Interactive practical

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

CCEA 1210 — matched to your course.

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At a glance

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Practical focuses1
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Available question marks4

Practical focus

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

Method

  1. Start with the capacitor discharged. Switch to charge it through 22 kΩ from 9.0 V and record V against time.
  2. When it is nearly fully charged, switch to discharge it through the same resistor and record V against time.
  3. For charging, V = V₀(1 − \(e^{-t/RC}\)); for discharging, V = V₀\(e^{-t/RC}\).
  4. 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 / sV charging / VV discharging / V
00.009.00
53.455.55
105.583.42
156.892.11
207.701.30

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

ProblemEffectImprovement
Capacitor not fully discharged at the startCharging curve starts above zeroShort it through a resistor first
Leakage currentNever quite reaches V₀Use a good-quality capacitor
Meter resistanceChanges effective RUse a high-resistance sensor

Risk assessment

HazardRiskControl measure
Stored chargeShockDischarge after use
Reversed polarityCapacitor damageObserve polarity markings

Common mistakes

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.

Assessment coverage

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2 questions available, worth 4 marks in total.

Practical 2
Foundation 1Standard 1
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Charging equation?

V = V₀(1 − \(e^{-t/RC}\)).

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