All subjects · Physics · WPH14 (Unit 4)

Core practical 11: capacitor charge and discharge

Core practical 11 uses an oscilloscope or data logger to display and analyse the p.d. across a capacitor as it charges and discharges through a resistor.

Method with a data logger

Build a circuit with a d.c. supply, a two-way switch, the capacitor and a resistor. In one switch position the capacitor charges; in the other it discharges through the resistor.
Connect a voltage sensor across the capacitor to a data logger.
Charge fully, then switch to discharge and record V against t at a high sampling rate.
Use a large R and C (RC of several seconds) or the logger for shorter times.

Method with an oscilloscope

For a small RC (milliseconds), replace the supply and switch with a square-wave signal generator. Each half-cycle charges, then discharges, the capacitor. Choose a period much longer than RC so it fully charges and discharges. Read V and t from the screen using the volts/div and time-base settings.

Analysis

Read the time for V to fall to 37% of V_0 to estimate RC.
Better: plot ln V against t. The gradient is -1/RC; using all the data reduces the effect of random errors.
Compare RC with the value from the labelled R and C, or find C from the gradient and a known R. Capacitors often have tolerances of ±10% or more.

Worked example

On an oscilloscope set to 5 ms/div, the p.d. falls to 37% of its initial value after 2.4 divisions. Find RC.

Time = 2.4 x 5 ms = 12 ms
At t = RC, V = 0.37V_0

Answer: RC = 12 ms

Key idea: Logger or oscilloscope across C; square wave for short RC; time to 37% gives RC; ln V against t has gradient -1/RC.

Check yourself

An oscilloscope time base is 5 ms/div. The p.d. across a discharging capacitor falls to 37% of V_0 after 2.4 divisions. Find RC in ms.
Show answer
12 ms. t = 2.4 x 5 = 12 ms, and the time to fall to 37% is RC.
A graph of ln V against t has a gradient of -0.050 s^-1 for a 100 kΩ resistor. Find C in µF.
Show answer
200 µF. RC = 1 / 0.050 = 20 s, so C = 20 / 100000 = 2.0 x 10^-4 F = 200 µF.
A data logger shows the p.d. halving in 7.0 s. Find RC in s, to 3 s.f.
Show answer
10.1 s. t_1/2 = RC ln 2, so RC = 7.0 / 0.693 = 10.1 s.
A graph of ln V against t, with V in volts, has a y-intercept of 2.48. Find the initial p.d. V_0 in V, to 3 s.f.
Show answer
11.9 V. ln V_0 = 2.48, so V_0 = e².48 = 11.9 V.
Why is a data logger used rather than a voltmeter and stopwatch?
- It makes the capacitor discharge more slowly
- It removes the need for a resistor
- It records many readings at precise, short intervals
- It increases the capacitance
Show answer
It records many readings at precise, short intervals. The p.d. changes quickly at first; a logger can sample many times per second.
When using a square-wave generator and an oscilloscope, how should the signal period be chosen?
- Much longer than RC, so the capacitor fully charges and discharges
- Much shorter than RC
- Exactly equal to RC
- It does not matter
Show answer
Much longer than RC, so the capacitor fully charges and discharges. Each half-cycle must last several time constants.
Why is a graph of ln V against t plotted?
- It makes the p.d. larger
- It removes the need to measure time
- It gives a straight line whose gradient gives RC using all the data
- It gives a curve that is easier to read
Show answer
It gives a straight line whose gradient gives RC using all the data. A straight line lets you use a best-fit gradient, -1/RC.

Practise with instant feedback

Sources

Last updated 7 November 2026. Spotted a mistake? Tell us.