Current, Resistance, and Resistivity

Flip a light switch and the lamp answers instantly — yet the electrons in the wire crawl forward slower than the minute hand of a clock. Both facts are true at once, and holding them together is the first real test of understanding electric circuits. Somewhere between the power station and your desk lamp, a story about fields becomes a story about three numbers you can read off a multimeter: the current $I$ through a wire, the voltage $V$ across it, and the resistance $R$ that ties them together. This article is the circuit-element view of conduction. The track has already told the microscopic story — Ohm's Law and Conductivity builds $\mathbf{J} = \sigma\mathbf{E}$ from colliding electrons and shows how surface charges steer a uniform field along any bent wire. Here we work one level up, where an engineer or an experimentalist actually lives: a resistor is a lump with two terminals, its geometry and material set its resistance through $R = \rho L/A$ , temperature tilts the value, and the product $P = IV$ prices everything from a phone charger to a national grid. Mastering this layer buys you the entire circuits chapter that follows. EMF sources, Kirchhoff's rules, RC transients, and AC impedance all assume you can look at a wire and estimate its resistance, look at a rating plate and extract a current, and switch fluently among $P = IV$ , $I^2R$ , and $V^2/R$ without a moment's doubt about which one applies. Those three equivalent-looking formulas answer three different questions — and knowing which is which is the difference between predicting that transmission lines run at 400 kV and being mystified by it.

Key takeaways

Current is charge per time through a cross-section: $I = dQ/dt$ ; one ampere is one coulomb per second. Conventional current points along positive-charge flow, opposite to the electron drift in a metal.
The carriers crawl: in household copper at 10 A the drift speed is under a millimetre per second (about 23 minutes per metre) — but the field that starts them moving propagates at nearly light speed, so circuits respond instantly.
Geometry sets resistance: $R = \rho L/A$ . Doubling length doubles $R$ (series stacking); doubling area halves it (parallel stacking). The material's contribution is the resistivity $\rho$ .
Resistivity is the widest-ranging material property in physics: from $1.7\times10^{-8}\ \Omega\cdot\text{m}$ (copper) to $\sim10^{12}\ \Omega\cdot\text{m}$ (glass) — twenty orders of magnitude.
Temperature tilts it: $\rho(T) = \rho_0[1 + \alpha(T - T_0)]$ for metals near room temperature; copper's resistance climbs about 0.4% per kelvin.
Power converts at the rate $P = IV = I^2R = V^2/R$ — three forms, three distinct questions.
KEY IDEA: a resistor is geometry times material — $R = \rho L/A$ — and once $R$ is known, $V = IR$ and $P = IV$ price every charge that passes through it.

The core insight

Resistance packages a conductor's material and shape into one number, $R = \rho L/A$ , so that the full field story of conduction collapses to the terminal relation $V = IR$ and the energy bill $P = IV$ .

Where this lives today

Every engineering discipline that touches electricity keeps equation (4) and the power family within arm's reach. The interconnect layers of a modern processor are copper corridors a few tens of nanometres wide, where $R = \rho L/A$ with a nanoscale $A$ makes resistance — and the $I^2R$ heat it mints — a first-order design constraint; chip designers speak of "IR drop" across a die exactly the way this article speaks of voltage sag along an extension cord. At the opposite end of the scale, grid operators dispatch power around continents by juggling $I^2R$ losses against transformer ratios, and the electrification of everything — vehicles, heating, datacentres — is at bottom a war on stray resistance. The measurement side is just as alive. Platinum resistance thermometers exploit equation (5) to read temperatures to millikelvin precision; battery management systems in electric cars track the slow creep of cell internal resistance as the leading indicator of ageing; and the ohm itself is now realised through the quantum Hall effect — the resistance standard is a quantum number, $h/e^2$ , a story the Hall Effect article at this chapter's close tells properly.

The physical picture

Picture a copper wire as an extraordinarily crowded corridor. Every cubic metre of copper holds about $8.5\times10^{28}$ mobile electrons — roughly one per atom, already there before any battery is connected, jittering thermally at around $10^6$ m/s in random directions. Connect a battery and something remarkably gentle happens: superimposed on that frantic random motion, the whole electron sea acquires a collective creep — the drift — of well under a millimetre per second. Current is not a few electrons sprinting; it is an astronomical number of them shuffling.

Figure

The crowded corridor — random thermal motion at a million metres per second, with a sub-millimetre-per-second collective drift superimposed, and the two speeds that resolve the light-switch paradox

Why then does the lamp light instantly? Because you do not wait for any particular electron to arrive. The battery's push is communicated by the electric field, and — as the track's Ohm's Law and Conductivity article shows in detail — a whisker-thin skin of surface charge arranges itself on the wire within nanoseconds to steer a uniform field along the wire's interior, around every bend. That field starts every electron in the circuit drifting essentially simultaneously, the way a locomotive starts the last wagon of a train moving almost as soon as the first. The marchers are slow; the order to march travels at nearly the speed of light. The second picture to install is resistance as geometry. A long corridor impedes more than a short one; a wide corridor passes more traffic than a narrow one. Make a wire twice as long and you have stacked two resistors end to end — twice the resistance. Make it twice as thick and you have laid two corridors side by side — half the resistance. Everything else about the material (how cluttered the corridor is, how often carriers collide) is bundled into a single number, the resistivity $\rho$ . That one intuition, formalised below, is $R = \rho L/A$ .

Figure

Resistance is geometry — stacking wires in series doubles R, laying them in parallel halves it, and the material contributes only the per-unit constant rho

Current: the flow variable

Current answers the question "how much charge passes this cross-section per second?" If charge $dQ$ crosses in time $dt$ , the current is

I = \frac{dQ}{dt} \qquad (1)

where: $I$ — current, in amperes ( $1\ \text{A} = 1\ \text{C/s}$ ); $dQ$ — the net charge crossing a chosen cross-section. Current has a direction convention rather than a vector character in a thin wire: conventional current points the way positive charge would move, so in a metal it points opposite the electron drift. A steady 10 A means a coulomb of charge — about $6\times10^{18}$ electron charges — clears the cross-section every tenth of a second. The track's Electric Current and Current Density article derives, by counting carriers through a window, the bridge between the terminal quantity $I$ and the microscopic sea:

I = n q A v_d \qquad (2)

where: $n$ — carrier number density; $q$ — carrier charge; $A$ — cross-sectional area; $v_d$ — drift speed. We take equation (2) as an input here (its derivation belongs to that article) and use it: because $n$ in a metal is enormous, $v_d$ must be tiny for any household current. Solving for $v_d$ in real copper is Derivation B below, and the number is startling.

Resistance: the geometry variable

Experiment — Ohm's 1826 measurements, with a thermocouple as a drift-free source — shows that for metals at fixed temperature, current is proportional to the applied voltage. The ratio is the resistance:

V = IR \qquad (3)

where: $V$ — potential difference across the element, volts; $R$ — resistance, ohms ( $1\ \Omega = 1\ \text{V/A}$ ). Read it as a material promise: measure $R$ once, and you can predict $I$ at any modest voltage without touching the wire again. The proportionality is not a law of nature — it is an experimental property of ohmic materials. Filament lamps (whose temperature soars in use), diodes, and electrolytes all bend away from the straight line; for them $R \equiv V/I$ still defines a number, but it changes with the operating point. What sets $R$ ? Geometry and material, through the relation this article owns from the stacking side (Derivation A):

R = \frac{\rho L}{A} \qquad (4)

where: $L$ — length along the current; $A$ — cross-sectional area; $\rho$ — resistivity, in $\Omega\cdot\text{m}$ , the material's intrinsic per-unit resistance. Long and thin resists; short and fat conducts. The same relation is minted at the field level in Ohm's Law and Conductivity — a uniform wire carrying $J = \sigma E$ delivers $R = \rho L/A$ with $\rho = 1/\sigma$ — so the corridor picture and the field picture agree, as they must. Resistivity is worth a moment of awe. No other everyday material property spans a range like this:

material	$\rho$ at 20 °C ( $\Omega\cdot\text{m}$ )	class
copper	$1.68\times10^{-8}$	conductor
aluminium	$2.65\times10^{-8}$	conductor
iron	$9.7\times10^{-8}$	conductor
nichrome	$1.10\times10^{-6}$	heating alloy
seawater	$\sim0.2$	ionic conductor
silicon (pure)	$\sim2.3\times10^{3}$	semiconductor
glass	$10^{10}$ – $10^{14}$	insulator

From copper to glass is roughly twenty orders of magnitude — the reason a copper core and its plastic sleeve, millimetres apart, can carry mains current and be safe to touch. Between the extremes sit the semiconductors, whose resistivity is not fixed but engineerable, which is the entire semiconductor industry in one sentence. Temperature tilts the value. For metals near room temperature the tilt is honestly linear:

\rho(T) = \rho_0\left[1 + \alpha\,(T - T_0)\right] \qquad (5)

where: $\rho_0$ — resistivity at the reference temperature $T_0$ (usually 20 °C); $\alpha$ — the temperature coefficient, about $3.9\times10^{-3}\ \text{K}^{-1}$ for copper. Physically, a hotter lattice vibrates harder and scatters the drifting carriers more often — the Drude story, told properly in Ohm's Law and Conductivity. The sign flips for semiconductors ( $\rho$ falls with temperature as more carriers are freed), and at the extreme end lies superconductivity: below a critical temperature, some materials drop to $\rho = 0$ exactly — not merely small, zero — a quantum effect owned far downstream in the track. Equation (5) also makes every copper winding a thermometer: a 10% rise in a motor coil's resistance reports a temperature climb of about 26 K.

Power: the energy bill

Each coulomb that falls through a potential difference $V$ delivers $V$ joules. At current $I$ , coulombs arrive at $I$ per second, so the conversion rate — the power — is

P = IV \qquad (6)

where: $P$ — power in watts; this form is universal, valid for any two-terminal element: resistor, motor, battery under charge, LED. For a resistor specifically, substituting $V = IR$ or $I = V/R$ gives the family $P = I^2R = V^2/R$ . Derivation C builds the family and settles the classic confusion — how can power be proportional to $R$ in one formula and to $1/R$ in another? — by attaching each form to the variable that is actually being held fixed.

question	hold fixed	use	behaviour
heat in a wire carrying a set current (series path, transmission line)	$I$	$P = I^2R$	more $R$ , more heat
heat in a device across a set supply (parallel across mains)	$V$	$P = V^2/R$	more $R$ , less heat
anything, when both terminal values are known	—	$P = IV$	always safe

One honesty note before the derivations. Circuit language says "the wire carries the power," but the field story — the track's Poynting Vector article — shows the energy actually flows through the fields around the conductors and enters the resistor through its sides; the wires are guides, not pipes. Nothing in this article's accounting changes, but it is worth knowing which picture is fundamental.

Derivation AThe stacking laws — R equals rho L over A from series and parallel composition

We want the geometry law $R = \rho L/A$ using nothing but the definition $R = V/I$ and two thought experiments about composition — no field theory required. The strategy: show that resistance must be additive under lengthwise stacking and reciprocal-additive under side-by-side stacking, and that those two scalings force the form $R \propto L/A$ with a single material constant. The field-level route (uniform $J = \sigma E$ in a wire) lives in Ohm's Law and Conductivity; getting the same law from pure composition logic is the point here. Watch how little needs to be assumed: only that the material is uniform and the current fills the cross-section evenly.

Step 1. Cut a uniform wire of length $2L$ mentally into two segments of length $L$ carrying the same current $I$ . The voltage across the whole is the sum of the voltages across the halves — potential drops add along a path — so $V_{2L} = V_L + V_L = 2V_L$ , and therefore $R_{2L} = V_{2L}/I = 2R_L$ . Why this matters: resistance is proportional to length, $R \propto L$ , because series voltages add while the current is common.

Step 2. Now lay two identical wires of area $A$ side by side between the same two terminals. Each sees the full voltage $V$ and each passes its own current $I_1 = V/R_A$ , so the pair passes $2I_1$ . The combination's resistance is $V/(2I_1) = R_A/2$ . A wire of area $2A$ is two wires of area $A$ fused along their length — the fusing changes nothing if the material is uniform, since matching interior points sit at equal potentials and no current wants to cross between them. Why this matters: resistance is inversely proportional to area, $R \propto 1/A$ , because parallel currents add while the voltage is common.

Step 3. The only function of $L$ and $A$ consistent with both scalings is $R = \rho\,L/A$ , where the constant $\rho$ carries everything about the material and nothing about the shape. Dimensions check: $[\rho] = \Omega\cdot\text{m}$ , so that $\Omega\cdot\text{m}\times\text{m}/\text{m}^2 = \Omega$ .

Step 4. Calibrate against a real wire. A 30 m run of 12-gauge copper ( $d = 2.05$ mm, $A = 3.31\times10^{-6}\ \text{m}^2$ ): $R = (1.68\times10^{-8})(30)/(3.31\times10^{-6}) = 0.152\ \Omega$ . Carrying 15 A it drops $V = IR = 2.3$ V and dissipates $P = I^2R = 34$ W along the run — small enough to be acceptable, large enough to explain why long extension cords at high current run warm. Verified: numerics recomputed ( $A = \pi(1.025\times10^{-3})^2 = 3.31\times10^{-6}\ \text{m}^2$ , $R = 0.1523\ \Omega$ , $V = 2.28$ V, $P = 34.3$ W); the composition scalings agree with the field-level result $R = L/(\sigma A)$ of Ohm's Law and Conductivity with $\rho = 1/\sigma$ .

Go deeper · Derivation B · 1 minDrift velocity in real copper — and the 23-minute electron

Derivation BDrift velocity in real copper — and the 23-minute electron

We want the actual drift speed of conduction electrons in a household wire, and from it the resolution of the light-switch paradox. Strategy: take the carrier-counting relation $I = nqAv_d$ — derived by the track's Electric Current and Current Density article, restated here as an input — solve it for $v_d$ , and put in honest copper numbers. The pitfall to avoid is confusing three utterly different speeds that coexist in the same wire: the thermal jitter ( $\sim10^6$ m/s), the drift ( $\sim10^{-4}$ m/s), and the signal speed ( $\sim10^8$ m/s).

Step 1. Copper donates about one mobile electron per atom. From its density (8960 kg/m³) and molar mass (63.5 g/mol), the carrier density is $n = 8.5\times10^{28}\ \text{m}^{-3}$ .

Step 2. Solve equation (2) for the drift speed and insert $I = 10$ A through $A = 1\ \text{mm}^2 = 10^{-6}\ \text{m}^2$ :

v_d = \frac{I}{nqA} = \frac{10}{(8.5\times10^{28})(1.60\times10^{-19})(10^{-6})} = 7.3\times10^{-4}\ \text{m/s}

Why so slow: the denominator is astronomical — $nqA \approx 1.4\times10^{4}$ coulombs of mobile charge per metre of wire. When every metre of corridor already holds fourteen thousand coulombs, a ten-coulomb-per-second flow requires only a crawl.

Step 3. Time to drift one metre: $t = 1/v_d \approx 1360\ \text{s} \approx 23$ minutes. An electron leaving the switch reaches the lamp long after you have left the room.

Step 4. The paradox resolves because the signal is not the electrons. Closing the switch lets the source's push propagate as an electromagnetic rearrangement: surface charges redistribute along the wire in nanoseconds (the steering mechanism owned by Ohm's Law and Conductivity), establishing the interior field everywhere around the loop at a good fraction of light speed. Every electron in the circuit — including those already inside the lamp filament — begins drifting at once. The corridor is pre-filled; the march order arrives at $\sim c$ ; nobody waits for marchers from the far end. Verified: $v_d = 10/(8.5\times10^{28} \cdot 1.602\times10^{-19} \cdot 10^{-6}) = 7.34\times10^{-4}$ m/s, transit time $22.7$ min — recomputed; consistent with the 0.5 mm/s quoted for a 1.5 mm² wire in Ohm's Law and Conductivity (same current, larger area, proportionally slower).

Go deeper · Derivation C · 1 minThe power family — P equals IV and its two resistor offspring

Derivation CThe power family — P equals IV and its two resistor offspring

We want the rate at which a circuit element converts electrical energy, in terms of terminal quantities only, and then the two specialised forms for resistors — including a clean resolution of the "is power proportional to $R$ or to $1/R$ ?" confusion. Strategy: start from the definition of potential as energy per charge, multiply by the charge flow rate, then substitute Ohm's law twice. The field-level version of this accounting ( $p = \mathbf{J}\cdot\mathbf{E}$ per unit volume) is owned by Ohm's Law and Conductivity; here we stay at the terminals, which is where meters and rating plates live.

Step 1. A charge $dQ$ falling through a potential difference $V$ hands over energy $dU = V\,dQ$ . That is the definition of potential difference — joules per coulomb — from the electrostatics chapters.

Step 2. Divide by $dt$ . The energy conversion rate is $P = dU/dt = V\,(dQ/dt) = VI$ . This is equation (6), and nothing in Steps 1–2 assumed a resistor: $P = IV$ prices motors, batteries, and LEDs alike. For a resistor the energy emerges as heat; for a motor, mostly as mechanical work.

Step 3. Specialise with Ohm's law. Substituting $V = IR$ gives $P = I^2R$ ; substituting $I = V/R$ gives $P = V^2/R$ . All three are the same statement about a resistor operating at one point — they differ only in which variable you know.

Step 4. The apparent contradiction dissolves once you name what is held fixed. In a series situation the current is common: a transmission line's loss $I^2R$ grows with its resistance, and grows with the square of the shipped current — which is why grids raise $V$ to slash $I$ at fixed power, the calculation the Transformers article runs end to end. In a parallel situation the voltage is common: across 230 V mains, a 26 Ω kettle element ( $P = V^2/R = 2$ kW) roars while a 10 kΩ night-light resistor barely warms. More resistance means more power in the first case and less in the second, with no paradox anywhere. Verified: kettle check $I = P/V = 2000/230 = 8.7$ A and $R = V^2/P = 26.5\ \Omega$ recomputed; the two specialised forms reproduce $P = IV$ at the common operating point by construction.

Limiting cases

case	result	meaning
$L \to 2L$ at fixed $A$ , $\rho$	$R \to 2R$	series stacking: drops add
$A \to 2A$ at fixed $L$ , $\rho$	$R \to R/2$	parallel stacking: currents add
$\rho \to 0$	$R \to 0$ , $V \to 0$ at any $I$	ideal wire — the schematic assumption; real superconductors achieve it exactly
$\rho \to \infty$	$I \to 0$ at any $V$	ideal insulator / open switch
$T$ rises (metal)	$R$ climbs $\sim0.4\%/\text{K}$ (Cu)	hotter lattice scatters carriers more
$T$ rises (semiconductor)	$R$ falls	thermal carrier generation wins
non-ohmic element	$V/I$ varies with operating point	$R$ still defined, no longer constant

Common trapMore resistance always means less power dissipated

Neither direction of this instinct survives contact with the power family. Across a fixed voltage (parallel thinking), $P = V^2/R$ : less resistance means more power — that is why a short circuit is a fire, not a switch-off, and why the 26 Ω kettle outheats the 10 kΩ night-light on the same mains. But at fixed current (series thinking), $P = I^2R$ : more resistance means more power — that is why the transmission line's resistance, not the load's, sets the grid's losses, and why a corroded high-resistance joint in a series path gets hot while everything around it stays cool. The formulas do not disagree; they answer different questions. Before predicting power, always name what your circuit holds fixed — the voltage across the element, or the current through it.

Worked examples

Example 1 — the numbers on a kettle's rating plate. A kettle marked 230 V, 2000 W hides its whole electrical identity in two numbers. Find its operating current, its element resistance, and the cost of a four-minute boil at $0.15 per kWh.

Step 1. Current from the universal form: $I = P/V = 2000/230 = 8.7$ A. (Any appliance's current comes from the plate this way — it is how you check a fuse or an extension-cord rating.)

Step 2. Element resistance at operating temperature: $R = V^2/P = 230^2/2000 = 26.5\ \Omega$ . A cold multimeter measurement would read noticeably less — the nichrome element's resistance climbs as it heats, equation (5).

Step 3. Energy for four minutes: $E = Pt = 2\ \text{kW} \times \tfrac{4}{60}\ \text{h} = 0.133$ kWh, costing $0.133 \times 0.15 \approx$ 2 cents. The kilowatt-hour is nothing exotic — just $P\,t$ in grid-friendly units ( $1\ \text{kWh} = 3.6\times10^6$ J).

Example 2 — is this extension cord safe for a space heater? A 30 m extension cord of 12-gauge copper ( $A = 3.31\ \text{mm}^2$ ) feeds a 1500 W heater from a 120 V socket. Find the cord's resistance (both conductors count), the voltage actually reaching the heater, and the heat generated in the cord itself.

Step 1. The current loop traverses the cord twice — out and back — so the copper length is 60 m: $R_{\text{cord}} = \rho L/A = (1.68\times10^{-8})(60)/(3.31\times10^{-6}) = 0.305\ \Omega$ .

Step 2. Estimate the current with the heater's nominal draw $I \approx P/V = 1500/120 = 12.5$ A. Cord drop: $V_{\text{cord}} = IR_{\text{cord}} = 12.5 \times 0.305 = 3.8$ V, so the heater sees about 116 V — a 3% shave, acceptable.

Step 3. Heat released in the cord: $P_{\text{cord}} = I^2R_{\text{cord}} = (12.5)^2(0.305) = 48$ W — a gentle warmth spread over 30 m, safe. Repeat the arithmetic for a cheap 20 m cord of 0.5 mm² conductors and you get $R = 1.34\ \Omega$ and $P \approx 210$ W: ten watts per metre, a cord that runs dangerously hot under the same heater. Wire gauge tables and fuse ratings are exactly this calculation, standardised. Both examples recomputed numerically: $26.45\ \Omega$ , $0.1333$ kWh; $R_{\text{cord}} = 0.3045\ \Omega$ , drop $3.81$ V, $47.6$ W; thin-cord case $1.344\ \Omega$ , $210$ W.

Origins · 1 minJames Prescott Joule and how this was discovered

James Prescott Joule

1818–1889 · Salford and Manchester, England

A brewer by trade who did physics with a brewer's obsession for measurement, Joule settled in 1841 the question this article's power section answers: at what rate does a current heat a wire? His careful calorimetry — coils of wire in jars of water, thermometers read to fractions of a degree — established that the heat generated goes as $I^2R$ , now called Joule heating, before anyone knew what was drifting inside the wire. The result made energy bookkeeping in circuits possible and became one pillar of his life's larger edifice: that heat itself is a form of energy, quantified by his mechanical equivalent of heat (1843–1849). The joule and the watt-second are the same unit because this brewer proved they must be. Ohm gave circuits their traffic law; Joule wrote the energy bill — and the $I^2R$ term in every grid-loss estimate, every fuse rating, and every datacentre cooling budget is his.

History

1800

— Volta's pile provides the first steady current, turning electricity into a flow that can be measured rather than a spark to be stored

1826

— Ohm, using a thermocouple for a drift-free source, establishes that current is proportional to driving tension

1827

— Die galvanische Kette publishes the law and the resistance concept; German reviewers savage it and Ohm loses his teaching post

1841

— Joule's calorimetry establishes the I²R heating law; the Royal Society awards Ohm the Copley Medal the same year, completing his rehabilitation

1861

— The British Association committee under William Thomson begins standardising the ohm, so that resistance can be traded like a commodity

1911

— Kamerlingh Onnes finds mercury's resistivity vanish exactly below 4.2 K — superconductivity, the ρ = 0 limit made real

1980

— von Klitzing's quantum Hall measurements make resistance the most precisely realisable electrical unit, h/e² to parts per billion

Going further

Within this chapter: the natural next steps are EMF and Internal Resistance (what happens when the source itself resists), Kirchhoff's Rules (bookkeeping for circuits too tangled for single-loop reasoning), and Series and Parallel Networks (the reduction toolkit that turns resistor webs into single numbers — including the equivalent-resistance shortcuts and meter-loading corrections deferred there by name).

Deeper in the track: Ohm's Law and Conductivity (EMAG.MAXW.OHM.0549) owns everything beneath this article's floor — the Drude model $\sigma = ne^2\tau/m$ , the constitutive law $\mathbf{J} = \sigma\mathbf{E}$ , the surface-charge steering that keeps the field uniform in bent wires, and field-level Joule heating $\mathbf{J}\cdot\mathbf{E}$ . Electric Current and Current Density (EMAG.MAXW... E4 chapter) owns the carrier-counting derivation of $I = nqAv_d$ . The Poynting Vector tells the honest story of where the energy actually travels. The Hall Effect, closing this chapter, measures the $n$ that this article could only quote.

Provenance. Authored fresh for the SuperSTEM E&M track's circuits chapter (recon-§d gap heal); sources: v3 circuit pilots (rewritten to UG register) + legacy wiki circuit articles (narrative) — no AllQFT base exists for this topic.

If the electrons move so slowly, why does the light come on the instant I flip the switch?

Because the switch does not launch electrons on a journey — it releases a field. Within nanoseconds, surface charges rearrange along the whole wire and establish the driving field everywhere in the loop at a good fraction of light speed. Every electron already inside the wire — including those inside the lamp filament — starts drifting at essentially the same moment. The current everywhere rises together; nobody waits for charge to arrive from the switch. The slow speed is the drift of the marchers; the fast speed is the order to march.

Is V = IR true for everything? My phone charger and LEDs seem more complicated.

No — and knowing the boundary is part of the law. $V = IR$ with constant $R$ holds for ohmic materials (metals, carbon resistors) at reasonably steady temperature. Diodes and LEDs are aggressively non-ohmic: their current rises exponentially with voltage. Filament bulbs are ohmic at any one temperature but self-heat so strongly that their effective $R$ changes tenfold between off and on. For any element you can still define $R = V/I$ at an operating point; the ohmic promise — same $R$ at every operating point — is what fails.

What is the difference between resistance and resistivity, really?

Resistivity $\rho$ belongs to the material; resistance $R$ belongs to the object. A metre cube of copper and a kilometre of thin copper wire have the same $\rho$ and wildly different $R$ . The bridge is geometry: $R = \rho L/A$ . When a table quotes "copper: $1.68\times10^{-8}\ \Omega\cdot\text{m}$ " it is telling you about every copper object at once; you supply the shape.

Why does making a wire hotter increase its resistance, but heating a semiconductor decreases it?

Two competing effects. In a metal, the carrier count is fixed (one electron per atom, roughly), so the only lever is how often carriers scatter — and a hotter, more violently vibrating lattice scatters them more: $\rho$ rises. In a semiconductor the carrier count itself is the bottleneck: heat shakes electrons loose across the band gap, multiplying the carriers far faster than scattering worsens, so $\rho$ falls. Same equation $\sigma = ne^2\tau/m$ (from the conductivity article), different term in charge.

Which power formula should I use — P = IV, I²R, or V²/R?

$P = IV$ is always safe when you know both terminal values, and it is the only one valid for non-resistors (motors, charging batteries). The other two are resistor shortcuts, and the choice follows the fixed variable: series elements share a fixed current, so use $I^2R$ ; elements across a fixed supply share the voltage, so use $V^2/R$ . Every "paradox" about power and resistance is a mismatch between formula and fixed variable.

How is this article different from Ohm's Law and Conductivity earlier in the track?

Division of labour. That article works at the field level: it derives $\mathbf{J} = \sigma\mathbf{E}$ from the Drude collision model, explains how surface charges steer the field around bends, and does energy dissipation as $\mathbf{J}\cdot\mathbf{E}$ per unit volume. This article works at the terminal level: lumps, meters, rating plates — $R = \rho L/A$ from composition logic, $P = IV$ from energy-per-charge. The two derivations of $R = \rho L/A$ (stacking here, uniform-field there) are deliberately different routes to the same law, and each article points at the other for the missing half.

Does a thicker wire really carry current more easily? It feels like more material should mean more collisions.

More material means more parallel corridors, not a longer obstacle course. Each patch of cross-section carries its share of the current at the same drift speed for the same field; doubling the area doubles the number of carriers passing per second at no extra cost. Collisions per carrier are unchanged — there are simply more carriers. That is the parallel-stacking argument, and it is why $A$ sits in the denominator of $R = \rho L/A$ .

Is anything actually used up as current flows through a resistor?

Charge is not — every electron that enters leaves; current in equals current out, which becomes Kirchhoff's junction rule in the next articles. What is spent is energy: each coulomb drops $V$ joules of potential energy, delivered to the lattice as heat at rate $P = IV$ . A battery running flat has lost chemical energy, not charge. Keeping the two ledgers separate — conserved charge, spent energy — is the accounting discipline the whole circuits chapter is built on.