Capacitor Charging/Discharging

 Purpose

         Capacitor are components that enable the storage of electrical energy via electric field. The purpose of this experiment was to design a capacitor circuit to store energy of 2.5mJ, charge in 20s and discharge in 2s.

Pre-lab

       As the output voltage is 9V, to store 2.5mJ energy, the capacitor should 62 microF.

       To charge in 20s, the time constant should one fifth of the total charge time. So R_charge should be  66k Ohms.

       To discharge in 2s, the time constant should one fifth of the total charge time. So R_discharge should be 6.48k Ohms.

Experiment





Data and Analysis

 

     The curve measured by "Logger Pro" during charging: 

 

    

    The capacitor was charged around 18 seconds, which is closed to 20 seconds. Final voltage was measured to be around 8v.

 

    According to the data,

 

    R_leak = R_charge /(V_s/V_final-1) = 264k Ohms

 

     The curve measured by "Logger Pro" during discharging:

 

 

  capacitor was discharged around 2 seconds.

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