Capacitor Charge and Discharge

The time-dependent behaviour of capacitors charging and discharging through resistors is one of the most important practical and theoretical topics in AQA A-Level Physics (specification 3.7.4). The exponential functions governing these processes appear frequently in exam questions.

Spec mapping (AQA 7408 §3.7.4 — Capacitor charge and discharge): This lesson covers the derivation and application of the exponential decay equations Q = Q₀e^(−t/RC), V = V₀e^(−t/RC) and I = I₀e^(−t/RC), the charging equations Q = Q₀(1 − e^(−t/RC)) etc., the time constant τ = RC and its role as the e-folding time, the half-life relation t½ = RC ln 2 ≈ 0.693RC, the graphical method of extracting RC from a ln V vs t plot (linearisation), and the energy dissipated during discharge equal to the initial stored energy. The Required Practical 9 (investigating capacitor discharge) is treated in full depth in Lesson 7. Refer to the official AQA 7408 specification document.

Synoptic links — where this resurfaces in 7408:

• §3.7.4 Capacitance (Lesson 4) — the C = Q/V relationship is the algebraic engine of every charging/discharging step.
• §3.5 Electricity — Kirchhoff's voltage law and Ohm's law are the underpinning of the RC differential equation; familiarity with V_R = IR and series circuits is assumed.
• §3.8.1.2 Radioactive decay — the same mathematical form N = N₀e^(−λt) governs nuclear decay; the parallel between the time constant τ = RC and decay constant λ is examined directly in synoptic Paper 3 questions. Half-life relations t½ = ln2 × τ for capacitors and t½ = ln2 / λ for nuclei are structurally identical.
• §3.6.1 Simple harmonic motion — RC circuits are first-order; LC oscillators are second-order and give SHM-type sinusoidal exchange between electric and magnetic energy (extension). The mathematical machinery of exponential decay vs sinusoidal oscillation is the synoptic stepping-stone.
• §3.7.5 Magnetic fields — inductor-resistor circuits (RL circuits) display exponential current growth/decay with τ = L/R, mathematically identical to RC.
• §3.4.1.7 Energy / §3.5.1.6 Power — energy dissipated in R during discharge ∫I²R dt equals the initial ½CV₀² stored — a direct application of energy conservation.

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The RC Circuit

An RC circuit consists of a resistor (R) and a capacitor (C) connected in series. The product RC is called the time constant and determines how quickly the capacitor charges or discharges.

τ = RC

where τ is the time constant (s), R is the resistance (Ω), and C is the capacitance (F).

Dimensional check: [Ω][F] = [V A⁻¹][C V⁻¹] = [C A⁻¹] = [A s A⁻¹] = [s] ✓

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Discharging a Capacitor

When a charged capacitor (initial charge Q₀, initial voltage V₀) is disconnected from the supply and connected across a resistor R, it discharges exponentially.

Derivation of the Discharge Equation

At any instant during discharge, Kirchhoff's voltage law gives:

V_C = IR → Q/C = IR

Since I = −dQ/dt (the negative sign indicates that Q is decreasing):

Q/C = −R(dQ/dt)

Separating variables: dQ/Q = −dt/(RC)

Integrating: ln Q = −t/(RC) + constant

At t = 0, Q = Q₀: constant = ln Q₀

ln(Q/Q₀) = −t/(RC)

Q = Q₀ e^(−t/RC)

Since V = Q/C: V = V₀ e^(−t/RC)

Since I = V/R: I = I₀ e^(−t/RC) where I₀ = V₀/R

All three quantities — charge, voltage, and current — decay exponentially with the same time constant τ = RC.

Meaning of the Time Constant

After one time constant (t = RC):

• Q = Q₀ e⁻¹ = Q₀ × 0.368 ≈ 37% of Q₀
• V = V₀ × 0.368
• I = I₀ × 0.368

After 5 time constants (t = 5RC):

• Q = Q₀ e⁻⁵ = Q₀ × 0.0067 ≈ 0.67% of Q₀

The capacitor is considered fully discharged after about 5RC.

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Charging a Capacitor

When an uncharged capacitor is connected to a supply of EMF ε through a resistor R, it charges exponentially.

Charging Equations

Q = Q₀(1 − e^(−t/RC)) where Q₀ = Cε V = V₀(1 − e^(−t/RC)) where V₀ = ε I = I₀ e^(−t/RC) where I₀ = ε/R

During charging:

• Charge and voltage rise exponentially towards their maximum values.
• Current decreases exponentially from its initial maximum value I₀ = ε/R (when the capacitor is uncharged and the full EMF appears across R) to zero (when the capacitor is fully charged and V_C = ε).

After one time constant during charging:

• Q = Q₀(1 − e⁻¹) = Q₀ × 0.632 ≈ 63% of Q₀
• V = 63% of V₀

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Graphical Representations

Described diagram — Discharge graphs: Three graphs are shown. (1) Q-t: Curve starts at Q₀ and decays exponentially towards zero. At t = RC, Q = 0.37Q₀. (2) V-t: Same shape as Q-t, starting at V₀ and decaying to zero. (3) I-t: Starts at I₀ = V₀/R and decays to zero. For discharge, the current flows in one direction only.

Described diagram — Charging graphs: Three graphs. (1) Q-t: Starts at zero and rises towards Q₀ = Cε, following Q = Q₀(1 − e^(−t/RC)). At t = RC, Q ≈ 0.63Q₀. (2) V-t: Same shape, rising from 0 to ε. (3) I-t: Starts at I₀ = ε/R and decays exponentially to zero — identical in shape to the discharge current graph.

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Worked Examples

Worked Example 1 — Discharge Calculation

Question: A 100 μF capacitor is charged to 12 V and then discharged through a 47 kΩ resistor. Calculate: (a) the time constant; (b) the initial discharge current; (c) the voltage after 8.0 s; (d) the time for the voltage to fall to 1.0 V.

Solution:

(a) τ = RC = 47 × 10³ × 100 × 10⁻⁶ = 4.7 s

(b) I₀ = V₀/R = 12 / (47 × 10³) = 2.55 × 10⁻⁴ A = 0.255 mA

(c) V = V₀ e^(−t/RC) = 12 × e^(−8.0/4.7) = 12 × e^(−1.702) = 12 × 0.1822 = 2.19 V

(d) V = V₀ e^(−t/RC) 1.0 = 12 × e^(−t/4.7) e^(−t/4.7) = 1.0/12 = 0.08333 −t/4.7 = ln(0.08333) = −2.485 t = 4.7 × 2.485 = 11.7 s

Worked Example 2 — Charging Calculation

Question: An uncharged 220 μF capacitor is charged through a 10 kΩ resistor from a 9.0 V supply. Calculate: (a) the time constant; (b) the voltage across the capacitor after 3.0 s; (c) the charge on the capacitor after 3.0 s.

Solution:

(a) τ = RC = 10 × 10³ × 220 × 10⁻⁶ = 2.2 s

(b) V = V₀(1 − e^(−t/RC)) = 9.0 × (1 − e^(−3.0/2.2)) V = 9.0 × (1 − e^(−1.364)) V = 9.0 × (1 − 0.2557) V = 9.0 × 0.7443 = 6.70 V

(c) Q = CV = 220 × 10⁻⁶ × 6.70 = 1.47 × 10⁻³ C = 1.47 mC

Alternatively: Q = Q₀(1 − e^(−t/RC)) = 220 × 10⁻⁶ × 9.0 × (1 − e^(−1.364)) Q = 1.98 × 10⁻³ × 0.7443 = 1.47 × 10⁻³ C ✓

Worked Example 3 — Determining C from a Discharge Experiment

Question: A capacitor is discharged through a 33 kΩ resistor. The voltage is recorded at regular intervals:

Time (s)	Voltage (V)	ln(V)
0	10.0	2.303
2	7.4	2.001
4	5.5	1.705
6	4.1	1.411
8	3.0	1.099
10	2.2	0.788

Use a ln(V)–t graph to determine the capacitance.

Solution:

Since V = V₀ e^(−t/RC), taking natural logs: ln V = ln V₀ − t/(RC)

This is a straight line of the form y = mx + c, where:

• gradient m = −1/(RC)
• y-intercept = ln V₀

From the data, the gradient is: m = (0.788 − 2.303) / (10 − 0) = −1.515 / 10 = −0.1515 s⁻¹

So: −1/(RC) = −0.1515 RC = 1/0.1515 = 6.60 s C = 6.60 / R = 6.60 / (33 × 10³) = 2.0 × 10⁻⁴ F = 200 μF

Exam Tip: Plotting ln(V) against t gives a straight line with gradient −1/(RC). This is the preferred method for finding RC (and hence C if R is known) because it uses all the data points and is more accurate than reading a single time constant value off a V-t curve. Always draw the best-fit straight line through all points and calculate the gradient using two points that lie on the line (not necessarily data points).

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Half-Life of Discharge

The half-life (t½) is the time taken for the charge (or voltage or current) to fall to half its original value during discharge.

From Q = Q₀ e^(−t/RC): Q₀/2 = Q₀ e^(−t½/RC) ½ = e^(−t½/RC) ln(½) = −t½/(RC) −0.693 = −t½/(RC)

t½ = 0.693 × RC = 0.693τ

Worked Example 4 — Half-Life

Question: A 470 μF capacitor is charged to 20 V and discharged through a 15 kΩ resistor. Calculate the half-life and the voltage after two half-lives.

Solution:

τ = RC = 15 × 10³ × 470 × 10⁻⁶ = 7.05 s

t½ = 0.693 × 7.05 = 4.88 s

After two half-lives, the voltage has halved twice: V = 20 × ½ × ½ = 5.0 V

Check: V = 20 × e^(−2 × 4.88/7.05) = 20 × e^(−1.384) = 20 × 0.2505 = 5.01 V ✓

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Energy Dissipated During Discharge

During discharge through a resistor, all the energy initially stored in the capacitor is dissipated as heat in the resistor:

Energy dissipated = initial energy stored = ½CV₀² = ½Q₀V₀ = Q₀²/(2C)

This is true regardless of the resistance R. A larger R slows the discharge but the same total energy is dissipated.

Worked Example 5 — Designing an RC Timing Circuit

Question: A windscreen-wiper delay circuit is required to switch a relay 8.0 s after a button is pressed. The relay activates when the voltage across a capacitor falls to 1.0 V, having started at 6.0 V. Calculate the value of the resistor needed if a 220 μF capacitor is used.

Solution:

Discharge equation: V = V₀ e^(−t/RC)

Substituting the target conditions: 1.0 = 6.0 × e^(−8.0/(R × 220 × 10⁻⁶))

e^(−8.0/(RC)) = 1.0/6.0 = 0.1667

−8.0/(RC) = ln(0.1667) = −1.792

RC = 8.0/1.792 = 4.464 s

R = 4.464 / (220 × 10⁻⁶) = 2.03 × 10⁴ Ω ≈ 20 kΩ

In practice, the designer would specify a 22 kΩ resistor (a standard E12 value), giving τ = 22 × 10³ × 220 × 10⁻⁶ = 4.84 s and a delay to reach 1.0 V of t = 4.84 × ln(6.0) = 4.84 × 1.79 = 8.67 s — within ~10% of the target, acceptable for a non-critical timing application.

Worked Example 6 — Energy Dissipated and Peak Current During Discharge

Question: A 4700 μF capacitor charged to 50 V is discharged through a 100 Ω resistor. Calculate (a) the energy initially stored; (b) the peak (initial) current during discharge; (c) the time for the voltage to drop to 10% of its initial value; (d) confirm by integration that all the stored energy is dissipated in the resistor.

Solution:

(a) U₀ = ½CV² = ½ × 4700 × 10⁻⁶ × 50² = ½ × 4700 × 10⁻⁶ × 2500 = 5.875 J ≈ 5.9 J

(b) I_peak = V₀/R = 50/100 = 0.50 A (occurs at t = 0)

(c) V/V₀ = 0.10 → e^(−t/RC) = 0.10 → t = −RC × ln(0.10) = RC × 2.303 τ = RC = 100 × 4700 × 10⁻⁶ = 0.47 s t = 0.47 × 2.303 = 1.08 s

(d) Energy dissipated = ∫₀^∞ I²R dt = ∫₀^∞ (I₀ e^(−t/RC))² R dt = I₀² R ∫₀^∞ e^(−2t/RC) dt = I₀² R × (RC/2) = (V₀/R)² × R × (RC/2) = V₀²C/2 = ½CV₀² ✓

The integrated dissipation exactly equals the initially stored energy 5.875 J, regardless of R. A smaller R would have given a larger peak current and a faster decay (smaller τ), but the integrated energy dissipated stays the same. This is the formal expression of energy conservation in the RC discharge.

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Specimen Exam Question (9-mark)

Specimen question modelled on the AQA 7408 Paper 2 / Required Practical 9 format.