par SoS-Math(33) » sam. 25 avr. 2020 11:17
Bonjour,
il faut utiliser \(A=\large \epsilon lc\)
\(\frac{\Large -dA}{\Large dt} = \frac{\Large-d(\Large \epsilon lc)}{\Large dt}\)
\(\frac{\Large -dA}{\Large dt} = \large \epsilon l\frac{\Large-dc}{\Large dt}\)
d'où \(\frac{\Large-dc}{\Large dt} = \frac{\large1}{\Large \epsilon l}\frac{\Large -dA}{\Large dt} \)
\(\frac{\Large-dc}{\Large dt} = \frac{\large1}{\Large \epsilon l}\frac{\Large A}{\Large k_\alpha + k_\beta A} \)
\(\frac{\Large-dc}{\Large dt} = \frac{\frac{\Large A}{\huge \epsilon l}}{\Large k_\alpha + k_\beta \Large \epsilon lc} \)
\(\frac{\Large-dc}{\Large dt} = \frac{\Large c}{\Large k_\alpha + k_\beta \Large \epsilon lc} \)
SoSmath
Bonjour,
il faut utiliser [TeX]A=\large \epsilon lc[/TeX]
[TeX]\frac{\Large -dA}{\Large dt} = \frac{\Large-d(\Large \epsilon lc)}{\Large dt}[/TeX]
[TeX]\frac{\Large -dA}{\Large dt} = \large \epsilon l\frac{\Large-dc}{\Large dt}[/TeX]
d'où [TeX]\frac{\Large-dc}{\Large dt} = \frac{\large1}{\Large \epsilon l}\frac{\Large -dA}{\Large dt} [/TeX]
[TeX]\frac{\Large-dc}{\Large dt} = \frac{\large1}{\Large \epsilon l}\frac{\Large A}{\Large k_\alpha + k_\beta A} [/TeX]
[TeX]\frac{\Large-dc}{\Large dt} = \frac{\frac{\Large A}{\huge \epsilon l}}{\Large k_\alpha + k_\beta \Large \epsilon lc} [/TeX]
[TeX]\frac{\Large-dc}{\Large dt} = \frac{\Large c}{\Large k_\alpha + k_\beta \Large \epsilon lc} [/TeX]
SoSmath