Exercice Corrige - Electrocinetique

[ RC \fracdV_Cdt + V_C = E ]

Thus ( B = 9.93 \ \textV ).

[ V_C(t) = E + A e^-t/RC ]

With ( i(t) = C \fracdV_Cdt ), we get:

[ V_C(t_1) = 10 \left(1 - e^-5\right) \approx 10 (1 - 0.0067) \approx 9.93 \ \textV ] The capacitor is almost fully charged. The source is disconnected, and the capacitor discharges through ( R ). The differential equation becomes: exercice corrige electrocinetique

[ \boxed\tau = 100 \ \textms ] At ( t_1 = 0.5 \ \texts ): [ RC \fracdV_Cdt + V_C = E ] Thus ( B = 9