Topic 11.8: Resistor Capacitor (RC) Circuits

What Topic 11.8 requires

Topic 11.8 of AP Physics C: Electricity and Magnetism Unit 11 has two learning objectives and eighteen essential-knowledge statements, second only to Topic 11.5's twenty. It prints no boundary statement, which turns out to be the most important fact on this page.

11.8.A, describe the equivalent capacitance of multiple capacitors.

• 11.8.A.1 A collection of capacitors in a circuit may be analyzed as though it was a single capacitor with an equivalent capacitance $C_{\text{eq}}$.
• 11.8.A.1.i The inverse of the equivalent capacitance of a set of capacitors connected in series is equal to the sum of the inverses of the individual capacitances. Relevant equation $\dfrac{1}{C_{\text{eq,s}}} = \sum_i \dfrac{1}{C_i}$.
• 11.8.A.1.ii The equivalent capacitance of a set of capacitors in series is less than the capacitance of the smallest capacitor.
• 11.8.A.1.iii The equivalent capacitance of a set of capacitors in parallel is the sum of the individual capacitances. Relevant equation $C_{\text{eq,p}} = \sum_i C_i$.
• 11.8.A.2 As a result of conservation of charge, each of the capacitors in series must have the same magnitude of charge on each plate.

11.8.B, describe the behavior of a circuit containing combinations of resistors and capacitors.

• 11.8.B.1 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. Derived equation $\mathcal{E} = \dfrac{dq}{dt}R + \dfrac{q}{C}$.
• 11.8.B.2 The time constant $(\tau)$ is a significant feature of an RC circuit.
• 11.8.B.2.i The time constant of an RC circuit is a measure of how quickly the capacitor will charge or discharge and is defined as $\tau = R_{\text{eq}} C_{\text{eq}}$.
• 11.8.B.2.ii For a charging capacitor, the time constant represents the time required for the capacitor's charge to increase from zero to approximately 63 percent of its final asymptotic value.
• 11.8.B.2.iii For a discharging capacitor, the time constant represents the time required for the capacitor's charge to decrease from fully charged to approximately 37 percent of its initial value.
• 11.8.B.3 The potential difference across a capacitor and the current in the branch of the circuit containing the capacitor each change over time as the capacitor charges and discharges, but both will reach a steady state after a long time interval.
• 11.8.B.3.i Immediately after being placed in a circuit, an uncharged capacitor acts like a wire, and charge can easily flow to or from the plates of the capacitor.
• 11.8.B.3.ii As a capacitor charges, changes to the potential difference across the capacitor affect the charge on the plates of the capacitor, the current in the circuit branch in which the capacitor is located, and the electric potential energy stored in the capacitor.
• 11.8.B.3.iii The potential difference across a capacitor, the current in the circuit branch in which the capacitor is located, and the electric potential energy stored in the capacitor all change with respect to time and asymptotically approach steady state conditions.
• 11.8.B.3.iv After a long time, a charging capacitor approaches a state of being fully charged, reaching a maximum potential difference at which there is zero current in the circuit branch in which the capacitor is located.
• 11.8.B.3.v Immediately after a charged capacitor begins discharging, the amount of charge on the capacitor and the energy stored in the capacitor begin to decrease.
• 11.8.B.3.vi As a capacitor discharges, the amount of charge on the capacitor, the potential difference across the capacitor, and the current in the circuit branch in which the capacitor is located all decrease until a steady state is reached.
• 11.8.B.3.vii After either charging or discharging for times much greater than the time constant, the capacitor and the relevant circuit branch may be modeled using steady-state conditions.

The suggested skills are 1.B, create quantitative graphs with appropriate scales and units, including plotting data; 2.A, derive a symbolic expression from known quantities; 2.D, predict new values or factors of change of physical quantities using functional dependence between variables; 3.A, create experimental procedures that are appropriate for a given scientific question; and 3.C, justify or support a claim using evidence from experimental data, physical representations, or physical principles or laws. Five skills, the joint largest in Unit 11 alongside Topic 11.3.

How far the exam takes the differential equation

This is the question the topic turns on, and it is a boundary-statement question rather than a memory question. Four sourced facts settle it.

One. AP Physics C Topic 11.8 prints no boundary statement at all. Nothing in the framework fences off the mathematics here. That is not true of the algebra-based course.

Two. AP Physics 2's Topic 11.8 does print one, and it is the exact prohibition AP Physics C lacks. Quoted whole: "Descriptions of charging/discharging RC circuits in AP Physics 2 are limited to qualitative descriptions and representations. While students should be able to mathematically describe initial and final states of RC circuits, students are not expected to mathematically model these behaviors with respect to time." Everything after "while" is the part that gets cut; it says the two limiting states are still fair game in the algebra-based course.

Three. The C framework adds exactly one statement on top of the Physics 2 set, and it is the differential equation. Compare the two 11.8.B trees and they are the same statements renumbered, except that AP Physics C inserts 11.8.B.1 at the front: 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. Everything from the time constant onward is identical wording in both courses, shifted down one number.

Four. That statement carries the Derived equation label. The CED's Required Equations page defines the label: not all equations in the framework appear on the equation sheet, many are provided for reference and guidance or to demonstrate the final results of derivations expected of students on the exam, and those are the Derived Equations. So the differential equation is a result you are expected to produce, and it is not on the sheet.

Put those together and the ceiling is legible. No solved exponential appears anywhere in Unit 11 of the AP Physics C framework, for charge, current or potential difference, and none is printed on the equation sheet. What the CED prints instead is the behaviour: 11.8.B.2.ii and 11.8.B.2.iii pin the 63 and 37 percent figures, and 11.8.B.3.iii says the quantities asymptotically approach steady state. Four statements describe an exponential without ever writing one.

The CED's own sample free-response question confirms it from the assessment side. Question 4, part B, instructs students to "Derive, but do not solve, a differential equation" for a rate of change, and part C asks them to justify a prediction "by referring to the differential equation you wrote for part B".

So the assessed skill is the setup and the reasoning from it, not the solution. That said, nothing forbids solving it, and the sheet quietly supplies the tools: its Calculus table prints $\int \dfrac{dx}{x+a} = \ln|x+a|$, $\dfrac{d}{dx}(e^{ax}) = ae^{ax}$ and $\int e^{ax} dx = \dfrac{1}{a}e^{ax}$, which is exactly what separating variables needs. Being able to produce the exponential in four lines is cheap insurance; treating it as the point of the topic is a misreading.

Equivalent capacitance, and why the rules look swapped

Objective 11.8.A is the capacitor mirror of Topic 11.5's equivalent resistance, and both printed rules are on the sheet:

The reciprocal is on the series rule for capacitors and on the parallel rule for resistors. Rather than memorising which flips, remember what each connection forces to be shared, and let the algebra follow.

Series capacitors share a charge. Statement 11.8.A.2 gives the reason: as a result of conservation of charge, each of the capacitors in series must have the same magnitude of charge on each plate. The plates between two series capacitors are isolated from the rest of the circuit, so whatever charge leaves one must arrive on the other. With $Q$ shared, the potential differences add, $\Delta V = \sum Q/C_i$, and dividing by $Q$ gives the reciprocal rule.

Parallel capacitors share a potential difference. Both plates of each are connected to the same two nodes, so $\Delta V$ is common and the charges add, $Q = \sum C_i \Delta V$, giving the direct sum.

Statement 11.8.A.1.ii supplies the sanity check that catches most errors: the equivalent capacitance of a set of capacitors in series is less than the capacitance of the smallest capacitor. If your series answer is bigger than any member, you added instead of adding reciprocals. The parallel counterpart is that the sum is larger than the largest member.

A physical way to see the series result: two capacitors in series is like one capacitor with the plate separation increased, and $C = \kappa \varepsilon_0 A / d$, printed on the sheet, falls as $d$ grows. Two in parallel is like one capacitor with a larger plate area, and the same equation rises with $A$.

Three relations from Unit 10 are printed on the sheet and needed constantly here: $C = Q/\Delta V$, $U_C = \tfrac{1}{2}Q\Delta V$ and $C = \kappa\varepsilon_0 A/d$. The exam's reference information adds that capacitors are air-filled with $\kappa = 1.0$ unless otherwise stated.

Deriving the loop equation, and reading it without solving

Statement 11.8.B.1 names the method: a fundamental differential equation derived from Kirchhoff's loop rule. Do it once carefully.

Take a battery of emf $\mathcal{E}$ in series with a resistor $R$, a capacitor $C$ and a switch, with the capacitor initially uncharged. Define the positive current direction as the one the battery drives, and traverse the loop in that direction.

1. Across the battery, from the short line to the long line, the potential rises by $\mathcal{E}$.
2. Across the resistor, travelling with the current, the potential falls by $IR$, from Ohm's law in the printed form $I = \Delta V / R$.
3. Across the capacitor, from the negative plate to the positive plate, the potential falls by $q/C$, from the printed $C = Q/\Delta V$.
4. Sum to zero around the loop, per 11.6.A.2.i: $\mathcal{E} - IR - \dfrac{q}{C} = 0$.
5. Substitute $I = dq/dt$, from 11.1.A.1. That single step turns an algebraic equation into a differential one, because the same unknown $q$ now appears both as itself and as its own derivative.

That is 11.8.B.1 exactly. Now read it rather than solve it, because reading it answers most questions.

At $t = 0$ with an uncharged capacitor, $q = 0$, so the $q/C$ term vanishes and the whole emf sits across the resistor: $R\,dq/dt = \mathcal{E}$, giving the initial current $I_0 = \mathcal{E}/R$ and the initial slope of the charge graph. That is 11.8.B.3.i in symbols, the statement that an uncharged capacitor acts like a wire.

Long after, the charge stops changing, so $dq/dt = 0$ and $\mathcal{E} = q/C$, giving the final charge $q_f = C\mathcal{E}$ and zero current. That is 11.8.B.3.iv.

In between, every volt the battery supplies is split between the resistor and the capacitor, and the split shifts continuously from all-resistor to all-capacitor. The current only ever decreases, because the capacitor's growing potential difference opposes it.

The time constant falls out of the coefficients. Divide through by $R$ and the coefficient of $q$ is $1/(RC)$, which has units of inverse time. That is where $\tau = RC$ comes from, and it is why you can read a time constant off a correctly written differential equation without solving anything.

The junction-rule route reaches the same place when the capacitor is in a branch rather than in the main loop. Topic 11.7 covers it: write $I_{\text{total}} = I_R + dq/dt$ and substitute. The CED's scoring guidelines accept either opening.

The two instants that need no calculus

Seven of Topic 11.8's essential-knowledge statements, 11.8.B.3.i through 11.8.B.3.vii, describe the circuit's behaviour at the two ends of time. Between them they let you answer most multiple-choice questions on this topic without solving anything, and they are the highest-value thing on this page.

Immediately after a switch closes on an uncharged capacitor (11.8.B.3.i): the capacitor acts like a wire, and charge can easily flow to or from its plates. So replace it with a plain wire and analyse the resulting resistor network. Consequences: the capacitor's potential difference is zero, anything in parallel with it is short-circuited and carries no current, and the current elsewhere is at its largest.

Long after a switch closes (11.8.B.3.iv): the capacitor approaches full charge, reaching a maximum potential difference at which there is zero current in its branch. So replace it with a break, an open circuit, and analyse the resulting network. Consequences: no current in that branch, no potential difference across any resistor in series with it inside that branch, and the capacitor's potential difference is whatever the rest of the circuit puts across its terminals.

The two most common errors both come from mixing up the row you are on.

• "A capacitor blocks current, so no current flows at $t = 0$." Backwards. At $t = 0$ an uncharged capacitor is the easiest path in the circuit, not the hardest.
• "The capacitor charges up to the battery's emf." Only if it is directly across the battery. Otherwise it charges to whatever the steady-state analysis says, which is usually the potential difference across whatever it is in parallel with.

Statements 11.8.B.3.v and 11.8.B.3.vi handle the discharge, and they are the mirror image: the charge and stored energy begin to decrease immediately, and the charge, potential difference and branch current all decrease until a steady state is reached. A discharging capacitor at $t = 0$ behaves like a battery of potential difference $Q_0/C$, which is what sets the initial discharge current.

Statement 11.8.B.3.vii is the licence for the whole approach: after either charging or discharging for times much greater than the time constant, the capacitor and the relevant circuit branch may be modeled using steady-state conditions. "Much greater" is the operative phrase, and the time constant is what "much" is measured against.

Statement 11.8.B.3.ii and 11.8.B.3.iii tie the three time-varying quantities together: the potential difference, the branch current and the stored energy all change with respect to time and all asymptotically approach steady state. Because $U_C = \tfrac{1}{2}Q\Delta V$ and $Q = C\Delta V$, the stored energy goes as $\Delta V^2$, so it approaches its final value more slowly than the charge does.

The time constant, 63 percent and 37 percent

Statement 11.8.B.2 says the time constant is a significant feature of an RC circuit, and 11.8.B.2.i defines it. The equation is printed on the sheet:

Read both subscripts. The CED writes $R_{\text{eq}}$ and $C_{\text{eq}}$, not $R$ and $C$, and in a circuit with more than one resistor or capacitor the values that matter are usually neither of the printed component values. This is the single most reliable way to lose marks in this topic, because getting $\tau$ wrong moves every point on every graph.

Finding $R_{\text{eq}}$: it is the resistance the capacitor sees. Turn off every ideal source, replacing batteries with wires, and compute the resistance between the capacitor's two terminals. In a circuit where a resistor $R_1$ feeds a node shared by $R_2$ and the capacitor, that gives $R_1$ in parallel with $R_2$, which is smaller than either. If you cannot see it, the coefficient of $q$ in a correctly derived differential equation gives $1/\tau$ directly.

Finding $C_{\text{eq}}$: combine the capacitors with the 11.8.A rules first.

Statements 11.8.B.2.ii and 11.8.B.2.iii give the two numbers, and both are about charge:

• Charging: $\tau$ is the time for the charge to rise from zero to approximately 63 percent of its final asymptotic value.
• Discharging: $\tau$ is the time for the charge to fall from fully charged to approximately 37 percent of its initial value.

Those two figures are $1 - e^{-1} = 0.632$ and $e^{-1} = 0.368$, and they add to 1, which is a useful memory hook and also a real statement: what a charging capacitor has gained after one time constant plus what a discharging one has left after one time constant is the whole.

Four more readings that follow and are worth having ready:

• The time constant has units of seconds. An ohm times a farad is a second. Checking that catches a prefix slip immediately.
• After $3\tau$ a circuit is about 95 percent of the way there, and after $5\tau$ about 99 percent. That is the practical content of 11.8.B.3.vii's "much greater than the time constant".
• $\tau$ is the initial slope's extrapolation time. If the charge kept rising at its starting rate, it would reach its final value in exactly one time constant. That is a fast way to sketch the curve correctly: draw the tangent at the origin and mark where it crosses the asymptote.
• Energy does not follow the 63 and 37 percent figures. Stored energy goes as the square of the potential difference and dissipated power as the square of the current, so both have an effective time constant of $\tau/2$.

The time constant glossary entry and the RC circuit entry are the short versions.

Solving it, if you want to, and what the sheet gives you

The framework never asks for the solved function in Unit 11 and never prints one. But the C: E&M sheet's Calculus table prints the three rules that make solving it a four-line job, so it is worth knowing how.

Start from the derived equation and separate variables. Writing $\mathcal{E} = R\,dq/dt + q/C$ as $R\dfrac{dq}{dt} = \mathcal{E} - \dfrac{q}{C}$ and rearranging:

The sheet prints $\int \dfrac{dx}{x+a} = \ln|x+a|$, which is precisely this integral. Integrating from $q = 0$ at $t = 0$, exponentiating, and rearranging gives the charging solutions:

For a capacitor discharging through a resistor there is no source, so the loop rule gives $R\,dq/dt + q/C = 0$ and the same method gives

Check each against the CED's statements rather than trusting the algebra. At $t = \tau$ the charging expression gives $1 - e^{-1} = 0.632$ of the final value, matching 11.8.B.2.ii, and the discharging one gives $e^{-1} = 0.368$ of the initial value, matching 11.8.B.2.iii. At $t = 0$ the charging current is $\mathcal{E}/R$, matching 11.8.B.3.i, and as $t \to \infty$ it goes to zero, matching 11.8.B.3.iv. Every asymptote the framework describes is reproduced.

The current always decays, whether charging or discharging. Only the charge behaves differently between the two cases. That asymmetry is worth holding, because a current-against-time graph looks the same shape either way and the sign is the only difference.

Linearisation is the lab version, and this is where skills 1.B and 3.A meet. Taking the natural logarithm of the discharge equation gives $\ln q = \ln Q_0 - t/\tau$, so a graph of $\ln q$ or $\ln \Delta V$ against $t$ is a straight line whose slope is $-1/\tau$. That is the linearisation the Experimental Design and Analysis free-response question asks for, and it turns a curve you cannot fit by eye into a line you can.

The same machinery reappears in Topic 13.5 with an inductor in place of the capacitor: the loop rule gives another derived differential equation, the time constant becomes $\tau = L/R_{\text{eq}}$, which is also printed on the sheet, and the two limiting instants return with the roles reversed.

The CED's own RC free-response question

Sample free-response Question 4 in the AP Physics C: Electricity and Magnetism course description is the single best guide to what this topic looks like on the exam. It is the Qualitative/Quantitative Translation question, worth 8 points, aligned to learning objectives 10.3.A, 11.1.A, 11.7.A, 11.3.B and 11.8.B, and assessed against skills 2.A, 2.D, 3.B and 3.C.

The circuit is deliberately unfamiliar: a closed switch S, a resistor $R$ and an initially uncharged parallel-plate capacitor $C$, all in parallel with an ideal constant current source generating a constant current $I_{\text{const}}$. Not a battery. At $t = 0$ the switch is opened and the source keeps supplying $I_{\text{const}}$.

Part A, 3 points. A long time after the switch has been opened, the rate of change $d(\Delta V)/dt$ of the potential difference across the capacitor approaches a constant value; indicate whether it is positive, negative or zero, and justify. The credited answer is Zero, with points for indicating that the charge builds up to a maximum amount and no longer changes, and that a constant charge corresponds to a constant potential difference. That is 11.8.B.3.iv restated.

Part B, 3 points. "Derive, but do not solve, a differential equation for the rate of change $d(\Delta V)/dt$ of the potential difference across the capacitor after the switch is opened. Express your answer in terms of $R$, $C$, $\Delta V$, $t$, and physical constants, as appropriate. Begin your derivation by writing a fundamental physics principle or an equation from the reference information." The three points are awarded for, in order: a multi-step derivation starting with a correct application of either Kirchhoff's junction rule or loop rule; substituting the resistor current as $\Delta V / R$ and the capacitor branch current as $dq/dt$; and correctly writing that capacitor branch current as $\dfrac{d(C\Delta V)}{dt}$.

Part C, 2 points. The resistor is removed, leaving a gap, and the process is repeated. Indicate whether $d(\Delta V)/dt$ shortly after opening will be zero, a nonzero constant, or a value with increasing magnitude, and justify by referring to the differential equation you wrote for part B. The credited answer is a nonzero constant value: removing the resistor is the same as letting $R$ go to infinity, which sends the $\Delta V / R$ term to zero and leaves $d(\Delta V)/dt = I_0 / C$.

Three lessons to take from the rubric.

The first line is a principle, not an answer. The stem says so and the rubric pays for it. Writing $\sum I_{\text{in}} = \sum I_{\text{out}}$ or $\sum \Delta V = 0$ for the named loop earns a point even if the algebra afterwards fails.

Reasoning from your own equation is a separate skill and separate points. Part C is answered by taking a limit inside the equation from part B, not by solving anything. The scoring note even says the point for using functional dependence does not require using it correctly.

A constant current source is fair game. Nothing in the framework restricts RC circuits to batteries. With a constant current source and no resistor, the capacitor's potential difference grows linearly forever instead of approaching an asymptote, and that is exactly the contrast part C is testing.

How Topic 11.8 is tested, and who each page is for

Checked against the Table of Information appendix rather than recalled:

Also printed and needed here: $C = Q/\Delta V$, $U_C = \tfrac{1}{2}Q\Delta V$, $C = \kappa\varepsilon_0 A/d$, $I = dq/dt$, $I = \Delta V/R$ and, on the Calculus table, the logarithm and exponential rules. Not printed: the differential equation, any solved exponential, both Kirchhoff rules, the squared power forms and terminal potential difference.

The skills tell you the three shapes to expect. 2.A and 2.D point at the derivation and the limit-taking of sample Question 4. 1.B and 3.A point at the Experimental Design and Analysis question, and the CED's own Topic 11.8 sample activity is precisely that: have students use a known capacitor charged and connected directly to a voltmeter to determine the voltmeter's high internal resistance, by taking voltage-versus-time data as the capacitor discharges through the meter and using the data to find the time constant $RC$, then $R$. Notice what that activity assumes, that a nonideal voltmeter can be treated quantitatively, which the AP Physics 2 boundary statement for Topic 11.5 does not permit. 3.C points at justifying a claim from a graph or from your own equation.

The [AP Physics 2 Topic 11.8 page](/ap-physics-2/unit-11-electric-circuits/11-8-resistor-capacitor-rc-circuits) is for AP Physics 2 students; this page is for AP Physics C students. The two courses share the equivalent-capacitance rules and every statement about the two limiting states and the time constant, so that page covers most of this ground for the algebra-based exam and stops exactly where its boundary statement stops. What is genuinely extra here is one statement and everything it implies: the differential equation, the loop-rule derivation that produces it, and the habit of reasoning by taking limits inside it.

From here the pattern repeats twice more in the course. Topic 13.5 swaps the capacitor for an inductor and gets $\tau = L/R_{\text{eq}}$, and Unit 13 closes with LC circuits, where the sheet prints $\omega_{LC} = 1/\sqrt{LC}$ and the same loop rule produces oscillation instead of decay. The full unit map is on the Unit 11 hub.