Problem — burnt-out bulb in series

Problem

Three identical light bulbs are connected in series to a battery. One of them burns out (infinite resistance). What happens to the other two? And what is the voltage measurable across the burnt-out bulb?

Three bulbs in series: $L_2$ burns out (broken filament).

Solution

What happens to the other two — in a series circuit the current has a single path. If a bulb burns out, its filament breaks and presents infinite resistance: the circuit is open. The current drops to zero everywhere: $i = \frac{\mathcal{E}}{R_1 + R_\text{burnt} + R_3} = \frac{\mathcal{E}}{\infty} = 0$

So the other two bulbs go out as well: all three stay off. This is the well-known fault of old series light strings, where a single blown bulb switched off the entire string.

Voltage across the burnt-out bulb — with zero current, the two still-intact bulbs have no voltage drop across them ($V = R\,i = 0$). Going around the loop, the entire e.m.f. of the battery therefore appears entirely across the break, i.e. across the burnt-out bulb: $V_\text{burnt} = \mathcal{E}$

A voltmeter placed across it reads the full battery voltage: this is the practical way to identify which bulb has blown.

$\ev{\text{all three off}; \quad V_\text{burnt} = \mathcal{E} \text{ (full battery voltage)}}$

Collegamenti

Argomenti: Circuiti elettrici Concetti: Resistenze in serie e parallelo