Résumé : stockage d’énergie

\[ \begin{gathered} \text{Condensateur :} \\[6pt] C = \varepsilon_0 \cdot \varepsilon_r \cdot \frac{A}{d} \text{ avec } \varepsilon_0 = 8{,}854 \cdot 10^{-12}\,\frac{\mathrm{As}}{\mathrm{Vm}} \\[6pt] u_C(t) = U_{C0} + \frac{1}{C} \cdot \int i_C(t)\,dt \\[6pt] i_C(t) = C \cdot \frac{du_C(t)}{dt} \\[6pt] W_C = \frac{1}{2} C \cdot U_C^2 \\[4pt] Q = C \cdot U \\[4pt] \tau = R \cdot C \\[6pt] \text{Charge du condensateur : } u_C(t = \tau) = U_{C0} + 0{,}63 \cdot \Delta U \\ u_C(t = 5\tau) = U_{C0} + 0{,}99 \cdot \Delta U \\[4pt] \text{Décharge du condensateur : } u_C(t = \tau) = U_{C0} - 0{,}63 \cdot \Delta U \\ u_C(t = 5\tau) = U_{C0} - 0{,}99 \cdot \Delta U \end{gathered} \]

\[ \begin{gathered} \text{Bobine :} \\[6pt] L = \mu_0 \cdot \mu_r \cdot N^2 \cdot \frac{A}{l} \text{ avec } \mu_0 = 4\pi \cdot 10^{-7}\,\frac{\mathrm{Vs}}{\mathrm{Am}} \\[6pt] i_L(t) = I_{L0} + \frac{1}{L} \cdot \int u_L(t)\,dt \\[6pt] u_L(t) = L \cdot \frac{di_L(t)}{dt} \\[6pt] W_L = \frac{1}{2} L \cdot I_L^2 \\[6pt] \tau = \frac{L}{R} \end{gathered} \]