diff --git a/exercices/circuits.md b/exercices/circuits.md
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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?
+
+{#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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