Quantitative Charging & Practical Investigation Revision Notes | A Level AQA Physics

1. Why Charging Is Never Instantaneous

Connect a capacitor to a battery through a resistor and something interesting happens: the capacitor does not charge instantly. The current starts high, then gradually falls to zero as the capacitor fills up. Understanding exactly how that happens — and being able to calculate the charge at any moment — is the heart of this lesson.

Think of it like filling a tank through a narrowing pipe. At first, the tank is empty and water rushes in quickly. As the tank fills, the back-pressure builds and the flow slows. The resistor is the narrow pipe; the capacitor is the tank. The fuller it gets, the harder it is to push more charge in.

This analogy captures the key physics: the rate of charging depends on how much charge is still to be added, which decreases over time. That self-referential dependence is exactly what produces an exponential curve — and the equation that describes it.

2. The Charging Equation

When a capacitor charges through a resistor from a supply of voltage $V_0$, the charge $Q$ on the capacitor at time $t$ is given by the charging equation. This equation comes directly from solving Kirchhoff's voltage law for the RC circuit, and it is one of the two key exponential forms you need at A level.

Notice the structure: the bracket $\left(1 - e^{-t/RC}\right)$ starts at zero when $t = 0$ (because $e^0 = 1$, so $1 - 1 = 0$) and approaches 1 as $t \to \infty$ (because $e^{-\infty} \to 0$, so $1 - 0 = 1$). Multiplying by $Q_0$ scales this to run from zero up to the maximum charge.

Because voltage across the capacitor is $V = Q/C$, dividing both sides by $C$ gives the equivalent voltage form: $V = V_0\left(1 - e^{-t/RC}\right)$. The current in the circuit follows $I = I_0 e^{-t/RC}$, which is a pure decay — current starts at its maximum and falls exponentially as the capacitor opposes further charging.

A larger $R$ means current is restricted more, so charging takes longer. A larger $C$ means more charge is needed to reach $V_0$, so again charging takes longer. Both effects multiply together in $RC$, which is why the time constant captures the combined influence of both components.

3. Reading the Charging Curve

The shape of the $Q$-$t$ graph during charging is a rising curve that starts steeply and levels off asymptotically towards $Q_0$. It never actually reaches $Q_0$ in finite time — it just gets arbitrarily close. Examiners frequently ask you to describe or sketch this curve, so the precise language matters.

Reading the graph: at $t = 0$ the charge is zero and the gradient (which equals the current divided by... well, it is proportional to the current) is at its steepest. As time increases, the gradient decreases continuously. The curve is always concave downwards during charging.

A common exam task is to read off the time constant directly from the graph. Locate $0.63 Q_0$ on the vertical axis, draw a horizontal line across to the curve, then drop a vertical line to the time axis — that time value is $RC$. Practise this graphical method because it appears in data-analysis questions.

4. Worked Calculation Using the Charging Equation

Exam questions on the charging equation typically give you $R$, $C$, $V_0$ (or $Q_0$), and a time $t$, then ask for $Q$ or $V$ at that moment. The method is always the same: identify your knowns, substitute carefully, and evaluate the exponential term first before subtracting from 1.

Worked example: Calculating Charge at a Given Time During Charging

A capacitor of capacitance $470\,\mu\text{F}$ is charged through a resistor of resistance $10\,\text{k}\Omega$ from a $12\,\text{V}$ supply. Calculate the charge stored on the capacitor $8.0\,\text{s}$ after charging begins. Give your answer to 3 significant figures.

1. $C = 470\,\mu\text{F} = 470 \times 10^{-6}\,\text{F} = 4.70 \times 10^{-4}\,\text{F}$ $R = 10\,\text{k}\Omega = 10 \times 10^{3}\,\Omega = 1.0 \times 10^{4}\,\Omega$ — Always convert to SI base units (F and $\Omega$) before substituting. Mixing $\mu$F with k$\Omega$ without conversion is the most common source of error in this type of question.
2. $RC = 1.0 \times 10^{4}\,\Omega \times 4.70 \times 10^{-4}\,\text{F} = 4.70\,\text{s}$ — Check units: $\Omega \times \text{F} = \text{s}$. The time constant is 4.70 s, so 8.0 s is about 1.7 time constants — the capacitor should be well past 63% but not yet fully charged.
3. $Q_0 = CV_0 = 4.70 \times 10^{-4}\,\text{F} \times 12\,\text{V} = 5.64 \times 10^{-3}\,\text{C}$ — $Q_0 = CV_0$ gives the charge when the capacitor is fully charged to the supply voltage. This is the ceiling the charging curve approaches.
4. $\frac{t}{RC} = \frac{8.0\,\text{s}}{4.70\,\text{s}} = 1.702...$ $\therefore -t/RC = -1.702$ — Keep extra decimal places here to avoid rounding errors in the final answer. The exponent is dimensionless — the seconds cancel.
5. $e^{-1.702} = 0.1827...$ — Use the $e^x$ function on your calculator with $x = -1.702$. Do not round this intermediate value yet.
6. $Q = Q_0\left(1 - e^{-t/RC}\right) = 5.64 \times 10^{-3} \times (1 - 0.1827)$ $= 5.64 \times 10^{-3} \times 0.8173$ $= 4.61 \times 10^{-3}\,\text{C}$ — The bracket gives 0.817, confirming the capacitor is at about 82% of maximum charge — consistent with our estimate in Step 2. Final answer to 3 s.f.: $4.61 \times 10^{-3}\,\text{C}$.

Answer: $Q = 4.61 \times 10^{-3}\,\text{C}$ (to 3 s.f.)

5. Required Practical 9 — Investigating Charge and Discharge

The quantitative equation is only half the story. Required Practical 9 asks you to collect real data from a charging (or discharging) capacitor and then use that data to determine the time constant $RC$ experimentally. The key analytical technique is log-linear plotting, which converts the exponential relationship into a straight line.

6. Log-Linear Analysis — Turning a Curve into a Straight Line

Here is where the practical and the mathematics connect. A raw $V$-$t$ graph during discharge is a decaying exponential curve — useful to look at, but hard to use for precise measurement. The log-linear technique transforms it into a straight line, which is far easier to analyse and from which you can extract $RC$ with a best-fit line.

Start from the discharge equation: $V = V_0 e^{-t/RC}$. Take the natural logarithm of both sides. Using the rule $\ln(AB) = \ln A + \ln B$ and $\ln(e^x) = x$, the equation becomes $\ln V = \ln V_0 - \frac{1}{RC} \cdot t$. This is now in the form $y = mx + c$, where $y = \ln V$, $x = t$, gradient $m = -1/RC$, and intercept $c = \ln V_0$.

To apply this: calculate $\ln V$ for each data point, then plot $\ln V$ against $t$. Draw a best-fit straight line through the points. Measure the gradient using a large triangle (at least half the length of the line) and include units in your gradient calculation: $\Delta(\ln V) / \Delta t$ has units of $\text{s}^{-1}$. Then $RC = -1/\text{gradient}$.

Worked example: Determining RC from a Log-Linear Graph

A student discharges a capacitor through a resistor and records the following data. Determine the time constant $RC$. $t$ (s): 0, 10, 20, 30, 40, 50 $V$ (V): 9.00, 6.07, 4.09, 2.76, 1.86, 1.25

1. $\ln(9.00) = 2.197$ $\ln(6.07) = 1.803$ $\ln(4.09) = 1.408$ $\ln(2.76) = 1.015$ $\ln(1.86) = 0.621$ $\ln(1.25) = 0.223$ — Use the natural logarithm (ln) button on your calculator, not log base 10. The values decrease linearly with time, which confirms the exponential decay.
2. Using points at $t = 0\,\text{s}$, $\ln V = 2.197$ and $t = 50\,\text{s}$, $\ln V = 0.223$ — Choose points that lie on the best-fit line, not necessarily data points. Using a large separation reduces the percentage uncertainty in the gradient.
3. $m = \frac{\Delta(\ln V)}{\Delta t} = \frac{0.223 - 2.197}{50 - 0} = \frac{-1.974}{50} = -0.03948\,\text{s}^{-1}$ — The gradient is negative, confirming the voltage is decaying. The units are $\text{s}^{-1}$ because $\ln V$ is dimensionless and $t$ is in seconds.
4. $RC = \frac{-1}{m} = \frac{-1}{-0.03948} = 25.3\,\text{s}$ — The time constant is the negative reciprocal of the gradient. Always check: $RC$ must be positive and have units of seconds.
5. $y\text{-intercept} = \ln V_0 = 2.197$ $V_0 = e^{2.197} = 9.00\,\text{V}$ — The intercept gives $\ln V_0$, so $V_0 = e^{\text{intercept}}$. This should match the initial voltage in the data (9.00 V here), providing a consistency check on the graph.

Answer: $RC = 25.3\,\text{s}$ (from the gradient of the $\ln V$ against $t$ graph)

7. Evaluating the Practical and Handling Uncertainty

Collecting good data from a capacitor charging experiment is harder than it looks. The voltage changes most rapidly at the start — exactly when you are also trying to start the stopwatch and take the first reading. This is a systematic timing error that affects early data points more than later ones.

Using a data logger with a voltage sensor eliminates the reaction-time problem entirely and allows readings to be taken every fraction of a second. If a data logger is unavailable, choosing a larger $RC$ (by increasing $R$ or $C$) slows the process down so that manual timing becomes more feasible.

• Uncertainty in $R$: resistors have a tolerance (typically $\pm 5\%$ or $\pm 1\%$). Measure $R$ with a multimeter rather than relying on the colour-code value.
• Uncertainty in $C$: electrolytic capacitors often have tolerances of $\pm 20\%$. The experimental $RC$ may differ noticeably from the calculated $R \times C$ for this reason alone.
• Uncertainty in the gradient: draw lines of maximum and minimum gradient through the error bars on the log-linear graph to estimate the uncertainty in $RC$.
• Leakage current: real capacitors are not perfect insulators. A small leakage current causes the voltage to decay slightly even when the circuit is open, introducing a systematic error in discharge measurements.
• Voltmeter resistance: an analogue voltmeter with low internal resistance draws current from the capacitor during measurement, altering the effective resistance in the circuit and therefore the time constant.

8. Exam Technique and Mark-Winning Strategies

Questions on capacitor charging fall into three categories: pure calculation, graph analysis, and practical evaluation. Each has a reliable method that picks up marks efficiently. The worked examples in this lesson have modelled all three — now here is how to transfer that to exam conditions.

For calculation questions, the most reliable approach is: write the equation, state your known values with units, calculate $RC$ explicitly as a separate step, evaluate the exponential term, then substitute. Skipping the $RC$ step is where most arithmetic errors creep in. Examiners award a method mark for a correct substitution even if the final arithmetic is wrong, so showing every step protects your marks.

For log-linear graph questions, the mark scheme almost always awards marks for: correctly calculating $\ln V$ values, plotting them accurately, drawing a best-fit line, calculating the gradient with a clearly shown triangle, and then extracting $RC$ as $-1/\text{gradient}$. Each of these is a separate mark — do not rush through them as a single step.