The potential difference between the plates of a leaky capacitor, C = 3.00 microfarads., drops from 6.80 V. to 3.40 V. in 1.00 s. What is the equivalent resistance (in ohms) between the capacitor plates?

Answer 1

c = q/v
q = c v
dq/dt = -i = -c dv/dt
v = i r
i = v/r
so
v/r = -c dv/dt
dv/dt = -v (1/rc)
let v = Vi e^-kt
then dv/dt = -k Vi e^-kt
-k Vi e^-kt = (-Vi e^-kt)(1/rc)
so
k = 1/rc
and
v = Vi e^-t/rc = 6.8 e^-t/(3*10^-6 r)
at t = 1
3.4 = 6.8 e^-10^6/3r
ln (3.4/6.8) = -10^6/3r
- .693 = -10^6 / 3r
r = 10^6 / 2.08 = 4.8*10^5 Ohms

Answer 2

wow thank you! book was fairly vague on this subject