Problem — short circuit

Problem

Explain in five lines what happens when a short circuit is created across the poles of a battery, and why it is dangerous even for small domestic circuits. Consider the power dissipated, the internal resistance and the effects on the materials.

Solution

What happens — a short circuit connects the two poles of the battery with an external resistance that is almost zero ($R_\text{ext} \to 0$). The only resistance left to limit the current is the battery’s internal resistance $r$ (and that of the wires), typically very small: $i_\text{short} = \frac{\mathcal{E}}{r}$

With $r$ of the order of fractions of an ohm, this current is enormous compared with normal operation.

Power dissipated — all the power $P = \mathcal{E}\,i = \mathcal{E}^2/r$ finds no useful load and is dissipated inside the battery and in the conductors of the short, through Joule heating: $P_\text{diss} = i_\text{short}^2\, r = \frac{\mathcal{E}^2}{r}$

Effects on materials — this concentrated power causes rapid heating: the wires can melt the insulation, the battery can overheat, swell, leak electrolyte or (in lithium batteries) go into thermal runaway with a risk of fire or explosion.

Why it is dangerous even on a small scale — what matters is not the voltage, which can be modest ($1{,}5\ \text{V}$, $9\ \text{V}$), but the current: a small battery with low internal resistance can deliver tens of amperes when shorted, enough to heat a thin wire red-hot. This is why fuses and circuit breakers exist.

$\ev{i_\text{short} = \mathcal{E}/r \text{ large} \Rightarrow P = \mathcal{E}^2/r \text{ dissipated internally} \Rightarrow \text{overheating}}$

Links

Topics: Circuiti elettrici Concepts: Forza elettromotrice · Potenza