Initial condition ( V_C(0) = 0 ): [ 0 = E + A \quad \Rightarrow \quad A = -E ]
[ RC \fracdV_Cdt + V_C = E ]
Solution:
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:
[ V_C(t) = E + A e^-t/RC ]
Initial condition ( V_C(0) = 0 ): [ 0 = E + A \quad \Rightarrow \quad A = -E ]
[ RC \fracdV_Cdt + V_C = E ]
Solution:
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
[ V_C(t) = E + A e^-t/RC ]
