Physics
posted by Bruce on .
A fully charged 90µF capacitor is discharged through a 60Ω resistor. How long will it take for the capacitor to lose 80% of its initial energy?
I know energy is given by E(t)=q(t)^2, but beyond that I really don't know how to solve this. Any suggestions are appreciated.
Thank you!

Q(t)=Q₀•exp(t/RC)
E~Q² =>
E=E₀•exp(2•t/RC)
10080=20%
0.2•E₀=E₀•exp(2•t/R•C)
ln0.2 = 2•t/R•C
t=  ln0.2• R•C/2 = (1.6) •60•90•10⁻⁶/2 =0.0043 s. 
E = (1/2) c v^2
c = 90 * 10^6
q = c v
so dq/dt = c dv/dt = i
v = i r = r c dv/dt
so
dv/dt = (1/rc) v
v = Vi e^kt
dv/dt = k Vi e^kt = (1/rc)Vi e^kt
k = 1/rc
v = Vi e^t/rc
E = constant * v^2
so v^2 at t = .8 v^2 at t = 0
so
v at t = sqrt (.8) v at t = 0
v at t = .64 Vi
.64 = e^(1/rc)t
ln .64 = .446 = (1/rc) t
t = .446 * 90 * 10^6 * 60 
sqrt .2 not .8
so
.04 = e^(1/rc) t
3.22 = (1/rc) t
t = 3.22 * 90 * 10^6 * 60 
thank you!