Electric current

AS · 14 min

An electric current is a flow of charged particles. This topic sets up the language of the whole electricity section: what a current actually is, how charge and current are related by $Q = It$ , why charge always comes in whole-number multiples of $e$ , and how fast the charge carriers really move inside a wire ( $I = nAvq$ ). It also introduces the circuit symbols you will use for the rest of the course. Expect a definition or a $Q = It$ calculation in Paper 2, and drift-speed ratios in Paper 1 multiple-choice questions.

What a current is

Matter contains charged particles: electrons (negative) and the positive nuclei of atoms. Usually they are locked in place or move randomly in all directions, so there is no overall movement of charge. A current exists when there is a net flow of charge in one direction.

The particles that move are called charge carriers. Which particles they are depends on the material.

Material	Charge carriers
Metal	Free (delocalised) electrons
Electrolyte (a solution or molten ionic compound)	Positive and negative ions
Ionised gas (a spark, a fluorescent tube)	Electrons and positive ions
Semiconductor	Electrons (and "holes", which behave like positive carriers)

In a metal, each atom gives up one or more outer electrons to a "sea" of free electrons that can wander through the lattice of positive ions. When a cell is connected, an electric field is set up along the wire and the free electrons drift towards the positive terminal.

Definition

An electric current is a flow of charge carriers. The current at a point is the rate of flow of charge past that point:

I = \frac{\Delta Q}{\Delta t}

Conventional current

Circuit diagrams show conventional current, which flows from the positive terminal of a supply, round the circuit, to the negative terminal. This convention was fixed before anyone knew about electrons. In a metal the electrons actually move the other way, from negative to positive. A flow of negative charge to the left is exactly equivalent to a flow of positive charge to the right, so every rule in this course works with conventional current and you rarely need to think about electron direction.

Watch out

Do not say that "current flows from negative to positive" when asked for the direction of the current. The current (conventional) is from $+$ to $-$ outside the supply; the electrons move from $-$ to $+$ . If a question asks about electron flow, say "electrons" explicitly.

Charge and the coulomb

The ampere is an SI base unit (see Physical quantities and SI units). The unit of charge, the coulomb, is derived from it.

Key result

Q = It

where $Q$ is the charge in coulombs (C), $I$ is the current in amperes (A) and $t$ is the time in seconds (s). This holds for a constant current.

Definition

One coulomb is the charge that passes a point in a circuit when a current of one ampere flows for one second: $1\ \text{C} = 1\ \text{A s}$ .

When the current varies, $Q = It$ applies over each short interval, and the total charge is the area under a current–time graph. This is exactly like finding displacement from the area under a velocity–time graph.

Charge is quantised

Every charge carrier carries a whole-number multiple of the elementary charge:

e = 1.60 \times 10^{-19}\ \text{C}

An electron carries $-e$ ; a proton carries $+e$ ; an ion such as $\ce{Cu^{2+}}$ carries $+2e$ . Nobody has ever isolated a free particle with a charge such as $1.5e$ . Because charge can only take values $\pm e,\ \pm 2e,\ \pm 3e, \ldots$ , we say that charge is quantised.

Key result

Q = Ne

where $N$ is a whole number of elementary charges.

The elementary charge is so small that ordinary currents involve enormous numbers of electrons. One coulomb is $1/(1.60 \times 10^{-19}) = 6.25 \times 10^{18}$ electrons.

Tip

Quarks carry charges of $\pm\tfrac{1}{3}e$ and $\pm\tfrac{2}{3}e$ , but they are never found on their own: they are always bound inside particles whose total charge is a whole multiple of $e$ (see Fundamental particles). So the statement "charge is quantised in units of $e$ " holds for every free particle and every charge carrier.

Charge and number of electrons

A current of $0.25\ \text{A}$ passes through a lamp for $3.0$ minutes. Calculate (a) the charge that flows through the lamp, (b) the number of electrons that pass through it.

Solution

(a) Convert the time to seconds: $t = 3.0 \times 60 = 180\ \text{s}$ .

Q = It = 0.25 \times 180 = 45\ \text{C}

(b)

N = \frac{Q}{e} = \frac{45}{1.60 \times 10^{-19}} = 2.8 \times 10^{20}

Is this charge possible?

In an experiment to measure the charge on small oil drops, a student records charges of $4.8 \times 10^{-19}\ \text{C}$ and $5.6 \times 10^{-19}\ \text{C}$ . Explain which of these could be a correct measurement.

Solution

Divide each charge by $e$ :

\frac{4.8 \times 10^{-19}}{1.60 \times 10^{-19}} = 3.0, \qquad \frac{5.6 \times 10^{-19}}{1.60 \times 10^{-19}} = 3.5

Charge is quantised: a drop can only carry a whole number of elementary charges. $4.8 \times 10^{-19}\ \text{C}$ is $3e$ (three extra electrons) and is possible. $5.6 \times 10^{-19}\ \text{C}$ would be $3.5e$ , which is not a whole number, so this measurement must be wrong.

Charge from a current–time graph

The current in a component rises steadily from zero to $40\ \text{mA}$ in $2.0\ \text{s}$ , stays at $40\ \text{mA}$ until $t = 5.0\ \text{s}$ , then falls steadily to zero at $t = 6.0\ \text{s}$ . Calculate the total charge that passes through the component.

Solution

(0, 0) -- (2, 40) (2, 40) -- (5, 40) (5, 40) -- (6, 0) fill 0 2 y = 20 x fill 2 5 y = 40 fill 5 6 y = 40 - 40 (x - 5)

The graph shows current in mA against time in s. The charge is the shaded area. Split it into a triangle, a rectangle and a triangle, working in amperes:

Q = \tfrac{1}{2}(2.0)(0.040) + (3.0)(0.040) + \tfrac{1}{2}(1.0)(0.040) = 0.040 + 0.120 + 0.020 = 0.18\ \text{C}

Converting mA to A before multiplying is the step most often missed.

Circuit diagrams and symbols

Cambridge expects you to recognise and draw the standard symbols below and to read circuit diagrams built from them. Draw wires as straight lines with right-angled corners, show a junction where wires join with a dot, and never draw a gap in a wire unless you mean a break in the circuit.

Circuit symbols used in this course. For the cell, the long thin line is the positive terminal. For the diode, conventional current can pass in the direction the triangle points.

Two measuring instruments matter from the start.

An ammeter measures the current through a component, so it is connected in series with it: the same charge must flow through both. An ideal ammeter has zero resistance, so it does not change the current it is measuring.
A voltmeter measures the potential difference across a component, so it is connected in parallel with it. An ideal voltmeter has infinite resistance, so it draws no current. (Potential difference is defined in Potential difference and power.)

A galvanometer is a very sensitive current meter whose zero is at the centre of the scale. It is used to detect when a current is exactly zero (see Potentiometers and null methods).

How fast do the charge carriers move?

When you switch on a light, it comes on almost at once. It is tempting to think the electrons race round the circuit, but they do not. They drift remarkably slowly. The equation that shows this links the current to what is going on inside the conductor.

Deriving $I = nAvq$

Consider a conductor of cross-sectional area $A$ in which every charge carrier has charge $q$ and drifts with average speed $v$ . Let $n$ be the number density of charge carriers: the number of carriers per unit volume (unit $\text{m}^{-3}$ ).

In a time $t$ , every carrier moves a distance $vt$ . So all the carriers in a length $vt$ of the conductor pass through a given cross-section.

Volume of that length of conductor: $A \times vt$ .
Number of carriers in it: $n \times Avt$ .
Charge passing the cross-section: $Q = nAvt \times q$ .

So the current is

I = \frac{Q}{t} = \frac{nAvtq}{t} = nAvq

Key result

I = nAvq

$n$ : number density of charge carriers ( $\text{m}^{-3}$ ); $A$ : cross-sectional area ( $\text{m}^2$ ); $v$ : average drift speed ( $\text{m s}^{-1}$ ); $q$ : charge on each carrier (C). For electrons, $q = e = 1.60 \times 10^{-19}\ \text{C}$ .

The speed $v$ is called the drift speed (or drift velocity). The electrons in a metal also have large random speeds (around $10^{5}$ to $10^{6}\ \text{m s}^{-1}$ ) as they bounce around the lattice, but these random motions have no overall direction and carry no net charge. The drift speed is the small average speed superimposed on that random motion by the electric field.

What the equation tells you

Rearranged, $v = \dfrac{I}{nAq}$ . For a given current:

A thinner wire (smaller $A$ ) has a faster drift speed. In a series circuit the current is the same everywhere, so where the wire narrows, the electrons speed up.
A material with fewer charge carriers per unit volume (smaller $n$ ) has a faster drift speed. Metals have $n \approx 10^{28}$ to $10^{29}\ \text{m}^{-3}$ ; a semiconductor such as silicon has far fewer, so its carriers drift much faster for the same current.
An insulator has almost no free carriers ( $n \approx 0$ ), so it cannot carry a measurable current at ordinary voltages.

So why does a lamp light instantly? The wire is already full of free electrons. When the switch is closed, the electric field is established along the whole circuit at nearly the speed of light, and electrons everywhere, including those already inside the filament, start drifting at once.

Drift speed in a copper wire

A copper wire of diameter $1.0\ \text{mm}$ carries a current of $2.0\ \text{A}$ . The number density of free electrons in copper is $8.5 \times 10^{28}\ \text{m}^{-3}$ . Calculate (a) the drift speed of the electrons, (b) the time an electron takes to drift $1.0\ \text{m}$ along the wire.

Solution

(a) Cross-sectional area, with radius $0.50 \times 10^{-3}\ \text{m}$ :

A = \pi r^2 = \pi \times (0.50 \times 10^{-3})^2 = 7.85 \times 10^{-7}\ \text{m}^2

v = \frac{I}{nAe} = \frac{2.0}{8.5 \times 10^{28} \times 7.85 \times 10^{-7} \times 1.60 \times 10^{-19}} = 1.9 \times 10^{-4}\ \text{m s}^{-1}

(b)

t = \frac{1.0}{1.87 \times 10^{-4}} = 5.3 \times 10^{3}\ \text{s}

That is about an hour and a half to drift one metre: drift speeds are a fraction of a millimetre per second.

Two wires in series

A copper wire X of diameter $1.2\ \text{mm}$ is joined end to end with a copper wire Y of diameter $0.60\ \text{mm}$ . A current of $3.0\ \text{A}$ passes through both wires. (a) State the current in Y. (b) Determine the ratio $\dfrac{\text{drift speed in Y}}{\text{drift speed in X}}$ . (c) Calculate the drift speed in Y. The number density of free electrons in copper is $8.5 \times 10^{28}\ \text{m}^{-3}$ .

Solution

(a) The wires are in series, so charge cannot build up at the join: the current in Y is also $3.0\ \text{A}$ .

(b) $I$ , $n$ and $e$ are the same in both wires, so $v \propto \dfrac{1}{A}$ , and $A \propto d^2$ :

\frac{v_Y}{v_X} = \frac{A_X}{A_Y} = \left(\frac{d_X}{d_Y}\right)^2 = \left(\frac{1.2}{0.60}\right)^2 = 4.0

v_Y = \frac{3.0}{8.5 \times 10^{28} \times 2.83 \times 10^{-7} \times 1.60 \times 10^{-19}} = 7.8 \times 10^{-4}\ \text{m s}^{-1}

Finding the number density

A metal strip has a rectangular cross-section $2.0\ \text{mm}$ by $0.50\ \text{mm}$ . When the current in it is $1.5\ \text{A}$ , the drift speed of the free electrons is $1.1 \times 10^{-4}\ \text{m s}^{-1}$ . (a) Calculate the number density of free electrons. (b) The metal has $8.5 \times 10^{28}$ atoms per cubic metre. Deduce how many free electrons each atom contributes.

Solution

(a) $A = 2.0 \times 10^{-3} \times 0.50 \times 10^{-3} = 1.0 \times 10^{-6}\ \text{m}^2$ .

n = \frac{I}{Ave} = \frac{1.5}{1.0 \times 10^{-6} \times 1.1 \times 10^{-4} \times 1.60 \times 10^{-19}} = 8.5 \times 10^{28}\ \text{m}^{-3}

(b) The number of free electrons per cubic metre equals the number of atoms per cubic metre, so each atom contributes one free electron.

Watch out

In $I = nAvq$ , $A$ is the cross-sectional area in $\text{m}^2$ . A diameter in millimetres must be halved and converted: $d = 0.60\ \text{mm}$ gives $r = 0.30 \times 10^{-3}\ \text{m}$ . Forgetting either step changes the answer by a factor of 4 or $10^6$ .
$n$ is a number per unit volume, not the total number of electrons in the wire. The length of the wire does not appear in the equation.
The drift speed is not the speed at which "the electricity" travels. The signal (the electric field) travels at nearly the speed of light; the electrons themselves drift slowly.

Exam tip

"Define electric current" or "what is meant by an electric current" earns its mark for "flow of charge (carriers)" or "rate of flow of charge". Writing only "flow of electrons" is often not accepted, because charge carriers are not always electrons.
"State what is meant by charge being quantised": charge exists only in discrete amounts, integer multiples of the elementary charge $e$ .
Ratio questions on $I = nAvq$ (very common in Paper 1) are fastest by writing which quantities are the same, then the proportionality: "same $I$ , $n$ , $e$ , so $v \propto 1/A \propto 1/d^2$ ".
Give answers to the number of significant figures of the data, usually two or three. Keep extra figures in intermediate steps.

Summary

An electric current is a flow of charge carriers: electrons in metals, ions in electrolytes.
Conventional current flows from $+$ to $-$ outside the source; electrons in a metal move the opposite way.
$Q = It$ ; one coulomb is one ampere for one second. For a varying current, charge is the area under the $I$ – $t$ graph.
Charge is quantised: $Q = Ne$ , with $e = 1.60 \times 10^{-19}\ \text{C}$ .
$I = nAvq$ , where $n$ is the number of charge carriers per unit volume and $v$ is the drift speed.
Drift speeds in metals are tiny (fractions of a $\text{mm s}^{-1}$ ) because $n$ is huge. For the same current, thinner wires and materials with smaller $n$ have larger drift speeds.
Ammeters go in series (ideally zero resistance); voltmeters go in parallel (ideally infinite resistance).

Practice

Question

A current of $1.5\ \text{A}$ flows for $5.0$ minutes. Calculate the charge that passes.
Calculate the number of electrons that make up a charge of $1.0\ \text{C}$ .
How long does it take $2.0 \times 10^{20}$ electrons to pass a point in a wire carrying a current of $0.40\ \text{A}$ ?
An electron beam carries a current of $25\ \mu\text{A}$ . Calculate the number of electrons striking the target each second.
A rechargeable battery is labelled $2400\ \text{mA h}$ (milliampere hours). (a) Calculate the charge it can deliver in coulombs. (b) For how long can it supply a constant current of $0.30\ \text{A}$ ?
Which of these charges cannot be the charge on an isolated object: $3.2 \times 10^{-19}\ \text{C}$ , $8.0 \times 10^{-20}\ \text{C}$ , $-1.6 \times 10^{-18}\ \text{C}$ ? Explain.
An aluminium wire of cross-sectional area $2.5\ \text{mm}^2$ carries a current of $5.0\ \text{A}$ . The number density of free electrons in aluminium is $1.8 \times 10^{29}\ \text{m}^{-3}$ . Calculate the drift speed.
A semiconductor strip measures $4.0\ \text{mm}$ by $0.20\ \text{mm}$ in cross-section and carries a current of $2.0\ \text{mA}$ . The number density of charge carriers, each of charge $e$ , is $5.0 \times 10^{22}\ \text{m}^{-3}$ . (a) Calculate the drift speed. (b) Explain why the drift speed is so much larger than in a metal wire carrying a similar current.
A wire of uniform material has a section P of diameter $d$ followed by a section Q of diameter $3d$ . The drift speed in P is $v$ . (a) Find the drift speed in Q in terms of $v$ . (b) The wire is replaced by one in which section Q is made of a metal with twice the number density of free electrons, keeping all diameters and the current the same. Find the new drift speed in Q in terms of $v$ .
A current of $4.0\ \text{mA}$ passes through a solution in which the charge carriers are $\ce{Cu^{2+}}$ ions moving one way and $\ce{SO4^{2-}}$ ions moving the other. In a certain region, the positive ions carry $60\%$ of the current. (a) Calculate the charge passing through the region in $10\ \text{min}$ . (b) Calculate the number of $\ce{Cu^{2+}}$ ions that pass through the region in this time. (c) Explain why ions moving in opposite directions both contribute to the current in the same direction.

Answers

$t = 5.0 \times 60 = 300\ \text{s}$ ; $Q = It = 1.5 \times 300 = 450\ \text{C}$ .
$N = \dfrac{1.0}{1.60 \times 10^{-19}} = 6.25 \times 10^{18}$ electrons.
$Q = Ne = 2.0 \times 10^{20} \times 1.60 \times 10^{-19} = 32\ \text{C}$ ; $t = Q/I = 32/0.40 = 80\ \text{s}$ .
In one second, $Q = 25 \times 10^{-6}\ \text{C}$ , so $N = \dfrac{25 \times 10^{-6}}{1.60 \times 10^{-19}} = 1.6 \times 10^{14}$ electrons per second.
(a) $Q = It = 2.400\ \text{A} \times 3600\ \text{s} = 8640\ \text{C} \approx 8.6 \times 10^{3}\ \text{C}$ . (b) $t = Q/I = 8640/0.30 = 2.88 \times 10^{4}\ \text{s} = 8.0\ \text{h}$ .
Divide by $e$ : $3.2 \times 10^{-19}\ \text{C} = 2e$ (possible); $8.0 \times 10^{-20}\ \text{C} = 0.5e$ (not possible: not a whole multiple of $e$ ); $-1.6 \times 10^{-18}\ \text{C} = -10e$ (possible).
$A = 2.5 \times 10^{-6}\ \text{m}^2$ (since $1\ \text{mm}^2 = 10^{-6}\ \text{m}^2$ ). $v = \dfrac{I}{nAe} = \dfrac{5.0}{1.8 \times 10^{29} \times 2.5 \times 10^{-6} \times 1.60 \times 10^{-19}} = 6.9 \times 10^{-5}\ \text{m s}^{-1}$ .
(a) $A = 4.0 \times 10^{-3} \times 0.20 \times 10^{-3} = 8.0 \times 10^{-7}\ \text{m}^2$ . $v = \dfrac{2.0 \times 10^{-3}}{5.0 \times 10^{22} \times 8.0 \times 10^{-7} \times 1.60 \times 10^{-19}} = 0.31\ \text{m s}^{-1}$ . (b) The number density of charge carriers in the semiconductor is about a million times smaller than in a metal. For the same current, $v = I/(nAq)$ , so the carriers must drift much faster.
(a) Same current, same $n$ and $e$ , so $v \propto 1/d^2$ . Q has $3^2 = 9$ times the area, so $v_Q = v/9$ . (b) Now $n$ is doubled as well, so $v_Q = \dfrac{v}{9 \times 2} = \dfrac{v}{18}$ .
(a) $Q = It = 4.0 \times 10^{-3} \times 600 = 2.4\ \text{C}$ . (b) Charge carried by the positive ions $= 0.60 \times 2.4 = 1.44\ \text{C}$ . Each $\ce{Cu^{2+}}$ ion carries $2e = 3.20 \times 10^{-19}\ \text{C}$ , so $N = \dfrac{1.44}{3.20 \times 10^{-19}} = 4.5 \times 10^{18}$ ions. (c) A flow of negative charge in one direction transfers charge in the same sense as a flow of positive charge in the opposite direction: both make one side more positive and the other more negative. So both kinds of ion add to the conventional current, which is in the direction of the positive ions.

What a current is

Conventional current

Charge and the coulomb

Charge is quantised

Circuit diagrams and symbols

How fast do the charge carriers move?

Deriving I=nAvqI = nAvqI=nAvq

What the equation tells you

Summary

Practice

Console

Deriving $I = nAvq$