Capacitor

Statement of Purpose
To see the affect of a discharging capacitor.

Procedure

First, we know that it takes 5 time constants to discharge a capacitor to near 0 levels, which depends on the resistance and capacitance of the RC circuit.

So to charge a capacitor in 20 seconds of 2.5mJ with a 9V power supply we make the following calculations.
The energy of a capacitor is U=1/2CV^2, so C=0.0617 mF
The time constant is 5T =RC -> R=64.8 kΩ
The power generated in the 1W resistor box is 1.25 mW, which will not fry the resistor box. 
The following charge curve is generated by LoggerPro,

To discharge a capacitor in 2 seconds of 2.5mJ, we make the following calculations.
The time constant is 5T =RC -> R=6.48 kΩ
The power generated in the 1W resistor box is 12.42 mW, which will not fry the resistor box. 
The following charge curve is generated by LoggerPro,

Questions:
Using the Thevenin equations,
1. R_cth = 59.4 kΩ, V_cth = 8.25V
 2. R_dth = 6.42 kΩ, V_dth = 0V

3. 0.6321*V_f = 5.215 V
5.215 V is around 4.6s
tau_c = RC = 4.6s
R = 4.6/C = 74.55 kΩ
Error = 15%

Practical Question:
1. U = (1/2)CV^2
C_eq = 2*U/V^2 = 2*160*10^6/(15*10^3)^2 = 1.42 F

2. 1/2C + 1/2C + 1/2C + 1/2C = 2C = C_eq
C = 1/2C_eq = 0.71 F


Conclusion
The time to charge and discharge a capacitor is fairly simple. What matters practically is the time constant, resistors, and power loading.