; it just cannot be collapsed into a single ReqR_{\text{eq}}.

Hold that distinction before topic 11.5, where the series and parallel shortcuts start to look like a separate rulebook to memorize. They are not one. Write the junction rule at each node and the loop rule around each loop and you can work the DC networks in this course without the shortcuts at all, which is worth being able to do when a diagram refuses to reduce.

The eight topics, in the order they build

TopicWhat it adds
11.1 Electric CurrentCurrent as the rate charge crosses a cross-section, I=Δq/ΔtI = \Delta q / \Delta t, and the direction convention
11.2 Simple CircuitsLoops, open and closed and short circuits, and the schematic symbol set
11.3 Resistance, Resistivity, and Ohm's LawR=ρ/AR = \rho \ell / A, ohmic versus non-ohmic behavior, and I=ΔV/RI = \Delta V / R
11.4 Electric PowerP=IΔVP = I \Delta V, its two derived forms, and bulb brightness
11.5 Compound Direct Current (DC) CircuitsSeries and parallel connections, equivalent resistance, internal resistance, ammeters and voltmeters
11.6 Kirchhoff's Loop RuleConservation of energy as ΔV=0\sum \Delta V = 0, plus potential-versus-position graphs
11.7 Kirchhoff's Junction RuleConservation of charge as Iin=Iout\sum I_{\text{in}} = \sum I_{\text{out}}
11.8 Resistor-Capacitor (RC) CircuitsEquivalent capacitance, the time constant τ=ReqCeq\tau = R_{\text{eq}} C_{\text{eq}}, and the two steady states

The sequence has a logic. Topics 11.1 and 11.2 hand you the vocabulary and the picture. Topics 11.3 and 11.4 attach numbers to a single element: how strongly it opposes charge, and how fast it converts energy. Topic 11.5 is the first topic about a circuit rather than an element, bringing in both connection types plus the two ways real hardware departs from the idealization, internal resistance and nonideal meters. Topics 11.6 and 11.7 then name the conservation laws that were quietly doing the work in 11.5, and topic 11.8 adds the one element whose behavior changes as time passes.

One practical note: the CED puts the shortcuts (11.5) before the laws they follow from (11.6 and 11.7). If the series and parallel formulas feel arbitrary the first time you meet them, read 11.6 and 11.7 first and come back. Topic 11.8 is also where the unit reaches back into Unit 10, since capacitance is introduced in 10.6 Capacitors.

Which Unit 11 equations are printed on the sheet

The AP Physics 2 equation sheet is organized into seven groups, and circuits live in the Electricity column, which holds 20 entries. Ten of those 20 are equations the CED names inside a Unit 11 essential knowledge statement:

I=ΔqΔtR=ρAI=ΔVRP=IΔVI = \frac{\Delta q}{\Delta t} \qquad R = \frac{\rho \ell}{A} \qquad I = \frac{\Delta V}{R} \qquad P = I \Delta V
Req,s=iRi1Req,p=i1RiΔUE=qΔVR_{\text{eq,s}} = \sum_i R_i \qquad \frac{1}{R_{\text{eq,p}}} = \sum_i \frac{1}{R_i} \qquad \Delta U_E = q \Delta V
1Ceq,s=i1CiCeq,p=iCiτ=ReqCeq\frac{1}{C_{\text{eq,s}}} = \sum_i \frac{1}{C_i} \qquad C_{\text{eq,p}} = \sum_i C_i \qquad \tau = R_{\text{eq}} C_{\text{eq}}

The other ten entries in that column belong to Unit 10, though two of them come straight back in topic 11.8 whenever a question asks for the charge or the stored energy on a capacitor at steady state: C=Q/ΔVC = Q / \Delta V and UC=12QΔVU_C = \frac{1}{2} Q \Delta V.

Now notice what is absent. Kirchhoff's two rules are named as relevant equations in EK 11.6.A.3 and EK 11.7.A.2, but neither ΔV=0\sum \Delta V = 0 nor Iin=Iout\sum I_{\text{in}} = \sum I_{\text{out}} is printed anywhere on the AP Physics 2 sheet. Neither is the terminal-voltage relation ΔVterminal=EIr\Delta V_{\text{terminal}} = \mathcal{E} - I r, which the CED lists as a derived equation under EK 11.5.B.3, and neither are the two rearrangements of power that it lists as derived under EK 11.4.A.1:

P=I2R=(ΔV)2RP = I^2 R = \frac{(\Delta V)^2}{R}

Four relationships, then, that you carry in your head. The power pair costs one substitution of ΔV=IR\Delta V = IR into P=IΔVP = I \Delta V and is worth drilling until it is automatic, because brightness questions in 11.4 usually hand you either the current or the potential difference and not both.

One notation point that has cost real marks. The sheet prints Ohm's law with a delta, as I=ΔV/RI = \Delta V / R, and the CED writes it the same way in EK 11.3.B.1. That symbol is a potential difference between two points, not a potential at a point, and when a resistor sits inside a network the question of which two points you are measuring across is the whole question. The Ohm's law guide works through the three rearrangements and the units.

The traps that run across the whole unit

The CED's exam-preparation note for Unit 11 singles out two student beliefs to dismantle: that batteries store charge, and that current is used up in a circuit.

  • A battery is not a charge tank. It maintains a potential difference between its terminals, supplying energy per unit charge, which is why it is rated in volts and not coulombs. The mobile charge is already in the wire.
  • Current is not consumed. The junction rule says so: whatever arrives at a junction leaves it, so the current entering a bulb equals the current leaving it. What the bulb consumes is energy, at a rate P=IΔVP = I \Delta V, and each coulomb leaves with less electric potential energy than it arrived with.

The same note flags a vocabulary problem, and it is a real one: five words that look interchangeable are not.

WordWhat it isWhat it depends on
CurrentRate at which charge passes a cross-sectionThe whole circuit
Potential differenceEnergy per unit charge between two pointsWhich two points you pick
ResistanceHow strongly an object opposes charge movementThe object's material and its shape
ResistivityHow strongly a material opposes charge movementThe material alone
CapacitanceCharge stored per voltThe capacitor's geometry and dielectric

Resistance against resistivity is the pair that gets tested, through R=ρ/AR = \rho \ell / A. Stretch a wire to twice its length at constant volume and its cross-section halves, so its resistance quadruples while ρ\rho does not move. EK 11.3.A.2.ii adds that the resistivity of a conductor typically increases with temperature, while EK 11.3.B.1.ii says the resistivity of an ohmic material is constant regardless of temperature. Those two together are why testing whether a bulb is ohmic is a real experiment rather than a formality.

Five more traps span topics:

  • Brightness tracks power, not current. EK 11.4.A.2 ties bulb brightness to power. In series the current is common, so P=I2RP = I^2 R makes the larger resistance brighter; in parallel the potential difference is common, so P=(ΔV)2/RP = (\Delta V)^2 / R makes the smaller resistance brighter. "More current means brighter" gets one of those backwards every time. The series versus parallel guide has the reduction routine behind the comparison.
  • Conventional current versus what actually moves. EK 11.1.A.2.i defines conventional current as the direction positive charge would move; EK 11.1.A.2.ii notes that in common circuits the carriers are electrons, moving the other way. EK 11.1.A.2 also insists current is not a vector quantity even though it has a direction, and that the direction is tied to no coordinate system, so never resolve a current into components.
  • Zero current does not mean nothing is moving. EK 11.1.A.1.ii: if the current in a section of wire is zero, the net motion of the charge carriers there is zero, but the individual carriers do not have zero speed.
  • Ideal wires are only ideal in company. EK 11.5.B.1.ii is worth reading twice: wire resistance may be neglected only if the circuit contains other elements that do have resistance. A source joined to itself through "ideal" wire is a short, which EK 11.2.A.2.iii defines as a circuit in which charges flow with no change in potential difference.
  • Meters are wired opposite ways. Ammeters go in series and are ideally of zero resistance, so they do not change the current they measure (EK 11.5.C.1). Voltmeters go in parallel and are ideally of infinite resistance, so no charge flows through them (EK 11.5.C.2). Swap them and you either open the branch or short the element. EK 11.5.C.3 confirms nonideal meters change the circuit being measured.

The four representations Unit 11 expects you to build

Every one of the eight topics carries at least one science practice 1 skill in the CED's Unit at a Glance table. Practice 1.A, creating diagrams, tables, charts or schematics, is attached to 11.1, 11.2, 11.5 and 11.7. Practice 1.B, creating quantitative graphs with scales and units including plotted data, is attached to 11.3 and 11.8. Practice 1.C, creating qualitative sketches of graphs, is attached to 11.4, 11.6 and 11.8. Four representations carry that load.

  1. The schematic. EK 11.2.A.4 makes schematics the representation used to describe and analyze circuits, and EK 11.2.A.4.ii fixes the symbol set: battery, bulb, switch, capacitor, resistor, ammeter and voltmeter, with a diagonal strikethrough arrow across a symbol marking a variable element. EK 11.2.A.4.i is why the drawing matters: a circuit's properties depend on the physical arrangement of its elements, so redrawing a tangled diagram as a clean set of loops is analysis, not tidying.
  2. Current as a function of potential difference. EK 11.3.B.1.iv says the resistance of an ohmic element can be determined from the slope of a graph of the current in it against the potential difference across it. Mind the direction: with II vertical and ΔV\Delta V horizontal the slope is 1/R1/R, so resistance is the reciprocal of the slope, not the slope. A straight line through the origin is the signature of an ohmic element; a curve answers the ohmic question in the negative.
  3. Electric potential as a function of position around a loop. EK 11.6.A.4 asks for this graph, and it is the loop rule drawn rather than written: the trace steps up across the source, holds flat along ideal wire, drops across each resistor, and lands back at its starting value after one circuit. If your plot does not close, a sign is wrong.
  4. RC quantities against time. Topic 11.8 carries both 1.B and 1.C. EK 11.8.B.2.iii says the charge, the potential difference and the current all approach steady state asymptotically, so a correct sketch needs the right starting value, final value and concavity. The curve flattens toward its limit; it never turns and never crosses.

One scoring note that should change how you practice: science practice 1 is not assessed at all on the multiple-choice section, and it carries 20 to 35 percent of the free-response section. Drawing and sketching earn nothing in Section I and between a fifth and a third of Section II.

The three boundary statements that cap what can be asked

Three boundary statements sit inside Unit 11. Each is a promise about what the exam will not do, and each is worth knowing before you spend study time on material that is out of scope.

Topic 11.2. Unless a question specifies otherwise, all circuit schematic diagrams are drawn using conventional current. You never have to work out which convention a diagram is using.

Topic 11.5. Three limits in one statement, and the third is the one that gets missed.

  • AP Physics 2 only expects a qualitative discussion of how a nonideal ammeter or voltmeter affects the results of measurements. You will not be asked to compute the error a real meter introduces.
  • Unless a question states otherwise, all batteries, wires and meters are assumed to be ideal.
  • Circuits with batteries of different potential differences connected in parallel will not be assessed.

That third line quietly removes a whole genre of multi-loop problem. Networks with two unequal sources in parallel branches are a staple of college circuit analysis and are off the table in this course.

Topic 11.8. Descriptions of charging and discharging RC circuits in AP Physics 2 are limited to qualitative descriptions and representations, and the statement carries its exception in the same breath: students should still be able to mathematically describe the initial and final states of RC circuits. What is excluded is mathematically modeling those behaviors with respect to time, so no exponential fitting and no Q(t)=Q0et/τQ(t) = Q_0 e^{-t/\tau}. What you do owe is the two endpoint states as numbers, the shape of the curve between them as a sketch, and the time constant τ=ReqCeq\tau = R_{\text{eq}} C_{\text{eq}} with the two figures the CED attaches to it: a charging capacitor reaches roughly 63 percent of its final charge in one time constant (EK 11.8.B.1.ii), and a discharging one falls to roughly 37 percent of its initial charge in one time constant (EK 11.8.B.1.iii).

How Unit 11 is assessed

The AP Physics 2 exam runs three hours. Section I is 42 multiple-choice questions in 85 minutes for 50 percent of the score; Section II is 4 free-response questions in 95 minutes for the other 50 percent. A four-function, scientific or graphing calculator is allowed on both sections.

The 15 to 18 percent figure is a multiple-choice weighting, so expect roughly 6 to 8 of those 42 questions to be about circuits. Unit 11 can also appear inside any of the four free-response questions, whose types are fixed and all four of which appear on every exam:

QuestionType
1Mathematical Routines, 10 points, suggested 20 to 25 minutes
2Translation Between Representations, 12 points, suggested 25 to 30 minutes
3Experimental Design and Analysis
4Qualitative/Quantitative Translation

The science practice weightings tell you what those 6 to 8 multiple-choice questions look like. Calculating an unknown quantity with units (practice 2.B) is 20 to 25 percent of the section, and applying a law, definition, relationship or model to make a claim (3.B) is another 20 to 25 percent. Deriving a symbolic expression (2.A) is 15 to 20 percent. Comparing quantities across scenarios (2.C) and predicting new values from functional dependence (2.D) are 10 to 15 percent each. Justifying a claim with evidence (3.C) is 5 to 10 percent. Practices 1.A, 1.B, 1.C and 3.A are not assessed there at all.

Read the per-topic skill codes with care. The CED calls them suggested skills and notes that required course content can be assessed with any skill, so pairing topic 11.4 with practice 1.C is a teaching suggestion, not a promise. And practice 3.A, creating experimental procedures, is attached to topics 11.3 and 11.8 yet scores nothing in Section I; the CED says the experimental-procedure and graphing work in this unit pays off on the free-response section. The unit's Progress Check in AP Classroom matches that shape, at about 24 multiple-choice questions plus 4 free-response questions covering the same four types.

How to study Unit 11

Work it in this order and the unit stays small.

  1. Learn the two rules before the shortcuts. Derive the series and parallel formulas from Iin=Iout\sum I_{\text{in}} = \sum I_{\text{out}} and ΔV=0\sum \Delta V = 0 once, by hand. After that they are recall rather than faith.
  2. Then make the shortcuts mechanical. The series versus parallel circuits guide has the reduce-and-unwind routine, and the Ohm's law calculator combines a list of resistors either way, so you can check a reduction before committing to it.
  3. Draw all four representations from memory: schematic, II against ΔV\Delta V, potential against position, and the RC curves.
  4. Learn the gaps in the sheet. Kirchhoff's rules, ΔVterminal=EIr\Delta V_{\text{terminal}} = \mathcal{E} - I r and P=I2R=(ΔV)2/RP = I^2 R = (\Delta V)^2 / R are not printed. Everything else you need is on the AP Physics 2 formula sheet, so practice with it open until the symbols stop needing translation.
  5. Do RC circuits as two snapshots, then join the dots with a curve.
  6. Rehearse the vocabulary in writing. Free-response points come from written justification, so write full sentences during practice instead of only chasing numbers.

Unit 11 leans on Unit 10 in two places: 10.5 Electric Potential for what a potential difference is, and 10.6 Capacitors for topic 11.8. If the energy bookkeeping in the loop rule feels shaky, the ideas underneath it come from AP Physics 1 Unit 3 Work, Energy, and Power, moved from a mass in a gravitational field to a charge in an electric one.

One loop, one junction, and a battery that is not ideal

A battery of emf E=18.0 V\mathcal{E} = 18.0 \text{ V} and internal resistance r=1.00 Ωr = 1.00\ \Omega drives a 3.00 Ω3.00\ \Omega resistor in series with a parallel pair, 12.0 Ω12.0\ \Omega and 24.0 Ω24.0\ \Omega. Using only Kirchhoff's two rules, find the current in the battery, the current in each parallel branch, and the terminal voltage. Then check the energy bookkeeping.

  1. Fix a convention and keep it. Take the loop clockwise, in the direction of conventional current. Crossing the source from its negative to its positive terminal counts as a rise; crossing any resistance in the direction of the current counts as a drop. Every sign below follows from that one choice.

  2. Name the currents. Let II be the current in the battery and in the 3.00 Ω3.00\ \Omega resistor, I1I_1 the current in the 12.0 Ω12.0\ \Omega resistor and I2I_2 the current in the 24.0 Ω24.0\ \Omega resistor. The junction rule where the branches split gives I=I1+I2I = I_1 + I_2.

  3. Apply the loop rule to the small loop containing only the two parallel branches: down through the 12.0 Ω12.0\ \Omega and back up through the 24.0 Ω24.0\ \Omega gives I1(12.0 Ω)+I2(24.0 Ω)=0-I_1 (12.0\ \Omega) + I_2 (24.0\ \Omega) = 0, so I1=2I2I_1 = 2 I_2. With the junction rule, I=2I2+I2=3I2I = 2I_2 + I_2 = 3I_2, so I2=I/3I_2 = I/3 and I1=2I/3I_1 = 2I/3.

  4. The potential difference across the parallel section is ΔVp=I2(24.0 Ω)=(I/3)(24.0 Ω)=I(8.00 Ω)\Delta V_p = I_2 (24.0\ \Omega) = (I/3)(24.0\ \Omega) = I (8.00\ \Omega). That 8.00 Ω8.00\ \Omega is the parallel formula arriving on its own. Nothing was memorized to produce it.

  5. Now the big loop, starting just inside the negative terminal: 18.0 VI(1.00 Ω)I(3.00 Ω)I(8.00 Ω)=018.0 \text{ V} - I(1.00\ \Omega) - I(3.00\ \Omega) - I(8.00\ \Omega) = 0, so 18.0 V=I(12.0 Ω)18.0 \text{ V} = I(12.0\ \Omega) and I=1.50 AI = 1.50 \text{ A}.

  6. Branch currents: I2=1.50/3=0.500 AI_2 = 1.50/3 = 0.500 \text{ A} and I1=2(0.500)=1.00 AI_1 = 2(0.500) = 1.00 \text{ A}. Junction check: 1.00+0.500=1.50 A1.00 + 0.500 = 1.50 \text{ A}. Equal-voltage check: (1.00 A)(12.0 Ω)=12.0 V(1.00 \text{ A})(12.0\ \Omega) = 12.0 \text{ V} and (0.500 A)(24.0 Ω)=12.0 V(0.500 \text{ A})(24.0\ \Omega) = 12.0 \text{ V}, as the loop rule required.

  7. Terminal voltage: ΔVterminal=EIr=18.0 V(1.50 A)(1.00 Ω)=16.5 V\Delta V_{\text{terminal}} = \mathcal{E} - I r = 18.0 \text{ V} - (1.50 \text{ A})(1.00\ \Omega) = 16.5 \text{ V}, which is what an ideal voltmeter across the terminals reads while the circuit runs. Loop check on the external circuit: the 3.00 Ω3.00\ \Omega drops (1.50 A)(3.00 Ω)=4.50 V(1.50 \text{ A})(3.00\ \Omega) = 4.50 \text{ V}, the parallel section drops 12.0 V12.0 \text{ V}, and 16.54.5012.0=016.5 - 4.50 - 12.0 = 0.

  8. Energy bookkeeping. The source converts energy at P=EI=(18.0 V)(1.50 A)=27.0 WP = \mathcal{E} I = (18.0 \text{ V})(1.50 \text{ A}) = 27.0 \text{ W}. Internal resistance dissipates I2r=(1.50)2(1.00)=2.25 WI^2 r = (1.50)^2 (1.00) = 2.25 \text{ W}, the 3.00 Ω3.00\ \Omega dissipates (1.50)2(3.00)=6.75 W(1.50)^2(3.00) = 6.75 \text{ W}, the 12.0 Ω12.0\ \Omega dissipates (1.00)2(12.0)=12.0 W(1.00)^2(12.0) = 12.0 \text{ W}, and the 24.0 Ω24.0\ \Omega dissipates (0.500)2(24.0)=6.00 W(0.500)^2(24.0) = 6.00 \text{ W}.

  9. Total dissipated: 2.25+6.75+12.0+6.00=27.0 W2.25 + 6.75 + 12.0 + 6.00 = 27.0 \text{ W}, matching the source exactly. That is the loop rule again, multiplied through by the current: conservation of energy per unit charge becomes conservation of energy per unit time.

I=1.50 AI = 1.50 \text{ A} in the battery, 1.00 A1.00 \text{ A} in the 12.0 Ω12.0\ \Omega branch and 0.500 A0.500 \text{ A} in the 24.0 Ω24.0\ \Omega branch, with a terminal voltage of 16.5 V16.5 \text{ V}. The battery converts 27.0 W27.0 \text{ W}, of which 2.25 W2.25 \text{ W} is dissipated inside it and 24.75 W24.75 \text{ W} reaches the external resistors.

An RC circuit at its two endpoint states

A 20.0 V20.0 \text{ V} source of negligible internal resistance is connected through a switch to a 5.00 kΩ5.00 \text{ k}\Omega resistor. A 15.0 kΩ15.0 \text{ k}\Omega resistor and an initially uncharged 4.00 μF4.00\ \mu\text{F} capacitor are connected in parallel with each other, and that combination completes the loop. The switch closes at t=0t = 0. Find the current in the source immediately afterwards and a long time later, then the charge and stored energy on the capacitor at steady state.

  1. Immediately after the switch closes, the capacitor is uncharged, so the potential difference across it is zero and it behaves like a wire (EK 11.8.B.2.i). A wire across the 15.0 kΩ15.0 \text{ k}\Omega resistor holds that resistor at zero potential difference too, so no current passes through it and the only resistance in the loop is the 5.00 kΩ5.00 \text{ k}\Omega.

  2. Loop rule at t=0t = 0: 20.0 V=I0(5.00×103 Ω)20.0 \text{ V} = I_0 (5.00 \times 10^3\ \Omega), so I0=4.00×103 A=4.00 mAI_0 = 4.00 \times 10^{-3} \text{ A} = 4.00 \text{ mA}. All of it goes onto the capacitor plates.

  3. A long time later the capacitor is fully charged and there is zero current in its branch (EK 11.8.B.2.iv). The remaining loop runs through the two resistors in series, so 20.0 V=I(5.00×103 Ω+15.0×103 Ω)20.0 \text{ V} = I_{\infty}(5.00 \times 10^3\ \Omega + 15.0 \times 10^3\ \Omega) and I=20.0/(2.00×104)=1.00×103 A=1.00 mAI_{\infty} = 20.0 / (2.00 \times 10^4) = 1.00 \times 10^{-3} \text{ A} = 1.00 \text{ mA}.

  4. Steady-state potential differences: (1.00×103 A)(5.00×103 Ω)=5.00 V(1.00 \times 10^{-3} \text{ A})(5.00 \times 10^3\ \Omega) = 5.00 \text{ V} across the first resistor and (1.00×103 A)(15.0×103 Ω)=15.0 V(1.00 \times 10^{-3} \text{ A})(15.0 \times 10^3\ \Omega) = 15.0 \text{ V} across the second. The capacitor is in parallel with the second resistor, so it sits at 15.0 V15.0 \text{ V}. Loop check: 20.05.0015.0=020.0 - 5.00 - 15.0 = 0.

  5. Charge, from C=Q/ΔVC = Q / \Delta V: Q=(4.00×106 F)(15.0 V)=6.00×105 C=60.0 μCQ = (4.00 \times 10^{-6} \text{ F})(15.0 \text{ V}) = 6.00 \times 10^{-5} \text{ C} = 60.0\ \mu\text{C}.

  6. Stored energy, from UC=12QΔVU_C = \frac{1}{2} Q \Delta V: UC=12(6.00×105 C)(15.0 V)=4.50×104 JU_C = \frac{1}{2}(6.00 \times 10^{-5} \text{ C})(15.0 \text{ V}) = 4.50 \times 10^{-4} \text{ J}.

  7. Stop there. The current falls from 4.00 mA4.00 \text{ mA} to 1.00 mA1.00 \text{ mA} along a curve that approaches its final value asymptotically, and the topic 11.8 boundary statement keeps that time dependence qualitative in AP Physics 2. Sketch the shape and label the two endpoints; do not fit an exponential to it.

I0=4.00 mAI_0 = 4.00 \text{ mA} immediately after the switch closes and I=1.00 mAI_{\infty} = 1.00 \text{ mA} a long time later. At steady state the capacitor holds 60.0 μC60.0\ \mu\text{C} at 15.0 V15.0 \text{ V} and stores 4.50×104 J4.50 \times 10^{-4} \text{ J}.

Frequently asked questions

What does AP Physics 2 Unit 11 cover?

Unit 11 is Electric Circuits, with eight topics: electric current, simple circuits, resistance and resistivity and Ohm's law, electric power, compound direct current (DC) circuits, Kirchhoff's loop rule, Kirchhoff's junction rule, and resistor-capacitor (RC) circuits. The College Board weights it at 15 to 18 percent of the AP Physics 2 multiple-choice section and estimates about 12 to 20 class periods for it. Every technique in the unit comes from one of two conservation laws: charge at a junction, energy around a loop.

How much of the AP Physics 2 exam is Unit 11?

Unit 11 carries a 15 to 18 percent weighting on the multiple-choice section of the AP Physics 2 exam, the top band in the course alongside Unit 9 Thermodynamics and Unit 10 Electric Force, Field, and Potential. On a 42-question multiple-choice section that is roughly 6 to 8 questions. Circuits can also appear inside any of the four free-response questions, which are worth the other 50 percent of the score.

Are Kirchhoff's rules on the AP Physics 2 equation sheet?

No. Neither the loop rule (the potential differences around a closed loop sum to zero) nor the junction rule (the current into a junction equals the current out of it) is printed on the AP Physics 2 equation sheet, even though the CED names both as relevant equations for topics 11.6 and 11.7. What the sheet does print for circuits is the definition of current, resistance from resistivity, Ohm's law, power, the series and parallel rules for resistors and for capacitors, the RC time constant, and the change in electric potential energy. The two rules themselves you have to know.

What is the difference between resistance and resistivity?

Resistivity is a property of a material; resistance is a property of an object made from that material. Resistivity, written as the Greek letter rho, depends on the atomic and molecular structure of the substance and measures how strongly it opposes the motion of electric charge. Resistance also depends on shape, through R = rho times length divided by cross-sectional area. Two wires drawn from the same metal have identical resistivity but different resistance if one is longer or thinner. The AP Physics 2 CED names this pair of words as one students confuse in written answers.

Do I need the exponential RC equations for AP Physics 2?

No. The boundary statement for topic 11.8 limits charging and discharging RC circuits to qualitative descriptions and representations, and says students are not expected to mathematically model those behaviors with respect to time. The same statement keeps one quantitative obligation: you should still be able to describe the initial and final states of an RC circuit mathematically. So compute the current and the capacitor's potential difference immediately after the switch closes and again a long time later, know that the time constant is the equivalent resistance times the equivalent capacitance, and sketch the curve between the two states instead of fitting an exponential to it.

Why doesn't current get used up in a circuit?

Because charge is conserved. Kirchhoff's junction rule says the charge arriving at a junction per unit time equals the charge leaving it per unit time, so the current entering a bulb equals the current leaving that bulb. What the bulb uses up is energy, not charge: each coulomb passes through and comes out at a lower electric potential, and the rate of that energy conversion is the power, current times the potential difference across the element. The AP Physics 2 CED names this belief, along with the idea that batteries store charge, as a misconception the unit should challenge directly.

Do I need Unit 10 before starting Unit 11?

In two specific places, yes. Kirchhoff's loop rule in topic 11.6 is built on the change in electric potential energy of a charge moving through a potential difference, which is the Unit 10 relationship. And topic 11.8 puts capacitors into circuits, so it needs the definition of capacitance and the stored-energy expression from topic 10.6. The rest of Unit 11, meaning current, resistance, power, and the series and parallel connections, can be learned on its own footing.