Problem — fuses and Joule heating

Problem

Explain in five lines why a fuse (a thin wire of a low-melting-point alloy) “blows” when the current exceeds a threshold value, protecting the rest of the circuit. Discuss the dependence of the power dissipated on the current and the thermal equilibrium of the wire with its surroundings.

Solution

Joule heating in the wire — the fuse wire has a small resistance $R$; carrying a current $i$, it dissipates power $P = R\,i^2$ which depends on the square of the current: doubling $i$ quadruples the power dissipated.

Thermal equilibrium — this heat produced warms the wire, which in turn transfers heat to the surroundings (by conduction, convection, radiation) at a rate that grows with its temperature. The wire reaches an equilibrium temperature when the heat produced by Joule heating equals the heat lost: $R\,i^2 = P_\text{lost}(T)$

Why it “blows” — as long as $i$ stays below the threshold, equilibrium is established at a low temperature and the wire holds. If $i$ exceeds the threshold value, the term $R\,i^2$ grows faster than the heat loss: the temperature rises until it reaches the melting point (low by design). The wire melts, the circuit opens and the current is interrupted, saving the appliances downstream.

The thinness of the wire (small cross-section $\Rightarrow$ high $R$ and little mass to heat) and the low-melting alloy are chosen specifically so that the fuse reacts before, and at a lower current, than the conductors it is meant to protect.

$\ev{P = R\,i^2 \text{ exceeds the heat loss} \Rightarrow T \to T_\text{melting} \Rightarrow \text{the wire melts and opens the circuit}}$

Links

Topics: Circuiti elettrici Concepts: Effetto Joule