Capacitor Charging and Discharging

We just learned that capacitors are electrical components that can store energy. They have the advantage over batteries in a way that they allow rapid extraction of energy. Thus making them very useful in places that a big amount of energy is needed to be delivered in a short amount of time.
To understand how a capacitor works first we went over a theoretical model of a capacitor circuit and then we conduct an experiment observing the capacitor behavior.
In the following circuit when switch is in the left position the capacitor stores energy delivered by the battery. later after switching to the right position, the capacitor delivers its energy to the resistor.

Capacitor is Charging 

Capacitor is Discharging

In first position the equation governing voltages is:

And the equation for the second part is:

solutions for these two equations are as follow:

In both cases current follows the ohm's law.

However these theory assume an ideal capacitor. A real capacitor loses energy due to contact losses and also leakage. In our experiment we ignore the contact losses since they are minimal but leakage we have to model as a parallel resistor.

As the equation show in order to control the charge or dischargee rate we have to change the values for Rcharge and Rdischarge. The bigger these two the bigger the rate will be.

For the lab we were asked to design a circuit in which a 9V DC voltage supply charges a capacitor in 20 seconds with total energy delivered of 2.5 mJ and then discharges completely in 2 seconds.

For our circuit if we assume an ideal capacitor, we can obtain the capacitance by using the energy formula. Assuming VC is 9V that means C=62.7μF.

It takes 4.6≈5 time constants for the capacitor to charge to 99% of 9V. Using this value and the equation for tau=4, Rcharge becomes 64.8 kΩ.Following the same procedure we obtain Rdischarge=6.48 kΩ.

Dividing VC by the Rcharge we see that the maximum current becomes 13.9 mA and Pmax=1.25 mW.

Repeating the same method yields that max discharge current  is 1.39 mA and 

Pmax=0.0125 W. Neither is big enough for the resistance box to handle.

Here is a picture of our setup:

In the actual experiment since we no longer could assume that the capacitor is ideal, we started with values calculated theoretically. However we observed that the charge and discharge times do not match the 20 and 2 seconds accordingly. As a result we recalculated the value for the capacitor but this time including the Rleak. C=64.5 μF. From there we calculated the values for Rleak and Vf which is the maximum charged voltage of the capacitor. 

Vf=8.8 V

Rleak=0.872 MΩ

with these values the charge and discharge times are 22.15s and 2.3 s respectively. 

The capacitor discharges in 2.3 s. This is due to the fact that we made some simplifying assumption like getting rid of contact losses and ignoring resistance in wires and the scope, also the fact that the meters might not be operating perfectly.

Vth for charging period is equal to Vc which was 8.8 V.

Vth in discharging mode is equal to 0 for absence of any power supply. and Rth is combination of Rleak and Rdischarge which is 6.432 kΩ.

As stated before charge and discharge times are 22.15s and 2.3 s respectively. Using these values we obtain tau to be 4.3. However in initial calculation we found that it takes 4.6 time steps to reach 99% of maximum charge but we rounded this number to 5. This might be the reason why we have a slightly different tau.

The above graphs are drawn in Freemat using the values measured in the circuit.