Part 3 · Chapter 26

Current and Resistance

The last five chapters held charge still. Now we let it move. A steady stream of charge through a wire is an electric current, and every material resists that flow to some degree — copper barely, glass almost totally. This chapter defines current and current density, ties the drift of electrons to the field that pushes them, builds resistance and resistivity from both the engineer's and the atom's point of view, and accounts for the energy a current dissipates as heat.

Fundamentals of Physics Prof. Mithun Mondal Reading time ≈ 55 min

i What you'll learn

Electric current as the rate of charge flow, $i = \dfrac{dq}{dt}$, measured in amperes (1 A = 1 C/s), and why it is the same through every cross section.
Current density $\vec{J}$ with $i = \displaystyle\int \vec{J}\cdot d\vec{A}$, and the drift speed of carriers, $\vec{J} = ne\vec{v}_{d}$.
Resistance $R = \dfrac{V}{i}$ in ohms, and the material property resistivity $\rho = \dfrac{E}{J}$, linked by $R = \dfrac{\rho L}{A}$.
How resistivity rises with temperature, $\rho = \rho_{0}\big[1 + \alpha(T - T_{0})\big]$, and what it means for a device to obey Ohm's law.
The microscopic free-electron model giving $\rho = \dfrac{m}{e^{2}n\tau}$, the power transferred $P = iV$, and the heat dissipated $P = i^{2}R = \dfrac{V^{2}}{R}$.

Section 26-1

What Is Physics?

The previous five chapters were about electrostatics — charges at rest. Here we turn to charges in motion, the physics of electric currents. The subject touches nearly every profession: meteorologists study lightning and the slow drift of charge through the atmosphere, physiologists trace the nerve currents that fire muscles, electrical engineers design power grids and music systems, and space scientists track the charged particles streaming from the Sun that can knock out satellites and even ground-based power lines.

Our questions are basic but far-reaching. What exactly is an electric current? Why does charge flow freely in some materials and hardly at all in others? And where does the energy go when a current runs through a wire? In this chapter we restrict ourselves to steady currents of conduction electrons drifting through metallic conductors such as copper — the bread and butter of everyday circuits.

Section 26-2

Electric Current

Moving charge alone is not enough; there must be a net flow through a surface. The conduction electrons in an isolated copper wire jiggle about at speeds near $10^{6}\,\mathrm{m/s}$, but as many cross any plane one way as the other, so the net transport — and the current — is zero. Connect a battery and the field biases that motion slightly in one direction, and now charge genuinely flows. If charge $dq$ passes a plane in time $dt$, the current is defined as:

Definition of current

\[ i = \frac{dq}{dt} \qquad\Longleftrightarrow\qquad q = \int_{0}^{t} i\,dt \]

The SI unit is the ampere: $1\,\mathrm{A} = 1\,\mathrm{C/s}$, an SI base unit. Current is a scalar; the arrow we draw shows only the sense of flow along a conductor, not a direction in space.

💧

Conservation of charge

Under steady-state conditions the current is the same through every cross section, no matter its size or orientation.

Charge cannot pile up or vanish, so for every electron entering one end of a wire, one must leave the other — like water through a hose. At a junction the branch currents add to the incoming current: $i_{0} = i_{1} + i_{2}$. By convention the current arrow points the way positive carriers would move, even though the actual carriers in a metal are electrons drifting the opposite way.

Section 26-3

Current Density

Sometimes we care not about the total current in a wire but about how the flow is distributed across a cross section, point by point. The current density $\vec{J}$ is the current per unit area, pointing along the carrier velocity (for positive carriers). The total current through a surface is the flux of $\vec{J}$ through it:

Current density

\[ i = \int \vec{J}\cdot d\vec{A}, \qquad\text{and for uniform } \vec{J} \parallel d\vec{A}: \quad J = \frac{i}{A} \]

The SI unit is the ampere per square metre (A/m²). Just as field lines picture $\vec{E}$, streamlines picture $\vec{J}$: where a conductor narrows, the same current must squeeze through less area, so the streamlines crowd together and $J$ rises — even though $i$ is unchanged.

Section 26-4

Drift Speed

With no current, the conduction electrons move randomly with no net direction. Apply a field and they acquire, on top of that random motion, a slow drift speed $v_{d}$ opposite the field. Counting $n$ carriers per unit volume, each of charge $e$, the charge in a length $L$ is $nALe$, and it clears a cross section in time $L/v_{d}$, giving $i = nAev_{d}$.

Current density and drift speed

\[ \vec{J} = (ne)\vec{v}_{d} \qquad\Longleftrightarrow\qquad v_{d} = \frac{J}{ne} = \frac{i}{nAe} \]

The product $ne$ is the carrier charge density (C/m³). For positive carriers $\vec{J}$ and $\vec{v}_{d}$ point the same way; for electrons they point opposite ways.

Slower than a snail, yet the lights turn on at once. In ordinary household wiring the drift speed is only about $10^{-5}\,\mathrm{m/s}$ — millimetres per hour. How, then, do the lights come on the instant you flip the switch? Because the electric field that sets all the electrons drifting propagates along the wire at nearly the speed of light. Electrons everywhere — including in the bulb — start moving almost simultaneously, even though each individual electron crawls.

Section 26-5

Resistance and Resistivity

Apply the same voltage across geometrically identical rods of copper and of glass, and wildly different currents flow. The property that decides this is the resistance. We measure it by applying a potential difference $V$ and reading the current $i$:

Definition of resistance

\[ R = \frac{V}{i} \]

The SI unit is the ohm: $1\,\Omega = 1\,\mathrm{V/A}$. Rearranged as $i = V/R$, it says that for a given voltage, more resistance means less current. A component built to provide a definite resistance is a resistor.

Resistance describes an object; the corresponding property of a material is its resistivity $\rho$, defined from the field and current density at a point rather than the voltage and current of a whole rod.

Resistivity and conductivity

\[ \rho = \frac{E}{J} \quad\Longleftrightarrow\quad \vec{E} = \rho\vec{J}, \qquad \sigma = \frac{1}{\rho} \]

The SI unit of resistivity is the ohm-metre ($\Omega\cdot\mathrm{m}$); its reciprocal $\sigma$ is the conductivity. These hold for isotropic materials, whose electrical behaviour is the same in every direction.

Table 26-1 · Resistivities at room temperature (20 °C)
Material	ρ (Ω·m)	α (K⁻¹)
Silver (metal)	1.62 × 10⁻⁸	4.1 × 10⁻³
Copper (metal)	1.69 × 10⁻⁸	4.3 × 10⁻³
Aluminum (metal)	2.75 × 10⁻⁸	4.4 × 10⁻³
Tungsten (metal)	5.25 × 10⁻⁸	4.5 × 10⁻³
Silicon, pure (semiconductor)	2.5 × 10³	−70 × 10⁻³
Glass (insulator)	10¹⁰ – 10¹⁴	—

Section 26-6

Calculating Resistance & Its Variation with Temperature

For a homogeneous wire of length $L$ and uniform cross section $A$, the field and current density are uniform, with $E = V/L$ and $J = i/A$. Substituting into $\rho = E/J$ and identifying $V/i = R$ gives the workhorse formula linking the object to its material:

Resistance from resistivity

\[ R = \rho\,\frac{L}{A} \]

A longer wire resists more; a fatter wire resists less. This applies only to a uniform, isotropic conductor with the voltage applied across its full cross section.

Resistivity also depends on temperature. For metals the relationship is very nearly linear over a wide range, well captured by an empirical formula about a reference temperature $T_{0}$ (usually 293 K).

Temperature variation of resistivity

\[ \rho = \rho_{0}\big[1 + \alpha(T - T_{0})\big] \]

Here $\alpha$ is the temperature coefficient of resistivity. For metals $\alpha > 0$ — they resist more when hot. Because only a temperature difference appears, Celsius and Kelvin work equally well in this equation.

📐

Object versus material

Resistance is a property of an object; resistivity is a property of a material.

A thin copper wire and a thick copper bar are made of the same material — same $\rho$ — but have very different resistances because of their differing $L/A$. Use the macroscopic trio $V, i, R$ when measuring a specific component; use the microscopic trio $E, J, \rho$ when studying the substance itself.

Section 26-7

Ohm's Law

It is tempting to call $V = iR$ "Ohm's law," but that equation is merely the definition of resistance and holds for every device. The real content of Ohm's law is a statement about whether $R$ stays constant:

The essence of Ohm's law

\[ \text{A device obeys Ohm's law} \iff R = \frac{V}{i} \text{ is independent of } V \]

Equivalently, a material obeys Ohm's law when its resistivity $\rho$ does not depend on the magnitude or direction of $\vec{E}$. A plot of $i$ versus $V$ is then a straight line through the origin.

What disobeys it. An ordinary resistor gives a straight $i$–$V$ line — Ohmic. A semiconducting pn junction diode does not: it carries almost no current until the voltage exceeds a threshold, then conducts strongly, and blocks current entirely in reverse. Its resistance changes with the applied voltage, so it violates Ohm's law — which is exactly what makes it useful as a one-way valve for charge.

Section 26-8

A Microscopic View of Ohm's Law

Why do metals obey Ohm's law? The free-electron model pictures conduction electrons rattling around like gas molecules, colliding only with the metal's atoms and losing all memory of their drift at each collision. Between collisions, separated on average by the mean free time $\tau$, the field $E$ accelerates each electron, building an average drift $v_{d} = eE\tau/m$. Combining this with $J = nev_{d}$ yields the resistivity from first principles:

Resistivity of a metal

\[ \rho = \frac{m}{e^{2}n\tau} \]

Here $n$ is the free-electron density and $m, e$ are the electron's mass and charge. Metals obey Ohm's law because $\tau$ is essentially constant: the drift speed is so minuscule next to the electrons' effective random speed (~$1.6\times10^{6}\,\mathrm{m/s}$) that the field barely changes how often they collide.

Section 26-9

Power in Electric Circuits

As charge $dq = i\,dt$ drops through a potential difference $V$ in a device, its electric potential energy falls by $dU = dq\,V$. The rate of this energy transfer is the electrical power:

Rate of electrical energy transfer

\[ P = iV \]

The unit is the volt-ampere, equal to the watt: $1\,\mathrm{V\cdot A} = 1\,\mathrm{W}$. This is the general rate at which a source delivers energy to any device — a motor (to mechanical work), a charging battery (to chemical energy), or a resistor (to heat).

When the device is purely resistive, the delivered energy turns entirely into thermal energy through electron–atom collisions. Combining $P = iV$ with $R = V/i$ gives two specialised forms:

Resistive dissipation

\[ P = i^{2}R = \frac{V^{2}}{R} \]

Caution: $P = iV$ applies to energy transfers of all kinds, but $P = i^{2}R$ and $P = V^{2}/R$ apply only to the conversion of electrical energy into thermal energy in a resistance. This dissipation heats everything from toasters and incandescent bulbs to the wiring in your walls.

Section 26-10

Semiconductors and Superconductors

Not every material fits the metal mould. A semiconductor such as silicon has far fewer charge carriers than a metal and a much higher resistivity, but — crucially — a negative temperature coefficient: it conducts better when heated, because rising temperature liberates many more carriers, swamping the increased collision rate. Looking again at $\rho = m/e^{2}n\tau$, in a metal $n$ is fixed and warming shortens $\tau$ (so $\rho$ rises); in a semiconductor $n$ grows rapidly (so $\rho$ falls).

The real power of semiconductors comes from doping — adding trace impurity atoms to inject controllable numbers of carriers. This is how transistors and diodes, the building blocks of the information age, are made. At the opposite extreme are superconductors: below a critical temperature their resistivity drops to exactly zero, so a current once started in a loop can persist for years without any driving voltage. Discovered in mercury in 1911, superconductivity now appears in ceramics at far higher, cheaper-to-reach temperatures.

Three classes, one equation. Conductors, semiconductors, and superconductors all read off the same physics of carriers and collisions. A good conductor has many carriers and modest scattering; an insulator has almost none free to move; a semiconductor sits between, its carrier count tunable by heat or doping; and a superconductor, below its critical temperature, scatters not at all. Resistivity spans more than twenty orders of magnitude across these materials.

Worked Examples

Putting It to Work

1 Charge and electrons in a current

Problem. A steady current of 5.0 A flows in a wire for 4.0 min. How much charge passes a cross section, and how many electrons?

Solution. For a steady current, $q = i\,t$; the number is $N = q/e$.

q = it, N = q/e

\[ q = (5.0)(240) = 1.2\times10^{3}\,\mathrm{C}, \qquad N = \frac{1.2\times10^{3}}{1.602\times10^{-19}} \approx 7.5\times10^{21} \]

A coulomb is an enormous amount of charge — over a sextillion electrons stream past in just four minutes.

2 Drift speed in copper

Problem. A copper wire of radius 0.90 mm carries 17 mA. With one conduction electron per atom, $n = 8.49\times10^{28}\,\mathrm{m^{-3}}$. Find the drift speed.

Solution. Use $v_{d} = i/(nAe)$ with $A = \pi r^{2}$.

v_d = i / (nAe)

\[ v_{d} = \frac{17\times10^{-3}}{(8.49\times10^{28})(1.6\times10^{-19})(2.54\times10^{-6})} \approx 4.9\times10^{-7}\,\mathrm{m/s} \]

That is under 2 mm/h — slower than a sluggish snail, yet the wire delivers its current essentially instantly.

3 Resistance depends on how you connect it

Problem. An iron block ($\rho = 9.68\times10^{-8}\,\Omega\cdot\mathrm{m}$) measures 1.2 cm × 1.2 cm × 15 cm. Find its resistance (a) end to end across the square faces, and (b) across two opposite long faces.

Solution. Apply $R = \rho L/A$, with $L$ the distance between the faces used.

R = ρL/A for each orientation

\[ R_{1} = \frac{(9.68\times10^{-8})(0.15)}{1.44\times10^{-4}} \approx 1.0\times10^{-4}\,\Omega, \qquad R_{2} = \frac{(9.68\times10^{-8})(0.012)}{1.80\times10^{-3}} \approx 6.5\times10^{-7}\,\Omega \]

Same block, same material — but the resistance differs by more than a factor of 100, set entirely by the ratio $L/A$.

4 Power dissipated by a heating wire

Problem. A Nichrome wire has resistance 72 Ω. Find the dissipation when 120 V is applied (a) across the full wire, and (b) across each half after cutting it in two.

Solution. Use $P = V^{2}/R$. Each half has resistance 36 Ω.

P = V²/R

\[ P_{\text{full}} = \frac{(120)^{2}}{72} \approx 200\,\mathrm{W}, \qquad P_{\text{two halves}} = 2\times\frac{(120)^{2}}{36} \approx 800\,\mathrm{W} \]

Halving the wire quadruples the heat output — but also draws far more current, which would quickly overload the wire. Convenient on paper, unwise in practice.

5 Cost of leaving a bulb on

Problem. A 100 W bulb runs continuously for a 31-day month at US$0.06 per kW·h. Find the cost, and the bulb's resistance and current on a 120 V supply.

Solution. Energy is $P\,t$; then $R = V^{2}/P$ and $i = P/V$.

Energy = Pt; R = V²/P; i = P/V

\[ Pt = (0.100\,\mathrm{kW})(744\,\mathrm{h}) \approx 74\,\mathrm{kW\cdot h} \Rightarrow \$4.5, \quad R = \frac{120^{2}}{100} = 144\,\Omega, \quad i = \frac{100}{120} \approx 0.83\,\mathrm{A} \]

The whole month costs only a few dollars — and the relations $P = V^{2}/R = iV$ tie the rating, resistance, and current together at a glance.

Review

Chapter Summary

✓ Current

$i = \dfrac{dq}{dt}$, in amperes; the same through every cross section (charge conserved).

✓ Current density

$i = \int \vec{J}\cdot d\vec{A}$; uniform case $J = i/A$, in A/m².

✓ Drift speed

$\vec{J} = ne\vec{v}_{d}$; tiny (~µm/s) even though the field signal travels near light speed.

✓ Resistance

$R = \dfrac{V}{i}$ in ohms; resistivity $\rho = E/J$, conductivity $\sigma = 1/\rho$.

✓ Geometry

$R = \dfrac{\rho L}{A}$; with temperature, $\rho = \rho_{0}[1 + \alpha(T - T_{0})]$.

✓ Ohm's law

A device is Ohmic when $R$ is independent of $V$ — a straight $i$–$V$ line.

✓ Microscopic ρ

$\rho = \dfrac{m}{e^{2}n\tau}$; metals obey Ohm's law because $\tau$ is field-independent.

✓ Power

$P = iV$ in general; resistive heating $P = i^{2}R = V^{2}/R$.

Practice

Problems

Take $e = 1.602\times10^{-19}\,\mathrm{C}$, copper's $n = 8.49\times10^{28}\,\mathrm{m^{-3}}$, and resistivities from Table 26-1. Keep straight which quantities are properties of the object ($i, V, R$) and which of the material ($J, E, \rho$).

A 5.0 A current flows for 4.0 min. How many coulombs, and how many electrons, pass a cross section?
A copper wire of diameter 2.5 mm carries a current of $1.2\times10^{-10}\,\mathrm{A}$. Find the current density and the electron drift speed.
A beam carries $2.0\times10^{8}$ doubly charged positive ions per cm³, all moving north at $1.0\times10^{5}\,\mathrm{m/s}$. Find the magnitude and direction of the current density.
The current density in a wire of radius 2.0 mm is uniform at $2.0\times10^{5}\,\mathrm{A/m^{2}}$. Find the current through the outer region between $R/2$ and $R$.
A wire 4.00 m long and 6.00 mm in diameter has resistance 15.0 mΩ across 23.0 V. Find the current, the current density, and the resistivity; identify the material from Table 26-1.
What is the resistivity of a wire of diameter 1.0 mm, length 2.0 m, and resistance 50 mΩ?
A wire has resistance $R$. A second wire of the same material is half as long with half the diameter. Find its resistance in terms of $R$.
At what temperature is a copper conductor's resistance double its value at 20 °C? Take $\alpha = 4.3\times10^{-3}\,\mathrm{K^{-1}}$.
A flashlight bulb runs at 0.30 A and 2.9 V; its tungsten filament has resistance 1.1 Ω at 20 °C. Estimate the filament temperature when lit ($\alpha = 4.5\times10^{-3}\,\mathrm{K^{-1}}$).
A space heater of resistance 14 Ω runs on 120 V. Find the rate of energy dissipation, and the cost of 5.0 h at US$0.05 per kW·h.
An unknown resistor dissipates 0.540 W across a 3.00 V battery. At what rate does it dissipate across a 1.50 V battery?
A copper wire of cross section $2.00\times10^{-6}\,\mathrm{m^{2}}$ and length 4.00 m carries 2.00 A. Find the field along the wire and the thermal energy produced in 30 min.
A wire of resistance 6.0 Ω is drawn out to three times its length (same material and density). Find the new resistance.
Starting from $v_{d} = eE\tau/m$ and $J = nev_{d}$, derive $\rho = m/e^{2}n\tau$, and explain why a constant $\tau$ implies that metals obey Ohm's law.

Tip: three habits keep current-and-resistance problems clean. First, distinguish object quantities ($i, V, R$, read off meters) from material quantities ($J, E, \rho$, defined point by point) — and use $R = \rho L/A$ to cross between them. Second, for power, reach for $P = iV$ for any device but only $P = i^{2}R$ or $V^{2}/R$ for pure resistive heating. Third, when a wire's shape changes (stretched, narrowed, cut), track how both $L$ and $A$ change — and remember that stretching at constant volume raises $L$ and lowers $A$ together.