Capacitor Charge and Discharge Experiment for H2 Physics Paper 4: RC Circuits, Time Constant, and Excel Analysis
14 Apr 2026, 00:00 Z
Reviewed by
Chee Wei Jie·Academic Advisor (Physics)
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> **TL;DR**\
> The capacitor charge and discharge experiment is a Paper 4 workhorse because it simultaneously tests uncertainty handling, exponential model recognition, graph linearisation, and the new 9478 Excel spreadsheet requirement introduced for the 2026 cohort.\
> Discharge through a resistor follows $V(t) = V_0 e^{-t/RC}$. Taking the natural log linearises this to $\ln V = \ln V_0 - t/(RC)$, so a plot of $\ln V$ against $t$ gives a straight line whose gradient is $-1/(RC)$. The time constant $\tau = RC$ is extracted from that gradient, and LINEST in Excel gives gradient uncertainty in one formula.\
> This guide covers full method, worked numerical example, Excel LINEST step-by-step, uncertainty propagation into capacitance, common student errors, and ACE evaluation points.\
> Pair with [H2 Physics Paper 4 Spreadsheet Skills (9478)](https://eclatinstitute.sg/blog/h2-physics-experiments/H2-Physics-Paper-4-Spreadsheet-Skills-9478-2026) and the [H2 Physics practicals hub](https://eclatinstitute.sg/blog/h2-physics-experiments) for the full practical landscape.
---
## 1 | Why this experiment is a Paper 4 workhorse
The capacitor charge and discharge experiment appears regularly in H2 Physics Paper 4 (9478) for good reasons. It packs four distinct skill demands into a single setup.
**Exponential model.** Most Paper 4 experiments fit a linear model, so examiners use the capacitor experiment to test whether candidates can recognise an exponential relationship and know to linearise it before plotting.
**Uncertainty handling.** The voltage readings drift slightly between repeat runs, the stopwatch has a reaction-time component, and the resistor has a marked tolerance. Candidates must identify and quantify all three sources and propagate them correctly into the final value of capacitance.
**Graph linearisation.** Taking a natural log transforms a curve into a straight line. The gradient and intercept of that line carry physical meaning. Extracting both — and knowing what units they carry — is an explicit Paper 4 skill.
**Excel spreadsheet skills (new for 9478).** The 9478 syllabus introduced LINEST and LOGEST as assessable tools for gradient extraction with uncertainty. The capacitor experiment is one of the cleanest contexts in which LINEST gives a physically meaningful output: gradient equals $-1/(RC)$ and the LINEST standard error propagates directly into an uncertainty in $\tau$. [1]
No other common Paper 4 experiment tests all four demands at once.
---
## 2 | The physics in one page
### 2.1 Charging
When a capacitor of capacitance $C$ is connected in series with a resistor of resistance $R$ and a DC supply of EMF $V_0$, the voltage across the capacitor rises from zero as:
$$V(t) = V_0\!\left(1 - e^{-t/RC}\right)$$
The current starts at its maximum value $V_0/R$ and falls exponentially as the capacitor charges.
### 2.2 Discharge
When the charged capacitor (initial voltage $V_0$) is connected across the resistor with the supply disconnected, it discharges as:
$$V(t) = V_0\, e^{-t/RC}$$
### 2.3 Time constant
The quantity $\tau = RC$ is the **time constant** of the circuit. After one time constant, a discharging capacitor has fallen to $e^{-1} \approx 0.368$ of its initial voltage — roughly 37% — and a charging capacitor has risen to $1 - e^{-1} \approx 0.632$ of its final voltage — roughly 63%. After five time constants, the capacitor is effectively fully discharged (to about 0.7% of $V_0$) or fully charged (to 99.3% of $V_0$).
| Time elapsed | Discharge voltage (fraction of $V_0$) | Charge voltage (fraction of $V_0$) |
| ------------ | ------------------------------------- | ----------------------------------- |
| $\tau$ | 0.368 | 0.632 |
| $2\tau$ | 0.135 | 0.865 |
| $3\tau$ | 0.050 | 0.950 |
| $4\tau$ | 0.018 | 0.982 |
| $5\tau$ | 0.007 | 0.993 |
In school labs, choosing $R$ and $C$ so that $\tau$ is between 20 s and 120 s gives a discharge curve that can be sampled every 10 s for about 5 minutes — manageable with a stopwatch and a digital voltmeter.
---
## 3 | Apparatus and circuit setup
**Capacitor.** Use an electrolytic capacitor in the range 100 µF to 1000 µF. Electrolytic capacitors produce time constants on the order of tens of seconds when paired with a 100 kΩ resistor, making the discharge visible and recordable by hand. Electrolytic capacitors are polarised: one lead must be connected to the positive terminal of the supply at all times.
**Resistor.** Carbon-film resistors in the range 100 kΩ to 1 MΩ are standard. The marked resistance has a ±5% tolerance (gold band) or ±10% tolerance (silver band). Record the marked value and tolerance before the experiment.
**DC supply.** A 6 V or 9 V regulated supply is typical. Higher voltage gives better resolution on the voltage readings but does not change the time constant.
**Voltmeter.** Use a **digital multimeter** set to the DC voltage range. A digital multimeter has a high input impedance (typically 10 MΩ), which means it draws negligible current from the circuit and does not significantly alter the discharge time constant. An analogue voltmeter has a much lower input impedance and will drain the capacitor faster than the resistor alone, giving a falsely short time constant — this is a common exam error discussed in section 8.
**Switch.** A single-pole double-throw (SPDT) switch allows the circuit to be toggled cleanly between the charge path (supply connected) and the discharge path (capacitor connected to the resistor only).
**Circuit description.** The supply positive terminal connects through the switch to one plate of the capacitor. The other plate of the capacitor connects through the resistor back to the supply negative terminal. The voltmeter is connected directly across the capacitor plates. In the charge position, the switch routes current from the supply through the resistor to the capacitor. Flicking the switch to the discharge position disconnects the supply and connects the capacitor terminals directly across the resistor.
---
## 4 | Full procedure
**(a) Pre-experiment checks.** Verify the electrolytic capacitor polarity. Connect the positive plate to the positive supply terminal. Set the voltmeter range to the next range above the supply voltage (e.g., 20 V range for a 9 V supply).
**(b) Charge the capacitor.** Flick the switch to the charge position. Wait until the voltmeter reading stabilises (typically 5τ, so about 5 minutes for a 60 s circuit). Record this stable voltage as $V_0$.
**(c) Begin discharge.** With the stopwatch ready, flick the switch to the discharge position and start the stopwatch simultaneously. Record the voltmeter reading at $t = 0$ (this should closely match $V_0$), then at regular intervals — every 10 s is typical for a 60 s time constant.
**(d) Continue for at least 5τ.** Record readings until the voltage has fallen to less than 5% of $V_0$. This gives enough data points in the tail of the curve to anchor the fit.
**(e) Repeat.** Recharge the capacitor and repeat the discharge at least twice more. You should have at least three complete discharge curves.
**(f) Average.** At each time point, average the voltage readings from the three runs. Note the spread of readings as an estimate of random uncertainty at that time point.
---
## 5 | Linearisation and time-constant extraction
### 5.1 The linearisation
Starting from the discharge equation:
$$V = V_0\, e^{-t/RC}$$
Take the natural logarithm of both sides:
$$\ln V = \ln V_0 - \frac{t}{RC}$$
This is of the form $y = c + mx$ where $y = \ln V$, $x = t$, $m = -1/(RC)$, and $c = \ln V_0$. Plot $\ln V$ on the y-axis and $t$ on the x-axis. A straight line confirms that the discharge is exponential. The gradient gives $-1/(RC)$ and the intercept gives $\ln V_0$.
### 5.2 Worked numerical example
The following readings are from a single discharge run with a 470 µF electrolytic capacitor and a 100 kΩ resistor (nominal $\tau = 47$ s) from a 9.0 V supply.
| $t$ (s) | $V$ (V) | $\ln V$ |
| ------- | ------- | ------- |
| 0 | 8.94 | 2.190 |
| 10 | 7.23 | 1.978 |
| 20 | 5.85 | 1.766 |
| 30 | 4.73 | 1.554 |
| 40 | 3.82 | 1.340 |
| 50 | 3.09 | 1.129 |
| 60 | 2.50 | 0.916 |
Plotting these points and drawing the best-fit line, the gradient is estimated by choosing two well-separated points on the line (not necessarily data points):
$$m = \frac{\ln V_2 - \ln V_1}{t_2 - t_1} = \frac{0.90 - 2.19}{60 - 0} = \frac{-1.29}{60} = -0.02150 \text{ s}^{-1}$$
The time constant is then:
$$\tau = RC = -\frac{1}{m} = \frac{1}{0.02150} = 46.5 \text{ s}$$
Dividing by the nominal resistance $R = 100 \times 10^3$ Ω:
$$C = \frac{\tau}{R} = \frac{46.5}{100 \times 10^3} = 4.65 \times 10^{-4} \text{ F} = 465 \text{ µF}$$
This lies within the ±20% tolerance of the marked 470 µF value, which is a reasonable agreement for an electrolytic capacitor.
The display-math summary of the extraction chain:
$$\tau = -\frac{1}{m} \quad \Longrightarrow \quad C = \frac{\tau}{R} = -\frac{1}{mR}$$
---
## 6 | Excel LINEST workflow
The 9478 syllabus requires candidates to use spreadsheet tools including LINEST for gradient extraction with uncertainty. The following workflow matches the examiner's expectation exactly.
**(a) Enter your data.** Open a new Excel sheet. Put the time values in column A (e.g., A2:A8) and the averaged voltage values in column B (e.g., B2:B8).
**(b) Compute ln V.** In column C, enter the formula `=LN(B2)` in cell C2 and fill down to C8. This gives your $\ln V$ column.
**(c) Run LINEST.** Select a 2-row by 2-column output range (e.g., E2:F3). Enter the array formula:
```
=LINEST(C2:C8, A2:A8, TRUE, TRUE)
```
Confirm with Ctrl+Shift+Enter (on older Excel) or just Enter (on Excel 365). The four cells return:
| E2: gradient $m$ | F2: intercept $c$ |
| ---------------- | ----------------- |
| E3: standard error of $m$ (se$_m$) | F3: standard error of $c$ (se$_c$) |
For the sample data above, you would see approximately:
| $m = -0.02148$ s$^{-1}$ | $c = 2.191$ |
| ----------------------- | ----------- |
| se$_m = 0.00004$ s$^{-1}$ | se$_c = 0.002$ |
**(d) Extract the time constant.** The time constant is $\tau = -1/m$. In cell H2, enter:
```
=-1/E2
```
The standard error of $\tau$ from LINEST propagates as $\Delta\tau = \tau \times (\text{se}_m / |m|)$. In cell H3:
```
=H2 * (E3 / ABS(E2))
```
For the sample data, this gives $\tau = 46.5 \pm 0.1$ s.
For a full explanation of all LINEST output rows and how to handle LOGEST for curves that cannot be linearised, see [H2 Physics Paper 4 Spreadsheet Skills (9478)](https://eclatinstitute.sg/blog/h2-physics-experiments/H2-Physics-Paper-4-Spreadsheet-Skills-9478-2026).
---
## 7 | Uncertainty handling
### 7.1 Sources of uncertainty
**Timing.** Human reaction time introduces a random uncertainty of approximately ±0.3 s at each data point. This is small compared with a 10 s sampling interval but becomes significant if readings are taken every 2 s.
**Voltage.** A typical 4.5-digit digital multimeter has a last-digit uncertainty of ±0.01 V on the 20 V range.
**Resistance.** A carbon-film resistor with a gold band has ±5% tolerance. The marked value $R = 100$ kΩ carries $\Delta R = 5000$ Ω. For a precision determination, measure $R$ with a four-terminal method before the experiment.
### 7.2 Propagating into capacitance
From $C = \tau/R$, the fractional uncertainty in $C$ combines the fractional uncertainty in $\tau$ (from LINEST) and the fractional uncertainty in $R$ (from the resistor tolerance or measurement):
$$\frac{\Delta C}{C} = \sqrt{\left(\frac{\Delta \tau}{\tau}\right)^2 + \left(\frac{\Delta R}{R}\right)^2}$$
Using $\tau = 46.5$ s, $\Delta\tau = 0.1$ s (from LINEST), and $\Delta R / R = 0.05$ (±5% tolerance):
$$\frac{\Delta C}{C} = \sqrt{\left(\frac{0.1}{46.5}\right)^2 + (0.05)^2} = \sqrt{(0.0022)^2 + (0.05)^2} \approx 0.050$$
The uncertainty in $C$ is dominated by the resistor tolerance. This is a pedagogically important result: refining the LINEST fit with more data points has almost no effect on the final uncertainty because $\Delta R/R$ is the bottleneck. To improve precision, measure $R$ directly with a multimeter (reducing $\Delta R/R$ to about 0.005 or less) rather than relying on the marked value.
The final result with correctly propagated uncertainty:
$$C = 465 \pm 23 \text{ µF}$$
---
## 8 | Common student errors
**Using an analogue voltmeter.** An analogue voltmeter typically has an input impedance of 20 kΩ per volt. On the 10 V range, this is 200 kΩ. Connected in parallel with a 100 kΩ discharge resistor, the effective resistance drops to about 67 kΩ, reducing $\tau$ by one third. The graph will appear linear but the extracted $C$ will be systematically low. Always use a digital multimeter with at least 10 MΩ input impedance.
**Offset timing error.** Starting the stopwatch after flicking the switch (or vice versa) introduces a constant offset into all time readings. This shifts the intercept of the $\ln V$ vs $t$ plot but does not change the gradient, so $\tau$ is unaffected. However, if the question asks for $V_0$ from the intercept, the offset error will make the fitted $\ln V_0$ disagree with the directly measured $V_0$.
**Electrolytic polarity reversed.** Connecting an electrolytic capacitor with reversed polarity damages it and causes rapid leakage. In a practical exam, this typically produces a non-exponential discharge curve. Check polarity before every charge cycle.
**Too few data points past 3τ.** If readings are only taken up to $t = 3\tau$, the tail of the exponential is underrepresented. LINEST fits are pulled toward the early-time data where voltage changes are largest. The gradient appears steeper than the true value, giving an underestimate of $\tau$. Collect data to at least $t = 5\tau$.
**Ignoring voltmeter input impedance in the effective resistance.** Even at 10 MΩ input impedance, the voltmeter is in parallel with the resistor. For a 100 kΩ resistor, the effective resistance is $100\,000 \times 10\,000\,000 / (100\,000 + 10\,000\,000) \approx 99\,010$ Ω — only 1% below nominal, which is within the resistor tolerance and can be ignored. But for a 1 MΩ resistor, the effective parallel resistance drops to about 909 kΩ, a 9% shift that matters if the resistor tolerance is being used to estimate uncertainty. State this assumption explicitly in your answer.
---
## 9 | ACE evaluation points
The ACE (Analysis, Conclusions, Evaluation) section awards marks for identifying specific limitations and proposing concrete improvements. The key is to convert a qualitative limitation into a quantitative fix.
**Temperature dependence of capacitance.** Electrolytic capacitor capacitance varies with temperature: a typical X5R-rated capacitor can shift by ±15% over the 0–85 °C range, and even within a 30-minute experiment at room temperature, a lamp or resistor heating the bench may shift $C$ by 1–3%. *Exam-ready evaluation point: The electrolytic capacitor's capacitance decreases by approximately 1% per 5 °C rise above 25 °C. The resistor dissipates $P = V^2/R \approx 0.8$ mW at 9 V into 100 kΩ — negligible self-heating — but ambient temperature drift over a 30-minute session could shift $C$ by up to 2%, introducing a systematic underestimate of the true room-temperature capacitance. Recording temperature at the start and end of the experiment and applying a manufacturer's temperature coefficient would convert this qualitative limitation into a quantitative fix.*
**Dielectric absorption.** When a charged electrolytic capacitor is briefly discharged and then disconnected, the voltage slowly recovers toward a fraction of the original charge voltage. This is caused by polarisation of the dielectric material and means the capacitor does not behave as an ideal component. In a discharge experiment, dielectric absorption makes the tail of the discharge curve slightly slower than the exponential model predicts, leading to an overestimate of $\tau$ if the fit is weighted toward the tail. *Improvement: use a polypropylene film capacitor instead of electrolytic. Film capacitors have negligible dielectric absorption and much lower leakage current, making the discharge follow the ideal exponential much more closely.*
**Leakage currents in long-τ circuits.** At very long time constants (τ greater than about 5 minutes), the self-discharge current of the capacitor — which flows through the dielectric even without an external path — becomes comparable to the discharge current through the external resistor. This causes the effective time constant to be shorter than $RC$, and $C$ calculated from $\tau/R$ will be an underestimate. The effect is negligible for well-chosen RC combinations but should be mentioned if τ exceeds 10 minutes.
**Meter loading effect.** Covered quantitatively in section 8. For the ACE section, the exam-ready version is: *The digital voltmeter's 10 MΩ input impedance is in parallel with the 100 kΩ discharge resistor. The effective resistance is 99 kΩ rather than 100 kΩ — a 1% deviation. Since the nominal resistor tolerance is ±5%, this loading error is within the uncertainty already present, so it is not the dominant source of error for this circuit. If the experiment were repeated with R = 1 MΩ, the voltmeter loading would reduce the effective resistance by 9%, making it the dominant source of systematic error.*
**Suggested refinement: data-logger.** Replacing manual stopwatch timing with a computer-based data-logger and voltage sensor removes the reaction-time offset, allows sampling every second or less, and extends the useful data range well past 5τ. A data-logger also removes observer-introduced correlation between successive readings (a student who misreads one point may compensate by reading the next differently). This is a genuine improvement rather than a minor tweak — state the mechanism and the quantitative benefit.
---
## 10 | Further reading
- [H2 Physics Paper 4 Spreadsheet Skills (9478)](https://eclatinstitute.sg/blog/h2-physics-experiments/H2-Physics-Paper-4-Spreadsheet-Skills-9478-2026) — full LINEST and LOGEST reference, error output interpretation, and common formula mistakes
- [H2 Physics Practical 2026 Lab Mastery Guide](https://eclatinstitute.sg/blog/h2-physics-experiments/H2-Physics-Practical-2026-Lab-Mastery-Guide) — MMO technique across all Paper 4 experiment families
- [Internal Resistance of Batteries: Beyond Textbook](https://eclatinstitute.sg/blog/h2-physics-experiments/Internal-Resistance-of-Batteries-Beyond-Textbook) — another linearisation experiment with a comparable uncertainty-propagation structure
- [H2 Physics practicals hub](https://eclatinstitute.sg/blog/h2-physics-experiments) — full guide to all Paper 4 technique families and past-paper experiment types
---
> **Preparing for Paper 4 without a well-equipped lab?**\
> Our H2 Physics programme runs structured practical sessions in a fully equipped laboratory with SEAB-aligned apparatus. [Find out more →](https://eclatinstitute.sg/blog/h2-physics-experiments)
---
## References
[1] SEAB. (2024). _Physics (Syllabus 9478) GCE A-Level 2026._ Singapore Examinations and Assessment Board. (Scheme of Assessment; Paper 4 practical assessment; spreadsheet skills including LINEST/LOGEST.)
