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A2 · Practice questions · Discharging a capacitor

Down to 37% every time constant.

Six original Cambridge-style questions on exponential discharge through a resistor, the time constant τ = RC, and finding times and values from x = x₀e−t/RC.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.
Keep these straight

One exponential for charge, p.d. and current.

01
Calculation
[2 marks]

A 1000 μF capacitor discharges through a 2.2 kΩ resistor. Calculate the time constant of the circuit, and estimate the time after which the capacitor can be taken as fully discharged.

  • τ = RC = 2200 × 1000 × 10⁻⁶ = 2.2 s. ✓
  • After about five time constants the charge is below 1% of its starting value, so roughly 5 × 2.2 = 11 s. ✓
02
Calculation
[3 marks]

A 100 μF capacitor is charged to 12 V and then discharged through a 150 kΩ resistor. Calculate the time constant, the potential difference across the capacitor 20 s after the discharge begins, and the current at that instant.

  • τ = RC = 150 × 10³ × 100 × 10⁻⁶ = 15 s. ✓
  • V = V₀e−t/RC = 12 × e−20/15 = 12 × 0.264 = 3.2 V. ✓
  • I = V / R = 3.16 / 150 × 10³ = 2.1 × 10⁻⁵ A (down from an initial 12 / 150 × 10³ = 8.0 × 10⁻⁵ A). ✓
03
Analysis
[3 marks]

Explain, in terms of the current and the charge remaining, why the charge on a discharging capacitor falls exponentially rather than at a steady rate. Show also that the product RC has the unit of time.

  • The discharge current is I = V / R = Q / RC, so the rate at which charge leaves the capacitor is proportional to the charge still on it. ✓
  • As Q falls the current falls with it, so the discharge slows continuously: equal fractions of the charge are lost in equal time intervals, which is exponential decay. ✓
  • Unit of RC: Ω × F = (V A⁻¹)(C V⁻¹) = C A⁻¹ = s. ✓
04
Calculation
[3 marks]

A 100 μF capacitor discharges through a 4.7 kΩ resistor. Calculate the time taken for the charge to fall to 10% of its initial value, and state the fraction of the initial charge remaining after two time constants.

  • τ = RC = 4.7 × 10³ × 100 × 10⁻⁶ = 0.47 s. ✓
  • Q / Q₀ = 0.10 = e−t/RC, so t = RC ln 10 = 0.47 × 2.30 = 1.1 s. ✓
  • After 2τ the fraction remaining is e−2 = 0.14 (about 14%, not 0.37² by coincidence but exactly because 0.37² = e−2). ✓
05
Analysis
[2 marks]

A student measures the potential difference V across a discharging capacitor at a series of times t and plots ln(V / V₀) against t. Explain why the points should lie on a straight line, and how the time constant is obtained from the graph.

  • Taking logs of V = V₀e−t/RC gives ln(V / V₀) = −t / RC, which is a straight line through the origin in t. ✓
  • The gradient is −1/RC, so the time constant is τ = −1 / gradient (and the line passing through the origin checks that V₀ was read correctly). ✓
06
Calculation
[4 marks]

In a timing circuit a 47 μF capacitor, initially at 9.0 V, discharges through a resistor of unknown resistance R. The potential difference falls to 3.0 V in 2.6 s.

(a) Calculate the time constant of the circuit. (b) Calculate R. (c) Calculate the charge on the capacitor and the discharge current at t = 2.6 s.

  • (a) 3.0 = 9.0 e−2.6/RC, so 2.6 / RC = ln(9.0 / 3.0) = ln 3 = 1.10. ✓
  • τ = RC = 2.6 / 1.10 = 2.4 s. ✓
  • (b) R = τ / C = 2.37 / 47 × 10⁻⁶ = 5.0 × 10⁴ Ω (50 kΩ). ✓
  • (c) Q = CV = 47 × 10⁻⁶ × 3.0 = 1.4 × 10⁻⁴ C and I = V / R = 3.0 / 5.0 × 10⁴ = 6.0 × 10⁻⁵ A. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.