Problem — bulbs in series or in parallel

Problem

Two identical bulbs, each of resistance $R$, are first connected in series to a battery of voltage $V$ (power $P_s$ on each), then in parallel to the same battery (power $P_p$ each). Calculate the ratio $P_p/P_s$ and explain why they shine more brightly in parallel.

(a) The two bulbs in series; (b) in parallel on the same battery.

Solution

Series configuration — the resistances add up, $R_\text{tot} = 2R$, so the current is $i_s = \frac{V}{2R}$ The power on one bulb (the same current flows through it): $P_s = R\,i_s^2 = R\left(\frac{V}{2R}\right)^2 = \frac{V^2}{4R}$

Parallel configuration — each bulb is subjected to the full voltage $V$: $P_p = \frac{V^2}{R}$

Ratio: $\frac{P_p}{P_s} = \frac{V^2/R}{V^2/(4R)} = 4$

Why they shine more brightly in parallel — in series the two bulbs share the voltage ($V/2$ each) and the current is halved; in parallel each receives the full voltage $V$. Since $P = V^2/R$, doubling the voltage across a bulb quadruples the power: each bulb in parallel dissipates four times the power it had in series.

$\ev{\dfrac{P_p}{P_s} = 4}$

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Topics: Circuiti elettrici Concepts: Resistenze in serie e parallelo · Potenza