Experiment 6

Charging and Discharging of a Capacitor




The objective of this experiment is to verify the exponential behavior of capacitors during charging and discharging processes.




A computer with the Internet connection, a calculator, paper, and pencil




A capacitor is a passive electric device that stores electric energy.  A parallel-plates capacitor is made of two parallel metallic surfaces, each of area A, separated by an insulation layer of thickness d, and it has a capacity of

where C is the capacity in Farads, A the area of each plate in m2,  d the insulation (dielectric) thickness in (m), and εo the permittivity of free space (vacuum) for electric field propagation expressed in F/m that reads Farads/meter.


The factor, κ , pronounced kappa denotes the dielectric constant, and depends on the material of the insulation layer.  The capacitance C does not depend on the material of the plates.  The dielectric constant εo is related to Coulomb's constantk by


Figure 1


   In Fig. 1, at t = 0, the capacitor is uncharged.  As soon as the key in the circuit is closed, electrons flow from the negative pole of the battery toward the lower plate of the capacitor.  They distribute over the lower plate, making it negative.  At the same time, the repelled free electrons of the upper plate flow toward the positive pole of the battery.  This causes the upper plate to become positively charged.  This process does not happen suddenly.  It takes some time.  The current is greatest to begin with, and decreases as charges accumulate on the plates.   At the beginning the capacitor is empty; therefore, the voltage across it is zero, but as more and more charges build up on its plates, its voltage keeps increasing.  The voltage across the capacitor VC  asymptotically approaches the battery voltage VBat .

   During the charging and discharging processes, the voltage across the capacitor and the current through it follow the following exponential equations:



Battery in Circuit


At t = 0,


VC = 0


IC = VB / R


As t → ∞

VC = VBat.


IC = 0




With battery removed, the initial capacitor voltage is  Vo= Qo/C making the initial current  Io = Vo/R


At t = 0,


VC = Qo /C


IC = Qo /(RC)


As t → ∞


VC = 0


IC = 0



It is a good  idea to examine the values in the third and fourth columns by once setting t = 0 and once t   in the appropriate equations.  Note that the charge-voltage formula for a capacitor is Q = CV.  These exponential variations will be observed in this experiment.



You may need to add the following Website to your Java exception list: http://www.phy.ntnu.edu.tw/.   To do this, follow the path (Windows operating system),

Start → All Programs → Java → Configure Java → Security (use High) → Edit Site List … → Add → Type in the site URL (http://www.phy.ntnu.edu.tw/).

You will notice circuits as shown in Figures 2 and 3 when you click on the following link. 


Figure 2


As shown above. the two-way switch must be put in position 1, for the charging process.  Doing this puts the battery in the circuit.


For discharging, as shown below, the two-way switch must be put in position 2 in which case the battery is excluded.



Figure 3


Now, click on http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=31.0 .


Start the applet, and with the mouse move the blue bar (on the left) up and down  to set a battery voltage.  You may then close the circuit for charging or discharging at the two-way switch.


Part I: Charging


1) As you see the resistance in the circuit is 100.0  that means R = 100,000Ω.  The capacitance is 100.0μF, or C = 100.0x10 -6F.  Calculate the time constant τ = RC of this circuit.  Record the value of the time constant, τ.


2) Now that you know τ = 10.00sec, go to the next step.


3) Click on the two-way switch a few times to observe the charging and discharging processes.  If you hold the left click down on any part of the figure, you can temporarily stop the experiment and take readings.  This eliminates the need for helpers as explained below.   You will see that at intervals of 5 seconds, you will need to take readings.   You can easily self-teach yourself the way it works.


4) Reset the circuit.  At the top, set the time to zero.  First choose a battery voltage of 10.00 volts by using the blue bar beside the battery.  Start the circuit for charging.  Have one helper to read VC , the voltage across the capacitor,  and the another helper read IC , the current through the capacitor.  While the circuit is on, VC can be read just beside the capacitor, and IC can be read at the top of the circuit below the moving yellow dots that represent the current.  Your job is to carefully announce "time" at equal intervals of 5 seconds.  Your helper job is to read the voltage or current as soon as he/she hears you announcing the word "time." and record it in the chart provided under "Data."


   Note that the values of VC and IC at the very beginning (t = 0) are known or can be easily calculated.  At t = 0, the capacitor is empty and therefore its voltage is zero.  This is already recorded in the chart.


You need to calculate IC at t = 0 and record it in the chart where it says VB /R .   At t = 0,  all of the battery voltage drop across the resistor.  Divide the battery voltage you have chosen by R = 100,000Ω to calculate the initial current (current att = 0). 


{If you choose VB = 6.00volts, for example, then the initial current will be (6/100,000)Amps or 60.0x10-6Amps, or 60.0μA.  This is just an example.} 


5) If you chose the battery voltage to be 10.00volts, then the initial current will be 100.0μA.  Calculate the initial current and record it.   When you are ready to turn the circuit on (t =0), you just say "Ready, set,  go."   Every 5s, when you say time, the helper that is watching the voltage, reads the voltage to his/her best judgment and writes it down, and the other helper does the same for current.  This should continue for 60 seconds After 50 seconds, some time between 50 to 60 seconds, you may want to click on the "stop" button and end the charging process because the capacitor is almost fully charged.


Part II: Discharging


6) After 50 seconds of charging, it means after 5 time-constants (if τ = 10.00s), the capacitor is almost fully (99 percent) charged and its voltage almost equates the battery voltage.  Now you need to prepare the group for thecapacitor discharging process.  The final VC of the charging stage becomes the initial VC of the discharging stage.  Record VC  in the chart.  You may use this voltage to calculate the initial current and write it in the chart as well.


When you announce "Ready, set, go", you already must have the values of VC and IC at t = 0 as in Part I.    Again at 5s intervals, announce "time" for your helpers to record the values they read.  Continue for another 60 seconds and record the values of capacitor voltage and current in the second chart.




Graph the following: For charging:  VC versus t    and   IC versus t, and

                                for discharging: VC versus t   and   IC versus t.






VB = the battery voltage (to be set at the starting time of charging)

VoC  = the initial capacitor voltage (to be read at the start of discharging)

                 R = 100.0kΩ and  C = 100.0μF.




Charging: Note that the value of IC at t = 0 is VBat /R. 


  t  (sec) 0 5 10 15 20 25 30 35 40 45 50 55 60
VC (volts) 0                        
IC (Amps) VB/R                        


Discharging: Note that the value of IC at t = 0 is VoC /R.


  t    (sec) 0 5 10 15 20 25 30 35 40 45 50 55 60
VC (volts) VoC                        
IC (Amps) VoC/R                        




For the charging part:  Io = VB /R.


For the discharging part Io = VoC /R.


Comparison of the Results:


The 4 graphs may be compared with the corresponding graphs in your text or course material.


Conclusion: To be explained by students.


Discussion:  To be explained by students.