Posted by **Bruce** on Wednesday, October 24, 2012 at 1:38pm.

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!

- Physics -
**Elena**, Wednesday, October 24, 2012 at 1:59pm
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.

- Physics -
**Damon**, Wednesday, October 24, 2012 at 2:02pm
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

- misread -
**Damon**, Wednesday, October 24, 2012 at 2:07pm
sqrt .2 not .8

so

.04 = e^-(1/rc) t

-3.22 = -(1/rc) t

t = 3.22 * 90 * 10^-6 * 60

- Physics -
**Bruce**, Wednesday, October 24, 2012 at 11:02pm
thank you!

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