Worked exercise — series vs parallel: total power

Problem

Three identical resistors of value $R$ are connected to the same source $\mathcal{E}$, first all in series, then all in parallel. Compare the total power dissipated in the two configurations.

Solution

In series. The equivalent resistance and the current are: $R_{\text{eq}} = 3R \qquad i_s = \frac{\mathcal{E}}{3R}$ The total power is $P_{\text{serie}} = R_{\text{eq}}\,i_s^2 = 3R\left(\frac{\mathcal{E}}{3R}\right)^2 = \frac{\mathcal{E}^2}{3R}$

In parallel. Now $R_{\text{eq}} = \frac{R}{3} \qquad i_p = \frac{\mathcal{E}}{R/3} = \frac{3\mathcal{E}}{R}$ The total power is $P_{\text{par}} = R_{\text{eq}}\,i_p^2 = \frac{R}{3}\left(\frac{3\mathcal{E}}{R}\right)^2 = \frac{3\,\mathcal{E}^2}{R}$

Ratio between the two configurations: $\frac{P_{\text{serie}}}{P_{\text{par}}} = \frac{\mathcal{E}^2/(3R)}{3\,\mathcal{E}^2/R} = \frac{1}{9}$

For the same source, the parallel combination dissipates nine times more than the series: its equivalent resistance is smaller, so it draws more current and more power.

$\ev{\dfrac{P_{\text{serie}}}{P_{\text{par}}} = \dfrac{1}{9}}$

Links

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