Question

A charged capacitor,C=10e-6 F, with an initial stored energy of 3.5J, is discharged across a resistor, R. The charge on the capacitor drops to 40% of its maximum value in time, t=8.2s, from the instant the switch is closed (t=0).
Find the current through the resistor at a time t=10s.
How much thermal energy is dissipated across the resistor after two time constants?
Determine the electrostatic energy on the capacitor after two time constants.

Answer 1

C = Q/V
V = (1/C) integral i dt
dV/dt = (1/C) i
but
i = -V/R
so
dV/dt = (1/C) -V/R
or
dV/V = (-1/RC) dt (I bet you knew that)
so
ln V = (-1/RC) t
V = Vi e^(-t/RC) (which you also know)
and Q = CV = CVi e^(-t/RC)
at t = 8.2
Q(8.2) = Cv = .4 CVi = CVi e^(-8.2/RC)
so
.4 = e^(8.2/{R*10^-5} )
ln .4 = -8.2*10^5/R
solve for R

but how much energy is stored at V ?
E = (1/2) C V^2
that should get you started