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