Lab 11: Capacitor

Introduction:
In this experiment, we are going to study the capacitors. Capacitors are component that can store energy via the electric field, and we can rapidly extract the energy from capacitors. The purpose of this experiment is to learn about charging and discharging a capacitor. The picture shown below illustrates how we will use simple circuit to analysis charging and discharging a capacitor.

Procedure:

Task 1: Calculate expressions for the Thevenin voltage and resistance for the charging and discharging circuits:

Charging                                              Discharging

R_th =(R_cR_leak)/(R­_c + R_leak)                 R_th =(R_disR_leak)/(R­_dis + R_leak)

V_th = V_sR_leak / (R_c + R_leak)                 V_th = V_c

Step1:  Design a charge and discharge system that utilizes a 9V DC power supply, is charging about 20s with a resulting stored energy of 2.5mJ, and then discharges that 2.5mJ in 2s.

E=V²C/2 = 2.5mJ

C= 2(2.5mJ)/9² =62µF.

Step2:  Estimate the value of charging resistance:

Charging time is 5s. 

R_c = 64.5kΩ.

Calculate the peak charge current and the peak power:

I_peak  =V/R = (9V)/(64.5kΩ) = 0.14mA

P= I²R = (0.14mA)²(64.5kΩ) = 1.26 mW 

Step 3:  Discharging time is 10 times less than charging times, so the R_dis =R_c/10 = 6.45kΩ.

Caculate the peak discharge current and the peak power: 

I_peak  =V/R = (9V)/(6.45kΩ) = 1.4mA

P= I²R = (1.4mA)²(6.45kΩ) = 12.6 mW 

Build the circuit:

Because the Voltage meter can measure maximum voltage of 7, we set the power supply to be 6V.

Measured V_f­inal  = 5.707 V

Solve for R_{leak :}

V_final  = V_s (R_leak)/(R_c + R_leak) ->    R_leak = V_final V_c /(V_s – V_final) = 1.256 MΩ

The graph of charging capacitor:

The charging process takes more than 20s to reach to 5.707V.

Discharging process:

The capacitor fully discharges about 2s.

Questions:

1 - Calculate the Thevenin equivalence voltage and resistance values seen by the capacitor during charging:

R_th = 61.4 kΩ

V_th  = 5.707 V

2 - Calculate the Thevenin equivalence voltage and resistance values seen by the capacitor during discharging:

R_th = 6.42 kΩ

V_th­ = 5.707 V

3 - When t equal one time constant, e^-t/T_­ = e^-1 =0.3679. The charging voltage equal  V_f(1-0.3679) = 0.6321*V_final = 0.6321*5.707V = 3.607V
Look at the graph, time for charging voltage to reach 3.607V is about 4s:
T= t = 4s =RC -> R= 4/C = 4/62µF = 64.5 kΩ

Practice Questions:
We want to scale our result to the rail gun problem we worked in previous exercise. The rail gun requires a stored energy of 16MJ, and the capacitor charging is 15kV.

1- Find the required equivalence capacitance:

E=V²C/2 = 160MJ

C= 2(160 MJ)/(15 kV)² =1.4F

2- If the capacitance will be achieved in the manner shown below, the required value of individual capacitance C= 0.7 F