A definition, a law, and a shape

Resistance is defined for any two-terminal device by

$$R \equiv \frac{V}{I}, \qquad 1 \ \Omega = 1 \ \text{V/A}.$$

That equation always applies, because it is a naming convention: put $V$ across the thing, measure the $I$ that results, and call the ratio $R$. It cannot be false and it predicts nothing on its own.

Ohm's law is a separate, much stronger statement: for some materials, over some range, that ratio comes out the same number no matter what $V$ you choose. Double the voltage and the current doubles too, leaving $R$ untouched. Materials for which this holds are called ohmic, and metallic resistors at fixed temperature are the standard example.

Plenty of devices are not ohmic and still have resistance:

1. A bulb filament heats up as more current runs through it, and metals grow more resistive when hot, so its $R$ climbs with $V$.
2. A diode passes almost nothing below a threshold voltage and then conducts hard, so its $R$ falls by orders of magnitude across a fraction of a volt.
3. A thermistor is engineered so $R$ swings with temperature, which is the whole point of the component.

The error this section exists to prevent is reading $R = V/I$ as a proportionality that lets $R$ track $V$. It does the opposite: for an ohmic resistor, doubling $V$ doubles $I$ and the ratio does not move. If you found yourself writing "the voltage doubled so the resistance doubled", the sentence has $I$ frozen when $I$ is exactly the thing that responded.

Take a uniform conductor of length $L$ and cross-section $A$. Inside it the field is uniform, $E = V/L$, and the material's response is $J = \sigma E$ with $\sigma = 1/\rho$ the conductivity. Then $I = JA = \sigma E A = \sigma A V / L$, so

$$R = \frac{V}{I} = \frac{L}{\sigma A} = \frac{\rho L}{A}.$$

Read the two halves separately. $\rho$ belongs to the material: copper's $\rho = 1.7\times10^{-8} \ \Omega\cdot\text{m}$ is the same number for a hair-thin strand and a bus bar. $R$ belongs to the object: it needs the material AND the dimensions. Quoting $1.7\times10^{-8}$ as "copper's resistance" mixes the two, and the units give it away, since ohm-meters are not ohms.

The scaling ladder, run one rung at a time:

1. Double $L$ at fixed $A$: $R$ doubles. Longer path, more collisions.
2. Double $A$ at fixed $L$: $R$ halves. Two lanes instead of one.
3. Double the radius at fixed $L$: $A = \pi r^2$ quadruples, so $R$ falls to $R/4$. This is the rung that gets skipped.
4. Stretch a wire to twice its length at constant volume: $L$ doubles AND $A$ halves, so $R$ becomes $4R$. Two changes, both raising $R$.

When two things change at once, write the ratio explicitly:

$$\frac{R_2}{R_1} = \frac{L_2}{L_1}\cdot\frac{A_1}{A_2}.$$

Both graphs get used and they are reciprocals of each other.

$$\text{On } V \text{ vs } I: \ \text{slope} = R. \qquad \text{On } I \text{ vs } V: \ \text{slope} = \frac{1}{R}.$$

The AP course uses the $I$-versus-$V$ orientation constantly, and there the reading is counterintuitive at first: a steeper line means more current per volt, which means a smaller resistance. Two resistors of $2 \ \Omega$ and $8 \ \Omega$ plot as lines with slopes $0.5 \ \text{A/V}$ and $0.125 \ \text{A/V}$, and the steep one is the small resistor.

For a straight line through the origin, the ratio at any point and the slope of the line are the same number, so an ohmic device lets you be careless. For a curved characteristic they are different quantities and you have to say which you want:

1. The chord from the origin to the operating point gives $V/I$, the resistance at that point. This is what $R$ means.
2. The tangent gives $dV/dI$, the dynamic or small-signal resistance, which is how much extra voltage a small extra current costs there.

They agree only for a line through the origin, which is another way of saying they agree exactly when the device is ohmic. A filament's $I$-$V$ curve bends toward the horizontal as it heats, so its chord resistance rises with voltage while its tangent resistance rises faster still.

In a metal, resistivity climbs with temperature because the lattice vibrates harder and scatters the carriers more often:

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

with $\alpha \approx 4\times10^{-3} \ \text{K}^{-1}$ for tungsten and similar metals. A lamp filament runs near $2500$ K and sits at room temperature when off, so a swing of roughly $2200$ K takes $\rho$ up by nearly an order of magnitude. Measure a $100$ W household bulb cold with an ohmmeter and you read something like $10 \ \Omega$; in operation it behaves like $144 \ \Omega$. Both readings are correct, and neither is the "true" resistance, because the device does not have one.

This is the practical face of the definition-versus-law distinction. $R = V/I$ still returns a number at every operating point, so the filament always has a resistance. What it does not have is a constant resistance, so it is not ohmic, so nothing that assumes constant $R$ may be used across a change in operating point.

Two habits follow. Never carry a cold resistance into a hot calculation. And when a problem says "assume the filament is ohmic", it is granting you permission you would not otherwise have, which is worth noticing rather than skimming.