Physics · Electric Fields · A2 Level

Discharging Capacitors

Let a charged capacitor discharge through a resistor and the current doesn't switch off cleanly — it dies away smoothly, fast at first and ever more slowly, never quite reaching zero. That shape, exponential decay, turns up everywhere in physics, and the capacitor-resistor circuit is the cleanest place to meet it properly.

8 min read

Discharging through a resistor

Connect a charged capacitor across a resistor and charge flows off the plates, driven by the potential difference across them. As charge $Q$ falls, the potential difference $V=Q/C$ falls too, reducing the discharge current $I=V/R$ — so the rate of discharge continuously slows. This self-limiting behaviour produces an exponential decay rather than a straight-line fall to zero.

DerivationV = V₀e^(−t/RC), from the discharge equation

At any instant during discharge, the current flowing off the capacitor is $I=-\dfrac{dQ}{dt}$ (negative because $Q$ is decreasing), and this current must equal the current through the resistor, $I=\dfrac{V}{R}=\dfrac{Q}{RC}$.

$V=Q/C$ from the definition of capacitance, applied at every instant.

Equating the two expressions for $I$: $-\dfrac{dQ}{dt}=\dfrac{Q}{RC}$.

This is a first-order differential equation whose solution (beyond A-level to derive fully, but standard to quote and use) is $Q=Q_0 e^{-t/RC}$, where $Q_0$ is the initial charge at $t=0$.

This is the same mathematical form as radioactive decay — both arise from a quantity whose rate of change is proportional to its own current value.

Since $V=Q/C$ and $C$ is constant, dividing both sides by $C$ gives the equivalent statement in terms of potential difference: $V=V_0 e^{-t/RC}$. The same form applies to the discharge current: $I=I_0 e^{-t/RC}$.

All three quantities — $Q$, $V$, $I$ — decay with the identical exponential shape and the identical time constant during discharge.

▪ derived

The time constant

The quantity $RC$ in the exponent has units of time (ohms times farads gives seconds) and is called the time constant, $\tau=RC$. It sets the timescale of the discharge — a larger resistance or capacitance means a slower discharge, since either limits how quickly charge can flow.

What the time constant actually means

Substituting $t=\tau=RC$ into $V=V_0e^{-t/RC}$ gives $V=V_0e^{-1}\approx0.37V_0$. So after one time constant, the potential difference (and charge, and current) has fallen to about 37% of its initial value — a 63% drop. This holds regardless of starting voltage: after a further time constant, whatever value $V$ has reached falls to 37% of that value, and so on. Since the decay never mathematically reaches zero, a capacitor is usually considered practically discharged after about five time constants, by which point under 1% of the original charge remains.

All three quantities — charge, potential difference and current — trace out the identical decay shape, just rescaled. Each graph below is marked at one time constant $\tau=RC$, where the quantity has fallen to about 37% of its starting value.

Animated graph of charge Q against time t during discharge. The curve starts at its maximum value and decays exponentially toward zero, with a dashed marker showing the value at one time constant. — $Q$ against $t$ during discharge: $Q=Q_0e^{-t/RC}$, reaching about 37% of its starting value after one time constant $\tau=RC$.

Animated graph of potential difference V against time t during discharge. The curve starts at its maximum value and decays exponentially toward zero, with a dashed marker showing the value at one time constant. — $V$ against $t$ during discharge: $V=V_0e^{-t/RC}$, the same exponential shape as $Q$, since $V=Q/C$.

Animated graph of current I against time t during discharge. The curve starts at its maximum value and decays exponentially toward zero, with a dashed marker showing the value at one time constant. — $I$ against $t$ during discharge: $I=I_0e^{-t/RC}$ — the discharge current itself decays with the same time constant as $Q$ and $V$.

Required Practical: Discharging a Capacitor

AQA Required Practical

This is AQA's Required Practical 9: investigating capacitor charge/discharge.

Equipment

Equipment	Purpose
Capacitor (known/measurable capacitance)	The component being charged and discharged
Resistor (known resistance)	Sets the time constant $\tau=RC$ together with the capacitor
Power supply	Charges the capacitor fully before discharge begins
Switch (or two-way switch)	Switches between charging and discharging the capacitor
Voltmeter, or data logger with a voltage sensor	Records voltage across the capacitor over time; a data logger removes manual reaction-time error
Stopwatch (if not using a data logger)	Times the readings at regular intervals

Method

MethodFinding τ = RC from ln V against t

Connect the capacitor in series with the resistor and a switch, with a voltmeter (or voltage sensor) connected across the capacitor.

Charge the capacitor fully by connecting it to the power supply, waiting until the voltmeter reading stabilises at the supply voltage.

At time $t=0$, switch the circuit so the capacitor discharges through the resistor alone, starting the stopwatch (or data logger) at the same instant.

Record voltage $V$ at regular time intervals as the capacitor discharges — either manually at fixed intervals, or continuously via the data logger.

Continue recording until the voltage has dropped to a small fraction of its initial value.

Take natural logarithms of both sides of $V=V_0e^{-t/RC}$ to linearise the decay: $\ln V = \ln V_0 - \dfrac{t}{RC}$.

Plot $\ln V$ against $t$ — a straight line, with $y$-intercept $\ln V_0$ and gradient $-\dfrac{1}{RC}$, from which the time constant $\tau=RC$ can be found.

▪ method

Sources of error and how to minimise them

Manual timing introduces reaction-time error

Reading a stopwatch and voltmeter by hand introduces reaction-time and timing-mismatch error, especially early in the discharge when the voltage is changing fastest. Use a data logger for continuous, precisely time-stamped recording wherever available.

Leakage current through the capacitor

A real capacitor is never a perfect insulator between its plates — a small leakage current flows internally even with no external circuit connected, so the capacitor loses charge that never appears in the resistor's current or the voltmeter's reading. Use a high-quality capacitor and a voltmeter/data-logger input with very high resistance to minimise this effect.

Starting the timing before the capacitor is fully charged

If discharge begins before the capacitor has reached the full supply voltage, $V_0$ in the analysis is wrong. Wait for the voltmeter reading to stabilise fully before switching to discharge and starting the timer.

Key takeaways

Discharge: $Q=Q_0e^{-t/RC}$, $V=V_0e^{-t/RC}$, $I=I_0e^{-t/RC}$ — all three decay with the same exponential shape and time constant.
The time constant $\tau=RC$ (units of seconds) is the time for the discharging quantity to fall to about 37% (i.e. by 63%) of its value at the start of that interval; a capacitor is practically discharged after about five time constants.
(AQA Required Practical 9) The time constant is found experimentally by discharging a capacitor through a known resistor and recording $V$ (or $I$) against $t$; plotting $\ln V$ against $t$ linearises the decay, giving a straight line of gradient $-1/RC$.
Sources of error include leakage current through the capacitor's imperfect insulation, and reaction-time error on manually timed readings, best avoided using a datalogger.

← Previous

Dielectrics

Charging Capacitors