Exercice Corrige Electrocinetique Site

Thus ( B = 9.93 \ \textV ).

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

Initial condition at ( t = t_1^+ ): ( V_C(t_1) = 9.93 \text V ) (continuity of capacitor voltage). exercice corrige electrocinetique

[ 0 = R i + V_C \quad \Rightarrow \quad RC \fracdV_Cdt + V_C = 0 ]

[ 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: Thus ( B = 9

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

Solution:

[ RC \fracdV_Cdt + V_C = E ]

[ E = V_R(t) + V_C(t) = R i(t) + V_C(t) ] The source is disconnected, and the capacitor discharges

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