Charging and Discharging Capacitors

When a capacitor charges or discharges through a resistor, the process follows exponential curves. Understanding these curves — and the mathematics behind them — is essential for the Edexcel A-Level specification and has wide practical applications in timing circuits, sensors, and signal processing.

Discharging a Capacitor

Consider a capacitor of capacitance C, initially charged to voltage V₀ (storing charge Q₀ = CV₀), connected across a resistor R. When the switch is closed, charge flows through the resistor and the capacitor discharges.

The charge, voltage, and current all decay exponentially:

Q = Q₀e^(−t/RC)

V = V₀e^(−t/RC)

I = I₀e^(−t/RC)

where I₀ = V₀/R is the initial current.

Why Exponential?

At any instant, the current through the resistor is I = V/R, and V = Q/C. As charge flows off the plates, Q decreases, so V decreases, so I decreases. The rate of discharge is proportional to the amount remaining — this is the defining property of exponential decay.

Mathematically: dQ/dt = −Q/RC, which has the solution Q = Q₀e^(−t/RC).

This is the same mathematical structure as radioactive decay (dN/dt = −λN), which is why capacitor discharge and radioactive decay share many features: exponential decay, a constant half-life, and the same ln graph method for analysis.

The Time Constant

The time constant τ (tau) is defined as:

τ = RC

where R is in ohms (Ω) and C is in farads (F), giving τ in seconds (s).

After one time constant:

Q = Q₀e^(−1) = Q₀ × 0.368

So after one time constant, the charge (and voltage and current) has fallen to 36.8% of its initial value — equivalently, 63.2% of the initial value has been lost.

Useful time constant benchmarks during discharge:

Time elapsed	Fraction remaining	Percentage remaining
1τ	e⁻¹ = 0.368	36.8%
2τ	e⁻² = 0.135	13.5%
3τ	e⁻³ = 0.050	5.0%
4τ	e⁻⁴ = 0.018	1.8%
5τ	e⁻⁵ = 0.007	0.7%

After 5τ, the capacitor is considered effectively fully discharged (less than 1% remaining).

Worked Example 1

A 2200 μF capacitor discharges through a 10 kΩ resistor. What is the time constant?

τ = RC = 10 000 × 2200 × 10⁻⁶ = 22 s

After 22 s, the voltage will have fallen to 36.8% of its initial value. After 5τ = 110 s, the capacitor is effectively fully discharged.

Worked Example 2: Finding Voltage at a Specific Time

A 500 μF capacitor is charged to 12 V and then discharged through a 20 kΩ resistor. Calculate the voltage after 15 s.

τ = RC = 20 000 × 500 × 10⁻⁶ = 10 s

V = V₀e^(−t/RC) = 12 × e^(−15/10) = 12 × e^(−1.5)

e^(−1.5) = 0.2231

V = 12 × 0.2231 = 2.7 V

Worked Example 3: Finding Time to Reach a Specific Voltage

A 1000 μF capacitor charged to 9.0 V discharges through a 47 kΩ resistor. How long does it take for the voltage to fall to 2.0 V?

τ = RC = 47 000 × 1000 × 10⁻⁶ = 47 s

V = V₀e^(−t/RC)

2.0 = 9.0 × e^(−t/47)

e^(−t/47) = 2.0/9.0 = 0.222

−t/47 = ln(0.222) = −1.506

t = 1.506 × 47 = 70.8 s ≈ 71 s

Half-Life of Discharge

The half-life (t½) is the time taken for the charge (or voltage or current) to fall to half its initial value:

Setting Q₀/2 = Q₀e^(−t½/RC):

1/2 = e^(−t½/RC)

ln(1/2) = −t½/RC

t½ = RC × ln 2 = 0.693 × RC = 0.693τ

Worked Example 4

For a circuit with τ = 22 s:

t½ = 0.693 × 22 = 15.2 s

After 15.2 s, the voltage has halved. After another 15.2 s (total 30.4 s), it has halved again to one quarter of its original value.

Charging a Capacitor

When an uncharged capacitor is connected to a DC supply of EMF ε through a resistor R, the capacitor charges up. The charging curves are:

Q = Q₀(1 − e^(−t/RC)) where Q₀ = Cε

V = V₀(1 − e^(−t/RC)) where V₀ = ε

I = I₀e^(−t/RC) where I₀ = ε/R

Note that:

Charge and voltage follow (1 − exponential) curves — they increase from zero towards their maximum values
Current still follows a simple exponential decay — it starts at its maximum value (I₀ = ε/R) and decreases to zero

Why Does the Current Decay During Charging?

Initially, the capacitor has zero voltage across it, so the full EMF drives current through the resistor: I₀ = ε/R. As the capacitor charges, its voltage increases, opposing the supply EMF. The net driving voltage (ε − V_C) decreases, so the current decreases. When V_C = ε, the current is zero and the capacitor is fully charged.

Worked Example 5: Charging Calculation

An uncharged 220 μF capacitor is charged through a 100 kΩ resistor from a 9.0 V supply. Calculate: (a) the time constant (b) the voltage across the capacitor after 30 s (c) the initial charging current (d) the current after 30 s

(a) τ = RC = 100 000 × 220 × 10⁻⁶ = 22 s

(b) V = 9.0 × (1 − e^(−30/22)) = 9.0 × (1 − e^(−1.364)) = 9.0 × (1 − 0.256) = 9.0 × 0.744 = 6.7 V

(d) I = 90 × 10⁻⁶ × e^(−30/22) = 90 × 10⁻⁶ × 0.256 = 23 μA

Logarithmic Graphs for Determining RC

Taking the natural logarithm of the discharge equation:

Q = Q₀e^(−t/RC)

ln Q = ln Q₀ − t/RC

This is in the form y = mx + c, where:

y = ln Q
x = t
m = −1/RC (gradient)
c = ln Q₀ (y-intercept)

A graph of ln Q against t (or ln V against t, or ln I against t) gives a straight line with gradient −1/RC.

This is the standard method for determining RC experimentally:

Measure V (or Q or I) at regular time intervals during discharge
Plot ln V against t
Draw the line of best fit
The gradient = −1/RC, so RC = −1/gradient

Worked Example 6

A student measures the voltage across a discharging capacitor and plots ln V against t. The straight line has a gradient of −0.045 s⁻¹. What is the time constant?

Gradient = −1/RC = −0.045

RC = 1/0.045 = 22.2 s

If the capacitance is known to be 470 μF, the resistance can be found:

R = RC/C = 22.2 / (470 × 10⁻⁶) = 47 200 Ω ≈ 47 kΩ

Shape of the Curves — Summary

Quantity	Discharging	Charging
Charge Q	Exponential decay from Q₀ to 0	Rises from 0 towards Q₀
Voltage V	Exponential decay from V₀ to 0	Rises from 0 towards ε
Current I	Exponential decay from I₀ to 0	Exponential decay from I₀ to 0
Current direction	Flows one way	Flows the opposite way

Current always decays exponentially, whether charging or discharging. During charging, the initial current direction is opposite to the discharge current direction.

Real-World Applications

Timing circuits: The time constant RC controls how quickly a capacitor charges or discharges. By choosing R and C, engineers can create precise time delays. A 555 timer IC uses RC circuits to generate square waves with a specific frequency.

Defibrillators: A capacitor (typically ~30 μF) is charged to about 5000 V, storing around 375 J. This is discharged through the patient’s chest in a few milliseconds to restart the heart. The doctor charges the capacitor (takes a few seconds) and then triggers a rapid discharge.

Smoothing in power supplies: After rectifying AC to DC, the output is a series of pulses. A large capacitor charges during the peaks and discharges slowly during the troughs, smoothing the output to near-constant DC. A larger time constant (larger RC) gives smoother output.

Windscreen wiper delay: An RC circuit controls the pause between wiper sweeps. Changing R (via a dial) changes the time constant and thus the delay.

Common Exam Mistakes

Using the wrong equation for charging vs discharging: Discharge uses Q = Q₀e^(−t/RC). Charging uses Q = Q₀(1 − e^(−t/RC)). The current always decays exponentially in both cases.
Forgetting to convert units: R must be in Ω and C in F. A common error is using kΩ or μF without converting.
Confusing time constant with half-life: τ = RC (falls to 36.8%), t½ = 0.693RC (falls to 50%).
Reading the ln graph incorrectly: The gradient is −1/RC, so RC = −1/gradient. Do not forget the negative sign.

Key Points

Discharge: Q = Q₀e^(−t/RC), V = V₀e^(−t/RC), I = I₀e^(−t/RC)
Charging: Q = Q₀(1 − e^(−t/RC)), V = V₀(1 − e^(−t/RC)), I = I₀e^(−t/RC)
Time constant τ = RC; after 1τ, the value has fallen to 36.8% (or risen to 63.2%)
Half-life: t½ = 0.693RC
A graph of ln V against t gives a straight line with gradient −1/RC
After approximately 5τ, the capacitor is effectively fully charged or discharged
Current always decays exponentially, whether charging or discharging

A-Level Deep Dive: Charging and Discharging Capacitors

Spec mapping

Edexcel 9PH0 specification Topic 7 — Electric and Magnetic Fields, capacitor sub-strand covers the exponential charge, discharge and current behaviour of capacitor–resistor circuits, the time constant $\tau = RC$ , the use of $Q = Q_0 e^{-t/RC}$ (and the corresponding voltage and current expressions), and the linearisation of decay data via a $\ln$ plot (refer to the official specification document for exact wording). Charging/discharging is examined in Paper 2 (Electricity, Magnetism, Nuclear and Particle Physics) and reappears synoptically in Paper 3 (General and Practical Principles), where Core Practical 15 — investigating capacitor discharge — is one of the most heavily assessed practicals on the entire course. The capacitor energy expressions $E = \tfrac{1}{2}QV = \tfrac{1}{2}CV^2 = \tfrac{Q^2}{2C}$ are listed in the Edexcel formula booklet; the exponential decay equations $Q = Q_0 e^{-t/RC}$ are also given, but candidates are still expected to derive $\tau = RC$ and the half-life relation $t_{1/2} = RC \ln 2$ on demand.

Worked example with full mark scheme

Question (8 marks):

A capacitor of capacitance $C = 220\ \mu\text{F}$ is charged to $V_0 = 12.0\ \text{V}$ and then discharged through a resistor of resistance $R = 47\ \text{k}\Omega$ .

(a) Calculate the time constant $\tau$ and the initial discharge current $I_0$ . (2)

(b) Calculate the charge $Q$ remaining on the capacitor after $t = 5.0\ \text{s}$ . (3)

(c) Show that the time taken for the voltage to fall to half of its initial value is independent of $V_0$ and calculate this half-time. (3)

Solution with mark scheme:

(a) Step 1 — time constant.

$\tau = RC = (47 \times 10^3) \times (220 \times 10^{-6}) = 10.34\ \text{s}$

A1 — $\tau \approx 10.3\ \text{s}$ with correct unit (seconds).

Step 2 — initial current. At $t = 0$ , the full $V_0$ sits across $R$ , so by Ohm's law

$I_0 = \frac{V_0}{R} = \frac{12.0}{47 \times 10^3} = 2.55 \times 10^{-4}\ \text{A} \approx 0.26\ \text{mA}$

A1 — correct $I_0$ to 2 s.f. with unit. Common slip: students write $I_0 = Q_0/\tau$ , which is dimensionally correct and gives the same answer ( $Q_0 = CV_0 = 2.64\ \text{mC}$ , divided by $10.34\ \text{s}$ gives $2.55 \times 10^{-4}$ A) — both routes earn the mark, but Ohm's law is faster.

(b) Step 1 — initial charge.

$Q_0 = CV_0 = (220 \times 10^{-6}) \times 12.0 = 2.64 \times 10^{-3}\ \text{C}$

M1 — calculation of $Q_0$ from $Q = CV$ .

Step 2 — apply the discharge equation.

$Q(5.0) = Q_0 e^{-t/RC} = 2.64 \times 10^{-3} \cdot e^{-5.0/10.34}$

The exponent: $-5.0 / 10.34 = -0.4836$ , so $e^{-0.4836} = 0.6166$ .

M1 — substitution into the correct exponential equation with consistent units.

$Q(5.0) = 2.64 \times 10^{-3} \times 0.6166 = 1.63 \times 10^{-3}\ \text{C}$

A1 — $Q \approx 1.6\ \text{mC}$ to 2 s.f. with unit.

$V = V_0 e^{-t/RC}$ , so $\tfrac{1}{2}V_0 = V_0 e^{-t_{1/2}/RC}$ .

M1 — $V_0$ cancels: this is the proof requested. The half-time depends only on $R$ and $C$ , not on the initial voltage.

Step 2 — solve for $t_{1/2}$ .

$\tfrac{1}{2} = e^{-t_{1/2}/RC}$ , take $\ln$ : $\ln(0.5) = -t_{1/2}/RC$ , so $t_{1/2} = RC \ln 2$ .

A1 — printed result $t_{1/2} = \tau \ln 2$ .

Step 3 — evaluate.

$t_{1/2} = 10.34 \times 0.6931 = 7.17\ \text{s}$

A1 — $t_{1/2} \approx 7.2\ \text{s}$ with unit.

Total: 8 marks (M3 A5, split as shown).

Specimen question modelled on the Edexcel 9PH0 Paper 2 format

Question (6 marks): A student investigates the discharge of a $C = 1000\ \mu\text{F}$ capacitor through a fixed resistor. They record voltage $V$ across the capacitor every 10 s and plot $\ln V$ against $t$ . The graph is a straight line with gradient $-0.0250\ \text{s}^{-1}$ and $y$ -intercept $\ln V = 2.40$ .

(a) Use the gradient to determine the resistance $R$ . (3)

(b) Use the intercept to determine the initial voltage $V_0$ to which the capacitor was charged, and explain how the linear graph confirms that the discharge is exponential. (3)

Mark scheme decomposition by AO:

(a)

M1 (AO2.1) — taking $\ln$ of $V = V_0 e^{-t/RC}$ gives $\ln V = \ln V_0 - t/RC$ , so the gradient equals $-1/RC$ .
M1 (AO1.2) — equating: $-1/RC = -0.0250$ , so $RC = 40.0\ \text{s}$ .
A1 (AO1.1) — $R = 40.0 / (1000 \times 10^{-6}) = 40\ 000\ \Omega = 40\ \text{k}\Omega$ .

(b)

M1 (AO2.1) — $\ln V_0 = 2.40$ , so $V_0 = e^{2.40} = 11.0\ \text{V}$ .
A1 (AO1.1) — $V_0 \approx 11\ \text{V}$ with unit.
B1 (AO3.2) — explanation: a straight line on the $\ln V$ vs $t$ plot means $\ln V$ is linear in $t$ , which is the algebraic signature of exponential decay; if the discharge were not exponential the points would deviate systematically from the line.

Total: 6 marks split AO1 = 3, AO2 = 2, AO3 = 1. Edexcel uses capacitor questions to test linearisation craft (AO2) as well as numerical fluency (AO1) — the explanation B1 is the discriminator for A* candidates.

Synoptic links

Connects to:

Capacitor energy storage (earlier in Topic 7): energy stored is $E = \tfrac{1}{2}CV^2$ . As $V$ decays exponentially, $E$ decays as $E_0 e^{-2t/RC}$ — twice the rate of voltage decay. The same time-constant framework governs both, but the energy time constant is $RC/2$ , a subtlety A* candidates note.
Exponential functions and natural logarithms (Pure Mathematics): $e^x$ , $\ln x$ and the relation $\ln(e^x) = x$ are the mathematical engine of capacitor decay. The linearisation $\ln V = \ln V_0 - t/RC$ is the same trick used for radioactive half-life and Newton's law of cooling.