Temperature dependence of resistivity

Touch the filament of a torch bulb the instant it lights up and, if you could measure it, you would find its resistance has jumped to several times its cold value — the same tungsten wire, the same length and thickness, simply hotter. Heat a coil of nichrome wire in a heater and its resistance barely budges. Heat a thermistor — a small semiconducting bead — and its resistance can fall to a fraction of what it was. Same word, "heating," three completely different electrical outcomes. That spread is not an accident; it is a direct fingerprint of how each material carries current. The free-electron picture built up earlier in this chapter already contains the answer. Resistivity was written as $\rho = m/(ne^2\tau)$ — a tug-of-war between how many charge carriers are available ( $n$ ) and how often they get knocked off course by collisions ( $\tau$ , the average time between collisions). Temperature does not touch this formula directly; it touches $n$ and $\tau$ separately, and which one moves depends on whether the material is a metal or a semiconductor. In a metal the carrier count is already saturated, so only $\tau$ shrinks as the lattice shakes harder — resistivity climbs. In a semiconductor, heat's main job is to free new carriers that were previously locked up, so $n$ explodes upward and resistivity falls, even though $\tau$ is also shrinking. This article turns that qualitative story into the one linear formula the CBSE board actually tests, $R_T = R_0(1+\alpha\Delta T)$ , shows how the temperature coefficient $\alpha$ is measured and used, and works through the board-level and competitive-level numbers — copper heated in a lab, platinum calibrated as a thermometer — that show up again and again in exam papers.

Key takeaways

Resistance changes with temperature; for a moderate range this is captured by the linear approximation $R_T = R_0(1+\alpha\Delta T)$ , and the same form holds for resistivity, $\rho_T = \rho_0(1+\alpha\Delta T)$ .
$\alpha$ , the temperature coefficient of resistance, has SI unit $\text{K}^{-1}$ (equivalently per degree Celsius, since a 1 K change equals a 1°C change).
Metals have a small positive $\alpha$ (order $10^{-3}\ \text{K}^{-1}$ ): carrier density $n$ is essentially fixed, but the relaxation time $\tau$ falls as lattice vibrations increase collisions, so $\rho \propto 1/\tau$ rises.
Alloys such as manganin and nichrome have a deliberately tiny $\alpha$ , almost temperature-independent resistance — ideal for standard resistors and heating elements.
Semiconductors have a negative $\alpha$ : heating frees many more charge carriers (a large rise in $n$ ) than the fall in $\tau$ can offset, so resistivity drops — the working principle of a thermistor.
Rearranging the linear law, $\alpha = (R_T - R_0)/(R_0\Delta T)$ , lets $\alpha$ , $R_0$ , $R_T$ or $\Delta T$ be solved for given the other three — the standard board and competitive numerical.
Key idea: over a moderate temperature range, resistance follows $R_T = R_0(1+\alpha\Delta T)$ , with metals showing a small positive $\alpha$ because collision frequency alone rises while carrier density stays fixed, and semiconductors showing a negative $\alpha$ because a temperature rise instead unlocks a flood of new charge carriers.

The core insight

Resistance follows the linear law $R_T = R_0(1+\alpha\Delta T)$ , with $\alpha$ positive for metals (rising collision rate, fixed carrier density) and negative for semiconductors (carrier density rises faster than collisions can slow it).

Where it fits in the syllabus

This topic sits inside the Current Electricity chapter of CBSE Class 12 Physics, directly after resistivity of materials and the drift-velocity/relaxation-time model of conduction, and just before electrical energy and power. The prescribed content is exactly the linear relation $R_T = R_0(1+\alpha\Delta T)$ together with a qualitative, drift-model explanation of why metals, alloys and semiconductors respond differently — no Debye-model or lattice-vibration mathematics is expected, and nothing beyond this linear law is examined at board level.

The physical picture

Begin from the microscopic expression for resistivity that the free-electron model supplies:

\rho = \dfrac{m}{ne^2\tau}

where: $m$ is the electron mass, $n$ is the number density of free (conduction) electrons, $e$ is the electron charge, and $\tau$ is the relaxation time — the average time between successive collisions of a free electron with the vibrating lattice ions. This formula is the hinge on which the entire temperature story turns: temperature cannot change $m$ or $e$ , so any change in $\rho$ with temperature must come from a change in $n$ , in $\tau$ , or in both, and reading which one dominates tells you whether a material's resistivity rises or falls when heated. For a wire of fixed length $L$ and cross-section $A$ , resistance and resistivity are proportional ( $R = \rho L/A$ ), so whatever linear law holds for $\rho$ also holds for $R$ . Over a moderate temperature range — the only range CBSE expects — both are captured by one equation:

R_T = R_0\big(1 + \alpha\Delta T\big), \qquad \rho_T = \rho_0\big(1+\alpha\Delta T\big)

where: $R_0$ (or $\rho_0$ ) is the resistance (or resistivity) at a reference temperature $T_0$ , commonly taken as $0^\circ\text{C}$ ; $R_T$ (or $\rho_T$ ) is the value at temperature $T$ ; $\Delta T = T - T_0$ is the temperature change, in kelvin or equivalently degrees Celsius since the two scales have identical-sized degrees; and $\alpha$ is the temperature coefficient of resistance, with SI unit $\text{K}^{-1}$ . Physically, $\alpha$ is just the fractional change in resistance per unit temperature rise — a small positive $\alpha$ means resistance creeps up slowly with heating; a negative $\alpha$ means it falls. The figure below is the qualitative graph every CBSE answer sheet is expected to be able to sketch: resistivity rising almost as a straight line for a metal, staying nearly flat for an alloy, and falling steeply for a semiconductor.

Figure

A graph of resistivity against temperature showing three curves, a metal rising almost linearly with a small positive slope, an alloy staying nearly flat, and a semiconductor falling steeply, all starting near the same origin region

Why metals rise. In a metal, essentially every atom's valence electron is already free at room temperature, so the carrier density $n$ is nearly fixed with temperature. What changes is $\tau$ : as $T$ rises, the lattice ions vibrate more vigorously about their fixed sites, so a drifting electron collides with them more often, and $\tau$ shrinks. Since $\rho \propto 1/\tau$ , resistivity increases, giving a small positive $\alpha$ , typically of order $10^{-3}\ \text{K}^{-1}$ for pure metals such as copper and platinum.

Why alloys stay flat. Alloys such as manganin (copper-manganese-nickel) and nichrome (nickel-chromium) are engineered specifically so that $\alpha$ is tiny — their resistance is nearly independent of temperature over a useful working range. This near-flatness is exactly why they are chosen for precision standard resistors (where drifting resistance would spoil a measurement) and for heating-element wire (which must survive large temperature swings without its current-carrying behaviour changing unpredictably).

Why semiconductors fall. In a semiconductor, at low temperature most valence electrons are still bound in covalent bonds and only a few are free to conduct. Raising the temperature breaks many more of these bonds, releasing a large number of new charge carriers — $n$ rises steeply, often by orders of magnitude over a modest temperature range. Although $\tau$ also falls (more lattice vibration means more collisions, just as in a metal), the surge in $n$ overwhelms it, and the net effect is that $\rho$ decreases sharply with rising temperature. This gives semiconductors a negative $\alpha$ , and it is the working principle of a thermistor — a semiconducting device whose large, predictable resistance change with temperature makes it a sensitive temperature sensor.

Derivation AGetting alpha and RT from the linear resistance-temperature law

The goal is to show, starting from nothing but the definition of a temperature coefficient as a fractional rate of change, how the single linear formula $R_T = R_0(1+\alpha\Delta T)$ is built, and how it is inverted to solve for $\alpha$ itself — the two directions every board numerical asks for. No calculus is needed; the "coefficient" is simply defined as an average fractional change per unit temperature over the range in question, which is exactly what the CBSE syllabus expects at this level.

Step 1. Define $\alpha$ as the fractional change in resistance per unit rise in temperature, taking $R_0$ at a reference temperature $T_0$ as the baseline: $\alpha = \dfrac{1}{R_0}\left(\dfrac{R_T - R_0}{\Delta T}\right)$ , where $\Delta T = T - T_0$ . This says $\alpha$ measures how big a fraction of the original resistance is added (or removed) for each kelvin of heating — a dimensionless-per-kelvin quantity, unit $\text{K}^{-1}$ .

Step 2. Rearrange for $R_T$ . Multiplying both sides of the Step 1 definition by $R_0\Delta T$ gives $R_T - R_0 = \alpha R_0 \Delta T$ , and adding $R_0$ to both sides gives the working form $R_T = R_0(1+\alpha\Delta T)$ . This is the equation quoted in every board answer, and it is nothing more than "start at $R_0$ , add a correction proportional to how far the temperature has moved."

Step 3. Note the linear approximation this hides. Physically $\alpha$ can itself depend a little on temperature (it is only exactly constant for an idealised material), so $R_T = R_0(1+\alpha\Delta T)$ is a first-order (linear) approximation valid over a moderate temperature range around $T_0$ — exactly the regime CBSE numericals stay within. Over huge temperature swings, or near a phase change, the straight-line approximation breaks down and $\alpha$ itself must be treated as temperature-dependent, which is beyond this syllabus.

Step 4. Invert the same formula the other way, for $\alpha$ given a measured $R_0$ and $R_T$ : divide through by $R_0$ , subtract 1, then divide by $\Delta T$ , giving back $\alpha = \dfrac{R_T-R_0}{R_0\Delta T}$ . This is exactly the experimental route: measure resistance at two known temperatures, and $\alpha$ falls out directly — precisely how a platinum resistance thermometer is calibrated (worked out numerically below).

Worked examples

Board level - resistance of a heated copper wire

A copper wire has resistance $R_0 = 10\ \Omega$ at $0^\circ\text{C}$ . Its temperature coefficient of resistance is $\alpha = 4.0\times10^{-3}\ \text{K}^{-1}$ . Find its resistance at $80^\circ\text{C}$ .

Step 1. Identify the given quantities and the formula to use. $R_0 = 10\ \Omega$ , $\alpha = 4.0\times10^{-3}\ \text{K}^{-1}$ , and $\Delta T = 80 - 0 = 80\ \text{K}$ (since the reference temperature is $0^\circ$ C). The relevant law is $R_T = R_0(1+\alpha\Delta T)$ .

Step 2. Compute the fractional correction term first: $\alpha\Delta T = (4.0\times10^{-3})(80) = 0.32$ . This means the resistance is expected to grow by 32 percent of its original value.

Step 3. Substitute back: $R_T = 10(1+0.32) = 10\times1.32 = 13.2\ \Omega$ . Answer: the copper wire's resistance rises from $10\ \Omega$ to $13.2\ \Omega$ , a $32\%$ increase, when heated from $0^\circ$ C to $80^\circ$ C.

Competitive level - calibrating a platinum resistance thermometer

A platinum resistance thermometer has resistance $5.00\ \Omega$ at $0^\circ\text{C}$ and $5.23\ \Omega$ at $100^\circ\text{C}$ . Find (a) the temperature coefficient of resistance of platinum, and (b) the temperature at which its resistance is $5.15\ \Omega$ .

Step 1. Use the two calibration points to find $\alpha$ first, via $\alpha = \dfrac{R_T-R_0}{R_0\Delta T}$ , with $R_0=5.00\ \Omega$ , $R_T = 5.23\ \Omega$ , $\Delta T = 100\ \text{K}$ : $\alpha = \dfrac{5.23-5.00}{5.00\times100} = \dfrac{0.23}{500} = 4.6\times10^{-4}\ \text{K}^{-1}$ .

Step 2. Use this $\alpha$ as the instrument's fixed calibration constant and apply it forward to the unknown reading, $R_T = 5.15\ \Omega$ : $5.15 = 5.00\big(1+(4.6\times10^{-4})\Delta T\big)$ , so $\dfrac{5.15}{5.00} = 1.03 = 1+(4.6\times10^{-4})\Delta T$ .

Step 3. Isolate $\Delta T$ : $0.03 = (4.6\times10^{-4})\Delta T$ , so $\Delta T = \dfrac{0.03}{4.6\times10^{-4}} \approx 65.2\ \text{K}$ . Since the reference temperature was $0^\circ$ C, this is directly the temperature reading. Answer: $\alpha \approx 4.6\times10^{-4}\ \text{K}^{-1}$ , and the resistance $5.15\ \Omega$ corresponds to a temperature of approximately $65.2^\circ\text{C}$ . Timing note: about ninety seconds. The chain is fixed and mechanical — find $\alpha$ from the two given calibration points, then reuse the same $\alpha$ as a constant to convert any further resistance reading into a temperature. This two-point-calibration-then-interpolate pattern is exactly how a real platinum resistance thermometer is used in the lab, and it is a favourite competitive-exam variant of the basic board numerical.

Common traps

Board scripts lose marks in the same few places every year on this topic. Assuming every material gets more resistive when heated is the single biggest one — true for metals, but the opposite for semiconductors and insulators, whose resistance falls (negative $\alpha$ ) as temperature rises. A related slip is explaining the metal case by saying "carrier density $n$ increases with temperature" — for a metal $n$ is essentially fixed; it is the collision rate (falling $\tau$ ) that changes, and mixing this up with the semiconductor mechanism is a common conceptual crossover error. Students also sometimes forget that $\Delta T$ in kelvin and in degrees Celsius are numerically identical (since both scales have equal-sized degrees), so no conversion is needed for $\Delta T$ itself — but the reference temperature $T_0$ at which $R_0$ was measured must always be tracked and stated. Finally, treating $\alpha$ as one universal number for "metals" is wrong: different pure metals have different $\alpha$ , and alloys like manganin and nichrome are deliberately formulated to make $\alpha$ nearly zero, unlike ordinary copper or platinum.

You should be able to…

Recall and apply $R_T = R_0(1+\alpha\Delta T)$ and $\alpha = (R_T-R_0)/(R_0\Delta T)$ in either direction.
Explain, using $\rho = m/(ne^2\tau)$ , why metals show a small positive $\alpha$ (fixed $n$ , falling $\tau$ ) and why semiconductors show a negative $\alpha$ (rising $n$ dominates).
Justify the choice of manganin or nichrome for standard resistors and heating elements.
Solve two-point calibration problems such as a resistance thermometer.
Sketch the qualitative resistivity-versus-temperature graph for a metal, an alloy, and a semiconductor on the same axes.

Ask the tutor: Want more practice turning resistance-versus-temperature data into $\alpha$ , or a deeper look at how a thermistor's characteristic curve is used in real temperature-sensing circuits?

Why is alpha positive for metals but negative for semiconductors if both have more collisions when heated?

Both do get more collisions when heated — that part is identical, and it always pushes resistivity up a little. The difference is what else happens to the carrier count $n$ . In a metal, almost all available electrons are already free at room temperature, so $n$ barely changes; the extra collisions are the only effect, and resistivity rises. In a semiconductor, heating also unlocks large numbers of new carriers that were previously bound in covalent bonds, so $n$ shoots up. That surge in $n$ is much bigger than the drop caused by extra collisions, so the two effects don't cancel — the carrier-count effect wins, and resistivity falls overall.

Is alpha really constant, or does it change with temperature too?

Strictly, $\alpha$ can drift a little with temperature for a real material — the linear formula $R_T = R_0(1+\alpha\Delta T)$ is a first-order approximation, accurate over a moderate range around the reference temperature $T_0$ . CBSE numericals always stay inside that safe range, so you can treat $\alpha$ as a fixed constant given in the problem. Over very large temperature swings, or near a phase change in the material, a single constant $\alpha$ would no longer describe the curve accurately, but that level of detail is beyond this syllabus.

Why does 0 degrees Celsius keep showing up as the reference temperature?

It is simply a conventional choice, not a physical requirement — $R_0$ could in principle be measured at any reference temperature $T_0$ , and the formula $R_T = R_0(1+\alpha\Delta T)$ still works with $\Delta T = T - T_0$ . $0^\circ$ C is used so often because it is easy to set up reliably in a lab (an ice-water bath) and because tabulated values of $\alpha$ for common materials are conventionally quoted relative to it, which keeps textbook problems and real calibration certificates consistent with each other.

How is this different from why a bulb filament glows brighter and brighter as it heats up during switch-on?

It is the same physics, just watched frame by frame instead of at two fixed points. A tungsten filament's resistance keeps climbing continuously as it heats from room temperature to its white-hot working temperature (often to several times its cold resistance), following the same underlying $\rho = m/(ne^2\tau)$ mechanism as the copper wire in the worked example — just over a far larger $\Delta T$ than a typical board numerical uses. This is also why a bulb draws a large surge current for an instant at switch-on: the filament is still cold and its resistance is at its lowest value.