CP11 – Capacitor Charge & Discharge | Edexcel A Level Physics

Theory — Capacitor Charge and Discharge

A capacitor charging and discharging through a resistor follows an exponential relationship with time.

Objectives

Measure how the potential difference across a capacitor changes with time during discharge
Plot V vs t and ln(V) vs t to confirm the exponential decay relationship
Determine the time constant τ = RC from the gradient of the ln(V) vs t graph
Investigate how changing R and C affects τ

Discharge Equation

When a charged capacitor discharges through a resistor, the voltage decreases exponentially:

V = V₀ e^(−t/RC) Taking natural log: ln(V) = ln(V₀) − t/RC

V₀ = initial voltage · t = time · R = resistance · C = capacitance · τ = RC = time constant

The time constant τ = RC is the time for the voltage to fall to V₀/e ≈ 0.368 V₀. A plot of ln(V) against t gives a straight line with gradient = −1/RC = −1/τ.

Charge Equation

V = V₀(1 − e^(−t/RC))

During charging, V rises from 0 toward V₀ exponentially. The time constant τ = RC is the time to reach V₀(1 − 1/e) ≈ 0.632 V₀.

Time Constant τ = RC

τ = RC (seconds, when R in Ω and C in Farads)

Example: R = 470 kΩ, C = 100 μF → τ = 470×10³ × 100×10⁻⁶ = 47 s
Example: R = 1 kΩ, C = 5 μF → τ = 1×10³ × 5×10⁻⁶ = 0.005 s = 5 ms

Larger τ means slower charging/discharging. After 5τ the capacitor is effectively fully charged or discharged (>99%).

Using the Oscilloscope

The oscilloscope displays voltage on the y-axis (vertical scale in V/div) and time on the x-axis (time base in s/div or ms/div). For slow RC circuits (τ ~ seconds), a stopwatch and voltmeter are used instead. For fast RC circuits (τ ~ ms), the oscilloscope directly displays the exponential curve and a square wave supply is used to repeatedly charge and discharge the capacitor.

Procedure

Equipment

100 μF capacitor · 470 kΩ resistor · 6 V battery or PSU · Two-way switch (or flying lead) · Oscilloscope or voltmeter · Stopwatch with lap timer · Connecting leads · Multimeter

1

Measure R with a multimeter

Use a multimeter on resistance mode to measure the actual value of the resistor. A 470 kΩ resistor may read anywhere from 423–517 kΩ (within 10% tolerance). Record the measured value.

💡 Always use the measured R value, not the nominal value, when calculating τ = RC.

2

Set up the oscilloscope

Connect the oscilloscope directly across the PSU. Adjust the time base until a steady horizontal line appears. Set vertical scale so 6 V sits near the top. Record the EMF V₀.

💡 In this simulation the oscilloscope shows a live green trace sweeping left to right, just like a real CRT oscilloscope.

3

Build the circuit

Connect: PSU → switch (position A: charge) → resistor R → capacitor C → back to PSU. Connect the oscilloscope (or voltmeter) across the capacitor only. Position B of the switch connects R across the capacitor to allow discharge.

💡 In this simulation use the two-way switch buttons: ⚡ Charge puts switch in position A, 📉 Discharge puts switch in position B.

4

Charge the capacitor

Set switch to position A (charge). Watch the voltage rise on the oscilloscope toward V₀. Wait until V reaches V₀ (fully charged). Record this as V₀.

💡 Full charging takes about 5τ. For R=470kΩ, C=100μF: τ=47s, so 5τ≈235s. In the simulation this is scaled for visibility.

5

Discharge and record readings

Set switch to position B (discharge) and start the stopwatch simultaneously. Use a data logger with automatic sampling to record V against t continuously until the capacitor is fully discharged (5τ). Repeat the full charge-discharge cycle at least 3 times for the same R and C to check repeatability.

💡 In this simulation, clicking Discharge automatically starts logging (V, t) points every few seconds — exactly like a real data logger. Recording auto-stops once the capacitor reaches 5τ (over 99% discharged). Complete at least 3 full charge-discharge runs for the same R, C — each run adds its own set of points to the same graph, letting you check that τ comes out consistently across independent trials.

6

Plot ln(V) vs t

Calculate ln(V) for each reading. Plot ln(V) (y-axis) against t (x-axis). Draw a best-fit straight line. Gradient = −1/τ = −1/RC. Calculate τ and compare with RC from your measured values.

💡 The y-intercept of the ln(V) vs t graph = ln(V₀), so V₀ = e^(y-intercept). This gives an independent check of V₀.

Simulation Speed

Points are auto-logged every 5 simulated seconds, so speed only affects how long you wait to see the full curve.

EMF (Battery / PSU)

Resistance R

Capacitance C

Time Constant

τ = RC47.0 s

5τ (full cycle)235 s

V at τ (0.368·V₀)2.21 V

Experimental Runs

Runs completed0 / 3 minimum

Auto-stop at5τ (≈99.3% discharged)

Two-Way Switch

Switch open — capacitor idle

0.00 V

Capacitor voltage V(t) — Run 1 not started

Elapsed time t0.00 s

Current I0.00 μA

Charge Q = CV0.00 μC

Auto-logged points: 0

OSCILLOSCOPE — V/div · — s/div

DATA LOGGER — V vs t live trace

Data Table

Raw readings from both charging and discharging (auto-logged). Calculate ln(V), ln(I), ln(Q), or ln(EMF−V) yourself for the graph you're investigating. τ = RC = — s

#	Run	Phase	t / s	V / V
No data yet — go to Simulation tab.

Graph & Analysis

Choose an investigation — collect readings in the Simulation tab first (charging and discharging both auto-record).

Uncertainty Estimate

%U in V (datalogger ADC ±0.01 V)—

%U in t (auto-logged, negligible)—

%U in τ (from gradient)—

Uncertainty in V matters most at low voltages late in the discharge, where ln(V) is most sensitive to small changes in V.

Discussion Questions

Write your answers and reveal model answers when ready.

Q1

Explain why the graph of ln(V) against t is a straight line during discharge, and state what the gradient and y-intercept represent.

During discharge, V = V₀e^(−t/RC). Taking the natural logarithm of both sides: ln(V) = ln(V₀) − t/RC. This is of the form y = c + mx where y = ln(V), x = t, c = ln(V₀) and m = −1/RC = −1/τ. Since both V₀ and RC are constants, the graph of ln(V) vs t is a straight line. The gradient = −1/τ = −1/RC (negative because V decreases with time). The y-intercept = ln(V₀), so V₀ = e^(y-intercept), providing an independent check of the initial voltage.

Q2

A capacitor discharges from 6.0 V through a 470 kΩ resistor. The voltage after 30 s is measured as 3.8 V. Calculate the capacitance C.

From V = V₀e^(−t/RC): 3.8 = 6.0 × e^(−30/(470×10³ × C)). Rearranging: e^(−30/(470×10³ × C)) = 3.8/6.0 = 0.6333. Taking ln: −30/(470×10³ × C) = ln(0.6333) = −0.4572. So 470×10³ × C = 30/0.4572 = 65.62. C = 65.62/(470×10³) = 1.396×10⁻⁴ F = 140 μF. (Close to the nominal 100 μF — the discrepancy could reflect a capacitor with higher than nominal capacitance, which is common, or experimental error in V.)

Q3

Explain what the time constant τ = RC represents physically, and state the voltage across the capacitor after one, two and three time constants.

The time constant τ = RC is the time for the capacitor voltage to fall to 1/e (≈ 36.8%) of its initial value during discharge. It represents the characteristic timescale of the RC circuit — a larger τ means the capacitor discharges more slowly. After 1τ: V = V₀/e ≈ 0.368 V₀. After 2τ: V = V₀/e² ≈ 0.135 V₀. After 3τ: V = V₀/e³ ≈ 0.050 V₀. After 5τ the voltage has fallen to less than 1% of V₀ and the capacitor is considered fully discharged.

Q4

In the experiment, a student charges the capacitor and then moves the flying lead to begin discharge. They notice their first few readings are higher than expected from the theory. Suggest a reason.

There is a time delay between moving the flying lead (starting the discharge) and starting the stopwatch. During this delay the capacitor continues to discharge, so when the student starts timing, the actual time elapsed is already greater than zero, but the stopwatch reads zero. This shifts all readings to the left on the t-axis — the measured voltage at t = 0 appears correct (V₀) but subsequent readings appear too high because the student's recorded t is less than the true t. To minimise this, the stopwatch and the switch change should be operated simultaneously. Using a data logger with an automatic trigger eliminates this error entirely.

Q5

Doubling R while keeping C fixed doubles τ. Explain why, in terms of the current flowing during discharge, and describe how this affects the shape of the V vs t curve.

During discharge, the current at any instant is I = V/R. Doubling R halves the current at every voltage. Since Q = CV, the rate of charge removal dQ/dt = I = V/R. A smaller current means charge is removed more slowly from the capacitor plates, so the voltage falls more slowly. This is why doubling R doubles τ = RC. On the V vs t graph, the curve still starts at V₀ and decays exponentially to zero, but it is "stretched" horizontally — it takes twice as long to reach the same fraction of V₀. On the ln(V) vs t graph, the straight line has half the magnitude of gradient (less steep), confirming |gradient| = 1/τ = 1/RC.

Capacitor Charge and Discharge Through a Resistor

Theory — Capacitor Charge and Discharge

Objectives

Discharge Equation

Charge Equation

Time Constant τ = RC

Using the Oscilloscope

Procedure

Equipment

Data Table

Graph & Analysis

Uncertainty Estimate

Interpretation

Discussion Questions