Capacitor Charging and Discharging

 Introduction
       
          Capacitor are components that enable the storage of electrical energy via electric field. In this lab, we will plot the graph of the charging and the discharging of capacitors. In this lab we will design, build and test a charge/discharge system that utilizes a 10V DC power supply, employs a charging interval of about 20 seconds with a resulting energy of 2.5 mJ in 2 seconds.

Experimental

The first task that is done is to do the calculation to obtain the required Capacitance
E = 0.5 Cq^2
C = 2E / q^2 = 61.7 µF

Since we don't have capacitor with such values, 66µF will be used.

Assuming an ideal capacitance, the value of the charging resistance can be calculated

5* time constant = 20s
time constant = 4


RC = 4
R = 60.6 k_ohm


From the value above, we can calculate peak current and peak power


I = 9V/606kohms = 0.149mA
P = 0.149 (9) = 1.341mW


We then calculate the discharge resistance the same way we calculate the charging resistance
R = 6.06kohms


Again the discharge peak current and peak power is calculated


I = 9/6.06 = 1.49mA
P= 1.49(9) = 13.4mW

Once everything is calculated, the circuit was built as follow:

The capacitor put in parallel

2 variable boxes one for charging and the other for discharging

graph of charging and discharging

By looking at the graph above, when its charging, the V final is 8.45V

The resistor leakage is calculated to be 1.03Mohms

Further it discharge in 2.5 seconds

Questions


Charging:
Vth = [Rl/(R+Rl)] Vs = 8.50V

Rth = (Rl*R)/(Rl+R) = 57.2kohms

discharging:

Vth = 8.50V

Rth = 6.025kohms


Vc(τ) = .6321*(8.45V)
Vc(τ) = 5.34V
 t = 21.05s
τcharge = 21.05s
Rth = τcharge/C
Rth = 57.8kΩ