In short

Ohm's law states that, for a metallic conductor at constant temperature, the current I through it is proportional to the voltage V across it:

\boxed{\;V \;=\; IR\;}.

The proportionality constant R is the resistance, measured in ohms (\Omega): one ohm is one volt per ampere. Resistance is a property of the specific piece of conductor — its material, length, cross-section, and temperature. For a uniform wire of length L and cross-sectional area A made of material with resistivity \rho,

\boxed{\;R \;=\; \rho\,\frac{L}{A}\;}.

Resistivity \rho is an intrinsic property of the material. Copper sits at 1.7\times 10^{-8}\ \Omega\cdotm; nichrome (the wire in your toaster) sits at 1.1\times 10^{-6}\ \Omega\cdotm — about 65 times higher — which is exactly why the copper carries current while the nichrome gets hot. Conductivity \sigma = 1/\rho and conductance G = 1/R are the reciprocals; use whichever is more convenient.

Ohm's law is not a fundamental law — it is an empirical statement that holds for ohmic conductors (most metals). Non-ohmic devices — diodes, filament lamps, semiconductors near breakdown — have nonlinear V–I curves. The slope of the V–I graph at any point is still called a "differential resistance", but the simple V = IR relation no longer holds.

Open the grey switchboard next to the entrance of a two-bedroom flat in a Delhi housing society. Behind the circuit breakers you see a rectangular copper bar about three millimetres thick — the bus-bar — feeding current to every socket in the flat. When the AC is running, the mixer is on in the kitchen, and both LED tube-lights in the living room are lit, something like 15 amps of current is flowing through that copper strip. The voltage drop across the entire metre-long length of the bus-bar is only a few millivolts. That tiny voltage, divided into 15 amps, is the bar's resistance — about 10^{-4}\ \Omega. Almost zero. The bar is there precisely because it refuses to drop voltage. It is a bad resistor and a good conductor.

Now walk into the kitchen. The pop-up toaster on the counter draws roughly 6 amps at 230 V. Inside the toaster is a wire coiled around a mica former — a thin nichrome wire, about 0.3 mm in diameter. That wire has a resistance of about 38\ \Omega, some 380,000 times higher than the bus-bar. When 6 amps flow through it, the voltage drop across it is 230 V and the power dissipated is I^{2}R = 36 \times 38 \approx 1.4 kilowatts, all converted to heat. That is what toasts your bread. The nichrome is there precisely because it does drop voltage. It is a good resistor and a very bad conductor.

Same current-carrying geometry, same circuit, two opposite roles. The physics that distinguishes them is Ohm's law and the concept of resistivity. This chapter makes both precise, derives R = \rho L/A from the drift-velocity picture you met in the previous chapter, and builds an interactive that lets you change a conductor's dimensions and watch the resistance respond.

The observation — current is proportional to voltage

Take a fixed piece of copper wire. Connect it across a variable DC supply. Measure the current through it as you vary the voltage. Plot the results.

What you see, for any moderate voltage, is a straight line through the origin. Doubling the voltage doubles the current. Tripling the voltage triples the current. Reversing the voltage reverses the current. The wire's response is linear.

V-I characteristic of an ohmic conductorStraight line through the origin on a voltage versus current plot, with slope labelled as resistance R. The line extends into the third quadrant (reversed current for reversed voltage).I (current, amperes)V (voltage, volts)123−1−224−2−4slope = Rohmic conductor
The $V$–$I$ characteristic of a copper wire (or any ohmic conductor) at fixed temperature is a straight line through the origin. The slope of that line is the resistance $R$. Doubling the voltage doubles the current; reversing the voltage reverses the current.

Georg Ohm established this empirically in 1827 by rolling his own batteries and wires. The modern statement, with the proportionality constant named R:

V \;=\; IR. \tag{1}

Equation (1) is the defining relation of resistance. In a moment you will derive it from the microscopic physics of electrons in a metal. But first, three observations about the formula.

Observation 1 — R is a property of the object, not the circuit

Plug the same copper wire into a battery of 3 V, it carries 3/R amps; into 9 V, it carries 9/R amps. The same R holds in both cases. Change the voltage, the current changes; R stays fixed.

Change the wire, though — use a longer copper wire, or a thinner one, or a piece of nichrome of the same shape — and R changes. Resistance is a property of the specific conductor, determined by its material and its geometry.

Observation 2 — the unit

From R = V/I, the unit of resistance is volts per ampere. This is named the ohm (\Omega) in honour of Ohm:

1\ \Omega \;=\; 1\ \text{V/A}.

An ordinary copper extension cord has about 0.1\ \Omega per metre. An LED resistor in an Arduino project is typically 220\ \Omega. The heating element of a hair dryer is about 15\ \Omega.

Observation 3 — temperature must be constant

Ohm's law holds at a fixed temperature. Heat a copper wire up and its resistance rises — metals get more resistive as they get hotter (the next chapter, Resistivity and Temperature Dependence, works this out). When a filament lamp's V–I curve bends upward at higher voltage, it is not a violation of Ohm's law; it is the filament heating up and R changing along the way.

Deriving V = IR from the drift-velocity picture

In the previous chapter you saw the drift-velocity formula: for electrons of density n per cubic metre drifting at speed v_{d} through a wire of cross-section A, the current is

I \;=\; n\,A\,e\,v_{d}. \tag{2}

where e is the electron's charge magnitude. Now ask: what determines v_{d} when you apply an electric field E along the wire?

Between collisions with lattice ions, an electron accelerates freely under the field's force eE. It picks up velocity v = (eE/m_{e})\,\tau over an average time \tau (the "mean free time" between collisions), where m_{e} is the electron mass. A collision randomises the velocity — after each collision the electron starts with zero average forward velocity again, and the cycle repeats. Averaging over many collisions gives the net drift velocity

v_{d} \;=\; \frac{eE}{m_{e}}\,\tau. \tag{3}

Why: the force on one electron is F = eE (field times charge), so acceleration is eE/m_{e}. Over time \tau between collisions, it gains extra speed a\tau = eE\tau/m_{e}. Collisions then scramble the direction randomly, so only this systematic drift accumulates — a net motion against the field (for the negative electron) or, equivalently, conventional current in the direction of the field.

Substitute (3) into (2):

I \;=\; n\,A\,e\cdot\frac{eE}{m_{e}}\,\tau \;=\; \frac{n e^{2} \tau}{m_{e}}\cdot A\cdot E.

Now use the field-voltage relation. For a uniform wire of length L with voltage V across it, the field inside the wire is uniform: E = V/L.

I \;=\; \frac{n e^{2} \tau}{m_{e}}\cdot A\cdot\frac{V}{L}.

Rearrange to isolate V:

V \;=\; I\cdot \frac{m_{e}L}{n e^{2}\tau A}. \tag{4}

Why: this is Ohm's law in disguise. The bracketed quantity m_{e}L/(ne^{2}\tau A) is exactly what we should call the resistance R — it depends on the wire's dimensions (L and A) and on material properties (n, m_{e}, e, \tau), but not on the voltage or current. So V is proportional to I, with a proportionality constant that is a property of the wire.

Identifying R in equation (4):

R \;=\; \frac{m_{e}L}{n e^{2}\tau\, A} \;=\; \left(\frac{m_{e}}{n e^{2}\tau}\right)\cdot\frac{L}{A}.

Define the resistivity \rho as the material-specific factor:

\boxed{\;\rho \;=\; \frac{m_{e}}{n e^{2}\tau}\;}, \qquad \boxed{\;R \;=\; \rho\,\frac{L}{A}\;}. \tag{5, 6}

Why: the factorisation separates geometry (L/A) from material (\rho). Two different pieces of the same material — a short fat one and a long thin one — have the same \rho but different R. Two pieces of different materials with the same L and A have the same geometric factor but different R, because their \rho's differ.

Dimensional check

Units of R should be ohms. Units of \rho:

[\rho] \;=\; \frac{[m_{e}]}{[n][e^{2}][\tau]} \;=\; \frac{\text{kg}}{(1/\text{m}^{3})\cdot\text{C}^{2}\cdot\text{s}} \;=\; \frac{\text{kg}\cdot\text{m}^{3}}{\text{C}^{2}\cdot\text{s}}.

Using 1\ \text{V} = 1\ \text{kg}\cdot\text{m}^{2}/(\text{C}\cdot\text{s}^{2}) (from energy-per-charge) and 1\ \text{A} = 1\ \text{C/s}:

\text{V}\cdot\text{s}/\text{C} \;=\; \text{kg}\cdot\text{m}^{2}/\text{C}^{2}/\text{s}.

So

\Omega\cdot\text{m} \;=\; (\text{V/A})\cdot\text{m} \;=\; \frac{\text{kg}\cdot\text{m}^{2}}{\text{C}^{2}/\text{s}}\cdot\text{m} \;=\; \frac{\text{kg}\cdot\text{m}^{3}}{\text{C}^{2}\cdot\text{s}}. \checkmark

Resistivity's SI unit is \Omega\cdotm. Resistance's SI unit is \Omega. Both consistent.

Interactive: shape the wire

Below, fix the material at copper (\rho = 1.72\times 10^{-8}\ \Omega\cdotm) and vary the wire length L from 0.1 m to 10 m. The cross-sectional area is fixed at A = 1\ \text{mm}^{2} = 10^{-6}\ \text{m}^{2} (the cross-section of typical household wiring in India). Watch resistance rise linearly with length.

Interactive: resistance grows linearly with lengthStraight line showing R equals rho times L divided by A for copper, with rho one point seven two times ten to the negative eight, A one square millimetre. A draggable red point on the length axis from zero point one to ten metres; a tracking dot rides the line. A readout panel shows the numerical resistance in ohms and milliohms.wire length L (metres)resistance R (ohms)00.050.100.150.2013579R = ρ L / Acopper, ρ = 1.72×10⁻⁸ Ω·mA = 1 mm² (fixed)drag the red dot along the L axis
The line shows how the resistance of a 1-mm² copper wire scales with length. Drag the red point to change $L$. Even at 10 m the resistance is only 0.17 ohms — which is why copper is used for wiring: long runs contribute almost no voltage drop.

Three things to notice as you drag:

Conductance and conductivity — the reciprocals

Resistance measures how much voltage it costs to push a given current through a conductor. Sometimes you want the opposite question: how much current flows per volt of applied voltage? That is called the conductance:

G \;=\; \frac{1}{R}, \qquad \text{unit: siemens (S)}, \quad 1\ \text{S} = 1\ \Omega^{-1} = 1\ \text{A/V}.

Likewise, the reciprocal of resistivity is the conductivity:

\sigma \;=\; \frac{1}{\rho}, \qquad \text{unit: S/m} = 1/(\Omega\cdot\text{m}).

Both formulations are equivalent. Circuit engineers use R (resistance) because voltages are what they measure; condensed-matter physicists often prefer \sigma (conductivity) because it is a more linear quantity under small perturbations. Ohm's law in conductivity form is

\vec{J} \;=\; \sigma\,\vec{E},

where \vec{J} is current density (current per area) and \vec{E} is the electric field. This is the local Ohm's law; it holds at every point inside the conductor, not just across the whole wire. Its integrated version over a uniform wire gives back V = IR.

Ohmic vs non-ohmic conductors

Most metals (copper, aluminium, silver, gold, iron, nichrome) obey Ohm's law accurately over many orders of magnitude of voltage, at constant temperature. Aqueous electrolytes also do, roughly.

But many important devices do not obey Ohm's law.

Ohmic vs non-ohmic V-I characteristicsThree V-I curves on the same axes. One straight line through the origin labelled ohmic, one upward-bending curve labelled filament lamp, one exponential forward-biased diode curve.IVohmic (copper)filament lampsilicon diode
On the same $V$–$I$ axes: an ohmic resistor (straight line), a filament lamp (upward bend as $R$ rises with temperature), and a silicon diode (exponential, only conducts beyond 0.7 V). Only the first obeys $V = IR$ with a constant $R$.

For non-ohmic elements, engineers define a differential resistance r_\text{d} = dV/dI that equals the local slope of the V–I curve. The differential resistance changes with the operating point — there is no single number R that describes the element.

The moral: Ohm's law is not a fundamental law of physics the way Newton's laws or Maxwell's equations are. It is an approximate material property of metals that happens to hold over a wide range. Whenever you assume "this device is ohmic", you are making a modelling decision — convenient, almost always correct for copper and nichrome, but broken for diodes, transistors, or any semiconductor.

Worked examples

Example 1: Nichrome element of a 1 kW toaster

An electric toaster draws 1000 W from the 230 V Indian mains. Its heating element is a nichrome wire of cross-section A = 0.075\ \text{mm}^{2} = 7.5\times 10^{-8}\ \text{m}^{2}, resistivity \rho = 1.1\times 10^{-6}\ \Omega\cdotm at operating temperature. Find (a) the current drawn, (b) the resistance of the element, and (c) the length of nichrome wire used.

Nichrome heating coil of a toasterSchematic of a 230 V mains connected across a coiled nichrome wire. Labels show 1000 watts total, 4.35 amps current, 53 ohms resistance, cross-section 0.075 square millimetres.230 Vnichrome coil, R = 53 ΩI = 4.35 AP = 1000 W dissipated as heat
A 1000 W toaster on 230 V mains pulls 4.35 A through a 53 Ω nichrome coil. The coil is about 3.6 m long; it is wound into a compact spiral on a mica former inside the toaster.

Step 1. Current from power.

P \;=\; VI \;\Rightarrow\; I \;=\; \frac{P}{V} \;=\; \frac{1000}{230} \;\approx\; 4.35\ \text{A}.

Why: the defining formula for electrical power dissipated in a resistor is P = VI, which follows directly from the definition of voltage (energy per unit charge) and current (charge per unit time). For a known appliance power rating and line voltage, this tells you the current draw.

Step 2. Resistance from Ohm's law.

R \;=\; \frac{V}{I} \;=\; \frac{230}{4.35} \;\approx\; 53\ \Omega.

Why: once you know two of the three quantities V, I, R, Ohm's law gives you the third. The 53 Ω element is what makes the current small enough (4.35 A) to safely go through household wiring yet high enough to dissipate 1 kW.

Step 3. Length from R = \rho L/A.

L \;=\; \frac{R A}{\rho} \;=\; \frac{(53)(7.5\times 10^{-8})}{1.1\times 10^{-6}} \;=\; \frac{3.975\times 10^{-6}}{1.1\times 10^{-6}} \;\approx\; 3.6\ \text{m}.

Why: rearrange R = \rho L/A to solve for L. About 3.6 m of wire coiled into a compact spiral is exactly what you see when you peek inside a disassembled toaster. The coiling does not change the resistance (coiled wire and straight wire of the same length have the same L), but it fits the element into a small volume.

Step 4. Cross-check with copper — the same length and cross-section.

If you substituted copper (\rho = 1.72\times 10^{-8}\ \Omega\cdotm) for nichrome at the same geometry:

R_\text{Cu} \;=\; \frac{(1.72\times 10^{-8})(3.6)}{7.5\times 10^{-8}} \;\approx\; 0.83\ \Omega.

Current at 230 V through copper: I_\text{Cu} = 230/0.83 \approx 277\ \text{A}. Power dissipated: P = V^{2}/R = 230^{2}/0.83 \approx 64 kW.

Why: putting 230 V across 0.83 Ω would either trip every circuit breaker in the building or melt the wire in a fraction of a second. This comparison makes concrete why nichrome, with its 64× higher resistivity than copper, is the right material for a heating element — it intentionally limits the current to a safe, controllable value.

Result: I = 4.35 A, R = 53\ \Omega, L = 3.6 m of nichrome.

What this shows: The same Ohm's law V = IR explains both the useful job of nichrome (limit current, generate heat) and the failure that would happen if you used copper in its place (short-circuit, trip breakers). Picking the material is picking \rho; the geometry is mostly determined by packaging constraints.

Example 2: Copper bus-bar in a Delhi flat's distribution panel

A rectangular copper bus-bar of length L = 1.2 m and cross-section A = 3\ \text{mm}\times 10\ \text{mm} = 30\ \text{mm}^{2} = 3\times 10^{-5}\ \text{m}^{2} carries the combined current of several household circuits. Find its resistance. If it carries 32 A on a busy evening, what is the voltage drop across its length, and how much power is dissipated as heat in the bus-bar itself?

Step 1. Resistance from R = \rho L/A, using copper's resistivity \rho = 1.72\times 10^{-8}\ \Omega\cdotm.

R \;=\; \frac{(1.72\times 10^{-8})(1.2)}{3\times 10^{-5}} \;=\; \frac{2.064\times 10^{-8}}{3\times 10^{-5}} \;\approx\; 6.9\times 10^{-4}\ \Omega.

Why: the bus-bar's cross-section is large (30 mm²) compared with the 1 mm² wiring that serves individual rooms, by design — it must carry the total current of the flat without warming up appreciably. A 0.69 mΩ resistance is about 150 times smaller than the 0.1 Ω of a typical wall-wiring run.

Step 2. Voltage drop at 32 A.

V \;=\; IR \;=\; (32)(6.9\times 10^{-4}) \;\approx\; 22\ \text{mV}.

Why: 22 millivolts is negligible compared with the 230 V line voltage. This is the key design criterion for bus-bars: voltage drops in the distribution panel must be far below the line voltage, so that appliances throughout the flat "see" essentially the same 230 V regardless of what else is running.

Step 3. Power dissipated in the bus-bar.

P \;=\; VI \;=\; (22\times 10^{-3})(32) \;=\; 0.7\ \text{W}.

Or equivalently P = I^{2}R = (32)^{2}(6.9\times 10^{-4}) \approx 0.7 W.

Why: less than one watt of heat in a 1.2-metre copper bar of 30 mm² cross-section is nothing — the bar can sink that much heat without a noticeable temperature rise. The electrician's rule of thumb (thick bars, short runs, good clamps) is exactly the rule that keeps these numbers manageable.

Step 4. Compare with the load — a 1 kW iron running at the other end.

Total circuit: 230 V source, 0.00069 Ω bus-bar in series with a 53 Ω iron. The bus-bar drops 22 mV, the iron drops the remaining 230 - 0.022 \approx 230.0 V. The bus-bar is effectively invisible to the iron — its 0.7 W of heating compared with the iron's 1000 W is 0.07% of the total.

Result: R = 0.69\ \text{m}\Omega, V_\text{drop} = 22 mV, P = 0.7 W.

What this shows: Distribution copper is sized so that its resistance is a thousandth or less of the load it feeds. The same Ohm's law that heats the toaster's nichrome refuses to heat the flat's bus-bar — precisely because \rho_\text{Cu}/\rho_\text{nichrome} \approx 1/64, and because the bus-bar is much thicker than the toaster coil. Material choice and geometric scaling between the two ends of the circuit is what lets the electrical network do its job.

Example 3: A silicon diode is non-ohmic

A silicon diode has the measured V–I curve: at V = 0.5 V, I = 1\ \muA; at V = 0.7 V, I = 1 mA; at V = 0.75 V, I = 10 mA. (a) Does it obey Ohm's law? (b) Compute the "resistance" V/I at each operating point and see what happens. (c) Compute the differential resistance between the last two points.

Step 1. Tabulate R = V/I at each point.

V (V) I R = V/I
0.5 1\ \muA 5\times 10^{5}\ \Omega
0.7 1 mA 700\ \Omega
0.75 10 mA 75\ \Omega

Why: a quick Ohm's-law-sanity-check. If the diode were ohmic, V/I would be the same constant at every point. It is not — V/I drops by four orders of magnitude as V rises by a factor of 1.5. The diode is emphatically non-ohmic.

Step 2. Check for linearity another way: does the V–I pass through the origin with constant slope?

From the data: going from V = 0.7 to V = 0.75 (a 0.05 V rise) changes I from 1 mA to 10 mA (a 9 mA rise). Going from V = 0.5 to V = 0.7 (a 0.2 V rise) changes I from 1\ \muA to 1 mA (a 1 mA rise, about 1000× bigger than at low voltage).

Why: for an ohmic conductor, equal voltage changes produce equal current changes. Here, equal voltage changes produce wildly different current changes — the response is exponential, not linear.

Step 3. Differential resistance between 0.7 V and 0.75 V.

r_{d} \;=\; \frac{\Delta V}{\Delta I} \;=\; \frac{0.75 - 0.70}{0.010 - 0.001} \;=\; \frac{0.05}{0.009} \;\approx\; 5.6\ \Omega.

Why: the differential resistance is the local slope of the V–I curve, computed by the change-over-change rule. A fully forward-biased silicon diode has a very small r_{d} — a few ohms — even though its V/I ratio is much larger (75 Ω in our table). This is a fingerprint of a non-ohmic device: differential and static resistances disagree.

Result: The diode is non-ohmic. Its V/I ratio drops 4 orders of magnitude across a modest voltage range; its differential resistance near 0.75 V is about 5.6 Ω, far smaller than the static 75 Ω.

What this shows: For any circuit element that is not a pure metal at constant temperature, you cannot simply write V = IR with a constant R. Diodes, transistors, LEDs, neon bulbs, and almost every semiconductor device are non-ohmic. They require more careful characterisation — often an equation like the Shockley diode equation or a SPICE model — instead of a single number.

Common confusions

If you came here to understand Ohm's law, compute resistances, and see the difference between ohmic and non-ohmic behaviour, you have what you need. What follows is for readers who want the mean-free-time estimate, the local-form derivation \vec{J} = \sigma\vec{E}, and the thermodynamic origin of why Ohm's law must break down at sufficiently high fields.

Estimating the mean free time for copper

The Drude model of metal conduction gives the conductivity

\sigma \;=\; \frac{n e^{2}\tau}{m_{e}}.

For copper, \sigma \approx 5.96\times 10^{7}\ \text{S/m}, n = 8.49\times 10^{28}\ \text{m}^{-3} (one conduction electron per atom, with atomic number density set by copper's density 8960 kg/m³ and atomic mass 63.5 g/mol), m_{e} = 9.11\times 10^{-31} kg, e = 1.6\times 10^{-19} C. Solving for \tau:

\tau \;=\; \frac{m_{e}\sigma}{n e^{2}} \;=\; \frac{(9.11\times 10^{-31})(5.96\times 10^{7})}{(8.49\times 10^{28})(1.6\times 10^{-19})^{2}} \;\approx\; 2.5\times 10^{-14}\ \text{s}.

The mean free time between collisions is about 25 femtoseconds. This is shorter than a single optical-photon period — copper's electrons collide trillions of times per second at room temperature.

Multiply by a typical thermal speed of an electron, v_\text{th} \approx \sqrt{2k_{B}T/m_{e}} \approx 10^{5} m/s (or, more correctly, the Fermi velocity v_{F} \approx 1.6\times 10^{6} m/s in copper), to get the mean free path between collisions:

\ell \;=\; v_{F}\tau \;\approx\; (1.6\times 10^{6})(2.5\times 10^{-14}) \;\approx\; 40\ \text{nm}.

An electron in copper travels about 40 nanometres — roughly 100 atomic spacings — between collisions. That is the microscopic length scale embedded inside the macroscopic resistivity 1.72\times 10^{-8}\ \Omega\cdotm.

The local form of Ohm's law

Equation (6), R = \rho L/A, is an integrated statement about a whole wire. Its pointwise equivalent is the local Ohm's law:

\vec{J} \;=\; \sigma\,\vec{E}, \tag{7}

where \vec{J} (amperes per square metre) is the current density and \vec{E} (volts per metre) is the electric field. This says: at every point inside the conductor, the current density vector is aligned with the field and is proportional to it, with conductivity \sigma as the constant of proportionality.

To go from (7) back to V = IR for a uniform wire: take the wire along the x-axis of length L, cross-section A, with uniform field E. Then J = I/A and

\frac{I}{A} \;=\; \sigma E \;=\; \sigma\,\frac{V}{L}.
V \;=\; I\cdot \frac{L}{\sigma A} \;=\; I\cdot\rho\,\frac{L}{A} \;=\; IR. \checkmark

The local form (7) is the version used in semiconductor device physics, where the field and current density vary from point to point inside the device.

Why Ohm's law must fail at high fields

Equation (3) derived the drift velocity v_{d} = eE\tau/m_{e}. If you keep increasing E, this formula predicts ever-larger drift velocities. But no electron in a metal can drift faster than its own thermal speed v_{F} — the Fermi velocity, about 1.6\times 10^{6} m/s in copper. Once the field is strong enough that v_{d} approaches v_{F}, the assumptions behind the derivation break down:

  • Collisions are no longer independent of E. A strongly field-biased electron doesn't "reset" to zero drift between collisions; the collisions themselves start to depend on the field.
  • Field emission and impact ionisation. At E \sim 10^{9} V/m (much higher than anything in ordinary circuits), electrons can be ripped out of bound states by the field alone — adding new carriers and making the current grow nonlinearly.
  • Velocity saturation. Even before those exotic regimes, in semiconductors the drift velocity saturates at \sim 10^{5} m/s for silicon because of optical-phonon emission. Current stops growing with field: V \propto I breaks, replaced by V rising steeply while I plateaus.

At high enough fields — either above \sim 10^{5} V/m in semiconductors or \sim 10^{8} V/m in metals — Ohm's law stops describing the physics. All engineering applications at human-scale voltages (wall socket to microelectronics) are safely below this threshold, but the theoretical point stands: Ohm's law is an empirical law with a domain of validity, not a fundamental truth.

The connection between \rho and material science

The formula \rho = m_{e}/(ne^{2}\tau) tells you that resistivity is controlled by two factors: the number of charge carriers per unit volume (n) and the mean time between their collisions (\tau). Different materials have wildly different values:

Material n (carriers/m³) \tau (s) \rho (\Omega\cdotm)
Copper (metal) 8.5\times 10^{28} 2.5\times 10^{-14} 1.7\times 10^{-8}
Nichrome (alloy) \sim 5\times 10^{28} \sim 2\times 10^{-16} 1.1\times 10^{-6}
Pure silicon (intrinsic) 1.5\times 10^{16} \sim 10^{-11} \sim 2000
Doped silicon (n-type, heavy) 10^{24} \sim 10^{-13} \sim 10^{-4}
Glass \sim 10^{10} variable 10^{10}–10^{14}

Alloys like nichrome have high resistivity not because they have fewer carriers (they have about as many as copper) but because the atoms of chromium, iron, and nickel that are mixed in create a disordered lattice that scatters electrons very frequently (\tau is tiny). Silicon has high resistivity because it has almost no free carriers at room temperature (n is tiny). Doping silicon with phosphorus or boron adds carriers and reduces \rho dramatically — the entire semiconductor industry is built on fine control of the carrier density inside silicon.

So when an engineer picks a material for a given job, they are really picking the two factors n and \tau that together give the target \rho. Copper for power delivery (high n, long \tau). Nichrome for heating (moderate n, short \tau). Silicon for logic (carefully controlled n in each transistor region, moderate \tau). The one formula \rho = m_{e}/(ne^{2}\tau) covers all of them.

Where this leads next