added stuff. must still add kirchoff law

05_courant_electrique.md

@@ -391,10 +391,10 @@ plt.title("Puissance instantanée et moyenne du courant alternatif.")
 plt.legend()
 ```
 
-La puissance moyenne, $\bar P$, est facilement calculée (on va le faire le calcul ici, mais on le voit bien 
+La puissance moyenne, $\bar{P}$, est facilement calculée (on va le faire le calcul ici, mais on le voit bien 
 sur l'illustration de @fig:puissance_alternatif) et est donnée par
 $$
-\bar P=\frac{1}{2}I_0^2R=\frac{1}{2}\frac{V_0^2}{R}.
+\bar{P}=\frac{1}{2}I_0^2R=\frac{1}{2}\frac{V_0^2}{R}.
 $$
 
 ---
@@ -453,7 +453,7 @@ De l'équation
 \end{equation*}
 on obtient
 \begin{equation*}
-I_\mathrm{rms}=\frac{\bar P}{V_\mathrm{rms}}=\frac{1000}{230}=4.35\A.
+I_\mathrm{rms}=\frac{\bar{P}}{V_\mathrm{rms}}=\frac{1000}{230}=4.35\A.
 \end{equation*}
 Ainsi le courant de pic est donné par
 \begin{equation*}
@@ -463,9 +463,9 @@ La résistance est donc donnée par
 \begin{equation*}
 R=\frac{V_\mathrm{rms}}{I_\mathrm{rms}}=\frac{V_0}{I_0}=\frac{325}{6.15}=52.8\Omega.
 \end{equation*}
-Aux USA, la tension étant de $120\V$, on obtient pour la puissance $\bar P$ disponible
+Aux USA, la tension étant de $120\V$, on obtient pour la puissance $\bar{P}$ disponible
 \begin{equation*}
-\bar P=\frac{V_\mathrm{rms}^2}{R}=\frac{120^2}{52.8}=272\W.
+\bar{P}=\frac{V_\mathrm{rms}^2}{R}=\frac{120^2}{52.8}=272\W.
 \end{equation*}
 On voit qu'on risque d'avoir un problème pour faire fonctionner notre sèche cheveux à plein régime. A l'inverse un sèche-cheveux américain va très probablement griller si on le branche en Europe.
 

06_cicruits_electriques.md

@@ -180,8 +180,74 @@ Deux résistances de $100\Omega$ sont connectées en parallèle ou en série sur
 
 ---
 
+On peut faire évidemment beaucoup plus compliqué. Et on va le faire immédiatement.
+
+---
+
+Exemple (Plein de résistances) #
+
+Soit le circuit de la @fig:complex_res. Si la tension $V=9\V$. 
+
+1. Quel est le courant sort de la source de tension? $
+2. Quel est le courant dans la résistance de $6\Omega$?
+
+![Un certain nombre de résitances...](figs/complex_res.svg){#fig:complex_res width=50%}
+
+---
+
+La méthodologie globale pour résoudre ce genre d'exercice, est de déterminer la résistance équivalente du circuit en le décomposant en sous-circuits série/parallèle puis en utilisant la loi d'Ohm.
+
+---
+
+Solution (Plein de résistances) #
+
+1. On commence par calculer la résistance équivalent des résistances $8\Omega$ et $4\Omega$. Ces résistances sont en parallèle. On a donc
+$$
+\frac{1}{R_\mathrm{eq1}}=\frac{1}{8}+\frac{1}{4}=\frac{3}{8}\frac{1}{\Omega}.
+$$
+On a donc que $R_\mathrm{eq1}=2.7\Omega$ (voir @fig:complex_res_1)
+
+![Un certain nombre de résitances, moins 1...](figs/complex_res_1.svg){#fig:complex_res_1 width=50%}
+
+Puis vient le temps de calculer la résistance équivalente des résistances de $6\Omega$ et $2.7\Omega$. Ces résistances sont en série, donc il suffit de les sommer et on a $R_\mathrm{eq2}=6+2.7=8.7\Omega$ (voir @fig:complex_res_2)
+
+![Un certain nombre de résitances, moins 2...](figs/complex_res_2.svg){#fig:complex_res_2 width=50%}
+
+On a à présent les résistances de $10\Omega$ et $8.7\Omega$ qui sont en parallèle et on calcule
+$$
+\frac{1}{R_\mathrm{eq3}}=\frac{1}{10}+\frac{1}{8.7}=0.21\frac{1}{\Omega},
+$$
+et il vient $R_\mathrm{eq3}=4.8\Omega$ (voir @fig:complex_res_3).
+
+![Un certain nombre de résitances, moins 3...](figs/complex_res_3.svg){#fig:complex_res_3 width=50%}
+
+Finalement, on a trois résistances en série qu'on peut simplement sommer et il vient
+$$
+R_\mathrm{eq}=0.5+5+4.8=10.3\Omega.
+$$
+
+On peut à présent calculer le courant dans le circuit avec la loi d'Ohm
+$$
+V=R_\mathrm{eq}I\Leftrightarrow I=\frac{9}{10.3}=0.87\A.
+$$
+
+2. Le courant dans la résistance de $6\Omega$ doit être la même que dans la branche correspondant à $\R_\mathrm{eq2}$. On doit encore calculer la tension $V_2$ aux bornes du sous-circuit du haut. On a que
+$$
+V=V_1+V_2+_3\Leftrightarrow V_2=V-V_1-V_3=V-(0.5+5)\cdot 0.87=4.3\V.
+$$
+Et finalement que 
+$$
+I=\frac{4.3}{8.7}=0.48\A.
+$$
+
+---
+
+
+
 Pour des circuits beaucoup plus compliqués que ceux vus ici, il faut noter qu'il y a un ensemble
-de règles qu'on peut appliquer pour déterminer le courant dans un circuit: les **lois de Kirchoff**. On ne va pas les lister ou les étudier ici.
+de règles qu'on peut appliquer pour déterminer le courant dans un circuit: les **lois de Kirchoff**.
+
+
 
 ## Les circuits contenant des capacités en parallèle ou en série
 
@@ -366,4 +432,31 @@ plt.xlabel('temps [s]')
 plt.ylabel('tension [V]')
 plt.title("Tension du condensateur")
 plt.legend()
-```
\ No newline at end of file
+```
+
+---
+
+Exemple (Décharge du condensateur) #
+
+Si un condensateur de $C=10\mu \mathrm{F}$ est connectée à une résistance de $R=100\Omega$ et possède une tension de $V_0$. Quand le condensateur se décharge, combien de temps faut-il pour que sa tension tombe à $10\%$ de sa valeur originale?
+
+---
+
+---
+
+Solution (Décharge du condensateur) #
+
+Comme la tension d'un condensateur se diminue comme
+\begin{equation*}
+V_C=V_0e^{-t/(RC)},
+\end{equation*}
+et qu'on veut connaître le temps nécessaire pour que $V_C=0.1V_0$, on peut écrire l'équation
+\begin{align*}
+0.1V_0&=V_0e^{-t/(RC)},\\
+0.1&=e^{-t/(RC)},\\
+0.1&=e^{-t/(10^{-5}\cdot 100)},\\
+\ln{0.1}&=-10^3t,\\
+t&=0.0023\s.
+\end{align*}
+
+---

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