Resistor–Capacitor (RC) Circuits

SECTION 1

Historical Context & Motivation

The study of RC circuits sits at the intersection of two foundational discoveries in electrical science: the capacitor and Ohm's law. The Leyden jar, invented in the 1740s, demonstrated that charge could be stored, while Georg Ohm's 1827 treatise established the proportional relationship between voltage and current through a resistor. Combining these ideas into a single circuit loop posed a rich question: what happens when a capacitor charges or discharges through a resistance? The answer—exponential time dependence—became one of the earliest examples of a first-order linear differential equation governing a physical system, linking circuit theory to the broader mathematical language of dynamics.

1745

The Leyden Jar

Pieter van Musschenbroek and Ewald Georg von Kleist independently invent the Leyden jar, the first device capable of storing electrical charge—a precursor to the modern capacitor.

1827

Ohm's Law Published

Georg Simon Ohm publishes 'Die galvanische Kette,' formalizing the linear relationship V = IR and enabling quantitative analysis of resistive elements in circuits.

1845

Kirchhoff's Circuit Laws

Gustav Kirchhoff introduces his voltage and current laws, providing the systematic loop and junction equations needed to analyze multi-element circuits, including RC combinations.

1876

Mica Capacitors in Telegraphy

Practical fixed-value capacitors appear in telegraph systems, and engineers begin characterizing RC time constants to optimize signal transmission along undersea cables.

1940s

RC Timing in Electronics

RC circuits become ubiquitous in radar, radio, and early computers—used as filters, timing elements, and coupling stages—cementing their role as a fundamental building block of modern electronics.

The central question that RC circuit theory answers is deceptively simple: how does the voltage across a capacitor change with time when current must flow through a resistor? Because the capacitor's voltage is proportional to stored charge and the resistor limits current, the resulting feedback loop produces the characteristic exponential approach to a final state—a pattern that recurs throughout physics and engineering whenever a system's rate of change is proportional to the distance from equilibrium.

SECTION 2

Core Principles & Definitions

An RC circuit contains at minimum a resistor (resistance R, measured in ohms) and a capacitor (capacitance C, measured in farads) connected in a loop with a voltage source or switch. The interplay between the resistor's Ohm-law behavior and the capacitor's charge-storage behavior gives rise to time-dependent voltages and currents that are governed by a single parameter: the time constant τ = RC. Understanding these core ideas is essential before we derive the governing equations.

Capacitor Equation

The voltage across a capacitor is V_C = Q/C, where Q is stored charge. Current into the capacitor equals I = dQ/dt, so I = C dV_C/dt.

Ohm's Law for the Resistor

The voltage drop across a resistor carrying current I is V_R = IR. This instantaneous relation constrains how fast charge can flow to or from the capacitor.

Kirchhoff's Voltage Law (KVL)

The sum of voltage gains and drops around any closed loop must equal zero. For an RC loop with EMF ε: ε − IR − Q/C = 0, which becomes the governing differential equation.

Time Constant τ = RC

The product RC has units of seconds (Ω × F = s). After one time constant, a charging capacitor reaches ≈ 63.2 % of its final voltage; a discharging one falls to ≈ 36.8 % of its initial voltage.

Energy Considerations

Energy stored in a capacitor is U = ½CV². During charging from a DC source, exactly half the energy delivered by the battery is dissipated in the resistor regardless of R—a surprising and frequently tested result.

✦ KEY TAKEAWAY

KEY TAKEAWAY

SECTION 3

Visual Explanation — RC Circuit Diagrams

A series RC charging circuit. When switch S closes at t = 0, current I(t) flows from the battery (EMF ε, yellow) through resistor R (violet zigzag) and into capacitor C (cyan plates). Kirchhoff's voltage law around the loop yields the first-order ODE whose solution is the exponential charging curve.

The diagram above shows the canonical series RC circuit used on the AP exam. Notice that when the switch is open, no current flows and the capacitor remains uncharged (assuming initial charge Q₀ = 0). The moment the switch closes, the full battery EMF appears across the resistor because V_C = 0 initially, so I(0) = ε/R—the maximum current. As charge accumulates on the capacitor plates, V_C rises and the voltage remaining across R decreases, slowing the current. This negative feedback is the physical origin of the exponential decay of current during charging.

SECTION 4

Mathematical Framework

Deriving the Charging Equations

Apply Kirchhoff's voltage law around the series loop containing ε, R, and C. With Q(t) representing the charge on the capacitor and I = dQ/dt, the loop equation is ε − IR − Q/C = 0. Substituting I = dQ/dt yields the first-order linear ODE: ε − R(dQ/dt) − Q/C = 0, which rearranges to dQ/dt = (Cε − Q)/(RC). Separating variables and integrating with the initial condition Q(0) = 0 produces the standard charging solutions.

CHARGE (CHARGING)

Q(t) = Cε(1 − e^(−t/RC))

Q(t) = charge on capacitor at time t; C = capacitance; ε = battery EMF; R = resistance; RC = τ (time constant). As t → ∞, Q → Cε = Q_max.

VOLTAGE ACROSS CAPACITOR (CHARGING)

V_C(t) = ε(1 − e^(−t/RC))

Dividing Q(t) by C gives the capacitor voltage. At t = 0, V_C = 0; at t = τ, V_C ≈ 0.632ε.

CURRENT (CHARGING)

I(t) = (ε/R) e^(−t/RC)

Current starts at its maximum value ε/R and decays exponentially. At t = τ, I ≈ 0.368(ε/R).

Deriving the Discharging Equations

For a capacitor initially charged to voltage V₀ and then disconnected from the battery (discharging through R alone), KVL gives Q/C + IR = 0, leading to dQ/dt = −Q/(RC). This is separable: dQ/Q = −dt/(RC). Integrating with Q(0) = CV₀ yields Q(t) = CV₀ e^(−t/RC), so V_C(t) = V₀ e^(−t/RC) and I(t) = −(V₀/R) e^(−t/RC), where the negative sign indicates current flows opposite to the charging direction.

VOLTAGE ACROSS CAPACITOR (DISCHARGING)

V_C(t) = V₀ e^(−t/RC)

The capacitor voltage decays from V₀ toward zero. After 5τ, V_C < 0.7 % of V₀, so the capacitor is considered 'fully discharged.'

AP Exam Tip

SECTION 5

Charging vs. Discharging — Graphical Analysis

Left: During charging, V_C (solid cyan) rises exponentially toward ε while current I (dashed pink) decays from ε/R toward zero. Right: During discharging, both V_C and |I| decay exponentially from their initial values toward zero. At t = τ, each quantity is at 36.8 % of its starting value.

The two graphs illustrate a crucial asymmetry that often appears on the AP exam. During charging, the capacitor voltage and the current behave as complements: V_C = ε(1 − e^(−t/τ)) rises while I = (ε/R)e^(−t/τ) falls, and at every instant their sum satisfies V_C + IR = ε. During discharging, both V_C and |I| share the same exponential decay shape, since V_C = V₀e^(−t/τ) and |I| = (V₀/R)e^(−t/τ). The practical benchmark is that after five time constants (5τ) the transient is effectively complete, since e⁻⁵ ≈ 0.0067.

Key voltage ratios at integer multiples of the time constant

Number of τ	Charging V_C / ε	Discharging V_C / V₀
0	0.000	1.000
1τ	0.632	0.368
2τ	0.865	0.135
3τ	0.950	0.050
5τ	0.993	0.007

SECTION 6

Worked Example — Charging and Energy Dissipation

A 12.0 V battery is connected in series with a 4.70 kΩ resistor and a 22.0 µF capacitor (initially uncharged). The switch closes at t = 0. Determine (a) the time constant, (b) the capacitor voltage at t = 150 ms, (c) the current at t = 150 ms, and (d) the total energy dissipated by the resistor as the capacitor fully charges.

Step 1 — Compute the Time Constantτ = RC = (4.70 × 10³ Ω)(22.0 × 10⁻⁶ F) = 0.1034 s ≈ 103 ms. This sets the timescale: at t = 103 ms the capacitor reaches about 63.2 % of 12.0 V.

τ ≈ 0.103 s

Step 2 — Capacitor Voltage at t = 150 msV_C(t) = ε(1 − e^(−t/τ)) = 12.0(1 − e^(−0.150/0.1034)) = 12.0(1 − e^(−1.451)). Evaluating: e^(−1.451) ≈ 0.2346, so V_C = 12.0 × 0.7654 ≈ 9.18 V.

V_C ≈ 9.18 V

Step 3 — Current at t = 150 msI(t) = (ε/R) e^(−t/τ) = (12.0/4700) e^(−1.451) = (2.553 × 10⁻³)(0.2346) ≈ 5.99 × 10⁻⁴ A. Alternatively, I = (ε − V_C)/R = (12.0 − 9.18)/4700 ≈ 6.00 × 10⁻⁴ A, consistent within rounding.

I ≈ 0.599 mA

Step 4 — Energy Dissipated in the ResistorTotal energy delivered by the battery is U_bat = Qε = Cε² = (22.0 × 10⁻⁶)(12.0)² = 3.168 × 10⁻³ J. Energy stored in the capacitor is U_C = ½Cε² = 1.584 × 10⁻³ J. By conservation of energy, heat dissipated in R equals U_bat − U_C = ½Cε² = 1.584 × 10⁻³ J. This confirms the famous result: exactly half the energy from the battery is dissipated in R, independent of R.

E_dissipated = ½Cε² ≈ 1.58 mJ

SECTION 7

Charging vs. Discharging — Key Differences

Side-by-side comparison of series RC charging and discharging

Property	Charging (ε present)	Discharging (no ε)
V_C(t)	ε(1 − e^(−t/τ)) — rises from 0 to ε	V₀ e^(−t/τ) — falls from V₀ to 0
I(t)	(ε/R) e^(−t/τ) — starts at ε/R, decays to 0	−(V₀/R) e^(−t/τ) — starts at −V₀/R, decays to 0
Initial current magnitude	ε/R (maximum)	V₀/R (maximum)
V_R(t)	ε e^(−t/τ) — decays from ε to 0	V₀ e^(−t/τ) — decays from V₀ to 0
Energy dissipated (total)	½Cε² (always half the battery's output)	½CV₀² (all stored energy becomes heat)
Current direction	Into positive plate of capacitor	Out of positive plate (reversed)

✦ KEY TAKEAWAY

KEY TAKEAWAY

SECTION 8

Connection to RL and RLC Circuits

The RC circuit is a member of a family of circuits governed by linear differential equations. Replacing the capacitor with an inductor produces the RL circuit, and including both gives the RLC circuit. While the RC and RL circuits are both first-order (one energy-storage element), the RLC circuit is second-order and supports oscillatory solutions—analogous to a driven, damped harmonic oscillator. Understanding how the RC time constant maps onto the RL time constant (τ = L/R) and the RLC natural frequency (ω₀ = 1/√(LC)) provides powerful conceptual bridges for the rest of the AP curriculum.

Comparison of RC, RL, and RLC circuits

Feature	RC Circuit	RL Circuit	RLC Circuit
Order of ODE	1st order	1st order	2nd order
Time constant / parameter	τ = RC	τ = L/R	ω₀ = 1/√(LC)
Energy storage	Electric field (½CV²)	Magnetic field (½LI²)	Both fields
Transient behavior	Exponential only	Exponential only	Damped oscillation
Mechanical analog	Dashpot + spring (overdamped)	Dashpot + inertia	Mass-spring-dashpot

On the AP Physics C exam, RC circuit questions often appear in the FRQ section as part of a multi-part problem that asks you to sketch graphs, derive equations from KVL, and compute energy. The same exponential form you master here reappears in RL circuits (Unit 11), so investing effort now in understanding the general solution technique—apply KVL, separate variables, integrate, apply initial conditions—pays dividends throughout the course.

SECTION 9

Practice Problems

A capacitor in a series RC circuit is charging from a battery of EMF ε. Which of the following correctly describes the voltage across the resistor V_R as a function of time?

A 10.0 µF capacitor, initially uncharged, is placed in series with a 50.0 kΩ resistor and a 9.00 V battery. What is the voltage across the capacitor at t = 0.500 s after the switch is closed?

A fully charged 47.0 µF capacitor (initial voltage 24.0 V) is discharged through a 2.00 kΩ resistor. At what time does the capacitor voltage first drop below 3.00 V?

PROBLEM 4 — APPLIED

An RC circuit is used in a cardiac defibrillator. A 64.0 µF capacitor is charged to 5000 V through a 1.00 MΩ resistance, then discharged through the patient's chest (modeled as 500 Ω). (a) How long does it take to charge the capacitor to 95.0 % of 5000 V? (b) What is the peak discharge current through the patient? (c) How much energy is delivered to the patient? Show all work.

PROBLEM 5 — CRITICAL THINKING

A student claims that doubling the resistance in a series RC charging circuit will cause the capacitor to store twice as much energy once fully charged. The battery EMF and capacitance remain fixed. (a) Evaluate this claim and justify your answer quantitatively. (b) Determine how doubling R affects the total energy dissipated by the resistor during the full charging process. (c) On a single set of axes, sketch V_C(t) for the original and doubled resistance, labeling both time constants. Explain how the graphs support your answers.