AP Physics 2 · Topic 11.8

Topic 11.8: Resistor-Capacitor (RC) Circuits

Q: What is the time constant of an RC circuit?

The time constant is a measure of how quickly a capacitor will charge or discharge, and the AP Physics 2 CED defines it in essential knowledge 11.8.B.1.i as the equivalent resistance multiplied by the equivalent capacitance. It is printed on the AP Physics 2 equation sheet, and an ohm times a farad gives a second. One time constant after charging begins, the capacitor's charge has reached approximately 63 percent of its final value; one time constant into a discharge, the charge has fallen to approximately 37 percent of its initial value.

Q: Does an uncharged capacitor act like a wire or like a break in the circuit?

Like a wire, but only at the start. Essential knowledge 11.8.B.2.i says that immediately after being placed in a circuit, an uncharged capacitor acts like a wire, and charge can easily flow to or from its plates. After a long time the opposite holds: 11.8.B.2.iv says a fully charged capacitor reaches a maximum potential difference at which there is zero current in the circuit branch in which it is located, so it behaves like a break. To solve either end, redraw the schematic with the capacitor replaced by a wire for the initial state, or with its branch deleted for the final state.

Q: Do you need the exponential equations for RC circuits on AP Physics 2?

No. The AP Physics 2 boundary statement for Topic 11.8 says that descriptions of charging and discharging RC circuits are limited to qualitative descriptions and representations, and that 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. So numerical work on the instant the switch closes and on the long-time steady state is fair game, and no exponential expression for charge, current, or potential difference as a function of time is required. None is printed on the AP Physics 2 equation sheet either.

Q: How do capacitors in series and parallel combine?

They combine the opposite way to resistors. Capacitors in parallel add directly, so the equivalent capacitance is the sum of the individual capacitances (essential knowledge 11.8.A.1.iii). For capacitors in series the inverses add, so the inverse of the equivalent capacitance is the sum of the inverses of the individual capacitances (11.8.A.1.i). Both formulas are printed on the AP Physics 2 equation sheet. Statement 11.8.A.1.ii gives a free check on series work: the equivalent capacitance of capacitors in series is less than the capacitance of the smallest capacitor in the group.

Q: Why do capacitors in series all carry the same charge?

Because of conservation of charge. Essential knowledge 11.8.A.2 says that as a result of conservation of charge, each of the capacitors in series must have the same magnitude of charge on each plate. The inner plate of one series capacitor, the wire, and the facing plate of the next form an isolated piece of conductor that started neutral and connects to nothing else in the circuit. Whatever charge is pulled off one of those plates has to appear on the other, so the two magnitudes are equal.

Q: What happens to the current in an RC circuit right after the switch closes and long afterward?

Right after the switch closes the current is at its largest, because the uncharged capacitor acts like a wire and offers nothing to push back (11.8.B.2.i). As the capacitor charges, the growing potential difference across it opposes further flow and the current in that branch falls. After a long time the current in the capacitor's branch is zero (11.8.B.2.iv). Charge and potential difference on the capacitor do the reverse: they start at zero and rise, approaching their final values asymptotically rather than reaching them.

Q: Is the RC time constant on the AP Physics 2 equation sheet?

Yes. The equation for the time constant as equivalent resistance times equivalent capacitance is the last entry in the Electricity group of the AP Physics 2 equation sheet, which holds 20 entries out of 129 across the whole sheet. That group also prints the definition of capacitance as charge divided by potential difference, the stored energy as half the charge times the potential difference, and the series and parallel rules for both resistors and capacitors. What it does not print is any expression for how charge or current varies with time in an RC circuit.

Unit 11: Electric Circuits15-18% of the multiple-choice section

In an RC circuit a capacitor charges through a resistor. Right after the switch closes, an uncharged capacitor acts like a wire and the current is largest. Long after, it is fully charged and no current flows in its branch. The time constant, resistance times capacitance, sets the pace.

AP Physics: Unit 11 (topics 11.8 Resistor-Capacitor (RC) Circuits). AP Physics 2 Unit 11, Topic 11.8. Two learning objectives: 11.8.A, describe the equivalent capacitance of multiple capacitors, and 11.8.B, describe the behavior of a circuit containing combinations of resistors and capacitors. Under 11.8.A: capacitors in series take reciprocals (11.8.A.1.i), the series equivalent is less than the smallest capacitance (11.8.A.1.ii), capacitors in parallel add (11.8.A.1.iii), and series capacitors carry equal magnitudes of charge as a result of conservation of charge (11.8.A.2). Under 11.8.B: the time constant is defined as the equivalent resistance times the equivalent capacitance (11.8.B.1.i), one time constant takes a charging capacitor to approximately 63 percent of its final charge (11.8.B.1.ii) and a discharging one down to approximately 37 percent of its initial charge (11.8.B.1.iii), an uncharged capacitor acts like a wire immediately after being placed in a circuit (11.8.B.2.i), a fully charged capacitor reaches a maximum potential difference at which there is zero current in its branch (11.8.B.2.iv), and for times much greater than the time constant the branch may be modeled using steady-state conditions (11.8.B.2.vii). The boundary statement reads in full: 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. Suggested skills 1.B, 1.C, 2.A, 2.D, 3.A, and 3.C. Unit 11 is weighted at 15 to 18 percent of the multiple-choice section and is allotted about 12 to 20 class periods.

What Topic 11.8 requires

Topic 11.8 carries two learning objectives, and they are doing quite different jobs.

11.8.A Describe the equivalent capacitance of multiple capacitors.
11.8.B Describe the behavior of a circuit containing combinations of resistors and capacitors.

Under 11.8.A sit two essential knowledge statements with three sub-statements. Under 11.8.B sit two, with three sub-statements under the first and seven under the second. That count is the shape of the topic: the largest block of it, the seven sub-statements under 11.8.B.2, is a careful description of how a capacitor behaves at the start of a charging process, during it, and long afterward.

The suggested skills are 1.B, create quantitative graphs with appropriate scales and units, including plotting data; 1.C, create qualitative sketches of graphs that represent features of a model or the behavior of a physical system; 2.A, derive a symbolic expression from known quantities by selecting and following a logical mathematical pathway; 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. Two of those six skills are about graphs.

Unit 11 is weighted at 15 to 18 percent of the multiple-choice section and the CED allots the unit about 12 to 20 class periods. Topic 11.8 is the last topic in it, and it answers two of the CED's essential questions for the unit directly: why lights on electronics dim slowly and then go out when unplugged, and how electrical energy can be stored effectively to use later.

What capacitance is, how plate geometry sets it, and how much energy a charged capacitor stores all belong to Topic 10.6. This page assumes those and covers what happens once you wire a capacitor to a resistor.

What the exam will and will not ask (the boundary statement)

Read this before anything else, because it decides how much mathematics you need. The CED's boundary statement for Topic 11.8 reads in full:

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.

Both halves matter and neither survives on its own.

The restriction is on time dependence. You are not expected to produce or manipulate an expression for charge, current, or potential difference as a function of $t$ . No exponential decay formula is required, and none is printed on the AP Physics 2 equation sheet: the sheet has 129 entries in seven groups, its Electricity group holds 20 of them, and the only exponential anywhere on it is the radioactive decay law $N = N_0 e^{-\lambda t}$ in the Modern Physics group.

The permission is easy to miss. The clause students should be able to mathematically describe initial and final states of RC circuits is not a footnote. Numerical work on the instant the switch closes and on the long-time steady state is squarely in scope, and those calculations are ordinary circuit analysis: Ohm's law, the loop rule, the junction rule, and $C = Q/\Delta V$ .

So the safe reading is: compute the two ends, describe the middle. If you have taken a calculus-based course or read a Physics C treatment, the derivative and the exponential you learned there are not wrong, they are simply not what this exam asks for, and time spent on them is time not spent on the graph sketching that skills 1.B and 1.C actually name.

Equivalent capacitance: series and parallel (11.8.A)

Learning objective 11.8.A treats several capacitors as one. Statement 11.8.A.1 says a collection of capacitors in a circuit may be analyzed as though it were a single capacitor with an equivalent capacitance $C_{\text{eq}}$ , and the three sub-statements give the rules.

Statement 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.

\frac{1}{C_{\text{eq,s}}} = \sum_i \frac{1}{C_i}

Statement 11.8.A.1.iii: the equivalent capacitance of a set of capacitors in parallel is the sum of the individual capacitances.

C_{\text{eq,p}} = \sum_i C_i

Both are printed on the equation sheet, in the Electricity group, immediately after the two resistor rules. That adjacency is exactly the trap. The capacitor rules are the resistor rules swapped. Resistors in series add and capacitors in parallel add; resistors in parallel take reciprocals and capacitors in series take reciprocals. Statement 11.8.A.1.ii gives you a check that costs nothing: 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 the smallest capacitor in the group, you used the wrong rule.

If skill 3.C asks why, the picture is geometric: connecting capacitors in parallel is like widening the plates, so the combination stores more charge per volt, while connecting them in series is like adding plate separation, so it stores less. Topic 10.6 owns that geometry.

Statement 11.8.A.2 is the one that makes series problems solvable: as a result of conservation of charge, each of the capacitors in series must have the same magnitude of charge on each plate. The reason is the same one behind the junction rule. The inner plates of two series capacitors, plus the wire joining them, form an isolated island of conductor that started neutral and is connected to nothing else. Charge pulled off one inner plate has to land on the other, so the magnitudes match.

The two states you can compute

The boundary statement licenses initial and final states, so learn them as two separate circuits that happen to share a diagram.

Immediately after the switch closes. Statement 11.8.B.2.i says that 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. Redraw the schematic with the capacitor replaced by a plain wire and solve the resulting resistor circuit. In that instant the potential difference across the capacitor is zero, the charge on its plates is zero, the energy stored is zero, and the current in its branch is at its largest value.

The reason the capacitor is not an obstacle yet is that $C = Q/\Delta V$ : with $Q = 0$ the potential difference across it is zero, and a component with no potential difference across it behaves like a piece of wire. Nothing is pushing back.

After a long time. Statement 11.8.B.2.iv says that 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. Redraw the schematic with the capacitor branch deleted, as an open circuit, and solve the resistor circuit that remains. Then find the potential difference across the capacitor from the circuit. The sheet prints $C = Q/\Delta V$ , which rearranges to $Q = C \Delta V$ , and it prints the stored energy directly as $U_C = \frac{1}{2} Q \Delta V$ .

A warning about the second redraw. Deleting the capacitor branch tells you the currents, but the potential difference across the capacitor is not automatically the source voltage. If there is a resistor in series with the capacitor inside its own branch, that resistor carries zero current at steady state, so it has zero potential difference across it, which means the capacitor takes the full potential difference of whatever it is in parallel with. Working out what the capacitor is in parallel with is the actual question.

Statement 11.8.B.2 covers the whole span: 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. Between the two ends, 11.8.B.2.ii and 11.8.B.2.iii say the charge on the plates, the current in that branch, and the stored energy all change with respect to time and asymptotically approach steady state conditions. Asymptotically is the operative word: nothing arrives, everything approaches.

The time constant, 63 percent and 37 percent

Statement 11.8.B.1 says the time constant $\tau$ is a significant feature of an RC circuit, and 11.8.B.1.i defines it: 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}}

That expression is printed on the AP Physics 2 equation sheet, as the last entry in the Electricity group. Check the units before you trust a number: an ohm times a farad is a second.

The CED then attaches two specific fractions, and they are about charge.

11.8.B.1.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.1.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.

Since $Q = C \Delta V$ and the capacitance does not change, the potential difference across the capacitor tracks the charge, so those percentages apply to $\Delta V$ as well. The current does not follow the same curve as the charge on a charging capacitor: it starts large and falls toward zero while the charge starts at zero and rises.

Skill 2.D, predicting new values using functional dependence, is the natural question here, and the dependence is a product. Double the resistance and the time constant doubles. Double the capacitance and it doubles. Double both and it quadruples. In a single loop with one resistor and one capacitor, changing $R$ does not change the final charge: at steady state that resistor carries no current, so it takes no potential difference, and the capacitor ends at the source voltage either way. Changing $C$ does change the final charge, because $Q = C \Delta V$ .

One honest caution about $R_{\text{eq}}$ . For a single loop with one resistor and one capacitor, $R_{\text{eq}}$ is that resistor and there is nothing to argue about. For a branched circuit, working out which combination of resistors the capacitor charges through is a piece of analysis the boundary statement does not require of you, since it declares the charging and discharging descriptions qualitative. Sketch the trend and say which change makes the circuit slower rather than inventing a number you cannot justify.

Discharging, and when steady state is a fair model

Discharging is the same physics run in reverse, and the CED gives it its own statements rather than leaving it as an exercise.

Statement 11.8.B.2.v says that immediately after a charged capacitor begins discharging, the amount of charge on the capacitor plates and the energy stored in the capacitor begin to decrease. Statement 11.8.B.2.vi says that 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.

Notice what is different from charging. On charging, charge and potential difference rise while the branch current falls. On discharging, all three fall together. That distinction is what a graph-sketching question comes down to, and it is easy to draw the right shape on the wrong axes.

A discharging capacitor is also the answer to the CED's essential question about why lights on electronics dim slowly and then go out when unplugged. Cut the supply and the stored charge keeps driving current through whatever resistance remains, so the light fades rather than stopping instantly, and the fade is slow enough to see when the time constant is a sizable fraction of a second.

Statement 11.8.B.2.vii tells you when you are allowed to stop worrying about time at all: 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. That is the license behind every problem that says after a long time and expects you to treat the capacitor branch as carrying no current. It also means that a question giving you $R$ , $C$ , and an elapsed time can be answered by comparing that time with $\tau$ and picking a state, with no formula to substitute into.

Sketching and reading the graphs (1.B and 1.C)

Two of the six suggested skills for this topic are about graphs, and the unit's Building the Science Practices note lists 1.B among the skills Unit 11 gives repeated practice in, saying those skills will be tested on the free-response section of the AP Physics 2 Exam. A correct sketch needs no time-dependent algebra at all.

For a capacitor charging through a resistor, four curves come up.

Quantity	Value at $t = 0$	Value after a long time	Shape
Charge on the capacitor	0	$Q_{\max} = C \Delta V_{\max}$	Rises, flattening toward a horizontal asymptote
Potential difference across the capacitor	0	$\Delta V_{\max}$	Same shape as the charge, since $Q = C \Delta V$
Current in the capacitor branch	Largest	0	Falls, flattening toward zero
Energy stored in the capacitor	0	$U_C = \frac{1}{2} Q \Delta V$	Rises toward a horizontal asymptote

For discharging, the charge, the potential difference, and the branch current all start at their maxima and fall toward zero (11.8.B.2.vi), so all three have the same falling shape.

Things that make a sketch wrong even when the trend is right:

A straight line. The approach is asymptotic (11.8.B.2.iii), so the curve must bend and flatten, never cross its asymptote, and never turn around.
A curve that touches the asymptote. Draw it approaching and leave a visible gap. The CED's own wording is that a charging capacitor approaches a state of being fully charged.
The current curve drawn as a mirror of the charge curve without a labeled axis. They are different quantities; label them.
No scale on a quantitative graph. Skill 1.B specifically asks for appropriate scales and units, so a plotted-data question wants numbers on both axes, not just a shape.

One useful marker to put on a sketch: at $t = \tau$ the charge on a charging capacitor is about 63 percent of its final value, and on a discharging capacitor about 37 percent of its initial value (11.8.B.1.ii and 11.8.B.1.iii). Two labeled points and a correct curvature make a sketch that reads as understood rather than guessed.

Designing the experiment, and the traps

Skill 3.A, create experimental procedures that are appropriate for a given scientific question, is a suggested skill for this topic, and the CED's Unit 11 overview says the unit provides opportunities for students to practice writing clear concise experimental procedures, as well as creating and using graphs of collected data. An RC question in that style asks you to design a measurement of $\tau$ , or of an unknown $C$ , or to test how $\tau$ depends on $R$ . The usable answer is built from the pieces the CED does give you: measure the potential difference across the capacitor against time with a voltmeter, which 11.5.C.2.i says must be connected in parallel with the element, and which 11.5.C.2.ii says has infinite resistance when ideal so no charge flows through it; identify the time at which the reading reaches about 63 percent of its final value; repeat for several known resistances and plot.

The traps on this topic are mostly confusions between the two states.

Treating a capacitor as a break at $t = 0$ . It is the other way round. Uncharged means it acts like a wire (11.8.B.2.i), and the current is largest then.
Treating a capacitor as a wire after a long time. Also backwards. Fully charged means zero current in that branch (11.8.B.2.iv).
Assuming the capacitor ends up at the source voltage. Only if it is directly across the source. Otherwise find what it is in parallel with once the branch current is zero.
Forgetting that a resistor in series with the capacitor drops nothing at steady state. Zero current means zero potential difference across it, by $I = \Delta V / R$ .
Using the resistor rules on capacitors. Series capacitors take reciprocals; check against 11.8.A.1.ii, that the series equivalent is smaller than the smallest one.
Mixing up which quantities rise and which fall. On charging, charge and potential difference rise while the branch current falls. On discharging, all three fall (11.8.B.2.v and 11.8.B.2.vi).
Reaching for an exponential. The boundary statement says students are not expected to mathematically model these behaviors with respect to time. If your working needs $e$ , you have left the course.

The unit overview also asks for precise vocabulary, naming the differences in meaning between current, potential difference, resistance, resistivity, and capacitance as a place where students inadvertently miscommunicate answers. On this topic that means saying the potential difference across the capacitor and the current in the branch, not the other way round.

Topic 11.8 closes Unit 11. Unit 12 picks up with magnetism, where the CED says students expand their investigations of the similarities and differences between electric and magnetic fields. The unit map for the whole course is on the AP Physics 2 hub.

Equivalent capacitance and the charge on each capacitor

A 6.0 microfarad capacitor is connected in series with a parallel pair made of a 3.0 microfarad capacitor and a 9.0 microfarad capacitor. A steady potential difference of 30 V is maintained across the whole combination. Find the equivalent capacitance, the charge on each capacitor, and the potential difference across each.

Reduce the parallel pair first, using 11.8.A.1.iii, that the equivalent capacitance of capacitors in parallel is the sum: $C_p = 3.0 + 9.0 = 12.0\ \mu\text{F}$ .
Now combine that with the 6.0 microfarad capacitor in series, using 11.8.A.1.i: $\frac{1}{C_{\text{eq,s}}} = \frac{1}{6.0} + \frac{1}{12.0} = \frac{2}{12.0} + \frac{1}{12.0} = \frac{3}{12.0}$ , so $C_{\text{eq,s}} = 4.0\ \mu\text{F}$ .
Check against 11.8.A.1.ii, which says the series equivalent is less than the smallest capacitance in the series group. The series group is 6.0 and 12.0 microfarads, and $4.0 < 6.0$ . The check passes.
Find the total charge from $C = Q/\Delta V$ rearranged: $Q = C_{\text{eq,s}} \Delta V = (4.0\ \mu\text{F})(30\ \text{V}) = 120\ \mu\text{C}$ .
By 11.8.A.2, capacitors in series carry the same magnitude of charge, so the 6.0 microfarad capacitor carries the full $120\ \mu\text{C}$ , and so does the parallel pair taken as one unit. Its potential difference is $\Delta V_1 = Q/C = 120/6.0 = 20\ \text{V}$ , and across the parallel pair $\Delta V_p = 120/12.0 = 10\ \text{V}$ .
Check the loop: $20 + 10 = 30\ \text{V}$ , the applied potential difference, as 11.6.A.3 requires.
Split the parallel pair. Both branches share $10\ \text{V}$ , so $Q_2 = (3.0\ \mu\text{F})(10\ \text{V}) = 30\ \mu\text{C}$ and $Q_3 = (9.0\ \mu\text{F})(10\ \text{V}) = 90\ \mu\text{C}$ . Their sum is $120\ \mu\text{C}$ , matching the charge that arrived, which is the junction-rule bookkeeping in charge form.

$C_{\text{eq}} = 4.0\ \mu\text{F}$ . The 6.0 microfarad capacitor holds $120\ \mu\text{C}$ at $20\ \text{V}$ ; the 3.0 microfarad capacitor holds $30\ \mu\text{C}$ and the 9.0 microfarad capacitor holds $90\ \mu\text{C}$ , both at $10\ \text{V}$ . The two checks are the useful part: the series equivalent came out smaller than the smallest series member (11.8.A.1.ii), and the two charges in the parallel pair added back to the charge that came through the series capacitor (11.8.A.2).

A single-loop RC circuit: both ends and the time constant

An ideal 9.0 V battery, a 200 kilohm resistor, an initially uncharged 10 microfarad capacitor, and an open switch are connected in one series loop. The switch is closed at t = 0. Find (a) the current immediately after closing, (b) the time constant, (c) the final charge, potential difference, and stored energy, and (d) the charge one time constant after closing.

(a) Statement 11.8.B.2.i says an uncharged capacitor acts like a wire immediately after being placed in a circuit, so redraw the loop with the capacitor replaced by a plain wire. The whole 9.0 V then sits across the 200 kilohm resistor: $I_0 = \Delta V / R = 9.0 / (2.0 \times 10^5) = 4.5 \times 10^{-5}\ \text{A}$ , which is $45\ \mu\text{A}$ . At this instant the capacitor's potential difference, charge, and stored energy are all zero.
(b) The time constant comes from 11.8.B.1.i, $\tau = R_{\text{eq}} C_{\text{eq}}$ . This is a single loop with one resistor and one capacitor, so $R_{\text{eq}} = 2.0 \times 10^5\ \Omega$ and $C_{\text{eq}} = 1.0 \times 10^{-5}\ \text{F}$ : $\tau = (2.0 \times 10^5)(1.0 \times 10^{-5}) = 2.0\ \text{s}$ . An ohm times a farad is a second, so the units are right.
(c) After a long time, 11.8.B.2.iv says the current in the capacitor branch is zero. This loop has only one branch, so the current everywhere is zero, and $I = \Delta V/R$ then gives zero potential difference across the resistor. The loop rule leaves the whole 9.0 V across the capacitor: $\Delta V_C = 9.0\ \text{V}$ .
The final charge follows from the sheet's $C = Q/\Delta V$ : $Q_f = (1.0 \times 10^{-5}\ \text{F})(9.0\ \text{V}) = 9.0 \times 10^{-5}\ \text{C}$ , which is $90\ \mu\text{C}$ .
The stored energy comes from $U_C = \frac{1}{2} Q \Delta V = \frac{1}{2}(9.0 \times 10^{-5})(9.0) = 4.05 \times 10^{-4}\ \text{J}$ , which is $0.405\ \text{mJ}$ .
(d) Statement 11.8.B.1.ii says that at one time constant the charge has risen from zero to approximately 63 percent of its final asymptotic value: $0.63 \times 90\ \mu\text{C} \approx 57\ \mu\text{C}$ , at $t = 2.0\ \text{s}$ . Because $Q = C \Delta V$ with fixed $C$ , the capacitor's potential difference at that moment is about $0.63 \times 9.0 = 5.7\ \text{V}$ . No exponential was needed, and the boundary statement says none is expected.

(a) $45\ \mu\text{A}$ . (b) $\tau = 2.0\ \text{s}$ . (c) $Q_f = 90\ \mu\text{C}$ at $\Delta V_C = 9.0\ \text{V}$ , storing $4.05 \times 10^{-4}\ \text{J}$ . (d) About $57\ \mu\text{C}$ , roughly $5.7\ \text{V}$ , at $t = 2.0\ \text{s}$ . Every number came from the two end states plus the CED's 63 percent figure, which is exactly the scope the boundary statement sets: mathematically describe the initial and final states, describe the journey between them qualitatively.

A branched RC circuit: what changes between the two ends

An ideal 24 V battery is connected in series with a 200 ohm resistor. That resistor feeds a node where a 400 ohm resistor and a 30 microfarad capacitor are connected in parallel with each other, and their far ends return to the battery. The capacitor is initially uncharged and the switch closes at t = 0. Find the current from the battery and the capacitor's charge and stored energy (a) immediately after closing and (b) after a long time. (c) Say what would change if the capacitance were doubled.

(a) Replace the uncharged capacitor with a wire (11.8.B.2.i). That wire connects the two ends of the 400 ohm resistor, so the 400 ohm resistor has zero potential difference across it and, by $I = \Delta V/R$ , carries no current. The battery therefore sees only the 200 ohm resistor: $I_{\text{batt}} = 24 / 200 = 0.12\ \text{A}$ .
Apply the junction rule at that node to see where the current goes: $0.12\ \text{A}$ in, $0$ through the 400 ohm resistor, so $0.12\ \text{A}$ into the capacitor branch. The capacitor's potential difference, charge, and stored energy are all zero at this instant, even though the current into it is at its largest.
(b) After a long time, 11.8.B.2.iv says there is zero current in the branch containing the capacitor. Delete that branch and the circuit is the 200 ohm and 400 ohm resistors in series: $I_{\text{batt}} = 24 / (200 + 400) = 0.040\ \text{A}$ .
The capacitor is in parallel with the 400 ohm resistor, so it takes that resistor's potential difference: $\Delta V_C = (0.040\ \text{A})(400\ \Omega) = 16\ \text{V}$ . Check the loop: the 200 ohm resistor drops $(0.040)(200) = 8.0\ \text{V}$ , and $8.0 + 16 = 24\ \text{V}$ .
Charge and energy: $Q = C \Delta V = (30 \times 10^{-6})(16) = 4.8 \times 10^{-4}\ \text{C} = 480\ \mu\text{C}$ , and $U_C = \frac{1}{2} Q \Delta V = \frac{1}{2}(4.8 \times 10^{-4})(16) = 3.84 \times 10^{-3}\ \text{J}$ , about $3.8\ \text{mJ}$ .
(c) Doubling $C$ to 60 microfarads changes nothing about either current, because the capacitor's role at both ends is set by whether its branch conducts, not by how much it holds. The steady-state potential difference stays at $16\ \text{V}$ , so $Q$ doubles to $960\ \mu\text{C}$ and $U_C = \frac{1}{2}Q\Delta V$ doubles to about $7.7\ \text{mJ}$ . The circuit also takes longer to get there, since 11.8.B.1.i makes the time constant proportional to $C_{\text{eq}}$ , but the boundary statement keeps that half of the answer qualitative: slower, with no number attached.

(a) $0.12\ \text{A}$ from the battery, all of it into the capacitor branch, with $Q = 0$ and $U_C = 0$ . (b) $0.040\ \text{A}$ from the battery, $\Delta V_C = 16\ \text{V}$ , $Q = 480\ \mu\text{C}$ , $U_C \approx 3.8\ \text{mJ}$ . (c) Doubling the capacitance leaves both currents and the 16 V unchanged, doubles the stored charge and the stored energy, and makes the charging slower. Notice that the capacitor did not end up at 24 V: it ended up at the potential difference of the resistor it is in parallel with, which is the step that is easiest to skip.

Frequently asked questions

What is the time constant of an RC circuit?

The time constant is a measure of how quickly a capacitor will charge or discharge, and the AP Physics 2 CED defines it in essential knowledge 11.8.B.1.i as the equivalent resistance multiplied by the equivalent capacitance. It is printed on the AP Physics 2 equation sheet, and an ohm times a farad gives a second. One time constant after charging begins, the capacitor's charge has reached approximately 63 percent of its final value; one time constant into a discharge, the charge has fallen to approximately 37 percent of its initial value.

Does an uncharged capacitor act like a wire or like a break in the circuit?

Like a wire, but only at the start. Essential knowledge 11.8.B.2.i says that immediately after being placed in a circuit, an uncharged capacitor acts like a wire, and charge can easily flow to or from its plates. After a long time the opposite holds: 11.8.B.2.iv says a fully charged capacitor reaches a maximum potential difference at which there is zero current in the circuit branch in which it is located, so it behaves like a break. To solve either end, redraw the schematic with the capacitor replaced by a wire for the initial state, or with its branch deleted for the final state.

Do you need the exponential equations for RC circuits on AP Physics 2?

No. The AP Physics 2 boundary statement for Topic 11.8 says that descriptions of charging and discharging RC circuits are limited to qualitative descriptions and representations, and that 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. So numerical work on the instant the switch closes and on the long-time steady state is fair game, and no exponential expression for charge, current, or potential difference as a function of time is required. None is printed on the AP Physics 2 equation sheet either.

How do capacitors in series and parallel combine?

They combine the opposite way to resistors. Capacitors in parallel add directly, so the equivalent capacitance is the sum of the individual capacitances (essential knowledge 11.8.A.1.iii). For capacitors in series the inverses add, so the inverse of the equivalent capacitance is the sum of the inverses of the individual capacitances (11.8.A.1.i). Both formulas are printed on the AP Physics 2 equation sheet. Statement 11.8.A.1.ii gives a free check on series work: the equivalent capacitance of capacitors in series is less than the capacitance of the smallest capacitor in the group.

Why do capacitors in series all carry the same charge?

Because of conservation of charge. Essential knowledge 11.8.A.2 says that as a result of conservation of charge, each of the capacitors in series must have the same magnitude of charge on each plate. The inner plate of one series capacitor, the wire, and the facing plate of the next form an isolated piece of conductor that started neutral and connects to nothing else in the circuit. Whatever charge is pulled off one of those plates has to appear on the other, so the two magnitudes are equal.

What happens to the current in an RC circuit right after the switch closes and long afterward?

Right after the switch closes the current is at its largest, because the uncharged capacitor acts like a wire and offers nothing to push back (11.8.B.2.i). As the capacitor charges, the growing potential difference across it opposes further flow and the current in that branch falls. After a long time the current in the capacitor's branch is zero (11.8.B.2.iv). Charge and potential difference on the capacitor do the reverse: they start at zero and rise, approaching their final values asymptotically rather than reaching them.

Is the RC time constant on the AP Physics 2 equation sheet?

Yes. The equation for the time constant as equivalent resistance times equivalent capacitance is the last entry in the Electricity group of the AP Physics 2 equation sheet, which holds 20 entries out of 129 across the whole sheet. That group also prints the definition of capacitance as charge divided by potential difference, the stored energy as half the charge times the potential difference, and the series and parallel rules for both resistors and capacitors. What it does not print is any expression for how charge or current varies with time in an RC circuit.