With ( i(t) = C \fracdV_Cdt ), we get:
Initial condition ( V_C(0) = 0 ): [ 0 = E + A \quad \Rightarrow \quad A = -E ] exercice corrige electrocinetique
[ 0 = R i + V_C \quad \Rightarrow \quad RC \fracdV_Cdt + V_C = 0 ] With ( i(t) = C \fracdV_Cdt ), we
[ RC \fracdV_Cdt + V_C = E ]
[ 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 ):
Thus ( B = 9.93 \ \textV ).