Electric Current and Resistance

Beginner

An electric current is simply moving electric charge.

Step 1: what moves. In a metal, each atom lets go of one or two outer electrons, which wander freely between the atoms left behind (the ions). The magnified wire in the animation shows this crowd: yellow ions marked + in neat rows, blue electrons darting between them.

Step 2: the push. With no battery, the electrons rush about at random and get nowhere on average. Connect a battery and they all get a gentle push the same way: they still zigzag and bump into ions, but they also creep slowly along the wire. Follow the ringed electron: a tangled path, yet it works its way to the right.

Step 3: counting. Current is measured in amperes. One ampere means one coulomb of charge, about six billion billion electrons, passing a point every second. If 10 coulombs pass in 5 seconds, the current is $10÷5=2\mathrm{A}$.

Step 4: which way. By an old convention, current is said to flow from the + end to the − end. Electrons are negative, so they actually drift the other way, from − to +.

Step 5: resistance. The bumps hold the electrons back and hand energy to the ions, which jiggle harder: the wire warms up. With the same battery, a wire twice as long lets through half the current; a wire twice as thick lets through twice as much. Metals conduct easily; glass and plastic hardly at all.

Electrons crawl slower than a snail, yet a lamp lights the instant you flip the switch: the wire is already full of electrons, and the push travels along it almost at the speed of light.

Check yourself: 30 coulombs pass in 10 seconds. What is the current? (Answer: 3 A.)

Intermediate

What flows, and which way

A metal wire carrying no current is already full of moving charges. In a metal each atom releases one or a few outer electrons, which wander freely between the positive ions fixed in the crystal lattice. In salt water and batteries the carriers are ions. The current is the charge that crosses a cross-section of the conductor per unit time:
$I=\frac{Q}{t}$

The unit is the ampere, $1\mathrm{A}=1C/s$. An electron carries $e=1.602\times 10^{-19}\mathrm{C}$, so one ampere means $1/e\approx 6.2\times 10^{18}$ electrons crossing every second.

Conventional current points the way positive charge would move: from + to − through the outside circuit, along the electric field $\vec{E}$ inside the wire. Electrons are negative, so they drift the other way. In the animation the green $E$ arrow and the red-orange current arrow $I$ point left, while the blue drift velocity $v_d$ points right: negative charge moving right carries the same current as positive charge moving left.

Ohm's law and the current–voltage graph

The push that keeps charge flowing is the voltage $V$ across the component, the energy each coulomb gives up between its ends (see Electric Field and Potential); in a uniform wire of length $L$ it sets up a field $E=V/L$. Resistance compares the push with the flow:
$R=\frac{V}{I}$

Its unit is the ohm, $1\Omega =1V/A$. For a metal at a fixed temperature the current is proportional to the voltage, so $R$ is a constant: this is Ohm's law, and a graph of $I$ against $V$ is a straight line through the origin with slope $1/R$. A lamp filament's line bends over as it heats, its resistance growing more than tenfold, and a diode conducts one way only: $V/I$ still defines their resistance at each point, but it is not a constant.

What sets the resistance of a wire

Picture a wire as segments in a row and as lanes side by side. Every extra metre is one more stretch to push the current through, so $R$ grows in proportion to the length $L$; every extra square millimetre opens another lane, so $R$ falls in inverse proportion to the area $A$. The material enters through its resistivity $\rho$:
$R=\frac{\rho L}{A}$

Resistivity, in ohm metres (Ω·m), is the resistance of a piece 1 m long with a cross-section of 1 m². The table in the animation shows its range: $1.7\times 10^{-8}\Omega \mathrm{m}$ for copper, a conductor, $2.3\times 10^3\Omega \mathrm{m}$ for pure silicon, a semiconductor, and $10^{10}$ to $10^{14}\Omega \mathrm{m}$ for glass, an insulator: more than twenty powers of ten. The heating alloy nichrome has about $1.1\times 10^{-6}\Omega \mathrm{m}$.

In a metal $\rho$ also rises with temperature, roughly as $\rho (T)\approx {\rho }_0(1+\alpha \Delta T)$, with ${\rho }_0$ the value at 20 °C and $\Delta T$ the rise above it. For copper $\alpha =3.9\times 10^{-3}{\mathrm{K}}^{-1}$, about 0.4 % per kelvin. In a semiconductor $\rho$ falls instead, because heat frees more carriers.

Worked example: the three wires in the animation

The three wires in the animation share the same 2.0 V. Suppose the top one, the reference wire, is nichrome 1.0 m long with a cross-section of $0.55\mathrm{m}{\mathrm{m}}^2=0.55\times 10^{-6}{\mathrm{m}}^2$, about 0.84 mm across. Then
$R=\frac{\rho L}{A}=\frac{1.1\times 10^{-6}\cdot 1.0}{0.55\times 10^{-6}}=2.0\Omega$
$I=\frac{V}{R}=\frac{2.0}{2.0}=1.0\mathrm{A}$

These are the 2 Ω and 1.0 A printed beside it. Each blue dot stands for one coulomb, so one dot leaves the wire every second, and over the 12 s loop its counter climbs steadily to $Q=It=1.0\cdot 12=12\mathrm{C}$, about $7.5\times 10^{19}$ electrons.

The middle wire is twice as long: $R=4.0\Omega$, $I=0.50\mathrm{A}$, and its counter reaches 6 C. The same 2.0 V is spread over 2.0 m, so the field halves to 1.0 V/m and its dots move half as fast. The bottom wire has twice the cross-section: $R=1.0\Omega$, $I=2.0\mathrm{A}$, 24 C after 12 s; its field is unchanged, so its dots move as fast as in the reference wire, but in two lanes.

How fast the electrons drift

Now count the carriers. With $n$ carriers per cubic metre, each of charge $q$ and drifting at speed $v_d$, every carrier within a distance $v_{d}t$ upstream of a cross-section gets through it in a time $t$: a volume $A{v}_{d}t$ holding $nA{v}_{d}t$ carriers. Their charge divided by $t$ is the current:
$I=nA{v}_{d}q$

Copper has about one free electron per atom, $n=8.5\times 10^{28}{\mathrm{m}}^{-3}$. For 1 A in a wire of 1 mm²,
$v_d=\frac{I}{nAe}=\frac{1.0}{8.5\times 10^{28}\cdot 10^{-6}\cdot 1.602\times 10^{-19}}\approx 7.3\times 10^{-5}m/s$
about 0.07 mm/s, or 26 cm per hour.

Yet the electrons dart about at some 1.6 million m/s; the drift is a tiny bias on that random motion. The ringed electron in the magnified wire shows both, a quick zigzag and a slow advance to the right, although in real copper the zigzag is over ten billion times faster than the drift, not seven. A lamp still lights at once, because the wire is already full of electrons and the field that sets them moving spreads at a large fraction of the speed of light.

Worked example: an extension cable

A 10 m extension cable has two copper cores of 1.5 mm²; the current goes out along one and back along the other, through 20 m of copper. A kettle plugged into it draws 10 A. With $A=1.5\times 10^{-6}{\mathrm{m}}^2$,
$R=\frac{\rho L}{A}=\frac{1.7\times 10^{-8}\cdot 20}{1.5\times 10^{-6}}\approx 0.23\Omega$
$V=IR=10\cdot 0.23\approx 2.3\mathrm{V}$
That is only 1 % of the 230 V mains, but at 10 A it becomes about 23 W of heat in the cable (power is voltage times current), harmless when spread along 20 m but dangerous in a cable left wound on its drum.

The drift speed is
$v_d=\frac{I}{nAe}=\frac{10}{8.5\times 10^{28}\cdot 1.5\times 10^{-6}\cdot 1.602\times 10^{-19}}\approx 4.9\times 10^{-4}m/s$
half a millimetre per second: a steady current would need about 11 hours to carry an electron along those 20 m. Mains current reverses 100 times a second (50 Hz), so each electron only shuffles back and forth by a couple of micrometres, while the energy reaches the kettle at once.

Common mistakes

• Current is not used up: the same current flows into a lamp and out of it, 1.0 A at both ends of the reference wire; what the lamp takes is energy.
• Electrons do not race to the lamp: they drift at about 0.07 mm/s for 1 A in 1 mm² of copper. Only the field that sets them moving travels fast.
• A thicker wire has less resistance, not more: doubling the cross-section halves $R$, from 2 Ω to 1 Ω in the animation.
• Writing $R=V/I$ does not make resistance depend on voltage: for an ohmic wire, doubling $V$ doubles $I$, and the reference wire stays at 2 Ω.
• Convert areas with $1\mathrm{m}{\mathrm{m}}^2=10^{-6}{\mathrm{m}}^2$, not $10^{-3}$: the wrong factor turns the nichrome wire's 2.0 Ω into 0.0020 Ω.

Where you meet it

Household wiring uses copper, while overhead power lines use aluminium, whose resistivity is about 1.6 times higher but which is far lighter. Toasters and heaters use nichrome, whose resistivity is high and changes little with temperature. Resistance thermometers read temperature through $\rho (T)$. Next, DC Circuits and Ohm's Law combines resistors in series and in parallel, Electric Power and Energy at Home follows the heat they produce, and Magnetic Fields of Magnets and Currents shows the magnetic field around every current.

Advanced

Current density and charge conservation

A current $I$ says how much charge crosses a whole cross-section each second, but not how the flow is spread over it. The current density $\vec{J}$ does. Its size is the current per unit area at right angles to the flow, in A/m², and it points the way positive charge flows. To connect it with the carriers themselves, we count how many of them cross one small patch of a surface in a short time.

Legend

$dA$	Area of the small patch, drawn edge-on.	${\mathrm{m}}^2$
$d\vec{A}$	The patch as a vector: length $dA$, at right angles to the patch.	${\mathrm{m}}^2$
$\vec{J}$	Current density, $nq{\vec{v}}_d$: current per unit area, along the flow of positive charge.	$A/{\mathrm{m}}^2$
$\theta$	Angle between $\vec{J}$ (and ${\vec{v}}_d$) and $d\vec{A}$.	$\mathrm{rad}$
${\vec{v}}_{d}dt$	How far the carriers drift in the time $dt$: the slanted edge of the slab.	$\mathrm{m}$
$v_{d}dt\cos \theta$	The same drift measured at right angles to the patch: the height of the slab.	$\mathrm{m}$
$dV$	Volume of the slab below the patch: the carriers in it cross the patch within $dt$.	${\mathrm{m}}^3$

Figure 1 shows such a patch edge-on: the line labelled $dA$, its area. To say which way the patch faces, we give it a vector $d\vec{A}$, of length $dA$ and at right angles to the patch. The carriers, $n$ of them per unit volume and each with charge $q$, are the dots. On average they move with the drift velocity ${\vec{v}}_d$, which makes the angle $\theta$ with $d\vec{A}$. Their fast random motions carry as many carriers across the patch one way as the other, so only the drift counts.

In a short time $dt$ each carrier moves by ${\vec{v}}_{d}dt$. The carriers that cross the patch in that time are therefore exactly those that start in the slanted slab below it: any carrier further back cannot reach the patch in time. Look at the two sides of the slab. Its slanted edge, along the drift, is $v_{d}dt$ long, but its height at right angles to the patch is only $v_{d}dt\cos \theta$. A slanted slab has the same volume as an upright one with the same base and height, so
$dV=dA{v}_{d}dt\cos \theta$
The slab holds the charge $nqdV$, and all of it crosses the patch within $dt$. Dividing by $dt$ gives the current through the patch:
$dI=nq{v}_d\cos \theta dA=\vec{J}\cdot d\vec{A},\vec{J}=nq{\vec{v}}_d$
The factor $\cos \theta$ is exactly what a dot product supplies. Adding up the patches of a whole surface gives the current through it:
$I=\int \vec{J}\cdot d\vec{A}$
For a uniform flow straight through a cross-section $A$, $\theta =0$ everywhere, the integral is simply $JA$, and
$I=nA{v}_{d}q$

Figure 1 is drawn for positive carriers, $q>0$, where $\vec{J}$ and ${\vec{v}}_d$ point the same way. For electrons $q=-e$, so $\vec{J}$ points against ${\vec{v}}_d$: the conventional current flows one way while the electrons drift the other.

Now take a closed surface, for example a box around a junction, with every $d\vec{A}$ pointing outwards. Charge is conserved: it cannot appear or vanish. So the current flowing out of the surface must come from the charge $Q$ inside it, which decreases at the same rate:
$\oint \vec{J}\cdot d\vec{A}=-\frac{dQ}{dt}$
In a steady state nothing piles up anywhere, so $dQ/dt=0$: as much current enters any closed surface as leaves it. Applied to a piece of wire between two cross-sections, this says the same current crosses every section, whatever the wire's shape, so current is never used up. Applied to a junction of wires, it is the junction rule for circuits.

Drift from random collisions

A steady field pushes each carrier with a steady force, yet the current is steady too: the carriers do not speed up for ever. Collisions are what stop them. A carrier of mass $m$ feels the force $\vec{F}=q\vec{E}$ from the field, plus brief random kicks from collisions. The mean free time, $\tau$, is the average time between collisions. Drude's model of 1900 describes this motion with three assumptions.

• Between collisions the carrier moves freely, accelerating at $\vec{F}/m$.
• Each collision sends it off in a random direction, with no memory of how it was moving before.
• Collisions strike at random, with the same chance $dt/\tau$ in every short time $dt$, whatever happened earlier.

Now look at all the carriers at one instant. Each carrier's velocity has two parts. The first is the velocity it left its last collision with: random, so it averages to zero over many carriers. The second is what the force has added since then, $\vec{F}/m$ times the time since that collision. So the average velocity, the drift, is $\vec{F}/m$ times the average time since the last collision.

What is that average? Because collisions have no memory, the chance of one in any short interval is $dt/\tau$ whether we run time forwards or backwards. Looking back from any instant, the time since the last collision therefore has the same statistics as the wait until the next one, and it also averages $\tau$. Hence
$v_d=\frac{F\tau }{m}$
and for an electron in a field $E$, with $F=eE$,
$v_d=\frac{eE\tau }{m}$

Why the long flights dominate

This result invites a doubt. During one flight the added velocity grows from zero to $Ft/m$, so surely the drift should be half of $F\tau /m$? It would be, if every flight lasted exactly $\tau$. They do not: because collisions strike at random, the duration $t$ of a flight is spread out, with the probability density $p(t)=e^{-t/\tau }/\tau$. A fraction $p(t)dt$ of all flights last between $t$ and $t+dt$.

Legend

$t$	Duration of one free flight, between two collisions.	$\mathrm{s}$
$\tau$	Mean free time: the average duration of a flight.	$\mathrm{s}$
$p(t)$	Density of the flight duration, $e^{-t/\tau }/\tau$: a fraction $p(t)dt$ of flights last between $t$ and $t+dt$.	${\mathrm{s}}^{-1}$
$t^{2}p(t)$	Forward distance contributed by flights of duration $t$: $t^2$ times how common they are, up to the factor $F/2m$.	$\mathrm{s}$

The top graph of Figure 2 is this density. It is highest for the shortest flights and falls steadily, and its mean is $\tau$. Yet most flights are shorter than the mean: those longer than $\tau$, the shaded part, are only $1/e$, about 37 %, of all flights.

Now weigh each flight by how far it carries the carrier forward. Starting with no added velocity, a flight of duration $t$ advances it by $\frac{1}{2}(F/m)t^2=(F/2m)t^2$, so a flight twice as long gains four times the distance. The lower graph shows $t^{2}p(t)$: how common flights of duration $t$ are, times how much each one contributes. It peaks at $2\tau$, well beyond the mean, and most of its area lies in the shaded part: the flights longer than $\tau$ supply about 92 % of the forward distance. The few long flights do almost all of the work, because they gather speed for longer. Taking half of $F\tau /m$ treats every flight as equally long and misses exactly this.

The drift is the mean forward distance per flight divided by the mean duration of a flight:
$v_d=\frac{F}{2m}\frac{\overline{t^2}}{\overline{t}}=\frac{F}{2m}\frac{2{\tau }^2}{\tau }=\frac{F\tau }{m}$
Here $\overline{t}=\tau$ is the mean of $p(t)$, and $\overline{t^2}=2{\tau }^2$ is the area under the lower curve: for random collisions the mean of $t^2$ is twice the square of the mean. The random velocities with which flights begin add nothing on average, so this is the whole drift, and it agrees with the result above.

The magnified wire in the animation obeys the same rule. Each free flight is a short parabola bending to the right; with a mean free time of 12/45 ≈ 0.27 s and an acceleration of 22.5 units/s², the drift is $a\tau =22.5\cdot 12/45=6$ units per second.

Ohm's law, resistivity and resistance

With the drift known, the current density follows. For carriers of charge $q$, the force is $F=qE$, so $v_d=qE\tau /m$ and
$J=nq{v}_d=\frac{n{q}^2\tau }{m}E$
The current density is proportional to the field. This is Ohm's law in its local form, true at each point of the metal. The constant of proportionality is the conductivity $\sigma$, and its inverse is the resistivity $\rho$; for electrons, with $q^2=e^2$,
$J=\sigma E$
$\rho =\frac{1}{\sigma }=\frac{m}{n{e}^2\tau }$
Because the charge enters squared, positive and negative carriers give the same conductivity: measuring $\sigma$ cannot reveal the sign of the carriers.

Check this against copper. With $n=8.5\times 10^{28}{\mathrm{m}}^{-3}$ and $\tau =2.5\times 10^{-14}\mathrm{s}$, the formula gives copper's measured $1.7\times 10^{-8}\Omega \mathrm{m}$. The agreement is less impressive than it looks: $\tau$ cannot be measured directly and was inferred from that very resistivity. The model explains the form of the law, a current proportional to the field, but not the size of $\rho$.

Now go from one point to a whole wire. In a uniform wire of length $L$ and cross-section $A$ the field is the same everywhere, so the voltage across the wire is $V=EL$ and the current is $I=JA$. Putting $E=V/L$ and $J=I/A$ into $J=\sigma E$ gives $I/A=V/(\rho L)$, that is $V=(\rho L/A)I$. So
$R=\frac{\rho L}{A}$
is derived, not guessed.

The two graphs in the animation's expert layer show its two faces, using the reference wire of the intermediate level: nichrome, 1.0 m long and 0.55 mm² in section, so 2 Ω, carrying 1.0 A at 2.0 V. Against length, $R$ is a straight line through the origin with slope $\rho /A$: the reference wire sits at $L_0$ and 2 Ω, the wire twice as long at $2L_0$ and 4 Ω. Against area, $R$ is a hyperbola, because $RA=\rho L$ stays constant: the reference wire sits at $A_0$ and 2 Ω, the thick wire at $2A_0$ and 1 Ω.

The hyperbola shows diminishing returns. Doubling the area saves 1 Ω, but going on to $3A_0$ saves only 0.33 Ω more, while shrinking the cross-section towards zero sends $R$ to infinity.

Where the energy goes

Between collisions the field does work on each carrier and speeds it up. At each collision the carrier hands that energy to the lattice of ions, which vibrates harder: the wire warms up. On average the field delivers the power $\vec{F}\cdot {\vec{v}}_d=q\vec{E}\cdot {\vec{v}}_d$ to each carrier, and there are $n$ carriers per unit volume, so the heat produced per unit volume per second is
$p=\vec{J}\cdot \vec{E}=\sigma E^2=\rho J^2$
Over a uniform wire, multiplying by its volume $AL$ gives $(JA)(EL)=IV$, the familiar electrical power.

For the reference wire, $E=2.0V/m$ and $J=1.0/(0.55\times 10^{-6})\approx 1.8\times 10^{6}A/{\mathrm{m}}^2$, so $p=JE\approx 3.6\times 10^{6}W/{\mathrm{m}}^3$. The wire holds 0.55 cm³ of metal, so it produces 2.0 W in all, the same as $IV=1.0\cdot 2.0\mathrm{W}$.

How the energy gets in

How does this energy reach each part of the wire? Not by travelling along the inside of it: the carriers pass their energy to the lattice right where they are. It flows in through the wire's surface, carried by the electric and magnetic fields in the space around it.

Legend

$I$	Current in the wire.	$\mathrm{A}$
$a$	Radius of the wire.	$\mathrm{m}$
$L$	Length of the wire.	$\mathrm{m}$
$\vec{E}$	Electric field just outside the surface: its part along the wire, the same as inside.	$V/m$
$\vec{B}$	Magnetic field of the current, circling the wire by the right-hand rule: towards you at the top, away from you at the bottom.	$\mathrm{T}$
$\vec{S}$	Energy flux $\vec{E}\times \vec{B}/{\mu }_0$: energy per second through unit area, here into the wire.	$W/{\mathrm{m}}^2$

Figure 3 shows a wire of radius $a$ and length $L$ carrying the current $I$ to the right. Two fields surround it. The first is the electric field $\vec{E}$ just outside the surface, pointing along the wire. It is there because the part of a field that is parallel to a surface is the same on both sides of it, so the field that drives the current inside continues just outside. Surface charges may also add a part at right angles to the surface, but that part moves energy along the wire, not into it. The second is the magnetic field $\vec{B}$ of the current, which circles the wire by the right-hand rule and has the size $B={\mu }_{0}I/(2\pi a)$ at the surface.

Where both fields are present, electromagnetic energy flows. Its flux, the energy per second through unit area, is $\vec{S}=\vec{E}\times \vec{B}/{\mu }_0$, at right angles to both fields. Read the directions off the figure. At the top of the wire $\vec{E}$ points along it and $\vec{B}$ points towards you, so $\vec{S}$ points down, into the wire. At the bottom $\vec{B}$ points away from you, and $\vec{S}$ points up, again into the wire. All round, the energy flows straight in through the surface.

It remains to find how much energy flows in. Since $\vec{E}$ and $\vec{B}$ are at right angles, and the surface has the area $2\pi aL$,
$S=\frac{EB}{{\mu }_0}=\frac{EI}{2\pi a},S\cdot 2\pi aL=ELI=VI$
because $EL=V$: the inflow is exactly the heat produced inside. For the reference wire $a\approx 0.42\mathrm{mm}$ and $S\approx 760W/{\mathrm{m}}^2$, which over its surface of about $2.6\times 10^{-3}{\mathrm{m}}^2$ is exactly the 2.0 W of heat. The battery's energy travels through the space around the circuit; the wires guide it and take it in.

What really scatters the electrons

Drude pictured electrons bouncing off every ion. That picture is wrong, and correcting it explains how resistance depends on temperature and purity. First, the electrons of copper are not slow. The exclusion principle allows only one electron in each quantum state, so they fill states up to the Fermi speed, about $1.6\times 10^{6}m/s$, even at absolute zero. Their mean free path, the distance between collisions, is then $\ell =v_F\tau \approx 39\mathrm{nm}$, some 150 ion spacings: an electron passes many ions without being scattered.

The reason is that an electron is a wave. In a perfectly periodic lattice, the waves scattered by the individual ions combine to carry the electron straight on. A perfect crystal would have no resistance at all.

Only departures from perfect periodicity scatter electrons, and their scattering rates add (Matthiessen's rule). Lattice vibrations grow with temperature. Impurities and defects do not, and leave a residual resistivity even at the lowest temperatures.

Near room temperature the vibrational part is roughly proportional to the absolute temperature $T$, and a resistivity proportional to $T$ changes by the fraction $1/T$ per kelvin. That is why copper's temperature coefficient $\alpha =3.9\times 10^{-3}{\mathrm{K}}^{-1}$ is close to $1/(293\mathrm{K})=3.4\times 10^{-3}{\mathrm{K}}^{-1}$, and $\rho (T)\approx {\rho }_0(1+\alpha \Delta T)$ is simply the linear term of $\rho (T)$ around 20 °C. On cooling, the vibrations die away and $\rho$ levels off at the residual value: very pure copper at 4 K conducts a hundred to a thousand times better than at room temperature.

Two other kinds of material behave quite differently. In a semiconductor, heat lifts electrons into conducting states, so $n$ climbs steeply and $\rho$ falls as it warms. A superconductor, below its critical temperature (about 4.2 K for mercury), has exactly zero resistance. Its electrons form one collective quantum state that carries a current without scattering: a new state of matter, not the limit of ever fewer collisions.

Worked problem: a wire that thickens

$R=\rho L/A$ holds for a uniform wire. What if the cross-section changes along the length? Take the reference wire (nichrome, 1.0 m, so $\rho /A_0=2.0\Omega /\mathrm{m}$) and let it widen steadily from $A_0$ at one end to $2A_0$ at the other, $A(x)=A_0(1+x/L)$, with 2.0 V across it. Its resistance must lie between the thick wire's 1 Ω and the reference wire's 2 Ω, but where?

Legend

$I$	Current: the same through every slice.	$\mathrm{A}$
$A_0$	Cross-section at the thin end; the thick end has $2A_0$.	${\mathrm{m}}^2$
$A(x)$	Cross-section at the distance $x$ from the thin end, $A_0(1+x/L)$.	${\mathrm{m}}^2$
$L$	Length of the wire.	$\mathrm{m}$
$x$	Distance from the thin end.	$\mathrm{m}$
$dx$	Thickness of one slice.	$\mathrm{m}$
$\rho /A(x)$	Resistance per unit length of the wire at $x$.	$\Omega {\mathrm{m}}^{-1}$
$\rho /A_0$	Resistance per unit length at the thin end.	$\Omega {\mathrm{m}}^{-1}$
$\rho /2A_0$	Resistance per unit length at the thick end: half that at the thin end.	$\Omega {\mathrm{m}}^{-1}$
$dR$	Resistance of the slice at $x$, $\rho dx/A(x)$: the area of the dark strip.	$\Omega$
$R$	Resistance of the whole wire: the area under the curve.	$\Omega$

Figure 4 cuts the wire into thin slices (wire and slice are drawn far thicker than they are). The slice at the distance $x$ from the thin end has the thickness $dx$ and the cross-section $A(x)$. So thin a slice is a short uniform wire, with the resistance $dR=\rho dx/A(x)$. The same current $I$ passes through one slice after another, so the slices are in series and their resistances add.

The graph below the wire turns this sum into an area. Its curve is the resistance per unit length, $\rho /A(x)$, which falls from $\rho /A_0$ at the thin end to $\rho /2A_0$ at the thick end. The dark strip under the curve has the height $\rho /A(x)$ and the width $dx$, so its area is the slice's $dR$. Adding up all the strips gives the whole area under the curve, which is the resistance $R$ of the wire. With $u=1+x/L$, so that $dx=Ldu$,
$R={\int }_0^L\frac{\rho dx}{A_0(1+x/L)}=\frac{\rho L}{A_0}{\int }_1^2\frac{du}{u}=\frac{\rho L}{A_0}\ln 2$
$R=2.0\cdot 0.693\approx 1.39\Omega$

So the current is $I=2.0/1.386\approx 1.44\mathrm{A}$. Two tempting shortcuts both fail. Averaging the two resistances gives 1.5 Ω, too high, because the curve sags below the straight line joining its ends. Using the average area, $1.5A_0$, gives 1.33 Ω, too low, because the slices add as $1/A$, and the mean of $1/A$ is not one over the mean of $A$. The slice picture itself is safe here because the wire widens gently: its diameter grows only from 0.84 mm to 1.18 mm over a whole metre, so the current in each slice flows almost straight along the wire.

The current is the same through every slice, so the field must follow the area: $E(x)=\rho I/A(x)$ is 2.9 V/m at the thin end and 1.4 V/m at the thick end, and 1.17 V of the 2.0 V, 58 %, falls across the thinner half. The drift is twice as fast at the thin end as at the thick end, and the heat there is twice as large per unit length and four times as large per unit volume. What shapes the field this way is a thin layer of surface charge, set up within nanoseconds of switching on. The same concentration of heat at a narrow section is why fuse elements have deliberate narrow necks: that is where they melt.

Limits of validity

$R=\rho L/A$ treats the metal as a continuum, with a definite $\rho$ at each point, so every dimension of the conductor must be much larger than the mean free path, about 40 nm in copper. In thin films and in chip interconnects a few tens of nanometres wide, electrons also scatter at the surfaces and at grain boundaries, and the resistivity rises two or three times above its bulk value. Conductors shorter than the mean free path are crossed without any collision, and resistance takes on a quantum character: a constriction a few atoms wide conducts in steps of $2e^2/h\approx 7.75\times 10^{-5}\mathrm{S}$, 12.9 kΩ per channel, whatever the metal.

The results also assume a steady current. In an alternating current, the changing magnetic field inside the wire induces an electric field that opposes the current near the axis and adds to it near the surface. The current crowds into a surface layer whose thickness, the skin depth, is
$\delta =\sqrt{\frac{2\rho }{\omega {\mu }_0}}$
where $\omega$ is the angular frequency. For copper that is 9.3 mm at 50 Hz, 66 µm at 1 MHz and 2.1 µm at 1 GHz, so at high frequency only a thin skin of the wire carries current. Near $1/(2\pi \tau )\approx 6\mathrm{THz}$ the field reverses within about one free time, and the inertia of the electrons, not their collisions, starts to set their response; that response to light is what makes metals shiny.

Ohm's law also needs the energy a carrier gains between collisions to be tiny. In copper carrying 1 A per mm², the field is only 0.017 V/m, and an electron gains about $7\times 10^{-10}\mathrm{eV}$ over one free path, against the thermal energy scale $kT\approx 0.026\mathrm{eV}$ at room temperature. In strong fields this fails: the electrons of a semiconductor heat up, and in silicon their drift speed saturates near $10^{5}m/s$ instead of growing with the field.

Where this leads

DC Circuits and Ohm's Law builds on this: resistors in series add like lengths of wire, and in parallel they act like extra lanes, so their conductances add. Electric Power and Energy at Home follows the heat $VI=I^{2}R$ into cables, fuses and kettles. A current makes a magnetic field (Magnetic Fields of Magnets and Currents), and a magnetic field deflects the drifting carriers (Magnetic Force on Moving Charges), producing a Hall voltage that reveals $n$ and the sign of $q$. Semiconductors, Diodes and LEDs controls $n$ to build devices, and Electromagnetic Waves meets the same skin depth when a radio wave enters seawater. Since the free electrons carry heat too, good electrical conductors are good conductors of heat (Conduction, Convection and Radiation).