Capacitance Measurement in R-C Circuits: Lab Report Insights

Unit 4 Lab Report: Week 8: Section 2 By: Loudenridge Sundling Abstract: In this lab section, we were required to experimentally find the capacitance of a capacitor in series with resistor in a circuit in which the battery is sending sinusoidal voltages. In order to do this, we had to first utilize our knowledge of the resistor and capacitor voltage amplitudes in order to find a relationship between the angular frequency and capacitance of the circuit. By utilizing assistance from the lab worksheet and algebra, we were able to come up with the formula, ( 1 R ) ∗ ( V Resistor amplitude V Capactior amplitude ) = C ∗ ❑ d , which allowed us to experimentally calculate capacitance by finding the slope between the relationship of ( 1 R ) ∗ ( V Resistor amplitude V Capactior amplitude ) and the angular frequency, ❑ d . Consequently, after we completed our graph, we calculated the slope/capacitance equal to 9.5 ∗ 10 − 8 . Furthermore, through comparison between this experimental capacitance value and our chosen capacitance which was equal to 10.0 ∗ 10 − 8 F , we were able to conclude that there is a direct relationship between the angular frequency of our voltage sine wave and the product of the inverse of the resistors resistance and the ratio of resistor amplitude voltage and capacitance amplitude voltage due to the high accuracy of our experiment results.

Procedure: In this section, we were charged with experimentally finding the capacitance of a capacitor by utilizing the amplitude values from the resistance and capacitance sine equations provided us during the pre-lab readings, V_Resistor amplitude= (R/Z)*V_Source and V_capacitor amplitude =(X_C/Z)*V_source. Therefore, using the fact that R is equal to the resistance, X_C is equal to the inverse ratio of angular frequency,  , and capacitance, C, and Z is equal to the square root of R^2+ X_C^2, we were able to create an equation which related the ratio of the capacitor and resistor amplitudes multiplied by the inverse of the resistance to a direct relationship between capacitance and angular frequency. Thus creating the equation ( 1 R ) ∗ ( V Resistor amplitude V Capactior amplitude ) = C ∗ ❑ d . Using this equation, we were able to create an experiment in which we set up a circuit that has constant resistance and capacitance. Then, using the Siglent function generator, we chose a specific angular frequency, ❑ d , and utilized the cursor feature on the oscilloscope to find the ratio between the resistor and capacitor voltage amplitude multiplied by our chosen resistance, in this case, 10,000 ohms. We repeated this procedure 7 times which provided us with 6 points on a graph of ( 1 R ) ∗ ( V Resistor amplitude V Capactior amplitude ) vs. ❑ d , which we could plot a line of best fit that would contain a slope equal to capacitance. Results:

After completing our experimental procedure in which we measured the ratio between the resistor amplitude voltage and the capacitor amplitude voltage all multiplied by the inverse of resistance, against the change in angular frequency we created a table that contained 7 different points: ( 1 R ) ∗ ( V Resistor amplitude V Capactior amplitude ) ❑ d 0.62 ∗ 10 − 4 1 Ω 200π Hz 0.92 ∗ 10 − 4 1 Ω 300π Hz 1.22 ∗ 10 − 4 1 Ω 400π Hz 1.52 ∗ 10 − 4 1 Ω 500π Hz 0.84 ∗ 10 − 4 1 Ω 600π Hz 2.10 ∗ 10 − 4 1 Ω 700π Hz 2.45 ∗ 10 − 4 1 Ω 800π Hz This table allowed us to plot 7 different points on a graph and then find a line of best fit which accurately depicted the data. Then, in order to experimentally find capacitance, we chose two different points on this line of best fit (200π, 0.62 ∗ 10 − 4 ) and (300π, 0.92 ∗ 10 − 4 ) and subtracted the y-axis values by each other and divided this by the x-axis values subtracted from each other, 0.92 ∗ 10 − 4 − 0.62 ∗ 10 − 4 300 π − 200 π , in which we found our slope/capacitance equal to 9.5 ∗ 10 − 8 F . Discussion:

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After completing our experimental procedure, we were able to find an experimentally measured capacitance and then compare this to the know capacitance value of the capacitor we chose to put in our circuit. Because our chosen capacitance value was 0.1µF which is approximately equal to 10.0 ∗ 10 − 8 F , and our experimental capacitance value was 9.5 ∗ 10 − 8 F which is 95% of the theoretical value, we can conclude that our experimental procedure was an capacitance calculator. Furthermore, because our experimental value was calculated under the assumption that there is a direct correlation between product of the ratio between resistor amplitude voltage and capacitor amplitude voltage and the inverse of the resistor’s resistance and the angular frequency, ❑ d , we can conclude that this assumption is validated due to our experimental capacitance being very close to our theoretical capacitance. Open Ended Discussion: Week 8 Section 4 In this open ended section, we were charged with attempting to find the K value, or dielectric constant, of the sandwiched cardboard between our capacitor that we created through foil and 3 cardboard pieces. In order to test for this, we had to first set up a circuit which included a resistor and our homemade capacitor in series. Then, in order to experimentally find capacitance, we used the same method that we used in the 2 nd section in which we created a direct relationship between ( 1 R ) ∗ ( V Resistor amplitude V Capactior amplitude ) and ❑ d and then find at least 5 different data points which we could then turn into a graph with a best fit line where the slope would be equal to our capacitance of our homemade capacitor. After completing this procedure and developing a

graph, we found our slope/capacitance of our homemade capacitor to be equal to 7.08 ∗ 10 − 10 F . Consequently, in order to find the dielectric constant, we must first find what the capacitance of the capacitor would be without the sandwiched cardboard piece. In order to do this, we had to utilize the equation C = ∈ 0 ∗ A d where A is the area of the two cardboard pieces, d is the distance between the two capacitors and ∈ 0 is equal to the permittivity of free space or 8.85*10^-12. Because the dimensions of our two cardboard pieces was 40 cm by 40 cm and the distance between the capacitors was 6 mm, we found our capacitance without the dielectric constant to be equal to 2.36 ∗ 10 − 10 F . Now that we have found the capacitance without the dielectric constant, we can use the fact that C’= K*C where C’ is our capacitance with the sandwiched cardboard piece, C is the capacitance without the sandwiched cardboard piece and K is the dielectric constant. Furthermore, when we divided 7.08 ∗ 10 − 10 F by 2.36 ∗ 10 − 10 F we found that our K value, or dielectric constant was equal to 3. We concluded that our experiment worked very well and that our dielectric constant was rather accurate. Moreover, the only errors that we could find in our experimental procedure was the calculation of the capacitance without the dielectric insulator could have been more accurate if we used our experimental method to calculate for capacitance rather than the theoretical formula because we did not consider experimental error or mismeasurement of distance between the two capacitor sheets in our dielectric calculation.

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Unit 4 lab Report

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