Physics › Capacitance › The time constant and exponential decay

The time constant and exponential decay

One number, R times C, sets the timescale for charging and discharging alike. After one time constant 37% is left, half is gone every 0.69 of one, and a log-linear plot flattens the whole exponential into a straight line you can measure.

Year 13AQA 3.7.4.4 CIE 19.3.1, 19.3.2, 19.3.3 OCR A 6.1.3(b), 6.1.3(c), 6.1.3(d), 6.1.3(e)EDEXCEL 118, 120

Builds on Charging and discharging and Capacitors and energy stored.

The maths behind it: Exponential functions and e on InkMaths.

IN THIS TOPIC

Calculate the time constant RC and read it from graphs.
Use the discharge and charging equations, and T½ = 0.69RC.
Model a discharge step by step from ΔQ/Δt = −Q/CR, and compare the model with the exponential.
Determine RC from a log-linear plot, as in required practical 9.

COMMON MISCONCEPTION

After one time constant the capacitor is empty.

RC, the circuit's own timescale

Multiply the resistance by the capacitance and the units come out as seconds. Volts per amp times coulombs per volt leaves coulombs per amp, and a coulomb per amp is a second. That product, RC, is the time constant of the circuit, the single number setting its pace. Big resistance limits the current, big capacitance means more charge to shift, and either way the process takes longer.

FIG. 1The two landmarks on the decay: half the charge gone by 0.69 RC, and 37% remaining at t = RC.

Two landmarks anchor every graph question. After one time constant, 37% of the charge remains, the fraction 1/e, so the capacitor is far from empty. And the charge halves in a fixed time,

T_{½} = 0.69RC

NOT ON THE AQA DATA SHEET: LEARN IT

the same halving again and again, exactly the constant-ratio behaviour you will meet once more in radioactive decay.

The equations

The discharge curve has an exact form.

Q = Q_{0}e^{-t/RC}

ON THE AQA DATA SHEET

and because V = Q/C and I = V/R, the pd and the current obey the same equation with their own starting values, V₀ and I₀ replacing Q₀. Charging mirrors it.

Q = Q_{0}(1 - e^{-t/RC})

ON THE AQA DATA SHEET

climbing to 63% of the final value after one time constant, the complement of the 37% left behind in discharge. In both directions, five time constants is the practical rule of thumb for a process effectively complete, with under 1% of the change still to run.

One notational point matters here. In the discharge equation Q₀ is the charge you start with, while in the charging equation it is the charge you finish with, the full value CV the capacitor is heading for.

WORKED EXAMPLE

How long to reach ten per cent?

A discharge has RC = 2.2 s. Find the time for the pd to fall to 10% of its starting value.

Set the fraction up first, V/V₀ = 0.10 = e^−t/RC.

Take the natural log of both sides. That gives −t/RC = ln 0.10, so t = RC ln 10 = 2.2 × 2.303 = 5.1 s.

Check it against the landmark ladder. Ten per cent sits between two time constants (13.5% left) and three (5% left), so the answer had to land between 4.4 s and 6.6 s. The logarithm agreed with the ladder, as it must.

Building the decay a step at a time

Where does that exponential come from, without calculus? Two facts, both already on the page. What is left on the plates sets the current, since I = V/R and V = Q/C give I = Q/CR, and that current is the rate at which charge leaves the plates. Put together, the discharge obeys $\frac{\Delta Q}{\Delta t} = -\frac{Q}{CR}$ , with the minus sign saying only that Q is on its way down.

As written that is a recipe rather than an answer, and a spreadsheet can follow a recipe. Choose a small interval Δt. Take the charge at the start of it, work out the loss over the interval as ΔQ = −(Q/CR)Δt, add it to the charge, and the result is the charge starting the next interval. Fill two columns down the page and the decay curve emerges from nothing but multiplication and addition, with no exponential assumed anywhere.

One error is built into that method, and naming it is half of what the modelling is for. Across each step the rate is held frozen at the value it had at the start, while the real capacitor's rate falls all the way through the step. Frozen at the largest value it takes, so every step removes slightly too much charge, and the modelled curve runs a little below the true exponential for the whole of its length.

WORKED EXAMPLE

A model and the equation, side by side

Model the discharge of a 100 μF capacitor through 10 kΩ from Q₀ = 1.0 mC, using steps of 0.1 s, and compare the charge after 1.0 s with the exponential's answer.

RC = 10 000 × 100 × 10⁻⁶ = 1.0 s, so each step takes ΔQ = −Q(0.1/1.0), a tenth of whatever is there. The model therefore multiplies the charge by 0.90 once per step.

Ten steps make 1.0 s, so it reports 0.90¹⁰ = 0.349 of the starting charge. The equation gives e⁻¹ = 0.368, and the model is about 5% low, exactly the direction expected.

Now cut the interval to 0.01 s. That second becomes a hundred steps of one per cent each, 0.99¹⁰⁰ = 0.366, and the shortfall drops below 1%. Halving Δt roughly halves the error, which is the test of whether a step is small enough; below about RC/100 the model sits inside the width of a plotted point.

The model is worth checking against the same straight line the measurements get. Take natural logs of the modelled charges, plot them against time, and the points should fall on a line whose gradient has magnitude 1/RC. When they do, the arithmetic in the spreadsheet and the exponential in the data booklet are describing the same decay.

Required practical 9, the straight-line test

FIG. 2Take the natural log of the discharge data and the exponential becomes a straight line, gradient minus one over RC.

The ninth required practical charges and discharges a capacitor while logging the decay, and its analysis is the standard straight-line test for an exponential. Take natural logs of the discharge equation and you have ln Q = ln Q₀ − t/RC, a straight line of ln Q against t with gradient −1/RC. Plot the logged data. Straightness confirms the decay is genuinely exponential, and the gradient gives the time constant far more reliably than reading a single point off a curve. The same log-linear move reappears, symbol for symbol, in the nuclear unit.

GUIDED PRACTICE

A capacitance from a gradient

An RP9 plot of ln V against t for a discharge through 4.7 kΩ gives a straight line of gradient −0.85 s⁻¹. Find the time constant, then the capacitance.

Show the working

The gradient is −1/RC, so RC = 1/0.85 = 1.2 s.

C = RC/R = 1.2/4700 = 2.5 × 10⁻⁴ F, or 250 μF. The graph has measured the component. One straight line, one gradient, and the printed label on the capacitor gets checked instead of trusted.

ASSESSMENT FOCUS

RC comes out in seconds when ohms and farads go in. Prefixes work too if you carry them deliberately, since kilo-ohms times microfarads give milliseconds; the danger is mixing prefixed and base units without noticing, so convert first, multiply second.
Learn the landmark fractions well enough to quote them cold. 37% left after one RC on discharge, 63% reached after one RC on charging, half gone every 0.69RC.
One equation serves Q, V and I on discharge; swap in the matching starting value and say that you have done so, since the sentence itself carries a mark.
From a graph you can take RC from T½ divided by 0.69, or from the time to fall to 37%. State which route you used.
A step-by-step model of ΔQ/Δt = −Q/CR scores on two sentences. Each row takes ΔQ = −(Q/CR)Δt from the charge the row before it held, and the rate is held fixed across a step during which it is really falling, so the model decays slightly too fast and needs Δt small compared with RC.
In the practical analysis, plot ln Q or ln V against t, quote the gradient as −1/RC, and offer the straightness of the line as your evidence that the decay is exponential.
Both exponentials are printed on the AQA data sheet, so no recall marks are available for them. T½ = 0.69RC is not printed, and nor is the log form ln Q = ln Q₀ − t/RC. Learn the halving time, and expect RP9 questions to want the log form unprompted.

CHECK YOURSELF

A 470 μF capacitor discharges through a 10 kΩ resistor. Find the time constant, the half-life, and the fraction of the charge remaining after 9.4 s.

Show a hint

Base units first; then notice what 9.4 s is in time constants.

Show the answer

$RC$ = 10 000 × 470 × 10⁻⁶ = 4.7 s.

$T_{½} = 0.69RC$ = 0.69 × 4.7 = 3.2 s.

9.4 s is two time constants, so the fraction left is e⁻² = 0.135, about 13.5%, which is two rounds of keeping 37%.

The time constant is RC: with R in ohms and C in farads, it comes out in seconds.

One time constant leaves 37%; every 0.69 of one halves the charge.

Log the data and the exponential plots as a straight line.

WORKBOOK

Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.

18 questions on this topicAnswer them one at a time and mark yourself against the mark scheme.Practise this topic

Or read them with their mark schemes on the time constant and exponential decay questions page.

6 flashcards on this topicDefinitions, off-sheet equations and a spot-the-error card, scheduled by spaced repetition in your browser.Revise with flashcards

WHERE TO GO NEXT

Required practical 9: charging and discharging a capacitor puts this topic in the lab, and the written papers ask about it.
Exponentials and logarithms is the maths this lesson leans on, worked through from GCSE.

CHECK YOUR PROGRESS

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Calculate the time constant RC and read it from graphs.
Use the discharge and charging equations, and T½ = 0.69RC.
Model a discharge step by step from ΔQ/Δt = −Q/CR, and compare the model with the exponential.
Determine RC from a log-linear plot, as in required practical 9.

Open the full revision checklist to track your progress across the whole unit.