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---
# author:
# - Orestis Malaspinas
title: Exercices circuits
subtitle: Courant et circuits électriques
autoSectionLabels: true
autoEqnLabels: false
eqnPrefix:
- "éq."
- "éqs."
chapters: true
numberSections: true
chaptersDepth: 1
sectionsDepth: 3
lang: fr
documentclass: article
papersize: A4
cref: false
pandoc-numbering:
- category: exercice
urlcolor: blue
---
\newcommand{\m}{\mathrm{m}}
\newcommand{\C}{\mathrm{C}}
\newcommand{\V}{\mathrm{V}}
\newcommand{\mm}{\mathrm{mm}}
\newcommand{\cm}{\mathrm{cm}}
# Préambule
Faites les exercices **SANS** lire les réponses.
# Problèmes divers {#unidim .unnumbered}
Exercice (Luminosité d'ampoules) #
Soit le circuit de la @fig:ampoules où les trois ampoules on la même résistance $R$.
1. Quand l'interrupteur est en position fermée, classer les ampoules en fonction de leur luminosité.
2. Que se passe-t-il quand l'interrupteur est ouvert? Que devient la luminosité des ampoules?
![Un circuit électrique avec trois ampoules de résistance $R$.](figs/ampoules.svg){#fig:ampoules width=50%}
Corrigé (Luminosité d'ampoules) #
1. Quand l'interrupteur est fermé, le courant passant dans $C$ est séparé en deux parties égales pour passer dans $A$ et $B$, car leurs résistances sont égales. Ainsi, $A$ et $B$ auront la même luminosité, mais brilleront moins que $C$ qui a un courant plus important.
2. Quand on ouvre l'interrupteur, l'ampoule $A$ ne sera plus alimentée en courant. On a donc que le courant passera uniquement par $B$ et le circuit sera équivalent à avoir uniquement deux ampoules en série. La résistance équivalente de ce circuit ($=2R$) sera plus grande que celle du circuit avec l'interrupteur fermé ($=3R/2$). Ainsi en fermant l'interrupteur on diminue le courant sortant de la batterie passant par $C$ qui sera moins lumineuse. En revanche, $B$ va recevoir plus de courant que dans le cas où l'interrupteur est fermé. Elle brillera donc plus. Aussi comme $C$ et $B$ ont la même résistance, elles auront la même luminosité.
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