Capacitor charging and discharging

CIE A-Level Physics· 45 min read

1. The Capacitor Charging Process★★☆☆☆⏱ 10 min

When an uncharged capacitor is connected in series with a resistor and a DC battery, charge starts to accumulate on the capacitor plates. Current flows in the circuit until the potential difference across the capacitor equals the emf of the battery, at which point current drops to zero.

📘 Definition

Capacitor Charging

Process where charge accumulates on capacitor plates when connected to a DC source, until the capacitor voltage matches the source emf

Example:

A 10 μF capacitor connected to a 12 V battery through a resistor will charge until it has 12 V across its plates.

📐 Worked Example

A 10 μF capacitor is charged through a 50 kΩ resistor from a 12 V battery. Calculate the charge on the capacitor after 1 time constant, and the initial current at the start of charging.

1
First calculate the time constant $τ = R C$ :
2
$\tau = (50 \times 10^3 \Omega)(10 \times 10^{-6} F) = 0.5 \text{ s}$
3
Final charge when fully charged is $Q_{0} = C V_{0} = 10 \times 1 0^{- 6} \times 12 = 1.2 \times 1 0^{- 4} C$ . For charging, charge at time $t$ is $Q = Q_{0} (1 - e^{- t / τ})$ . At $t = τ$ :
4
$Q = 1.2 \times 10^{-4} (1 - e^{-1}) ≈ 0.63 \times 1.2 \times 10^{-4} = 7.56 \times 10^{-5} C$
5
At the start of charging ( $t = 0$ ), all voltage is across the resistor, so initial current is:
6
$I_0 = \frac{V_0}{R} = \frac{12}{50 \times 10^3} = 2.4 \times 10^{-4} A = 0.24 mA$

2. The Capacitor Discharging Process★★☆☆☆⏱ 10 min

When a charged capacitor is connected across a resistor, charge flows from one plate to the other through the resistor until the potential difference across the capacitor is zero. The rate of discharge is proportional to the remaining charge, leading to exponential decay.

📘 Definition

Exponential Discharge

Discharge of a capacitor follows exponential decay, where the charge, voltage and current all decrease exponentially from their initial values to zero.

📐 Worked Example

A 20 μF capacitor is charged to 10 V, then discharged through a 100 kΩ resistor. Calculate the voltage across the capacitor after 2 seconds, and the current after 1 second.

1
Calculate the time constant:
2
$\tau = RC = (100 \times 10^3 \Omega)(20 \times 10^{-6} F) = 2 \text{ s}$
3
For discharge, voltage follows $V = V_{0} e^{- t / τ}$ . Substitute $t = 2$ s:
4
$V = 10 e^{-2/2} = 10 e^{-1} ≈ 3.68 V$
5
Initial current is $I_{0} = V_{0} / R = 10/ (100 \times 1 0^{3}) = 1 \times 1 0^{- 4} A$ . At $t = 1$ s:
6
$I = I_0 e^{-t/\tau} = 1 \times 10^{-4} e^{-0.5} ≈ 6.07 \times 10^{-5} A = 60.7 \mu A$

Exam tip:

Always check whether the question asks for charging or discharging before selecting the correct equation.

3. Time Constant and Graph Interpretation★★★☆☆⏱ 15 min

The time constant $τ = R C$ determines how fast charging or discharging occurs. A larger $τ$ (bigger R or C) gives a slower process, while a smaller $τ$ gives faster charge/discharge.

📘 Definition

Time Constant

\tau = RC

For discharge: time taken for charge to fall to $1/ e \approx 37%$ of initial value. For charging: time taken for charge to rise to $\approx 63%$ of final value.

Quantity	Charging (vs time)	Discharging (vs time)
Charge $Q$	Rises exponentially from 0 to $Q_{0}$	Falls exponentially from $Q_{0}$ to 0
Capacitor Voltage $V_{C}$	Rises exponentially from 0 to $V_{0}$	Falls exponentially from $V_{0}$ to 0
Circuit Current $I$	Falls exponentially from $I_{0}$ to 0	Falls exponentially from $I_{0}$ to 0

📐 Worked Example

The half-life of discharge of a capacitor is 10 s. Calculate the time constant of the circuit.

1
Half-life $T_{1/2}$ is the time when $Q = Q_{0} /2$ . Substitute into the discharge equation:
2
$\frac{Q_0}{2} = Q_0 e^{-T_{1/2}/\tau} → \frac{1}{2} = e^{-T_{1/2}/\tau}$
3
Take natural logarithms of both sides:
4
$\ln(1/2) = -\frac{T_{1/2}}{\tau} → -\ln 2 = -\frac{T_{1/2}}{\tau} → \tau = \frac{T_{1/2}}{\ln 2}$
5
Substitute $T_{1/2} = 10$ s:
6
$\tau = \frac{10}{0.693} ≈ 14.4 \text{ s}$

4. Derivation of the Discharge Equation★★★★☆⏱ 15 min

CIE regularly asks for the derivation of the exponential discharge equation from first principles, so it is important to remember all steps.

🔬 Derivation

Goal:

Derive the exponential equation for charge during capacitor discharge

Starting from:

Kirchhoff's Voltage Law for a discharging RC circuit

1
For a charged capacitor discharging through a resistor, sum of potential differences around the loop is zero. $V_{C} = Q / C$ , $V_{R} = I R = - R \frac{d Q}{d t}$ (negative because Q decreases with time).
2
$\frac{Q}{C} - R \frac{dQ}{dt} = 0 → \frac{dQ}{Q} = -\frac{dt}{RC}$
3
Integrate both sides, with initial condition $Q = Q_{0}$ at $t = 0$ :
4
$\int_{Q_0}^{Q} \frac{dQ}{Q} = -\frac{1}{RC} \int_{0}^{t} dt$
5
Evaluate the integrals:
6
$\ln Q - \ln Q_0 = -\frac{t}{RC} → \ln\left(\frac{Q}{Q_0}\right) = -\frac{t}{RC}$

Result:

Exponentiating both sides gives the final discharge equation: $Q = Q_{0} e^{- t / R C}$

Exam tip:

Always state the initial condition and show full integration steps to get full marks for derivation questions.

5. Common Pitfalls

Wrong move:

Using the discharge equation $Q = Q_{0} e^{- t / τ}$ for charging calculations

Why:

Confusing the functional form for charging and discharging

Correct move:

Use $Q = Q_{0} (1 - e^{- t / τ})$ for charging, which approaches $Q_{0}$ from 0 as $t$ increases.

Wrong move:

Calculating time constant as $R + C$ instead of $R \times C$

Why:

Mistaking the definition of time constant, mixing units

Correct move:

Time constant $τ = R C$ , which has units of ohm × farad = second, as required.

Wrong move:

Assuming half-life equals the time constant

Why:

Confusing the definition of half-life (time to halve) and time constant (time to fall to 1/e)

Correct move:

Use the relationship $T_{1/2} = τ ln 2 \approx 0.69 τ$ to convert between the two values.

Wrong move:

Assuming voltage across the resistor is constant during charging/discharging

Why:

Treating the RC circuit as a steady-state DC circuit

Correct move:

Voltage across the resistor changes as current changes, proportional to the instantaneous current in the circuit.

Wrong move:

Forgetting that current decays exponentially during charging, just like discharge

Why:

Only focusing on the change in charge/voltage, not current

Correct move:

Current starts at maximum and decays exponentially to zero for both charging and discharging processes.

6. Quick Reference Cheatsheet

Process	Charge Q	Voltage V	Current I	Time Constant
Discharge	$Q_{0} e^{- t / R C}$	$V_{0} e^{- t / R C}$	$I_{0} e^{- t / R C}$	$τ = R C$
Charging	$Q_{0} (1 - e^{- t / R C})$	$V_{0} (1 - e^{- t / R C})$	$I_{0} e^{- t / R C}$	$τ = R C$
Half-Life				$T_{1/2} = τ ln 2 \approx 0.69 τ$
Key Rule	1τ discharge: $Q \approx 0.37 Q_{0}$	1τ charge: $V \approx 0.63 V_{0}$		5τ ≈ 99% complete

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2022 · 2
Charging capacitor graph analysis
2023 · 1
Time constant calculation
2024 · 4
Derive exponential discharge equation

What's Next

Understanding capacitor charging and discharging is core to working with RC circuits, which appear across CIE A-Level Physics in alternating current topics, electromagnetic induction, and pulse circuits. The exponential behaviour you learn here also prepares you for other common exponential processes in physics, such as radioactive decay. Now that you have mastered this sub-topic, you can move on to related topics in capacitance, or extend your knowledge to alternating current circuits with capacitors.