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Learn9702Paper 4 · A Level Structured QuestionsCapacitanceDischarging a capacitor
Capacitance · Subtopic 3 of 3

Discharging a Capacitor

When a capacitor discharges through a resistor, the current is proportional to the charge remaining, so charge, p.d. and current all decay exponentially: x=x0e−t/RCx=x_0e^{-t/RC}. The time constant τ=RC\tau=RC is the time for any of them to fall to 1e\dfrac1e (37%37\%) of its starting value. A graph of ln⁡V\ln V against tt is a straight line with gradient −1RC-\dfrac1{RC}.

Cambridge 9702 Paper 4Edexcel YPH11 WPH14AQA 7408 Paper 2≈ 40 minRevision 7 · updated 2026-10-27
Section 1 of 14

Learning objectives

  • Explain why discharge is exponential.
  • Use x=x0e−t/RCx=x_0e^{-t/RC} for QQ, VV and II.
  • Use τ=RC\tau=RC and the half-life 0.69RC0.69RC.
  • Find RCRC from a ln⁡V\ln V against tt graph.
Pause and think

A capacitor loses half its charge in 55 seconds. How long until it has lost three quarters?

Section 2 of 14

Core idea

Start hereA fading light

Discharge a capacitor through a lamp: it starts bright and fades, quickly at first, then ever more slowly. The less charge is left, the more gently it drains. That's the signature of exponential decay.

The big idea

x=x0e−t/RCx=x_0e^{-t/RC} for QQ, VV, II. τ=RC\tau=RC: time to fall to 37%. t1/2=RCln⁡2t_{1/2}=RC\ln2.

1. Where the exponential comes from

Q=Q0e−t/RCQ=Q_0e^{-t/RC}
  1. I=VR=QRCI=\frac VR=\frac Q{RC}
    p.d. across RR equals the capacitor's.
  2. I=−dQdtI=-\frac{dQ}{dt}
    Current drains the charge.
  3. dQdt=−QRC\frac{dQ}{dt}=-\frac Q{RC}
    Rate of loss ∝ what's left.
  4. ∫dQQ=−∫dtRC⇒ln⁡QQ0=−tRC\int\frac{dQ}Q=-\int\frac{dt}{RC}\Rightarrow\ln\frac Q{Q_0}=-\frac t{RC}
    Separate and integrate.
  5. Q=Q0e−t/RCQ=Q_0e^{-t/RC}
    Exponentiate.
V=Q/CV=Q/C and I=V/RI=V/R follow the same curve.
Discharge curve: V = V₀ e^(−t/RC)
246810246810120time (s)V (V)dischargeτ = RC37% of V₀
Bigger RR or bigger CC means slower discharge. After 5RC5RC it is essentially empty (under 1%).
Worked exampleC = 470 μF, R = 10 kΩ, from 10 V

τ\tau; VV after τ\tau; half-time.

Goal
Three values.
Strategy
Formulas.
  1. τ=470×10−6×104=4.7 s\tau=470\times10^{-6}\times10^4=4.7\text{ s}
    RCRC.
  2. V=10e−1=3.7 VV=10e^{-1}=3.7\text{ V}
    After one time constant.
  3. t1/2=4.7ln⁡2=3.3 st_{1/2}=4.7\ln2=3.3\text{ s}
    e−t/τ=12e^{-t/\tau}=\frac12.
Answer4.74.7 s; 3.73.7 V; 3.33.3 s.
✓ Checkt1/2<τt_{1/2}<\tau since 50% > 37% ✓.

2. Finding RC from data

ln V against t is a straight line
246810−3−2−11230time (s)ln(V / V)gradient = −1/RC
Taking logs turns the curve into a line: gradient −1RC=−0.5-\dfrac1{RC}=-0.5, so RC=2RC=2 s. This is how RCRC is found experimentally.
Linearising
  1. ln⁡V=ln⁡V0−tRC\ln V=\ln V_0-\frac t{RC}
    Take logs.
  2. gradient=−1RC\text{gradient}=-\frac1{RC}
    Plot ln⁡V\ln V against tt.
Common misconceptionLinear discharge It never "runs out" at a fixed time; use 5RC5RC as "practically empty" (0.7% left).

Summary

  • x=x0e−t/RCx=x_0e^{-t/RC}.
  • t1/2=RCln⁡2t_{1/2}=RC\ln2.
Key words
Time constant
RCRC.
HistoryEuler's number. The constant e=2.718…e=2.718\ldots was found by Jacob Bernoulli in 16831683 studying compound interest, and named by Leonhard Euler. It appears whenever a quantity changes at a rate proportional to itself: capacitors, radioactive nuclei, cooling coffee and bank savings.
Fun factWhy you should wait to open a TV. Old CRT televisions and microwave ovens contain high-voltage capacitors that can hold a dangerous charge long after unplugging. With RCRC of tens of seconds to minutes and a huge starting voltage, technicians always discharge them deliberately before working.
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