Resistor–Capacitor (RC) Circuits - AP Physics 2

SECTION 1

Historical Context & Motivation

The study of RC circuits sits at the intersection of two foundational discoveries in electrical science: the capacitor's ability to store charge and the resistor's role in limiting current flow. Long before engineers designed the timing circuits and signal filters that permeate modern electronics, physicists wrestled with the basic question of how electrical charge moves through materials and accumulates on conductors separated by insulators. Understanding this history illuminates why the exponential time dependence of RC circuits is not merely a mathematical curiosity but a direct consequence of the physical laws governing charge, voltage, and current.

1745 • The Leyden Jar

Pieter van Musschenbroek and Ewald Georg von Kleist independently invent the Leyden jar, the first device capable of storing appreciable electric charge — effectively the earliest capacitor.

1827 • Ohm's Law Published

Georg Simon Ohm establishes that current through a conductor is proportional to the voltage across it and inversely proportional to its resistance, giving us V = IR — the governing equation for resistors.

1831 • Faraday's Capacitance Experiments

Michael Faraday systematically studies how different dielectric materials affect the charge-storing capacity of capacitors, laying the groundwork for the definition of capacitance (C = Q/V).

1845 • Kirchhoff's Circuit Laws

Gustav Kirchhoff formalizes the junction rule and loop rule, providing the algebraic framework needed to analyze circuits with both resistors and capacitors in series and parallel.

1890s–1920s • Rise of RC Filter Theory

As telephony and radio communication emerge, engineers such as Oliver Heaviside develop the mathematics of RC time constants to design filters that separate signals of different frequencies — establishing the RC circuit as a cornerstone of electrical engineering.

The central question an RC circuit answers is deceptively simple: if a capacitor is connected through a resistor to a voltage source, how does the charge on the capacitor change with time? The answer — an exponential approach to equilibrium — reveals a deep interplay between the energy stored in the electric field of the capacitor and the energy dissipated as heat in the resistor. Mastering this behavior is essential for the AP Physics 2 exam and for any further work in electronics, signal processing, or biomedical instrumentation.

SECTION 2

Core Principles & Definitions

An RC circuit is any closed loop containing at least one resistor and one capacitor, often connected to a DC voltage source (a battery or power supply) or initially charged and then allowed to discharge. The time-dependent behavior of these circuits emerges from two fundamental relationships: Ohm's law (V_R = IR) and the capacitor voltage–charge relation (V_C = Q/C). When combined with Kirchhoff's loop rule, these produce a first-order differential equation whose solution is an exponential function of time.

Capacitance (C)

A measure of a capacitor's ability to store charge per unit voltage, defined as C = Q/V. The SI unit is the farad (F). Typical capacitors range from picofarads (pF) to millifarads (mF).

Resistance (R)

The opposition a resistor presents to current flow, measured in ohms (Ω). A larger resistance limits the rate at which charge can flow onto or off of the capacitor.

Time Constant (τ = RC)

The product of resistance and capacitance, measured in seconds, that characterizes how quickly the circuit responds. After one time constant, the voltage has changed by about 63% toward its final value.

Kirchhoff's Loop Rule

The algebraic sum of all voltage changes around any closed loop must equal zero. This rule connects the EMF of the battery, the voltage drop across the resistor, and the voltage across the capacitor at every instant.

Exponential Behavior

During charging, the capacitor voltage rises as (1 − e^−t/RC); during discharging, it decays as e^−t/RC. Both processes are asymptotic — the capacitor never quite reaches full charge or fully discharges in finite time.

✦

KEY TAKEAWAY

Think of charging a capacitor through a resistor like filling a water tank through a narrow pipe: at first, water flows quickly because the pressure difference is large, but as the tank fills, the back-pressure reduces the flow rate. Similarly, as V_C rises toward the battery voltage, the driving voltage across the resistor shrinks, slowing the current to an exponential trickle. The time constant τ = RC tells you how wide the pipe is: a large τ means a slow fill.

SECTION 3

Visual Explanation — RC Charging Circuit

A series RC charging circuit with a battery of EMF ε, a switch S that closes at t = 0, a resistor R (purple), and a capacitor C (cyan). The Kirchhoff loop equation ε − IR − Q/C = 0 governs the circuit at every instant. The current arrow shows conventional current direction.

In the diagram above, notice that the three circuit elements — battery, resistor, and capacitor — form a single closed loop. When the switch S closes at t = 0, charge begins to accumulate on the capacitor plates. As charge builds, the voltage across the capacitor (V_C = Q/C) increases, which reduces the net voltage driving current through the resistor. This self-limiting feedback is responsible for the exponential character of the charging process. The current starts at its maximum value I₀ = ε/R (when Q = 0 and the full battery voltage appears across R) and decays asymptotically toward zero as V_C approaches ε.

SECTION 4

Mathematical Framework

The quantitative description of RC circuits follows directly from applying Kirchhoff's loop rule and recognizing that current is the rate of change of charge. Although AP Physics 2 does not require you to solve differential equations, you should understand where the exponential solutions come from and be able to use them fluently. Below are the key equations for both charging and discharging scenarios.

Charging Equations (Battery + R + C)

CHARGE ON CAPACITOR (CHARGING)

Q(t) = Cε(1 − e^(−t/RC))

Q(t) = charge at time t; C = capacitance; ε = battery EMF; R = resistance; τ = RC = time constant (seconds). The maximum charge is Q_max = Cε.

VOLTAGE ACROSS CAPACITOR (CHARGING)

V_C(t) = ε(1 − e^(−t/RC))

Since V_C = Q/C, this follows directly. At t = 0, V_C = 0; as t → ∞, V_C → ε.

CURRENT (CHARGING)

I(t) = (ε/R) e^(−t/RC)

The current starts at I₀ = ε/R and decays exponentially. At t = τ, I has fallen to about 37% of its initial value (since e⁻¹ ≈ 0.368).

Discharging Equations (No Battery, Charged C through R)

CHARGE ON CAPACITOR (DISCHARGING)

Q(t) = Q₀ e^(−t/RC)

Q₀ = initial charge on the capacitor at t = 0. The charge decays toward zero as the capacitor releases its stored energy through the resistor.

VOLTAGE ACROSS CAPACITOR (DISCHARGING)

V_C(t) = V₀ e^(−t/RC)

V₀ = initial voltage = Q₀/C. Both voltage and charge decay with the same exponential factor.

💡

Significance of the Time Constant

After one time constant (t = τ = RC), a charging capacitor has reached ≈ 63.2% of its final voltage, and a discharging capacitor has fallen to ≈ 36.8% of its initial voltage. After five time constants (t = 5τ), the circuit is considered to be effectively at its steady-state value (>99% complete). These benchmarks appear frequently on the AP exam.

SECTION 5

Graphical Analysis — Charging vs. Discharging

The AP Physics 2 exam places significant emphasis on interpreting and sketching graphs of voltage, charge, and current as functions of time for RC circuits. Being able to translate between the mathematical expressions and their graphical representations is a core skill tested in both multiple-choice and free-response questions. The two diagrams below compare the charging and discharging cases side by side, highlighting the exponential growth toward a limit (charging) and exponential decay toward zero (discharging).

Left: Charging — V_C rises exponentially toward ε (cyan curve) while I decays from ε/R toward zero (violet curve). Right: Discharging — V_C decays from V₀ toward zero (pink curve) while I similarly decays (orange curve). In both cases, the dot marks the value at t = τ.

Time Elapsed	Charging: $V_C$ / ε	Discharging: $V_C$ / V₀
t = 0	0%	100%
t = 1τ	63.2%	36.8%
t = 2τ	86.5%	13.5%
t = 3τ	95.0%	5.0%
t = 5τ	99.3%	0.7%

Percentage of final (charging) or initial (discharging) voltage reached at multiples of the time constant τ = RC.

💡

AP Exam Tip

The AP exam often asks you to sketch or identify the correct graph shape. Remember: charging V_C is a rising curve that flattens out (concave down), while discharging V_C is a falling curve that flattens toward zero (concave up). Current always decays exponentially in both cases.

SECTION 6

Worked Example — Charging an RC Circuit

A 12.0 V battery is connected in series with a 5.00 kΩ resistor and a 2.00 μF capacitor. The capacitor is initially uncharged and the switch is closed at t = 0. Find: (a) the time constant, (b) the voltage across the capacitor at t = 15.0 ms, (c) the current at t = 15.0 ms, and (d) the energy stored in the capacitor after a very long time.

Charging an RC Circuit

1
Step 1 — Identify Given Values

ε = 12.0 V, R = 5.00 kΩ = 5.00 × 10³ Ω, C = 2.00 μF = 2.00 × 10⁻⁶ F. The capacitor is initially uncharged, so Q₀ = 0.
2
Step 2 — Calculate the Time Constant τ

τ = RC = (5.00 × 10³ Ω)(2.00 × 10⁻⁶ F) = 1.00 × 10⁻² s = 10.0 ms. This means the circuit's characteristic response time is 10.0 milliseconds.

τ = 10.0 ms
3
Step 3 — Find $V_C$ at t = 15.0 ms

Using V_C(t) = ε(1 − e^−t/τ), we compute t/τ = 15.0 ms / 10.0 ms = 1.50. Therefore V_C = 12.0 V × (1 − e^−1.50) = 12.0 V × (1 − 0.2231) = 12.0 V × 0.7769 = 9.32 V.

$V_C$ = 9.32 V
4
Step 4 — Find the Current at t = 15.0 ms

Using I(t) = (ε/R) e^−t/τ, the initial current is I₀ = 12.0 V / 5.00 × 10³ Ω = 2.40 × 10⁻³ A = 2.40 mA. At t = 15.0 ms: I = 2.40 mA × e^−1.50 = 2.40 mA × 0.2231 = 0.535 mA.

I = 0.535 mA
5
Step 5 — Find Energy Stored at t → ∞

After a very long time (t >> 5τ), the capacitor is fully charged to V_C = ε = 12.0 V. The energy stored in a capacitor is U = ½CV². Therefore U = ½ × (2.00 × 10⁻⁶ F) × (12.0 V)² = ½ × 2.00 × 10⁻⁶ × 144 = 1.44 × 10⁻⁴ J = 0.144 mJ.

U = 1.44 × 10⁻⁴ J = 0.144 mJ
6
Step 6 — Consistency Check

We can verify Step 3 using Kirchhoff's loop rule: V_R = IR = (0.535 × 10⁻³ A)(5.00 × 10³ Ω) = 2.68 V. Then V_R + V_C = 2.68 V + 9.32 V = 12.0 V = ε. ✓ The loop rule is satisfied.

SECTION 7

Charging vs. Discharging — Key Comparisons

While charging and discharging share the same time constant τ = RC and both involve exponential functions, they differ in important ways regarding initial conditions, boundary values, and the direction of current flow. The table below provides a systematic comparison that is useful for avoiding common AP exam mistakes.

Property	Charging (Battery Connected)	Discharging (No Battery)
$V_C$ at t = 0	0 V (capacitor initially uncharged)	V₀ (initial voltage from prior charge)
$V_C$ as t → ∞	ε (battery EMF)	0 V
$V_C$ equation	ε(1 − $e^{−t/RC}$ )	V₀ $e^{−t/RC}$
Current at t = 0	ε / R (maximum)	V₀ / R (maximum)
Current as t → ∞	0 A	0 A
Energy source	Battery supplies energy	Capacitor's stored energy
Energy dissipated in R	½Cε² (equal to energy stored!)	½CV₀² (all stored energy)

Comparison of charging and discharging behavior in a series RC circuit.

✦

KEY TAKEAWAY

A remarkable result of RC charging is that exactly half the energy supplied by the battery is dissipated as heat in the resistor, regardless of the resistance value. The other half is stored in the capacitor's electric field. This is analogous to dragging a box at constant velocity across a rough surface: the work done by you is split between kinetic energy of the box and thermal energy from friction. In the RC circuit, τ determines how quickly the process occurs, but not the 50/50 energy split, which is dictated by the circuit topology alone.

SECTION 8

Connections to Advanced Topics

The RC circuit you study in AP Physics 2 is a first-order linear system — the simplest type of time-dependent circuit. In more advanced physics and engineering courses, this foundation extends naturally into AC analysis, higher-order circuits, and even quantum phenomena. The table below maps your current knowledge to the territory that lies ahead, giving you a sense of how this material connects to broader topics.

AP Physics 2 Concept	Advanced Extension
τ = RC (DC time constant)	Cutoff frequency $f_c$ = 1/(2πRC) for AC low-pass and high-pass filters
Series RC circuit	RLC circuits with oscillatory (underdamped) and overdamped behavior
Exponential decay of current	Impedance and phasor analysis in AC circuits (complex exponentials)
Energy stored: U = ½CV²	Energy exchange between C and inductor L in LC oscillations (electromagnetic analog of a spring-mass system)
Capacitor blocks DC at steady state	Coupling and decoupling capacitors in electronic circuits; high-pass filter behavior

How AP Physics 2 RC concepts map to more advanced circuit theory.

Perhaps the most striking extension is the RLC circuit, in which an inductor is added to the RC loop. The inductor stores energy in a magnetic field, and when combined with the capacitor's electric field energy, the circuit can exhibit oscillatory behavior analogous to a mass on a spring. The RC circuit's purely exponential response is the critically damped or overdamped limit of this more general oscillatory system. Understanding the RC case thoroughly thus prepares you to appreciate the richer dynamics of resonant circuits.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

A capacitor is being charged through a resistor by a battery. At the instant the capacitor voltage equals half the battery EMF, which of the following correctly describes the current in the circuit?

A. The current is zero because the capacitor has reached half its maximum charge.
B. The current is half of its initial value ε/(2R).
C. The current is still at its maximum value ε/R because only half the charge has accumulated.
D. The current is increasing because the capacitor is demanding more charge to reach its maximum.

Show answer (B)

The correct answer is B. By Kirchhoff's loop rule, ε = IR +

V_C

. When

V_C

= ε/2, the voltage across the resistor is

V_R

= ε − ε/2 = ε/2, so I =

V_R

/R = ε/(2R), which is exactly half the initial current. Choice A is wrong because the current is zero only when

V_C

= ε (fully charged). Choice C is wrong because the current decreases continuously from its initial value as

V_C

increases. Choice D is wrong because the current in an RC charging circuit is always decreasing, never increasing.

Question 2: BASIC CALCULATION

A 10.0 kΩ resistor is connected in series with a 4.70 μF capacitor and a 9.00 V battery. What is the time constant of the circuit, and what is the current at t = 0?

A. τ = 47.0 ms; I₀ = 0.900 mA
B. τ = 47.0 ms; I₀ = 9.00 mA
C. τ = 4.70 ms; I₀ = 0.900 mA
D. τ = 470 ms; I₀ = 0.0900 mA

Show answer (A)

The correct answer is A. The time constant τ = RC = (10.0 × 10³ Ω)(4.70 × 10⁻⁶ F) = 4.70 × 10⁻² s = 47.0 ms. At t = 0 the capacitor is uncharged so

V_C

= 0, meaning the full battery voltage appears across the resistor: I₀ = ε/R = 9.00 V / 10.0 × 10³ Ω = 9.00 × 10⁻⁴ A = 0.900 mA. Choice B uses the wrong power of ten for the current (off by a factor of 10). Choice C has the wrong τ — likely a decimal place error in the exponent. Choice D multiplies by an extra factor of 10 in both τ and inverts the current error.

Question 3: INTERMEDIATE

A fully charged 100 μF capacitor (initial voltage 20.0 V) is discharged through a 50.0 kΩ resistor. How long does it take for the voltage across the capacitor to drop to 5.00 V?

A. 3.47 s
B. 5.00 s
C. 6.93 s
D. 10.0 s

Show answer (C)

The correct answer is C. First, find τ = RC = (50.0 × 10³)(100 × 10⁻⁶) = 5.00 s. The discharging equation is

V_C

= V₀

e^{−t/τ}

. Setting

V_C

= 5.00 V and V₀ = 20.0 V: 5.00 = 20.0

e^{−t/5.00}

. Dividing: 0.250 =

e^{−t/5.00}

. Taking the natural log: ln(0.250) = −t/5.00, so −1.386 = −t/5.00, giving t = 6.93 s. This is about 1.39τ. Choice A equals one τ times ln(2), which would correspond to half the voltage (10.0 V), not one-quarter. Choice B is exactly one time constant, at which

V_C

would be 0.368 × 20.0 = 7.36 V, not 5.00 V. Choice D is 2τ, at which

V_C

would be about 2.71 V.

Question 4: APPLIED

5 pts

A biomedical engineer designs a heart-rate monitor that uses an RC circuit to generate a timing pulse. The circuit must reach 90% of its full voltage within 200 ms. If the capacitor available has a capacitance of 22.0 μF, design an experiment to determine the maximum resistance that can be used, and explain what measurements you would need.

Show answer

Point 1 (Setup): Assemble a series circuit with the 22.0 μF capacitor, a variable resistance box (or set of known resistors spanning a range such as 1 kΩ to 10 kΩ), a DC power supply of known voltage ε, and a switch. Connect a voltage probe or digital oscilloscope across the capacitor to monitor

V_C

as a function of time after the switch is closed. Point 2 (Procedure): For each resistance value R, ensure the capacitor is fully discharged, then close the switch and record the time t₉₀ at which

V_C

first reaches 0.90ε. Repeat each trial at least three times and average the results to reduce random error. Use at least five different resistance values across the chosen range. Point 3 (Analysis): From the charging equation

V_C

= ε(1 −

e^{−t/RC}

), setting

V_C

= 0.90ε gives 0.90 = 1 −

e^{−t/RC}

, so

e^{−t/RC}

= 0.10, and therefore t₉₀ = RC × ln(10) ≈ 2.303 × RC. Plot t₉₀ (vertical axis) vs. R (horizontal axis); the relationship is linear with slope equal to C × ln(10). The slope can be used to verify the capacitance value. To find the maximum resistance directly, set t₉₀ = 0.200 s and solve: R_max = t₉₀ / (C × ln(10)) = 0.200 s / (22.0 × 10⁻⁶ F × 2.303) ≈ 3.95 kΩ. Alternatively, read R_max from the graph at t₉₀ = 200 ms. Point 4 (Conclusion with reasoning): The maximum allowable resistance is approximately 3.95 kΩ. Since t₉₀ = 2.303 × RC, increasing R beyond this value increases the time constant τ = RC, which causes the capacitor to take longer than 200 ms to reach 90% of ε, failing the design requirement. Point 5 (Sources of error and refinement): Potential sources of error include: (1) internal resistance of the power supply, which adds to the total series resistance and makes the effective R larger than the nominal resistor value; (2) capacitor tolerance, which for electrolytic capacitors can be ±10–20%, shifting the actual capacitance from the labeled 22.0 μF; and (3) limited time resolution of the voltage-measuring instrument, which introduces uncertainty in t₉₀. To improve accuracy, use a digital oscilloscope with high sampling rate, fit the full

V_C

(t) curve to an exponential function to extract τ directly, and measure the actual capacitance with an LCR meter before the experiment.

Question 5: CRITICAL THINKING

5 pts

Two identical capacitors (each with capacitance C) are connected in series with a single resistor R and a battery of EMF ε. (a) Determine the time constant of this circuit. (b) Find the voltage across each capacitor after a very long time. (c) Now suppose the two capacitors are instead connected in parallel with the same R and ε. Determine the new time constant and compare the total energy stored at steady state in both configurations.

Show answer

Point 1 — Series equivalent capacitance: For two identical capacitors in series, 1/C_eq = 1/C + 1/C = 2/C, so C_eq = C/2. The time constant is τ_series = R × C_eq = RC/2. Point 2 — Voltage across each capacitor (series): At steady state, I = 0 and the full battery voltage ε appears across the series combination. By symmetry (identical capacitors carry identical charge Q), each capacitor has V = Q/C = ε/2. Point 3 — Parallel equivalent capacitance and time constant: For two identical capacitors in parallel, C_eq = C + C = 2C. The time constant is τ_parallel = R × 2C = 2RC. Point 4 — Energy comparison: In the series case, U_series = ½C_eq × ε² = ½(C/2)ε² = Cε²/4. In the parallel case, U_parallel = ½C_eq × ε² = ½(2C)ε² = Cε². Therefore the parallel configuration stores four times as much energy as the series configuration. Point 5 — Physical reasoning: The parallel configuration has a larger equivalent capacitance (and thus a longer time constant), allowing it to accumulate more total charge and store more energy. In series, the reduced equivalent capacitance means less total charge flows from the battery, each capacitor only reaches ε/2, and less energy is stored overall.

SUMMARY

Summary — Resistor–Capacitor (RC) Circuits

Summary

An RC circuit combines a resistor and a capacitor to produce time-dependent voltages and currents governed by exponential functions. The time constant τ = RC sets the characteristic time scale: after one τ, a charging capacitor reaches 63% of the battery voltage, and a discharging capacitor retains only 37% of its initial voltage. After approximately five time constants, the circuit is effectively at its steady-state value.

During charging, V_C = ε(1 − e^−t/RC) and the current decays as I = (ε/R)e^−t/RC. During discharging, both voltage and current decay as pure exponentials toward zero. Kirchhoff's loop rule provides the fundamental constraint at every instant: the sum of voltage changes around the loop is zero. Energy stored in the capacitor is U = ½CV², and during charging, exactly half the battery's energy output is dissipated as heat in the resistor. These principles form the basis for timing circuits, filters, and sensor systems across all branches of electrical engineering.