Capacitor Charging & Discharging: Graphs & the Time Constant Revision Notes | A Level AQA Physics

1. Why Capacitors Don't Charge Instantly

Picture filling a bath through a tap that gets progressively weaker as the water level rises. At first, water rushes in quickly. As the bath fills, the back-pressure slows the flow. The bath never fills in a straight line — it curves. A capacitor charging through a resistor behaves in exactly the same way, and understanding why unlocks every graph and equation in this topic.

When a capacitor charges, the growing voltage across it opposes the supply voltage, reducing the current. Less current means charge accumulates more slowly, which means the voltage rises more slowly, which reduces the current further. This self-limiting feedback is the physical origin of the exponential curve.

2. Charging Graphs: Q, V, and I Against Time

When a capacitor charges through a resistor from a supply of e.m.f. $\varepsilon$, three quantities change with time: charge $Q$, voltage across the capacitor $V_C$, and current $I$ in the circuit. Each has a distinct graph shape, and the AQA specification requires you to recognise and describe all three.

Both $Q$ and $V_C$ start at zero and rise towards their maximum values ($Q_0 = C\varepsilon$ and $V_C = \varepsilon$) along a curve that gets progressively shallower. After one time constant $RC$, they have reached approximately 63% of their final value. After five time constants, they are within 1% of the maximum — effectively fully charged for practical purposes.

The current graph is the mirror image in behaviour: $I$ starts at its maximum value $I_0 = \varepsilon / R$ and decays exponentially towards zero. At $t = 0$, the uncharged capacitor offers no opposition, so the full supply voltage drives current through the resistor. As $V_C$ rises, the net voltage driving current falls, and $I$ decreases accordingly.

3. Discharging Graphs: Q, V, and I Against Time

Now remove the supply and connect the charged capacitor directly across the resistor. The stored charge drives a current through the resistor, and all three quantities — $Q$, $V_C$, and $I$ — decay from their initial values towards zero. This is the discharging scenario, and it produces the classic exponential decay curves that AQA tests most frequently.

All three quantities follow the same exponential decay shape during discharging. The charge $Q$, voltage $V_C$, and current $I$ each fall to $1/e \approx 0.368$ (about 37%) of their initial value after one time constant $RC$. The curves never actually reach zero mathematically, but after five time constants the capacitor is considered fully discharged in practice.

One subtlety worth noting: during discharging, the current flows in the opposite direction to the charging current. If you define the charging current direction as positive, the discharging current is negative. AQA questions sometimes ask you to account for this sign difference when describing or sketching graphs.

4. The Exponential Equations

The graphs have precise mathematical descriptions. For discharging, all three quantities share the same exponential form. For charging, $Q$ and $V$ follow a complementary rising exponential, while $I$ still decays. You need to be comfortable applying these equations and rearranging them to find time or the time constant.

The time constant $\tau = RC$ is the single most important quantity in this topic. A large $RC$ means the circuit charges and discharges slowly; a small $RC$ means it responds quickly. Physically, a large resistance limits the current (slowing charge flow), and a large capacitance means more charge must flow before the voltage changes significantly — both effects increase $\tau$.

5. Calculating with the Time Constant

Exam questions on this topic fall into two categories: calculating $RC$ from given component values, and using the exponential equations to find a quantity at a specific time (or the time at which a quantity reaches a specific value). The second type requires rearranging the exponential equation using natural logarithms.

Worked example: Finding the time for voltage to fall to a given fraction during discharge

A capacitor is charged to $12 \, \text{V}$ and then discharged through a $22 \, \text{k}\Omega$ resistor. The capacitance is $100 \, \mu\text{F}$. Calculate the time taken for the voltage to fall to $4.0 \, \text{V}$. Give your answer to 2 significant figures.

1. $V = V_0 \, e^{-t/RC}$ — This is the standard discharging equation. $V_0 = 12 \, \text{V}$, $V = 4.0 \, \text{V}$. [M1]
2. $RC = 22 \times 10^3 \, \Omega \times 100 \times 10^{-6} \, \text{F} = 2.2 \, \text{s}$ — Convert units first: $22 \, \text{k}\Omega = 22{,}000 \, \Omega$ and $100 \, \mu\text{F} = 100 \times 10^{-6} \, \text{F}$. Product gives seconds. [M1]
3. $\frac{V}{V_0} = e^{-t/RC} \Rightarrow \frac{4.0}{12} = e^{-t/2.2}$ — This isolates the exponential term. $4.0 / 12 = 0.3333...$
4. $\ln\left(\frac{4.0}{12}\right) = -\frac{t}{2.2}$ $\ln(0.3333) = -1.0986$ — $\ln(e^x) = x$, so the right side becomes $-t/RC$. [M1]
5. $t = -2.2 \times \ln(0.3333) = -2.2 \times (-1.0986) = 2.417 \, \text{s}$ — The two negatives cancel to give a positive time, as expected. [A1]
6. $t = 2.4 \, \text{s}$ — Sanity check: $4.0 \, \text{V}$ is less than $12/e \approx 4.4 \, \text{V}$, so the time should be slightly more than one time constant (2.2 s). Our answer of 2.4 s is consistent with this. [A1]

Answer: $t = 2.4 \, \text{s}$

Notice the sanity check in step 6: $12/e \approx 4.4 \, \text{V}$, and since $4.0 \, \text{V}$ is slightly below this, the time must be slightly more than one time constant. Performing this kind of order-of-magnitude check takes seconds and catches errors before they cost marks.

6. Gradients and Areas Under the Graphs

The AQA specification explicitly requires you to interpret gradients and areas under capacitor graphs. This is where the connection between $Q$, $V$, $I$, and $C$ becomes visible on the graph itself, and examiners frequently test it with 'describe' or 'explain' command words.

• Gradient of a Q–t graph: The gradient $dQ/dt$ equals the current $I$ at that instant. A steeper gradient means a larger current. During charging, the gradient is steepest at $t = 0$ (maximum current) and approaches zero as the capacitor becomes fully charged.
• Gradient of a V–t graph: Since $Q = CV$ and $C$ is constant, $dV/dt = I/C$. The gradient of the V–t graph equals $I/C$, so it is also steepest at $t = 0$ and flattens as charging progresses.
• Area under an I–t graph: The area under the current–time graph equals the total charge transferred, $Q = \int I \, dt$. This follows from the definition of current as the rate of flow of charge.
• Area under a V–t graph: This does not directly equal a simple physical quantity in the same way, but the area under a Q–V graph (not Q–t) equals the energy stored: $E = \frac{1}{2}QV = \frac{1}{2}CV^2$.

When a question asks you to 'interpret the gradient' of a Q–t graph, the expected answer is that the gradient represents the current at that point in time. Stating this with the correct unit (amperes) and linking it to the physical meaning — rate of flow of charge — is what earns the mark.

7. Determining the Time Constant from a Graph

In a required practical or data-analysis question, you will often need to extract the time constant from a graph rather than calculate it from component values. There are two reliable methods, and knowing both gives you flexibility depending on what the question provides.

Method 1 — The 37% method: Read off the initial value $x_0$ from the graph. Calculate $0.37 \times x_0$. Find the time at which the curve reaches this value. That time is the time constant $RC$. This works because $e^{-1} \approx 0.368$, so after one time constant the quantity has fallen to approximately 37% of its starting value.

Method 2 — Linearising the graph: Take natural logarithms of the discharging equation. Starting from $V = V_0 e^{-t/RC}$, taking $\ln$ of both sides gives $\ln V = \ln V_0 - \frac{t}{RC}$. Plotting $\ln V$ against $t$ produces a straight line with gradient $-1/RC$ and y-intercept $\ln V_0$. The time constant is then $RC = -1/\text{gradient}$.

Worked example: Extracting RC from a linearised ln V vs t graph

A student plots $\ln V$ against $t$ for a discharging capacitor and draws a best-fit straight line through the points. The line passes through $(0, \, 2.77)$ and $(6.0 \, \text{s}, \, 0.77)$. Determine the time constant $RC$.

1. $\text{gradient} = \frac{\Delta(\ln V)}{\Delta t} = \frac{0.77 - 2.77}{6.0 - 0} = \frac{-2.00}{6.0}$ — Use two well-separated points on the best-fit line, not individual data points. Show the triangle explicitly. [M1]
2. $\text{gradient} = -0.333 \, \text{s}^{-1}$ — $-2.00 \div 6.0 = -0.333$. The negative sign confirms exponential decay, as expected.
3. $\text{gradient} = -\frac{1}{RC}$ $\therefore RC = -\frac{1}{\text{gradient}}$ — This comes directly from the linearised equation $\ln V = \ln V_0 - t/RC$. [M1]
4. $RC = -\frac{1}{-0.333} = 3.0 \, \text{s}$ — The two negatives cancel. Always state the unit (seconds). [A1]
5. $\ln V_0 = 2.77 \Rightarrow V_0 = e^{2.77} \approx 16 \, \text{V}$ — The y-intercept gives the initial voltage. This is a useful cross-check: if the initial voltage is physically reasonable for the circuit described, the gradient calculation is likely correct.

Answer: $RC = 3.0 \, \text{s}$

8. Exam Technique and Common Mistakes

This topic rewards careful attention to which equation applies (charging or discharging), correct unit conversion before substitution, and precise use of natural logarithms. The following guidance consolidates the most frequently penalised errors seen in AQA mark schemes.