Define the temperature coefficient of resistance.

II. Short Answer Questions · Q8

Q. Define the temperature coefficient of resistance.

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Concept understanding — Temperature Dependence of Resistance

Temperature Dependence of Resistance

Imagine you're trying to walk through a crowded market. When the market is cool and calm, people move slowly and you can weave through easily. Now imagine the same market on a hot, chaotic day — everyone is jostling, moving faster, bumping into each other. Getting from one end to the other becomes much harder.

That's exactly what happens inside a metal wire when you heat it up.

The Intuition

In a metal, electric current is carried by free electrons drifting through a fixed lattice of positive ions. At room temperature, these ions are vibrating slightly around their positions. When you heat the metal, the ions vibrate more vigorously — they shake faster and with larger amplitude.

Think of the vibrating ions as a row of swinging doors. At low temperature, the doors barely move, so electrons slip through easily. At high temperature, the doors swing wildly, and electrons get knocked off course constantly. Each collision with a vibrating ion scatters the electron, making it harder for the current to flow.

The result: resistance increases as temperature increases — for most conductors.

The Precise Statement

For a metallic conductor over a moderate temperature range (not too close to absolute zero), the resistance changes linearly with temperature:

R(T) = R_0 [1 + \alpha (T - T_0)]

Where:

$R(T)$ is the resistance at temperature $T$
$R_0$ is the resistance at a reference temperature $T_0$ (often $0^\circ$ C or $20^\circ$ C)
$\alpha$ is the temperature coefficient of resistance (units: per °C or per K)

R = R_0 (1 + \alpha \Delta T)

The coefficient $\alpha$ tells you how sensitive the material is to temperature changes. For copper, $\alpha \approx 0.0039 /^\circ\text{C}$ — meaning for every 1°C rise, resistance increases by about 0.39%.

What About Other Materials?

Not everything behaves like metals.

Semiconductors (like silicon, germanium) do the opposite: their resistance decreases sharply as temperature rises. Why? Because heating frees more electrons from their bonds, creating many more charge carriers. Even though the lattice vibrates more, the huge increase in available carriers overwhelms that effect, so resistance drops.

Insulators also show decreasing resistance with temperature, but the effect is much smaller than in semiconductors.

Alloys like constantan (copper-nickel) have a very small $\alpha$ — their resistance barely changes with temperature. This is useful for making precision resistors that stay stable.

Superconductors are a special case: below a critical temperature, resistance drops to exactly zero. …

Why this formula?

Temperature Dependence of Resistance — Why the Formula Holds

Let’s build this from the ground up. The key formula you’ll see in exams is:

R_T = R_0 (1 + \alpha T)

But why does resistance change with temperature? It’s not magic — it’s about what happens inside the wire.

1. What determines resistance?

Resistance $R$ of a conductor depends on three things:

Length $L$ (longer → more resistance)
Cross-sectional area $A$ (thicker → less resistance)
Resistivity $\rho$ — a material property

The formula is:

R = \rho \frac{L}{A}

When temperature changes, $L$ and $A$ change very slightly (thermal expansion), but the big effect is on $\rho$ .

2. Why does resistivity change with temperature?

Resistivity $\rho$ depends on how easily electrons can move through the material.

In metals: Atoms vibrate more as temperature rises. These vibrations scatter electrons, making it harder for them to flow. So $\rho$ increases.
In semiconductors: More electrons get enough energy to jump into the conduction band. So $\rho$ decreases.

For most metals (and many conductors), the change is linear over a moderate temperature range.

3. Deriving the linear formula

Let $\rho_0$ be resistivity at a reference temperature $T_0$ (often $0^\circ$ C or $20^\circ$ C).

For a small change $\Delta T = T - T_0$ , the change in resistivity is proportional to $\Delta T$ and to $\rho_0$ :

\Delta \rho \propto \rho_0 \Delta T

Introduce the temperature coefficient of resistivity $\alpha$ :

\Delta \rho = \alpha \, \rho_0 \, \Delta T

So the new resistivity is:

\rho = \rho_0 + \Delta \rho = \rho_0 (1 + \alpha \Delta T)

Now, since $R = \rho \frac{L}{A}$ , and $L$ and $A$ change negligibly (for small $\Delta T$ ), we get:

R = \rho \frac{L}{A} = \rho_0 (1 + \alpha \Delta T) \frac{L}{A} = R_0 (1 + \alpha \Delta T)

That’s the formula:

\boxed{R_T = R_0 (1 + \alpha \Delta T)}

Where:

$R_T$ = resistance at temperature $T$
$R_0$ = resistance at reference temperature $T_0$
$\alpha$ = temperature coefficient of resistance (unit: $^\circ$ C $^{-1}$ or K $^{-1}$ )
$\Delta T = T - T_0$

--- …

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Step 1. Starting from $R_T=R_0[1+\alpha(T-T_0)]$ , rearranging gives $R_T-R_0=\alpha R_0(T-T_0)$ .

Step 2. Writing $\Delta R=R_T-R_0$ and $\Delta T=T-T_0$ , this becomes $\Delta R=\alpha R_0\Delta T$ , i.e. $\alpha=\dfrac{1}{R_0}\cdot\dfrac{\Delta R}{\Delta T}$ . …

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Q.Define the temperature coefficient of resistance.

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