Physics 9702 · Paper 4

Discharge & the Time Constant

Let a capacitor push its charge through a resistor and every quantity — charge, p.d., current — dies away along the same curve. This is the most-examined idea in the topic: it has appeared in every session for three years.

Why the decay is exponential (three marks)

The chain, each link creditable (9702/41/M/J/24 Q6(a)) (9702/42/M/J/25 Q7(c)): the capacitor's p.d. is proportional to its charge ( $V = Q/C$ ); that p.d. sits across the resistor, so the current is proportional to it ( $I = V/R$ ); and the current is the rate at which charge leaves. Put together: the rate of loss of charge is proportional to the charge remaining — the signature of exponential decay.
Less charge → less push → slower loss: the decay keeps slowing but never quite finishes.

x = x_0\, e^{-t/RC}, \qquad \tau = RC

Symbols

$x$ = charge Q, p.d. V or current I — all decay identically (—)
$x_0$ = the value at t = 0 (—)
$\tau = RC$ = the time constant of the circuit (s)

Discharge sampled from Q = Q₀e^(−t/RC). After each time constant τ = RC the value falls to 37% of what it was: 0.37Q₀ at τ, 0.14Q₀ at 2τ. Charge, p.d. and current all follow this same curve.

The $\tau = RC$ sets the pace: after each $\tau$ the value falls to $1/e \approx 37\%$ of what it was — not 50%. A 2024 paper read $\tau$ off an $I$ – $t$ graph exactly this way: find where the current has fallen to $I_0/e = 0.048$ mA, read $\tau = 4.3$ s, then $C = \tau/R$ (9702/41/M/J/24 Q6(b)).

Worked example

An isolated sphere leaks to earth (2025 paper)

A sphere of capacitance 69 pF holding 83 pC discharges to earth through $120\ \text{M}\Omega$ . How long until the charge is 26 pC? (9702/42/O/N/25 Q6(c))

$\tau = RC = 120\times10^{6} \times 69\times10^{-12} = 8.3\times10^{-3}\ \text{s}$ .
$26 = 83\,e^{-t/\tau}$ , so $t = \tau \ln(83/26) = 9.6\times10^{-3}\ \text{s}$ .

Answer

t \approx 9.6\ \text{ms}

Common mistake

Two recurring slips. First, percentages need the ln step: for a fall to 15%,

t = \tau\ln(1/0.15) = 1.9\tau

— it is not a neat number of “half-lives” (9702/42/F/M/25 Q5(c)(ii)). Second, when a network discharges,

\tau

uses the combined capacitance — the same 2025 question needed

\tau = R \times \tfrac{3}{2}C = 3.6

s, and using the single capacitor's 44

\mu\text{F}

instead of the network's 66 lost the marks.

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