Suppose a capacitor of C=2.0 uF with no excess charge on it is suddenly connected in series with a resistor of R=447 ohms , and an ideal battery of voltage V=25 V . How long will the capacitor take to gain 80% of the charge it will eventually have when current stops flowing through the battery?

Answer 1

C = Q/Vc
so Q = CVc = integral i dt
so Vc = (1/C) int i dt
Vc = 0 at start

25 = i R + Vc

25 = i R + (1/C) int i dt

0 = R di/dt + i/C

di/dt = -i/(RC)
di/i =-dt/(RC)
ln i = -t/(RC)
i = e^[-t/(RC)]
when t = 0, Vc = 0 and i = V/R
i = (V/R) e^[-t/(RC)]
Remember from above
Vc =(1/C) int i dt
so
Vc = (V/RC) int e^[-t/(RC)] dt
Vc = (V/RC) (-RC) e^[-t/(RC)] + c
Vc = -V e^[-t/(RC)] + c
at t = 0 Vcapacitor = 0 so C = V
Vc = V (1 - e^[-t/(RC)] )
.8 = (1 -e^[-t/(RC)])
.2 = e^[-t/(RC)]
-1.61 = -t/(RC)
put in R and C to get t