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Capacitor charge, discharge and time constant

Record capacitor voltage during charge and discharge, linearise the discharge data and determine the RC time constant and capacitance.

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A-Level Physics

Capacitor charge, discharge and time constant

Pearson Edexcel 9PH0 — matched to your course.

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Practical focus

Record capacitor voltage during charge and discharge and determine the RC time constant.

Practical method

Aim

Investigate the charging and discharging of a capacitor and compare both with the time constant RC.

Variables

Independent variable: time
Dependent variable: capacitor p.d. during charging and discharging
Control variables: R and C; supply p.d..

Apparatus

470 μF capacitor
22 kΩ resistor
9.0 V supply
Two-way switch
Data logger with voltage sensor

Method

Start with the capacitor discharged. Switch to charge it through 22 kΩ from 9.0 V and record V against time.
When it is nearly fully charged, switch to discharge it through the same resistor and record V against time.
For charging, V = V₀(1 − \(e^{-t/RC}\)); for discharging, V = V₀\(e^{-t/RC}\).
Find the time to reach 63% (charging) and fall to 37% (discharging) of V₀; both equal RC.

Sample results

C = 470 μF, R = 22 kΩ, V₀ = 9.0 V.

t / s	V charging / V	V discharging / V
0	0.00	9.00
5	3.45	5.55
10	5.58	3.42
15	6.89	2.11
20	7.70	1.30

Analysis

Charging reaches 0.63 × 9.0 = 5.7 V at about 10.3 s and discharging falls to 3.3 V at the same time: both give RC ≈ 10.3 s. At each time the charging and discharging p.d.s add to 9.0 V.

Uncertainty

The data logger samples every 0.1 s, so the 63% and 37% times can be read to about ±0.2 s (2%). The capacitor tolerance (±10%) affects comparison with the nominal value.

Errors and improvements

Problem	Effect	Improvement
Capacitor not fully discharged at the start	Charging curve starts above zero	Short it through a resistor first
Leakage current	Never quite reaches V₀	Use a good-quality capacitor
Meter resistance	Changes effective R	Use a high-resistance sensor

Risk assessment

Hazard	Risk	Control measure
Stored charge	Shock	Discharge after use
Reversed polarity	Capacitor damage	Observe polarity markings

Common mistakes

Using the discharge equation for charging.
Reading 50% instead of 63% for the time constant.
Starting with a partly charged capacitor.

Worked analysis

Charging: V = 9.0(1 − \(e^{-10/10.34}\)) = 9.0 × 0.62 = 5.58 V at 10 s. Discharging: V = 9.0 \(e^{-10/10.34}\) = 3.42 V. They add to 9.0 V.

Common misconception

Charging current is largest at the start and falls to zero, even though the p.d. is still rising.

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