Question

A fully charged 90-µF capacitor is discharged through a 60-Ω 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!

Answer 1

Q(t)=Q₀•exp(-t/RC)
E~Q² =>
E=E₀•exp(-2•t/RC)
100-80=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.

Answer 2

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

Answer 3

sqrt .2 not .8
so
.04 = e^-(1/rc) t
-3.22 = -(1/rc) t
t = 3.22 * 90 * 10^-6 * 60

Answer 4

thank you!