Topic 9 — Temperature Dependence of Resistance

Part 1 · The idea, told simply

Heat a copper wire and it conducts worse. Heat a piece of silicon and it conducts better. Same heating, opposite results. Why?

Story track · the crowded corridor, revisited

Go back to the corridor full of people from Topic 4. The electrons are trying to get through, and the metal ions are obstacles standing in the way.

Now warm the metal up. The ions do not move out of the way — they are locked in place — but they start jiggling harder. A jiggling obstacle is much easier to bump into than a still one. So the electrons collide more often, travel for less time between collisions, and make less progress.

Less time between collisions means a smaller relaxation time τ. And from Topic 3, ρ = m/(ne²τ). Smaller τ, larger ρ. The metal resists more.

Story track · why a semiconductor does the opposite

In a semiconductor, something else happens as well — and it is far more dramatic.

At low temperature, almost all the electrons in silicon are tied down in bonds and cannot move at all. There are hardly any charge carriers. Heat it, and the thermal energy starts ripping electrons free. The number of available carriers n shoots up, and it shoots up very steeply.

The jiggling-obstacle effect is still there, so τ still falls. But the flood of new carriers completely overwhelms it. In ρ = m/(ne²τ), the n in the denominator grows far faster than the τ shrinks. The semiconductor resists less.

The whole topic in one comparison — memorise this shape, not just the answers

	Metal	Semiconductor / insulator
What happens to n	essentially unchanged	rises steeply
What happens to τ	falls	falls
Which effect dominates	τ (the only one acting)	n, overwhelmingly
Result for ρ	RISES	FALLS
Sign of α	positive	negative

NCERT states the metal case exactly: "In a metal, n is not dependent on temperature to any appreciable extent and thus the decrease in the value of τ with rise in temperature causes ρ to increase." An assertion–reason question that blames a change in n for a metal's rising resistance is false — the reason is τ.

Animation 1 · Heat the metal, watch the collisions

The ions jiggle harder. The electron's free run between collisions gets shorter.

temperature =

Animation 2 · The semiconductor's flood of carriers

Same heating, but here electrons are being freed from bonds. Count them as the temperature rises.

temperature =

Part 2 · The formula

Maths track · NCERT §3.8

Over a limited range of temperature that is not too large, the resistivity of a metallic conductor is approximately

ρ_T = ρ₀ [1 + α (T − T₀)]

where ρ_T is the resistivity at temperature T and ρ₀ that at a reference temperature T₀. The constant α is the temperature coefficient of resistivity. From the equation, its dimension is (Temperature)⁻¹, so its unit is °C⁻¹ or K⁻¹. For metals α is positive.

Since R = ρl/A and the geometry barely changes, the same relation carries over to resistance:

R_T = R₀ [1 + α (T − T₀)]

Rearranged to find α from two measurements:

α = (R₂ − R₁) / [R₁ (T₂ − T₁)]

Note carefully that the denominator carries R₁, the resistance at the reference temperature — not R₂ and not their average.

The word "approximately" is doing real work

NCERT is careful here, and so is NEET. The linear relation implies that a graph of ρ_T against T is a straight line. That holds well over a limited range around the reference temperature. But at temperatures much lower than 0 °C the graph deviates considerably from a straight line (Fig. 3.8). The formula is a local approximation, not a law.

Animation 3 · Three materials, three curves

Copper rising and curving at low temperature, nichrome almost flat, a semiconductor falling away.

Why standard resistors are made of alloys

Nichrome — an alloy of nickel, iron and chromium — has a very weak dependence of resistivity on temperature. Manganin and constantan behave similarly. NCERT gives the practical consequence directly: these materials are widely used in wire-bound standard resistors, since their resistance values would change very little with temperature.

Copper is a far better conductor than nichrome, but that is not what a standard resistor needs. A standard resistor needs to be the same tomorrow as it was today, in a warm room or a cold one. Stability beats conductivity here.

Animation 4 · Why you would not build a standard resistor from copper

The same 50 °C temperature swing, applied to two 100 Ω resistors.

temperature swing =

Part 3 · The two standard numericals

Maths track · the heating element (NCERT Example 3.3)

A toaster's nichrome element measures 75.3 Ω at 27.0 °C with a negligible current. Connected to 230 V, the current settles at 2.68 A. Find the steady temperature. Take α = 1.70 × 10⁻⁴ °C⁻¹.

The logic matters as much as the arithmetic. With a negligible current there is no heating, so the element sits at room temperature. Once connected, it heats up, its resistance rises, and the current falls slightly — until a steady state is reached where the heat generated equals the heat lost.

R₂ = 230 V / 2.68 A = 85.8 Ω T₂ − T₁ = (85.8 − 75.3) / (75.3 × 1.70 × 10⁻⁴) = 820 °C T₂ = 820 + 27.0 = 847 °C

The same structure solves Exercise 3.6, where a nichrome element draws an initial 3.2 A settling to 2.8 A on a 230 V supply: R₁ = 71.9 Ω, R₂ = 82.1 Ω, giving a rise of about 840 °C and a steady temperature near 867 °C.

Maths track · the platinum resistance thermometer (NCERT Example 3.4)

A platinum wire reads 5 Ω at the ice point and 5.23 Ω at the steam point. In a hot bath it reads 5.795 Ω. What is the bath temperature?

Here the two fixed points define the scale, so no value of α is needed at all:

t = (R_t − R₀)/(R₁₀₀ − R₀) × 100 = (5.795 − 5)/(5.23 − 5) × 100 = (0.795/0.23) × 100 = 345.65 °C

This is the entire principle of resistance thermometry: resistance is easy to measure precisely, and it maps onto temperature through two calibration points.

Animation 5 · The platinum resistance thermometer

Two fixed points calibrate the scale. Any resistance reading between them becomes a temperature.

reading R_t =

Trap · the six that cost marks

Blaming a metal's rising resistance on a change in n. In a metal n barely changes; the cause is the falling τ.
Dividing by R₂ instead of R₁ when finding α. The denominator carries the resistance at the reference temperature.
Forgetting to add the reference temperature back at the end. The formula gives a temperature DIFFERENCE.
Treating the linear relation as exact. It fails badly well below 0 °C.
Taking α as positive for semiconductors. It is negative.
Assuming alloys are chosen for standard resistors because they conduct well. They are chosen because α is nearly zero.

Part 4 · Formula sheet

The core relation

ρ_T = ρ₀ [1 + α (T − T₀)] R_T = R₀ [1 + α (T − T₀)] α = (R₂ − R₁) / [R₁ (T₂ − T₁)] ← denominator uses R₁ T₂ − T₁ = (R₂ − R₁) / (R₁ α) Unit of α: °C⁻¹ or K⁻¹ · dimension (Temperature)⁻¹ Numerically α is the same in °C⁻¹ and K⁻¹ (interval sizes match)

Sign of α by material

Metals α positive, small (≈ 10⁻³ to 10⁻⁴ °C⁻¹) → ρ rises with T Alloys (nichrome, manganin, constantan) α ≈ 0, very weak dependence → used for standard resistors Semiconductors α negative → ρ falls with T Insulators α negative → ρ falls with T

The mechanism — the reason, not just the result

ρ = m/(n e² τ) Metal: n ~ constant, τ falls → ρ RISES Semiconductor: n rises steeply, τ falls; the rise in n dominates → ρ FALLS The competing quantity is always n versus τ

Resistance thermometry

Platinum resistance thermometer (two fixed points, no α needed): t = (R_t − R₀)/(R₁₀₀ − R₀) × 100 where R₀ is at the ice point and R₁₀₀ at the steam point

The heating-element problem

1. Negligible current → element sits at room temperature → R₁ 2. Connected to supply → steady current I → R₂ = V/I 3. T₂ − T₁ = (R₂ − R₁)/(R₁ α) 4. Add T₁ back: T₂ = (T₂ − T₁) + T₁ Steady state = heat generated equals heat lost

Sanity checks

For a metal, R must come out LARGER at the higher temperature For a semiconductor, SMALLER α for a metal is small: a 1000 °C rise roughly doubles R when α ≈ 10⁻³ The linear formula is a local approximation, not a law Graph of ρ against T for a metal: rising, straight over a limited range, curving away well below 0 °C

Part 5 · 50 NEET-pattern questions with full solutions

Includes 4 graph-based questions, 3 assertion–reason questions, and 8 questions tagged Mechanism — the n-versus-τ reasoning that Tier 3 questions are built on. Year labels are not attached: the patterns are authentic, the wording is mine.