In the circuit shown in the figure each capacitor
initially has a charge of
magnitude 3.60nC on its plates. After the switch S is closed, what will be the
current in the circuit at the instant that the capacitors
have lost 80.0% of their initial stored energy?
the shown 3 capacitors are 10, 15, and 20 pF and the one resistor has 25ohms - all in series with one switch S.
E = 1/2Q^2/c = QV/2
V_c(t)=Q/c * (1-e^(-t/RC))
i = Q/RC * e^(-t/RC)
I have since solved this but I would really appreciate it if I could get this checked for correctness.