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| -rw-r--r-- | main.tex | 22 |
1 files changed, 11 insertions, 11 deletions
@@ -117,11 +117,11 @@ \subsection{Mise en équation par la méthode des n\oe uds} \begin{gather} - I_{\textrm{total}} = I_1 + I_2 + I_3 \label{eq:sumOfI} \\ - \frac{V_{out}}{R_{16}} = \frac{V_1-V_{out}}{R_{13}} + - \frac{V_2-V_{out}}{R_{14}} + \frac{V_3-V_{out}}{R_{15}} \\ - V_{out} = \frac{\frac{V_1}{R_{13}}+\frac{V_2}{R_{14}}+\frac{V_3}{R_{15}}} - {\frac{1}{R_{13}}+\frac{1}{R_{14}}+\frac{1}{R_{15}}+\frac{1}{R_{16}}} \label{eq:vOut-symb} + I_{\textrm{total}} = I_1 + I_2 + I_3 \label{eq:sumOfI} \\ + \frac{V_{out}}{R_{16}} = \frac{V_1-V_{out}}{R_{13}} + + \frac{V_2-V_{out}}{R_{14}} + \frac{V_3-V_{out}}{R_{15}} \\ + V_{out} = \frac{\frac{V_1}{R_{13}}+\frac{V_2}{R_{14}}+\frac{V_3}{R_{15}}} + {\frac{1}{R_{13}}+\frac{1}{R_{14}}+\frac{1}{R_{15}}+\frac{1}{R_{16}}} \label{eq:vOut-symb} \end{gather} @@ -165,12 +165,12 @@ Cela permet de trouver une valeur de $V_{out}$ en fonction des diffentes combina \begin{gather} I_{\textrm{out}} = \SI{0}{\ampere},\quad R_{13} = \SI{10}{\kilo\ohm},\quad R_{14} = \SI{20}{\kilo\ohm}, - \quad R_{15} = \SI{40}{\kilo\ohm} \\ - V_{out} = \frac{\frac{V_1}{\SI{10}{\kilo\ohm}}+ - \frac{V_2}{\SI{20}{\kilo\ohm}}+\frac{V_3}{\SI{40}{\kilo\ohm}}} - {\frac{1}{\SI{10}{\kilo\ohm}}+\frac{1}{\SI{20}{\kilo\ohm}}+ - \frac{1}{\SI{40}{\kilo\ohm}}+\frac{1}{\SI{80}{\kilo\ohm}}} \\ - V_{out} = \frac{8V_1}{15}+\frac{4V_2}{15}+\frac{2V_3}{15} \label{eq:vOut} + \quad R_{15} = \SI{40}{\kilo\ohm} \\ + V_{out} = \frac{\frac{V_1}{\SI{10}{\kilo\ohm}}+ + \frac{V_2}{\SI{20}{\kilo\ohm}}+\frac{V_3}{\SI{40}{\kilo\ohm}}} + {\frac{1}{\SI{10}{\kilo\ohm}}+\frac{1}{\SI{20}{\kilo\ohm}}+ + \frac{1}{\SI{40}{\kilo\ohm}}+\frac{1}{\SI{80}{\kilo\ohm}}} \\ + V_{out} = \frac{8V_1}{15}+\frac{4V_2}{15}+\frac{2V_3}{15} \label{eq:vOut} \end{gather} |