Problem — chains of light bulbs

Problem

Four circuits with identical light bulbs of resistance $R$ powered by a battery $V$: (A) one bulb; (B) two bulbs in series; (C) two bulbs in parallel; (D) four bulbs in parallel. Rank the circuits by total power delivered by the battery, from smallest to largest.

Solution

Strategy — the total power delivered by the battery depends on the equivalent resistance seen at its terminals: $P = \frac{V^2}{R_\text{eq}}$ The smaller $R_\text{eq}$, the greater the power. Let’s compute $R_\text{eq}$ and $P$ for each circuit.

(A) one bulb — $R_\text{eq} = R$: $P_A = \frac{V^2}{R}$

(B) two in series — $R_\text{eq} = 2R$: $P_B = \frac{V^2}{2R}$

(C) two in parallel — $R_\text{eq} = R/2$: $P_C = \frac{V^2}{R/2} = \frac{2V^2}{R}$

(D) four in parallel — $R_\text{eq} = R/4$: $P_D = \frac{V^2}{R/4} = \frac{4V^2}{R}$

Ranking — factoring out $V^2/R$: the coefficients are $\tfrac{1}{2}$ (B), $1$ (A), $2$ (C), $4$ (D). From smallest to largest: $P_B < P_A < P_C < P_D$

Bulbs in series increase the resistance and reduce the power; bulbs in parallel lower it and increase total consumption.

$\ev{B < A < C < D}$

Collegamenti

Argomenti: Circuiti elettrici Concetti: Potenza Competenze: Ragionamento di ranking