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The time constant and exponential decay

The time constant of a resistor-capacitor circuit is the product RC, in seconds, and after one time constant of discharge 37% of the charge is left on the plates. Discharge follows Q = Q₀e^(−t/RC), halving every 0.69RC. Taking natural logs straightens the curve, which is how RC is measured.

Time constant and exponential decay

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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 the dielectric.

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.

After one time constant the charge has fallen to 1/e, about 37%, of its initial value, not zero. Each further time constant loses the same fraction, so it never quite empties.

RC, the circuit's own timescale

Multiply the resistance by the capacitance and the unit is the second: volts per ampere multiplied by coulombs per volt gives coulombs per ampere, and a coulomb per ampere is a second. That product, RC, is the time constant of the circuit, and it sets the timescale of the process. A large resistance limits the current and a large capacitance means more charge to transfer, and either lengthens the process.

The discharge curve with two landmarks dashed in: the half-life at 0.69 RC where half the charge remains, and the time constant RC where 37 percent remains, the curve approaching zero but never arriving. — FIG. 1The two landmarks on the decay: half the charge gone by 0.69 RC, and 37% remaining at t = RC.

Two values are used in most graph questions. After one time constant, 37% of the charge remains, the fraction 1/e, so the capacitor is far from fully discharged. The charge also halves in a fixed time,

T_{½} = 0.69RC

NOT IN THE AQA DATA BOOKLET: LEARN IT

so equal time intervals give equal fractional decreases. The same constant-ratio behaviour appears in radioactive decay.

The equations

The discharge curve has an exact form.

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

IN THE AQA DATA BOOKLET

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})

IN THE AQA DATA BOOKLET

rising to 63% of the final value after one time constant, the complement of the 37% remaining in a discharge. In both directions, five time constants is the usual working rule for a process being effectively complete, with under 1% of the change remaining.

Note the difference in meaning: in the discharge equation Q₀ is the initial charge, whereas in the charging equation it is the final charge, the value CV the capacitor is approaching.

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.

Write the fraction 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 the answer against the landmarks at whole time constants: 10% lies between the values at two time constants (13.5% remaining) and three (5% remaining), so the answer must lie between 4.4 s and 6.6 s.

Building the decay a step at a time

The exponential can be obtained without calculus from two facts already established. The charge remaining 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. Together these give $\frac{\Delta Q}{\Delta t} = -\frac{Q}{CR}$ , the minus sign indicating that Q is decreasing.

That relation gives a step-by-step method a spreadsheet can follow. Choose a small interval Δt. Take the charge at the start of the interval, calculate the loss over it as ΔQ = −(Q/CR)Δt, add it to the charge, and use the result as the charge starting the next interval. Two columns of multiplication and addition then generate the decay curve, with no exponential assumed.

The method carries a systematic error, and describing it is part of what the modelling assesses. Across each step the rate is held at its start-of-step value, whereas the real rate falls throughout the step. Because that is the largest value the rate takes in the step, every step removes slightly too much charge, and the modelled curve lies a little below the true exponential throughout.

A step-by-step model of a discharge runs below the true exponential, and shrinking the step shrinks the gap without ever closing it — FIG. 2The true exponential with two step-by-step models running over it, one taking steps of 0.5 s and one of 0.1 s on a circuit whose time constant is 1.0 s. Both sit below the truth from their first step onwards, because freezing the rate at its start-of-step value takes slightly too much charge every time. At t = RC the coarse model has 0.2500 left, the fine one 0.3487, and the equation 0.3679: shrinking the step shrinks the error without ever removing it.

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 gives 0.90¹⁰ = 0.349 of the initial charge. The equation gives e⁻¹ = 0.368, so the model is about 5% low, which is the expected direction.

Now reduce the interval to 0.01 s. That second becomes a hundred steps of one per cent each, 0.99¹⁰⁰ = 0.366, and the shortfall falls below 1%. Halving Δt approximately halves the error, so a step is small enough when halving it no longer changes the result noticeably; below about RC/100 the difference is smaller than the width of a plotted point.

The model can be checked against the same straight line used for the measurements. Take natural logs of the modelled charges and plot them against time; the points should lie on a line of gradient magnitude 1/RC. If they do, the step-by-step arithmetic and the exponential in the data booklet describe the same decay.

Required practical 9, the straight-line test

Required practical 9, capacitor discharge on a log scale: ln of the measured p.d. against time is a straight line of gradient minus one over RC; ln Q shares that gradient — FIG. 3Take 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. Taking natural logs of the discharge equation gives ln Q = ln Q₀ − t/RC, a straight line of ln Q against t of gradient −1/RC. Plot the logged data: a straight line confirms that the decay is exponential, and the gradient gives the time constant more reliably than reading a single point off a curve. The same log-linear method is used 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. One straight line and one gradient give the capacitance of the component, which can then be compared with its printed value.

ASSESSMENT FOCUS

RC comes out in seconds when ohms and farads are used. Prefixes can be carried through consistently, since kilohms × microfarads give milliseconds; the risk is mixing prefixed and base units, so convert first and multiply second.
Learn the standard fractions: 37% remaining after one RC on discharge, 63% reached after one RC on charging, and half gone every 0.69RC.
One equation serves Q, V and I on discharge; substitute the matching initial value and state that you have done so, since that statement carries credit.
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 requires two statements: each row takes ΔQ = −(Q/CR)Δt from the charge held by the row before it, 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. T½ = 0.69RC is not printed, and nor is the log form ln Q = ln Q₀ − t/RC, so learn the halving time and expect RP9 questions to require the log form without prompting.

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

Convert to base units first, then express 9.4 s as a number of 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 remaining is e⁻² = 0.135, about 13.5%, which is 37% of 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.

19 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 step by step.

USED LATER IN

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.