Experiment 10

The RC Circuit with an AC Source



To study the voltage variations across a capacitor in an RC circuit in series with an ac source, as well as obtaining the time constant of the circuit from oscilloscope readings




A resistance box or two appropriate resistors, a capacitance box or two appropriate capacitors, an oscilloscope, appropriate connecting wires, and a function generator




In a series circuit with a dc source, the voltage increase or decrease during the charging and discharging processes (Experiment 6) are given below:


Charging needs a battery in the circuit.  VBat. = the battery voltage.


Discharging requires an initially charged capacitor.  Vo = Qo/C.



Table 1


The corresponding graphs are:



According to the charging equation (Table 1), after a time of one time constant (RC), the capacitor voltage increases to 0.63 VB .  For the discharging process (Table 1), at t = RC, the capacitor voltage drops by 0.63Vo and becomes 0.37Vo.  Note that Vo means Vmax.


This information is used in this experiment to estimate the circuit´┐Żs time constant (RC) from the graph of capacitor voltage versus time on the oscilloscope screen.


When an AC source (in rectangular mode) is used instead of a battery, as in Fig. 1, the capacitor keeps getting charged in every half cycle and discharged in the next half cycle.  The extent of charge flow through the capacitor and the voltage across it is a function of the time constant RC of the circuit.  The voltage  across the capacitor, VC, is plotted in Fig. 2.



From the 10 horizontal divisions on the oscilloscope, the sweep frequency, and the number of full cycles generated on the screen, it is possible to measure RC, the time constant of the circuit.


If the sweep frequency is set at 100 per second, then each of the 10 horizontal divisions on the screen represents 1/1000 of a second.  If there is one full cycle on the screen as shown in Fig. 3, and the 0.63Vmax point corresponds to 1.5 horizontal divisions, for example, then RC = 1.5(1/1000)s = 0.0015s.





If, with a sweep frequency of 100 per second, for example, 2 full cycles are formed, the period of each cycle is (1/200) s.  Each cycle is then using half of the width of the screen or 5 horizontal divisions.  Each horizontal division is still (1/200)/5 =  (1/1000) s.  Now, if for a different RC combination, the 0.63Vmax point corresponds to 0.75 divisions of the horizontal axis, then the time constant is RC= 0.75(1/1000) s or  RC = 0.00075 s.  (Fig. 4)





1) Set the function generator frequency to 10 Hz and to the square-wave option, and connect it to the oscilloscope.


2) Tune it to a stationary trace of one cycle, and adjust its amplitude very close to maximum as shown in Fig. 5.

    Your circuit should look like Fig. 1.



3) Calculate the sweep time, as well as the sweep time per horizontal division.


4) Place a capacitor and a resistor (values determined by your instructor) into the circuit as shown in Fig. 1.  The shape of the graph should change and become similar to the one shown in Fig. 3.  Make adjustments as needed.


5) Carefully, estimate the location for 0.63Vmax point on the vertical axis, and find the corresponding value of time on the horizontal axis as shown in Fig. 3.  This will be your measured value for RC of the circuit.


6) Use the values of R and C as determined by your instructor to calculate the accepted value of RC.


7) Compare the two values for RC and calculate a percent error.


8) Repeat Steps 4 through 7 for another set of resistance and capacitance (determined by your instructor).







                                                Case I)                                     Case II)




                                                Case I)                                     Case II)




Perform as necessary.


Comparison of the Results:


Calculate a percent error on the value of RC in each case.


Conclusion:  To be explained by students.


Discussion:   To be explained by students.