In short

For a metal, the resistivity rises roughly linearly with temperature:

\boxed{\;\rho(T) \;=\; \rho_{0}\bigl[\,1 + \alpha\,(T - T_{0})\,\bigr]\;}

where \rho_{0} is the resistivity at a reference temperature T_{0} (usually 20 °C) and \alpha is the temperature coefficient of resistance. For copper \alpha \approx 3.9\times 10^{-3}\ \text{K}^{-1}; for nichrome \alpha \approx 4\times 10^{-4}\ \text{K}^{-1} (ten times smaller, which is exactly why nichrome makes a stable heating element). The microscopic reason is that hotter atoms vibrate more strongly and scatter drifting electrons more often — a shorter \tau in \rho = m_{e}/(ne^{2}\tau).

Semiconductors behave oppositely. In silicon or germanium, raising the temperature frees more charge carriers (the carrier density n rises exponentially), and the extra carriers overwhelm the shorter \tau. So semiconductor resistivity falls with temperature, often by more than a factor of two for every 10 K rise near room temperature.

Superconductors are the sharp limit. Below a critical temperature T_{c} the resistivity of certain materials drops to exactly zero — not small, zero — and currents circulate without dissipating any energy. Mercury below 4.2 K was the first example (Onnes, 1911); modern ceramic superconductors work up to 138 K, well above the boiling point of liquid nitrogen.

Colour coding lets you read a resistor's nominal value off the four or five painted bands on its body. The first two (or three) bands encode significant digits, the next band is the decimal multiplier, and the last is the tolerance. A resistor with bands brown-black-red-gold is 10 \times 10^{2} = 1\ \text{k}\Omega at \pm 5\%.

Ride the old Kolkata tram from Esplanade to Ballygunge on a wet August afternoon and look up at the overhead catenary wire — a single stretched copper conductor, a couple of centimetres thick, strung between masts. In January that wire sits at about 18 °C; in May it bakes at 45 °C. Same wire, same route, same 550 volts DC on the supply. But the current the motorman draws from the supply to maintain the same tractive force is measurably higher in January than in May. The reason is the wire itself. Its resistance has gone up by roughly 10% between the two seasons, and the tram's motors now see a slightly stiffer conductor feeding them. Copper gets more resistive as it gets hotter.

Now open the bonnet of a Maruti Alto and find the coolant temperature sensor — a small ceramic bead with two wires, screwed into the coolant channel. That bead is a thermistor: a piece of semiconductor whose resistance falls from about 9 k\Omega at 0 °C to about 200 \Omega at 100 °C. The engine-control unit measures this resistance, reads off the temperature, and decides how much fuel to inject. Semiconductor, not metal. Opposite sign. Same underlying physics, different dominant mechanism.

This chapter works out why these two behaviours exist and gives you the linear formula that handles the metal case for JEE-style problems. It also gives the full exponential picture for semiconductors, the story of superconductors, and the practical trick of reading a resistor's value off its coloured bands.

The metal picture — hotter lattice, more scattering

The previous chapter's going-deeper section derived the resistivity from the Drude model:

\rho \;=\; \frac{m_{e}}{n\,e^{2}\,\tau}. \tag{1}

Here n is the density of conduction electrons, e and m_{e} are the electron's charge and mass, and \tau is the mean time between collisions of an electron with the lattice. Equation (1) is the bridge between microscopic physics and macroscopic \rho: everything that changes resistivity has to change one of those four quantities.

In a metal, n, e, and m_{e} are fixed. The conduction-electron density does not depend on temperature — copper has one free electron per atom at every temperature it exists as a solid. So the only knob that temperature can turn in equation (1) is \tau, the mean free time.

What is \tau? It is the time, on average, that a drifting electron travels between one scattering event and the next. The scattering events in a pure metal are collisions with phonons — the quantised vibrations of the atomic lattice. Heat the metal and the atoms vibrate harder (larger amplitude), present a bigger cross-section to the passing electron, and scatter it more often. \tau goes down. Look at (1): if \tau falls, \rho rises. That is the physics of metal resistivity rising with temperature, in one microscopic sentence.

From microscopic \tau(T) to linear \rho(T)

For the mathematics, you only need two facts:

So 1/\tau \propto T, or \tau \propto 1/T. Put this into (1):

\rho \;=\; \frac{m_{e}}{n\,e^{2}\tau} \;\propto\; T.

In words, a metal's resistivity is approximately proportional to absolute temperature, over any range where the phonon contribution dominates. Over the restricted range that engineers and students actually care about — room temperature plus or minus a few hundred kelvin — it is convenient to expand around a reference T_{0} using a first-order Taylor series:

\rho(T) \;\approx\; \rho_{0} + \left.\frac{d\rho}{dT}\right|_{T_{0}}(T - T_{0}).

Pull out \rho_{0} from the derivative term:

\rho(T) \;=\; \rho_{0}\biggl[\,1 + \underbrace{\frac{1}{\rho_{0}}\,\frac{d\rho}{dT}\bigg|_{T_{0}}}_{\alpha}\,(T - T_{0})\,\biggr].

The bracketed coefficient in front of (T - T_{0}) is given a special name: the temperature coefficient of resistance \alpha, defined as the fractional change of resistivity per kelvin of temperature change, evaluated at the reference point T_{0}.

Why: \alpha is not some new physical mechanism. It is just a compact numerical label for "how sharply does \rho change with T near T_{0}", normalised by the value of \rho_{0} itself. Dividing by \rho_{0} makes \alpha dimensionally clean — it has units of 1/K — and lets you compare the temperature sensitivity of copper and tungsten directly, even though their resistivities differ by factors of ten.

The final working formula — the one you will use to answer JEE problems, and the one you will see on Indian physics-lab scripts — is

\boxed{\;\rho(T) \;=\; \rho_{0}\bigl[\,1 + \alpha\,(T - T_{0})\,\bigr]\;} \tag{2}

and by dividing each side by \rho_0 times the geometric factor L/A, the same linear form holds for resistance itself:

R(T) \;=\; R_{0}\bigl[\,1 + \alpha\,(T - T_{0})\,\bigr]. \tag{3}

Equation (3) assumes the wire's length and cross-section don't change appreciably with temperature — a reasonable approximation for metals, where linear thermal expansion is only \sim 10^{-5}\ \text{K}^{-1}, a hundred times smaller than \alpha itself.

A table of \alpha for common materials

Material \rho_{0} at 20 °C (\Omega\cdotm) \alpha at 20 °C (K^{-1})
Silver 1.6\times 10^{-8} +3.8\times 10^{-3}
Copper 1.7\times 10^{-8} +3.9\times 10^{-3}
Aluminium 2.8\times 10^{-8} +3.9\times 10^{-3}
Tungsten (bulb filament) 5.6\times 10^{-8} +4.5\times 10^{-3}
Iron 1.0\times 10^{-7} +5.0\times 10^{-3}
Nichrome (heater) 1.1\times 10^{-6} +4\times 10^{-4}
Manganin (precision resistor) 4.8\times 10^{-7} \pm 2\times 10^{-5}
Carbon (graphite) 3.5\times 10^{-5} -5\times 10^{-4}
Silicon (pure) \sim 2300 -70\times 10^{-3}
Germanium (pure) \sim 0.5 -50\times 10^{-3}

Three things jump out. First, the pure metals at the top all sit near \alpha \approx 4\times 10^{-3}\ \text{K}^{-1} — about a 0.4% rise per kelvin. Second, nichrome's \alpha is ten times smaller than copper's: this is a deliberate feature of the nickel-chromium alloy, because an Indian 1 kW room heater or a toaster element needs a resistance that does not change much as the coil glows at 800 °C. A nichrome heater stays at a stable operating point; a copper heater (with \alpha ten times higher) would run away — as it heats, resistance rises, current falls, heating drops, and the thermal equilibrium becomes extremely sensitive to mains voltage. Nichrome avoids this.

Third, manganin's \alpha is a hundred times smaller still, at the level of a few parts per million per kelvin. This is why every precision laboratory resistance standard, every strain-gauge bridge, every four-wire ohmmeter shunt inside an IIT strain-testing setup is made of manganin — an accidental alloy of copper, manganese, and nickel whose temperature coefficient is almost zero over a wide range. A 1 k\Omega manganin resistor changes by less than 0.01 Ω over a 50 K temperature swing. That is stability no pure metal can offer.

Finally, notice the negative signs at the bottom of the table. Carbon, silicon, and germanium all have negative \alpha: their resistance falls as they get hotter. That is the semiconductor regime, and it requires a different mechanism entirely.

Interactive — rho versus T for metals and semiconductors

Interactive: normalised resistivity versus absolute temperature Two curves: copper resistivity (linear, rising with T, normalised to 1 at 300 K) and silicon resistivity (decreasing exponentially with T, normalised to its 300 K value). Drag the temperature slider to see how each responds. absolute temperature T (K) ρ(T) / ρ(300 K) 250 300 350 400 450 500 1 2 3 copper (metal) silicon (semiconductor) drag the red dot along T
Drag the red marker along the temperature axis. The copper curve (red) rises gently and linearly — a 200 K rise from 300 K to 500 K about doubles its resistivity. The silicon curve (dark) drops steeply — a modest 100 K rise can cut resistivity by a factor of several. The sign of $\alpha$ is the difference between a power-line conductor and a thermistor.

The silicon curve uses a simplified \rho(T) \propto \exp(E_{g}/2k_{B}T), the Arrhenius-like form derived in the going-deeper section. Over the range 250 K to 500 K, it changes by about a factor of twenty — which is exactly what makes silicon-based temperature sensors so sensitive near room temperature.

Semiconductors — why the sign flips

For a semiconductor, all three of the factors on the right of equation (1) can change with temperature: n, \tau, and (in a subtle way) the effective mass too. The big one is n.

At absolute zero, a pure (intrinsic) semiconductor is a band insulator: the top filled band of electrons (the valence band) is completely full, and the next available band (the conduction band) is empty, separated by an energy gap E_{g}. Silicon has E_{g} \approx 1.1 eV; germanium has E_{g} \approx 0.67 eV. No free carriers, so n = 0, so \rho = \infty. Semiconductors at T = 0 are perfect insulators.

At any temperature above zero, thermal fluctuations occasionally kick an electron across the gap into the conduction band, leaving behind a positively charged hole in the valence band. Both the electron and the hole can now drift in response to a field, so the carrier density is no longer zero. Statistical mechanics (the Maxwell-Boltzmann distribution applied to the energy gap) gives

n(T) \;=\; n_{0}\,\exp\!\left(-\,\frac{E_{g}}{2\,k_{B}T}\right), \tag{4}

where n_{0} depends on the density of available states in each band and is only weakly temperature-dependent. The factor of 2 in the denominator is there because both the electron and the hole contribute to conduction, and the gap is crossed symmetrically.

Why: the exponential factor is the probability that a thermal fluctuation has enough energy to excite an electron across the gap. The gap E_{g} plays the role of the "activation energy" in the Boltzmann factor \exp(-E/k_{B}T) — if the gap is small or the temperature is high, excitations are common; if the gap is large or the temperature is low, excitations are rare.

At room temperature (T = 300 K), k_{B}T \approx 0.026 eV. For silicon, E_{g}/(2k_{B}T) \approx 1.1/0.052 \approx 21, so e^{-21} \approx 10^{-9}. Compared with the density of atoms in the crystal (\sim 5\times 10^{28}\ \text{m}^{-3}), this gives about 10^{16} carriers per m³ in intrinsic silicon at 300 K — ten trillion times fewer than copper's 10^{29}. That huge deficit is why pure silicon's resistivity is \sim 2000\ \Omega\cdotm while copper's is \sim 10^{-8}\ \Omega\cdotm: a factor of 10^{11} apart, which matches the ratio of their carrier densities exactly.

Now increase the temperature. The exponential in (4) is the dominant factor: n grows very fast with T. Specifically, for silicon at room temperature, raising T by 10 K increases n by about a factor of 2. Even though \tau shortens slightly (for the same phonon-scattering reason as in metals), the exponential growth in n swamps it. Put (4) into (1):

\rho(T) \;\propto\; \frac{1}{n\,\tau} \;\propto\; \exp\!\left(+\,\frac{E_{g}}{2\,k_{B}T}\right)\cdot (\text{weaker } T \text{ dependence from } \tau). \tag{5}

Take the log of both sides and you get the operational form used to characterise a thermistor:

\ln \rho \;=\; \ln A + \frac{E_{g}}{2\,k_{B}\,T}.

A plot of \ln \rho against 1/T is a straight line, whose slope lets you read the band gap directly. The NTC thermistor in the Maruti's engine control unit is exactly a device whose slope is calibrated and used as a thermometer: measure the resistance, take the log, convert to 1/T, solve for T.

The sign of \alpha for a semiconductor is therefore negative, and \alpha is not even constant — it depends strongly on T itself. Quote the linear formula (2) for a semiconductor only over a narrow temperature window; for any serious calculation, use the exponential (5) directly.

Superconductors — the sharp limit

At very low temperatures, certain metals and compounds show a behaviour that no classical model can explain: their resistivity drops not gradually, not asymptotically, but discontinuously, to exactly zero. This is superconductivity, and the temperature at which it switches on is the critical temperature T_{c}.

Resistivity of mercury versus temperature showing the superconducting transitionCurve of resistivity versus temperature. Above 4.2 K, rho is a small positive value rising gently with T. Below 4.2 K, rho collapses to zero. The critical temperature Tc is marked with a dashed vertical line.temperature T (K)resistivity ρ24681012T_c = 4.2 Kρ = 0 (superconducting)ρ rising with T (normal metal)
The resistivity of mercury as a function of temperature. Above $T_{c} = 4.2$ K, mercury behaves like an ordinary metal — small $\rho$, rising gently with $T$. Below $T_{c}$, $\rho$ drops to exactly zero. This is not a steep extrapolation of metal behaviour; it is a thermodynamic phase transition to a fundamentally different state of matter.

Heike Kamerlingh Onnes discovered superconductivity in mercury in 1911, shortly after he succeeded in liquefying helium (boiling point 4.2 K). He was measuring the resistance of pure mercury down to very low temperatures, expecting the phonon-scattering story to predict a smooth decrease toward some small residual value. Instead the resistance collapsed to a value he could not distinguish from zero. A superconducting current set up in a loop of niobium-titanium wire has been observed to persist, with no measurable decay, for years.

The mechanism was not explained until 1957, in the BCS theory of Bardeen, Cooper, and Schrieffer: at low temperatures, electrons in certain materials pair up into Cooper pairs via a subtle phonon-mediated attraction, and these pairs behave as a single quantum condensate that flows without scattering. The details are beyond the scope of this chapter (see the full article on superconductivity), but three facts are JEE-relevant and useful to remember:

An Indian example: the Variable Energy Cyclotron Centre in Kolkata operates a superconducting cyclotron whose coils use Nb-Ti at 4.5 K. The zero-resistance coils carry 800 A of current to generate a field of 6 T, and the only electrical power needed is what leaks through the cryogenic refrigerator — not ohmic heating, because there isn't any.

Reading resistor colour codes

Flip over any small through-hole resistor you pull out of a project box. On the body are four or five coloured bands, separated by gaps. Those bands are the resistor's value — painted on, because the resistors are too small to fit a numerical label.

The colour-to-digit mapping is a mnemonic every physics lab in India teaches on day one:

Colour Digit Multiplier Tolerance
Black 0 \times 10^{0} —
Brown 1 \times 10^{1} ±1%
Red 2 \times 10^{2} ±2%
Orange 3 \times 10^{3} —
Yellow 4 \times 10^{4} —
Green 5 \times 10^{5} ±0.5%
Blue 6 \times 10^{6} ±0.25%
Violet 7 \times 10^{7} ±0.1%
Grey 8 \times 10^{8} —
White 9 \times 10^{9} —
Gold — \times 10^{-1} ±5%
Silver — \times 10^{-2} ±10%

Reading rule for a four-band resistor:

Example: reading a four-band resistor as 10 k-ohmSchematic of a resistor with four coloured bands: brown, black, red, gold. These encode 1, 0, times ten to the two, and five-percent tolerance, giving a nominal value of 1 kilo-ohm.brown1black0red×10²gold±5%10 × 10² Ω = 1 kΩ (±5% tolerance)leads
A four-band resistor: brown (1), black (0), red (×10²), gold (±5%). Reading left to right: the first two digits are "10", the multiplier is $10^{2}$, so the nominal value is $10 \times 10^{2} = 1\ \text{k}\Omega$, with a tolerance of 5%. The actual resistance is guaranteed to be between 950 Ω and 1050 Ω.

Five-band resistors (used for higher precision, 1% or better) give three significant digits instead of two: band 1, band 2, band 3 are digits; band 4 is the multiplier; band 5 is the tolerance. A five-band resistor with bands red-red-black-orange-brown is 220 \times 10^{3} = 220\ \text{k}\Omega at ±1%.

The mnemonic Indian students memorise — brown, red, orange, yellow, green, blue, violet, grey, white for 1-9 — is "B B ROY of Great Britain has a Very Good Wife". Black is 0; gold and silver are multiplier-only.

Worked examples

Example 1: Tungsten filament of an incandescent bulb

An old-style 100 W tungsten-filament bulb runs at 230 V at its hot operating temperature of about 2500 K. At room temperature (300 K), the filament is a coiled tungsten wire whose cold resistance is measured as 40 Ω. Taking \alpha_\text{W} = 4.5\times 10^{-3}\ \text{K}^{-1}, predict the hot resistance and the cold inrush current when the bulb is first switched on. Compare the hot-to-cold resistance ratio with the simple linear formula, and say where the discrepancy comes from.

Tungsten bulb cold vs hot filamentA lightbulb shown in two states: at switch-on, the cold 40-ohm filament draws a large inrush current of 5.75 A. At steady state, the hot filament has grown to about 435 ohms and draws 0.53 A.cold: 300 KR = 40 ΩI = 5.75 A (inrush)filament heatshot: 2500 KR = 435 ΩI = 0.53 A (steady)
Cold filament at 300 K with 40 Ω draws a large inrush current when first connected to 230 V. Within milliseconds it heats to 2500 K, and the resistance grows by about an order of magnitude to 435 Ω, limiting the steady current to 0.53 A.

Step 1. Apply the linear formula.

With T_{0} = 300 K and T = 2500 K, \Delta T = 2200 K. Equation (3) predicts

R_\text{hot} \;=\; R_\text{cold}\bigl[\,1 + \alpha\,\Delta T\,\bigr] \;=\; 40\bigl[\,1 + (4.5\times 10^{-3})(2200)\,\bigr].
R_\text{hot} \;=\; 40\,[1 + 9.9] \;=\; 40 \times 10.9 \;\approx\; 436\ \Omega.

Why: the linear formula predicts R rises by a factor of nearly 11 between room temperature and operating temperature. The coefficient \alpha\,\Delta T \approx 10 is dominated by the enormous \Delta T — 2200 K is eight times the room-temperature reference. The small \alpha gets multiplied by a large \Delta T and the product is no longer small.

Step 2. Check the formula's range of validity.

The Taylor expansion behind equation (3) assumes \alpha\,\Delta T \ll 1. At \alpha\,\Delta T \approx 10, you are far outside that regime. The actual resistance of a tungsten filament at 2500 K, measured experimentally, is about 14 times its room-temperature value — i.e. R_\text{hot}\approx 560\ \Omega, not 435 Ω.

Why: the linearised equation (3) is a local approximation. For tungsten over a range as extreme as 300 K → 2500 K, the second-order (and higher) terms in the Taylor series start to contribute. The physical reason is that \tau does not actually fall as 1/T at all temperatures — at very high T, more phonon modes get thermally populated, and \rho grows slightly faster than linearly. For JEE problems, the linear formula is still the right tool, but the reader should know it breaks down at extreme \Delta T.

Step 3. Compute the currents, first cold then hot, using the linear-formula value.

Cold: I_\text{cold} = V/R_\text{cold} = 230/40 = 5.75 A.

Hot: I_\text{hot} = V/R_\text{hot} = 230/436 \approx 0.53 A.

Power dissipated when hot: P = VI = 230 \times 0.53 \approx 122 W. (Roughly consistent with a 100 W bulb; the rated 100 W is typically measured at a lower operating temperature and with more precision.)

Why: the bulb's rated current is just over half an amp, but for the first few tens of milliseconds after switch-on it draws almost six amps — more than ten times steady state. This is exactly why old incandescent bulbs fail at the moment you switch them on, not during normal running. The fuse wire in the filament survives 0.53 A for years but is pushed close to its limit by repeated 5.75 A surges.

Step 4. Quick consistency check with the cold resistance itself.

The filament is a tungsten wire of length L and cross-section A with cold resistivity \rho_{0} = 5.6\times 10^{-8}\ \Omega\cdotm. For R_\text{cold} = 40 Ω: L/A = R_\text{cold}/\rho_{0} = 40/5.6\times 10^{-8} \approx 7.1\times 10^{8} m^{-1}. Typical filament: A = \pi (25\,\mu\text{m})^{2} = 2\times 10^{-9} m², so L \approx 1.4 m — a metre-long tungsten wire coiled into a spiral a few millimetres across.

Result: Hot resistance ≈ 435 Ω (linearised) or ≈ 560 Ω (measured). Inrush current ≈ 5.75 A; steady-state current ≈ 0.53 A.

What this shows: At extreme \Delta T, the linear formula underestimates the true resistance. The qualitative picture — tungsten's resistance rises roughly tenfold when it glows — is correct, and this is precisely the mechanism that protects a filament from thermal runaway: as it heats, it becomes more resistive, drawing less current, limiting further heating.

Example 2: NTC thermistor as a car temperature sensor

An NTC (negative-temperature-coefficient) thermistor in a Maruti coolant sensor has resistance 9.0 k\Omega at 0 °C and 200 Ω at 100 °C. Model its resistivity as R(T) = R_{\infty}\exp(B/T) with T in kelvin. (a) Find the constants B and R_{\infty}. (b) Predict the resistance at engine idle, 85 °C. (c) The ECU reads the thermistor as a voltage divider, using a 2 k\Omega fixed series resistor and a 5 V supply. Compute the sensor's output voltage at 85 °C.

Thermistor voltage-divider circuit for a coolant temperature sensorSchematic: 5 V supply at top, fixed 2 k-ohm series resistor labelled R_fixed, then the thermistor R_T whose resistance depends on temperature, then ground. The output voltage V_out is measured across the thermistor.+5 VR_fixed = 2 kΩV_out (to ECU)R_T (thermistor)GND
The thermistor sits in the lower half of a voltage divider, in series with a fixed 2 k$\Omega$ resistor. As the coolant heats up, $R_{T}$ drops from kilo-ohms to hundreds of ohms, and $V_\text{out}$ falls. The ECU maps $V_\text{out}$ back to temperature through the calibrated $R(T)$ curve.

Step 1. Set up two equations from the two calibration points.

T = 273 K: 9000 = R_{\infty}\,\exp(B/273).

T = 373 K: 200 = R_{\infty}\,\exp(B/373).

Step 2. Take the ratio to eliminate R_{\infty}.

\frac{9000}{200} \;=\; \frac{\exp(B/273)}{\exp(B/373)} \;=\; \exp\!\left(B\,\left(\frac{1}{273} - \frac{1}{373}\right)\right).
45 \;=\; \exp\!\left(B\,\frac{373 - 273}{273\times 373}\right) \;=\; \exp\!\left(B\cdot \frac{100}{101{,}829}\right) \;=\; \exp(B \cdot 9.82\times 10^{-4}).
\ln 45 \;=\; B \cdot 9.82\times 10^{-4} \;\;\Rightarrow\;\; B \;=\; \frac{3.807}{9.82\times 10^{-4}} \;\approx\; 3875\ \text{K}.

Why: the ratio cancels R_{\infty} and leaves a single equation in B. The constant B is in kelvin because E_{g}/(2k_{B}) has units of temperature — it is the gap in temperature units, not in eV. For silicon, E_{g}/(2k_{B}) \approx (1.1\,\text{eV})/(2\cdot 8.6\times 10^{-5}\,\text{eV/K}) \approx 6400 K; our fitted B = 3875 K corresponds to an effective gap of about 0.67 eV — consistent with a commercial thermistor that uses a metal oxide, not pure silicon.

Step 3. Solve for R_{\infty}.

From 9000 = R_{\infty}\exp(3875/273) = R_{\infty}\exp(14.20) = R_{\infty}\cdot 1.47\times 10^{6}:

R_{\infty} \;=\; \frac{9000}{1.47\times 10^{6}} \;\approx\; 6.1\times 10^{-3}\ \Omega.

Why: R_{\infty} is the extrapolated resistance as T \to \infty, when the exponential approaches 1. It is not physically meaningful as a measurable quantity (the thermistor would melt long before T \to \infty), but it sets the scale for the R(T) curve.

Step 4. Predict R at 85 °C = 358 K.

R(358) \;=\; 6.1\times 10^{-3}\,\exp(3875/358) \;=\; 6.1\times 10^{-3}\,\exp(10.82) \;=\; 6.1\times 10^{-3}\,\cdot 50{,}180 \;\approx\; 306\ \Omega.

Why: 85 °C is close to 100 °C, so the resistance is close to (but a bit higher than) the 100 °C calibration point of 200 Ω. Exponential dependence on 1/T means a 15 K swing near 100 °C changes R by a factor of about 1.5, which matches our 306 Ω vs 200 Ω result.

Step 5. Compute the divider output.

With R_\text{fixed} = 2000\ \Omega and R_{T} = 306\ \Omega:

V_\text{out} \;=\; V_\text{supply}\,\frac{R_{T}}{R_\text{fixed} + R_{T}} \;=\; 5\,\cdot\,\frac{306}{2306} \;\approx\; 0.66\ \text{V}.

Why: as the coolant heats from 0 °C to 85 °C, V_\text{out} falls from 5 \cdot 9000/11000 = 4.09 V down to 0.66 V — a clean, monotone, wide-dynamic-range signal for the ECU's analog-to-digital converter. A ±0.01 V resolution on the ADC corresponds to about ±1 °C resolution on temperature near the idle point.

Result: B \approx 3875 K, R_{\infty} \approx 6 m\Omega; predicted R(85\,°\text{C}) \approx 306\ \Omega; divider output ≈ 0.66 V.

What this shows: A two-point calibration is enough to fix the exponential R(T) curve completely. Once B and R_{\infty} are known, the thermistor becomes a thermometer over the full operating range. The negative \alpha (actually: the full exponential, not a constant \alpha) is what makes the temperature reading sharp.

Example 3: Reading a nichrome heater's resistance change in service

A 1 kW nichrome room-heater element runs at 230 V. Cold (at 20 °C), the coil has resistance 48 Ω. When running, the coil glows at about 820 °C. (a) Predict the hot resistance using \alpha_\text{nichrome} = 4.0\times 10^{-4}\ \text{K}^{-1}. (b) Compute the operating current and power, and compare with the rated 1 kW. (c) Explain why this small change in R is exactly what makes nichrome useful for the job.

Step 1. Apply the linear formula.

\Delta T = 820 - 20 = 800 K. With \alpha = 4.0\times 10^{-4}\ \text{K}^{-1} and \alpha\,\Delta T = 0.32:

R_\text{hot} \;=\; R_\text{cold}(1 + \alpha\,\Delta T) \;=\; 48 \times 1.32 \;=\; 63.4\ \Omega.

Why: because nichrome's \alpha is ten times smaller than copper's, the fractional rise in R over this enormous 800 K range is only 32%, not 300%. The linear approximation is entirely adequate for this range.

Step 2. Compute hot current and power.

I_\text{hot} = V/R_\text{hot} = 230/63.4 = 3.63 A.

P_\text{hot} = V I_\text{hot} = 230 \times 3.63 \approx 834 W.

Compare with cold: I_\text{cold} = 230/48 = 4.79 A, P_\text{cold} = 230 \times 4.79 = 1102 W.

Why: the rated "1 kW" is the power at the hot operating point. At 834 W I compute a slightly low number because I'm assuming a coil temperature of 820 °C; a commercial heater might run at 870 °C where R is a little lower. Either way, the steady power is lower than the cold-switch-on power of 1100 W — the opposite of the tungsten-bulb case — but only by 25%, not by factor 10.

Step 3. Why nichrome, and not copper?

If this element had been made of copper of the same initial geometry (same 48 Ω at 20 °C), then at 820 °C its resistance would be

R_\text{Cu,hot} \;=\; 48\bigl[1 + (3.9\times 10^{-3})(800)\bigr] \;=\; 48 \times 4.12 \;=\; 198\ \Omega.

Current: 230/198 \approx 1.16 A. Power: 267 W. The heater would dissipate a quarter of its rated wattage, and never stabilise in temperature (because as R keeps rising, the heating keeps falling, and the coil might cool back down before reaching equilibrium).

Why: this is the negative feedback that makes nichrome a poor choice for a thermal regulator but an excellent choice for a heater. The heater wants a fixed thermal output against mains voltage variation. Nichrome's small \alpha delivers that — R is nearly constant over the operating range, so P = V^{2}/R is nearly constant too. Copper's large \alpha would give wildly voltage-dependent power, unusable for a domestic appliance.

Result: Hot R \approx 63\ \Omega, I \approx 3.6 A, P \approx 834 W.

What this shows: The small temperature coefficient of nichrome — 10× smaller than copper — is not a defect. It is the exact property that makes nichrome the material for heating elements. When engineers pick a material, they pick \alpha as carefully as \rho itself.

Common confusions

If you came here for the formula R(T) = R_{0}(1 + \alpha\,\Delta T), its physical origin, and the applications to heaters, thermistors, and bulbs, you have what you need. What follows is the statistical-mechanics derivation of the semiconductor \rho(T), Matthiessen's rule for impurity scattering, and a word on resistance-thermometry standards.

Matthiessen's rule — adding scattering rates

When an electron travels through a metal, it can scatter off two independent things:

  • Thermal phonons (lattice vibrations): rate 1/\tau_\text{ph}(T), which grows with T.
  • Static defects (impurities, dislocations, grain boundaries): rate 1/\tau_\text{imp}, which does not depend on T.

If the two scattering mechanisms are independent, the total scattering rate is the sum:

\frac{1}{\tau(T)} \;=\; \frac{1}{\tau_\text{ph}(T)} + \frac{1}{\tau_\text{imp}}.

Since \rho \propto 1/\tau, the resistivities add:

\rho(T) \;=\; \rho_\text{ph}(T) + \rho_\text{imp}.

This is Matthiessen's rule. It predicts that a metal's resistivity never drops below \rho_\text{imp} — even at T = 0, impurity scattering remains. This residual resistivity is the reason an ordinary copper wire does not become a superconductor at low T: its phonon scattering drops to zero, but its impurity scattering does not.

Superconductivity is a fundamentally different physical state, not a limit of Matthiessen. That is why it requires pair formation (Cooper pairs) and cannot be achieved merely by cooling a pure enough metal.

Deriving the semiconductor exponential from Fermi-Dirac statistics

The density of electrons thermally excited across the gap follows from integrating the Fermi-Dirac distribution over the conduction-band states. For an intrinsic semiconductor at moderate temperature (when the chemical potential sits near the middle of the gap), this integral gives

n(T) \;=\; \sqrt{N_{c}\,N_{v}}\,\exp\!\left(-\,\frac{E_{g}}{2\,k_{B}T}\right).

Here N_{c} and N_{v} are the effective density-of-states of the conduction and valence bands respectively, each approximately proportional to T^{3/2}. So the full temperature dependence of n is not a pure exponential — it has a pre-factor that grows as T^{3/2} — but over the narrow range of temperatures relevant to a thermistor, the exponential dominates.

Substituting into \rho = 1/(n e \mu), with the mobility \mu = e\tau/m^{*} of the carriers (only weakly T-dependent), you get

\rho(T) \;\approx\; \rho_{0}\,\exp\!\left(\frac{E_{g}}{2\,k_{B}T}\right).

This is the form the worked example fits. The constant B in the fit is E_{g}/(2k_{B}) — you can read the energy gap directly off the thermistor's calibration data.

Resistance thermometry and the ITS-90 scale

Pure platinum has one of the most reproducible resistance-vs-temperature curves of any metal. A 100 Ω platinum RTD (PT100) in an ISRO-grade sensor drifts by less than 1 part in 10^{5} per year, over a range from -200 °C to +850 °C. For this reason, platinum resistance thermometers define the International Temperature Scale of 1990 between the triple point of hydrogen (13.8 K) and the freezing point of silver (961.78 °C).

The calibration equation for a platinum RTD, used down to liquid-nitrogen temperatures, is the Callendar-Van Dusen equation:

R(T) \;=\; R_{0}\bigl[1 + A\,T + B\,T^{2} + C\,T^{3}(T - 100)\bigr],

with A = 3.91\times 10^{-3}\ \text{K}^{-1}, B = -5.78\times 10^{-7}\ \text{K}^{-2}, C = -4.18\times 10^{-12}\ \text{K}^{-4} (for T < 0 °C), and T measured in degrees Celsius. Below 0 °C the cubic correction switches in; above, the quadratic is enough. The leading coefficient A is essentially the \alpha of platinum near room temperature. The higher-order corrections are what distinguish a laboratory temperature standard from a bulk resistor.

If you ever wire up a four-wire resistance bridge at a physics lab in IIT Kanpur or IISc Bangalore, using a PT100 to read the temperature of a cryogenic sample, the formula above is what the software is evaluating as you watch the readout update. The linear formula (3) of this chapter is the first term; the polynomial corrections are the price of high-precision thermometry.

Superconducting transitions in applied magnetic fields

One subtlety beyond "\rho = 0 below T_{c}": a superconductor also has a critical magnetic field H_{c}(T) above which superconductivity is destroyed even below T_{c}. The curve H_{c}(T) starts at some finite value at T = 0 and falls to zero at T = T_{c}, typically as H_{c}(T) \approx H_{c}(0)[1 - (T/T_{c})^{2}]. A superconducting MRI magnet at IIT or at Tata Memorial Hospital in Mumbai operates comfortably below both T_{c} (the Nb-Ti transition at 9.5 K) and its H_{c} (about 15 T at liquid-helium temperature). A stray field above H_{c} quenches the superconductor back to the normal state, suddenly — a failure mode that requires special protection circuitry in every high-field magnet.

So the simple statement "below T_{c}, \rho = 0" is not quite enough. The full statement is: below T_{c} and below H_{c}(T) and below a critical current J_{c}(T, H), \rho = 0. Miss any of the three and the material reverts to its normal-state resistivity. Designing practical superconducting devices is largely the art of staying inside this three-variable safe region.

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