diff --git a/06_cicruits_electriques.md b/06_cicruits_electriques.md
index 405f7a4e88bf38eaa46801401de7dcdf43be7673..4fa21d7adef233b1ec5f610b3e34bfd422733941 100644
--- a/06_cicruits_electriques.md
+++ b/06_cicruits_electriques.md
@@ -53,4 +53,64 @@ $$
 et donc qu'on peut remplacer les trois résistances par une résistance équivalente (ou nette), où
 $$
 R_\mathrm{eq}=R_1+R_2+R_3.
-$$
\ No newline at end of file
+$$
+
+---
+
+Exemple (Trois résistances) #
+
+Soit une source de tension qui produit un courant continu de $12\V$ et trois
+résistances en série de $3\Omega$ chacune connectées au circuit comme sur la @fig:three_res. Quelle est le courant total dans le circuit?
+
+---
+
+---
+
+Solution (Trois résistances) #
+
+Les trois résistances étant connectées en série, ce circuit est équivalent à
+un circuit avec une seule résistance de $R=R_1+R_2+R_3=9\Omega$. En utilisant ensuite
+la loi d'Ohm on obtient
+$$
+I=V/R=12 / 9=4\A.
+$$
+
+---
+
+Une autre façon simple de connecter eds résistances sur un circuit est en **parallèle** (voir @fig:three_res_par). Les charges dans ce circuit suivent
+trois chemins différents et donc le courant total, $I$, est séparé en trois parties (pas forcément égales), $I_1$, $I_2$, et $I_3$.
+$$
+I=I_1+I_2+I_3.
+$${#eq:isum}
+Dans ce type de circuit le courant n'est pas interrompu si une des résistances est
+déconnectée (dans le cas où un des appareils que la dite résistance représente arrête de fonctionner par exemple).
+
+![Les résistances $R_1$, $R_2$, $R_3$ sont connectées en parallèle à la source $V$ et traversées par les courants $I_1$, $I_2, $I_3$.](figs/three_res_par.svg){#fig:three_res_par width=50%}
+
+Quand les résistances sont en parallèle, le voltage qui les traverse doit être le même.
+En effet, étant donné qu'on néglige la résistance des fils, deux points connectés
+directement entre eux ont le même voltage. Ainsi, on a pour les courants
+$$
+I_1=\frac{V}{R_1},\quad 
+I_2=\frac{V}{R_2},\quad 
+I_3=\frac{V}{R_3}. 
+$${#eq:i3}
+De plus on sait de la loi d'Ohm, qu'on doit pouvoir remplacer les trois résistances
+par une unique résistance équivalent, $R_\mathrm{eq}$, avec
+$$
+I=\frac{V}{R_\mathrm{eq}}.
+$${#eq:ipar}
+En mélangeant les 3 équations ci-dessus (voir @eq:isum, @eq:i3, et @eq:ipar), on obtient
+$$
+\frac{V}{R_\mathrm{eq}}=
+\frac{V}{R_1}+
+\frac{V}{R_2}+
+\frac{V}{R_3},
+$$
+et on déduit
+$$
+\frac{1}{R_\mathrm{eq}}=
+\frac{1}{R_1}+
+\frac{1}{R_2}+
+\frac{1}{R_3}.
+$$
diff --git a/figs/three_res_par.svg b/figs/three_res_par.svg
new file mode 100644
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