Problem — light bulbs in series and in parallel

Problem

Light bulbs in series and in parallel. Two identical light bulbs of resistance $R$ are connected to an ideal battery of EMF $\mathcal{E}$. (a) If you put them in series, is their total brightness greater or smaller than that of a single one of them? (b) If you put them in parallel, how does the brightness of each bulb change compared to the case where only one were connected? (c) If one of the two “burns out” (open circuit), what happens in the two cases? Answer without explicit calculations, reasoning about how voltage and current are shared.

Solution

(a) In series: SMALLER total brightness. The two bulbs in series present a resistance $2R$: the current supplied is $\mathcal{E}/2R$, half that of a single bulb ($\mathcal{E}/R$). The total power dissipated is $P_\text{series} = \frac{\mathcal{E}^2}{2R}$ i.e. half the power $\mathcal{E}^2/R$ of a single bulb. Adding a bulb in series obstructs the current and reduces the overall light.

(b) In parallel: each EQUAL to when it is alone. Each branch of a parallel combination sees the full voltage $\mathcal{E}$ of the ideal battery, independently of the other branch. So each bulb dissipates $\mathcal{E}^2/R$ exactly as if it were the only one connected: the brightness of each is unchanged (however, the total current supplied by the battery doubles, and so does the overall light).

(c) If one burns out (open branch):

• In series, the single current path is interrupted: both go out (the intact bulb no longer has current flowing through it).
• In parallel, the other branch stays closed and still sees the same $\mathcal{E}$: the other bulb keeps lighting normally.

$\ev{\text{(a) smaller } (P/2)\;;\quad \text{(b) equal to the single bulb}\;;\quad \text{(c) series: both off} - \text{parallel: the other stays on}}$

Links

Topics: Electric circuits Concepts: Resistors in series and parallel