Capacitor Charging/Discharging

We learned that capacitor is the element that store energy in electrical field. We are able to set the time to charge and discharge by using the desire value of resistors

In this experiment, we will consider charging and discharging a capacitor

we will have a circuit that like the one below
 

In real life, the capacitor will have a 'leak resistance' that is parallel to the capacitor due to the imperfectness of the world.
We will have two circuits. One for charging the capacitor and one for discharging the capacitor
First, we will need to calculate the expressions for the Thevenin voltage and resistance for the circuits

Charging:
V_Th=Vs* R_leak /(R_leak+R_charge)
R_TH=((R_charge+R_Leak)/R_charge*R_leak))^-1

Discharging:
V_th=Vcap*(R_leak/Rdischarge*Rleak)
R_TH=R_leak*R_discharge/(R_leak+R_discharge)

Next, we need to build and the circuit with a power supply of 9V, and it needs to employ a charging interval of about 20s with a energy of 2.5mJ and discharges with the same amount of energy in 2S

The require capacitance will be
0.5*CV^2=2.5*10^-3 with V=9V
we get  C= 61.7 uF

charging resistance
4=R_charge*C where C= 61.7uF
R_charge=64.8kohm

peak current
I=V*e^(-t/4) /R_charge
I_peak=0.139mA
P=1.251mW, and the resistance box is able to handle the power

Again, we need to find the R_discharge value
since we know the time requires is 2s
and tau=2/5
2/5 = R_discharge*C
R_discharge=6.47 kohm

for the peak current:
I_max=1.39mA
P_max=1.39mA*9=12.5mW, and the box is able to handle the power

Next, we obtain the desire value of elements
since there is no capacitor has the exact value we want, we combine three 22uF capacitors in parallel to a close value we want, which we measured is 63.6uF

Next, we assemble the elements on a bread board with multimeter and resistor box hooked up to the circuits

We were able to use logger pro to measure the change in voltage on the capacitor by connecting a probe to the capacitor

This is our result of charging capacitor

This is our result of discharging capacitor

The V_source is 9.01V

Charging:
The final voltage is 8.562V=V_source*R_leak/(R_leak+64.8K)
solve for R_leak, we obtain 1.27Mohm
The V_TH seen by the capacitor is V_th=V_source*(R_leak/R_charge*R_leak)=9.01*(1.27M/(64.8k*1.27M)=8.57 which is voltage seen by the capacitor

Discharging:
V_th=V_source*R_leak/(R_leak+R_discharge)=9.01*1.27M/(1.27M+6.47K)
=8.96V




Follow-up Questions
1. Calculate the Thevenin equivalent voltage and resistance values "seen" by the capacitor during charging

V_TH=8.57V
R_TH= (1.27M+64.8K)/(1.27M*64.8K)=61654ohm

2. Calculate the Thevenin equivalent voltage and resistance values "seen" by the capapcitor during discharging

V_TH=8.96V
R_TH=(1.27M+6.47K)/(1.27M*6.47K)=6437ohm

3. At t equal to one time constant, e^(-t/tau)=e^-1=0.3679: So, when t=tau_c during "charging" the voltage should equal 0.6321*(final voltage). Use this to estimate the value of tau_c from your "charging" waveform. Compute the corresponding value o resistance(this value should be close to the value found in question1)

8.562*0.6321=5.41V
5.41=-10.56e^(-0.2293t)+8.533
t=5.313
5.313=R*C
R=86K ohm

Practical Question
Given: 160MJ , 15k V
1. Determine the required equivalent capacitance
.5*C(15000)^2=160*10^6
C=1.42F

2.If we assume that the capacitance will be achieved in the manner shown below, via a series and parallel combination, calculate the required value of individual capacitance C

1.42=(C^2/2C)*4
C=0.71F