Charging/Discharing a Capacitor

PROBLEM
Design, build, and test a charge/discharge system that utilizes a 9V DC power supply, employs a charging interval of about 20s with stored energy of 2.5mJ, and then discharges that energy energy in 2s.

METHOD
We consider charging and discharging a capacitor. We will toggle a cable connected to the capacitor's "+" terminal from the CHARGE position to the DISCHARGE position. In the CHARGE position, as long as Vs > Vcap, energy is transferred to the capacitor. In the DISCHARGE position, the energy is absorbed by the resistance Rdischarge. In either case, C, Rcharge, and Rdischarge control the rate at which energy is transferred.
Additionally, we will add a leak resistance in parallel with the capacitor to model a "real" storage capacitor and not the ideal case. A circuit is constructed similar to the figure. We compute Thevenin voltage and resistance as "seen" by the capacitor for each case.

"Real" capacitor circuit

Thevenin charge circuit

Thevenin discharge circuit

We first compute the required capacitance. W=1/2CV^2 >> C=2W/V^2 >> 2*2.5/81 = 62 uF
Next, we compute the charging resistance. T=capacitive time constant. T=charge_time/num_charge_intervals =RC >> 20/5=62R >> R=4/62uF >> Rcharge = 64.8 kOhm
Now we can compute the peak current in the charge resistance and the peak power. Where, i=(Vs/R)*e^(-t/T) = 0.936 uA  & P=Vs*i=9*0.936=8.4 uW

We use a similar technique to compute the discharge resistance. T=capacitive time constant. T=charge_time/num_charge_intervals =RC >>2/5=R62uF >> Rdischarge = 6.48 kOhm.
We calculate the peak discharge current and power. V=IR>> I =V/R >> 9/6.48k = 1.389 mA, P=R*I^2 >> (6.48k)*(1.389)^2 >> P= 0.0125 W

charging

charging & discharging