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.
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.
Answers
Damon
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
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