Chapitre 4¶

Exercice à accomplir 5 :

P_{g}=P_{i}+P_{j} \Leftrightarrow E_{i}=\frac{d}{d t}\left(\frac{1}{2} c u_{c}^{2}\right)+r i^{2}

De plus d’après les questions précédentes $u_{C}(t)=E\left(1-e^{-i / c}\right)$

D’après la loi de comportement

\begin{aligned} i & =c \frac{d u c}{d t} \\ \Rightarrow i & =c \times \frac{E}{\tau} e^{-\frac{E}{\tau} u} \frac{-t}{\tau} \\ \Rightarrow i(t) & =\frac{E}{\tau} e^{\frac{-t}{\tau}} \\ -\varepsilon_{g} & =\int_{0}^{+\infty} P_{g} d t=\int_{0}^{+\infty} \frac{\varepsilon^{2}}{r} e^{-\frac{t}{\tau}} d t \\ \Rightarrow \varepsilon_{g} & =\frac{E^{2}}{r}\left[-\tau e^{-t / \tau}\right]_{0}^{\infty} \\ & =\frac{E^{2}}{r}(0-(-\tau)) \\ & =\frac{E^{2}}{r} \tau=c E^{2} \end{aligned}

\begin{aligned} -\varepsilon j & =\int_{0}^{+\infty} p_{j} d t \\ & =\int_{0}^{+\infty},\left(\frac{E}{r_{2}} e^{-\frac{1}{t}}\right)^{2} d t \\ & =\int_{0}^{+\infty} \frac{e^{2}}{r} e^{-\frac{2 t}{t}} \end{aligned}

\begin{aligned} & =\frac{E^{2}}{r}\left[-\frac{\bar{\tau}}{2} e^{-\frac{2 t}{\tau}}\right]_{0}^{\infty} \\ & =\frac{E^{2}}{r}\left(0-\left(-\frac{\tau}{2}\right)\right)=\frac{E^{2}}{r} \times \frac{\tau}{2}=\frac{c E^{2}}{2} \\ & \Rightarrow \frac{c}{\varepsilon}-\varepsilon_{c}=\int_{0}^{+\infty} p_{c} d t=\int_{0}^{+\infty} \frac{d}{d t}\left(\frac{1}{2} c u^{2}\right) d t \\ & \varepsilon_{c}=\left[\frac{1}{2} c u_{c}^{2}\right]_{0}^{+\infty} \\ & =\frac{1}{2} c E^{2}-0 \\ & =\frac{c E^{2}}{2} \end{aligned}