Capacitors and RC Decay Lab Report PDF

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Summary

The mesaurement of capacitance will be tested and related to the laws that determine the charging and discharging of a capacitor....

Description

Danny Stein PHYS 1440L – 812 Robert Martin 6 March 2018 Partners: Louis St Onge and Bobby Austin Capacitors and RC Decay

Objective: The measurement of capacitance will be tested and related to the laws that determine the charging and discharging of a capacitor.

Introduction A capacitor is a type of device used to store electric charge. The potential difference, V, between two plates is proportional to the charge, Q, if one plate receives a positive charge, while the other receives a negative charge. This can be expressed by: Q = CV

(1)

where C is the capacitor’s capacitance and is measured in farad (F). Volts are used to measure the potential, while the charge is expressed in coulombs. When capacitors are connected in parallel, as shown in Figure 1, the total capacitance is the sum of the individual capacitance values. This can be represented by: Cp = C1 + C2

(2)

Figure 1: Capacitors Connected in Parallel

However, if the capacitors are connected in a series (figure 2), then the total capacitance can be determined by: C C

Cs = C 1+C2 1

2

(3)

Figure 2: Capacitors Connected in Series

Figure 3: Resistance-Capacitance Circuit

Once the switch connects to terminal “a”, the DC voltage supply connects to the resistor and capacitor, and a current charges the capacitor. The output voltage throughout the circuit can be determined from the resistor voltage, Vr, and the capacitor voltage, Vc by: V0 = Vr + Vc As the capacitor begins to charge, the rate of charge exponentially increases and can be calculated by: −t

Q = CV0 (1 - e RC ) which can be written in terms of the capacitor voltage by:

−t

Q

Vc = = V0 (1 - e RC ) C

(4)

This can be rewritten in terms of n, where n equals 1, 2, 3,… etc, and can be expressed by Vc = V0 e-n Time can also be determined from the product of n and RC, The time constant, t’, of the circuit can be calculated by: t’ = RmC1

(5)

where Rm is the internal resistance. Similarly, Rm can be determined by:

Rm =

t′ −t′′ t′′

Rx

For this experiment, Rx was given as 10 MΩ. Apparatus and Procedure Equipment 

DC Regulated Power Supply



Resistor



Switch



Voltmeter



Two Capacitors



Stopwatch



Voltmeter Probes

(6)

Procedure To begin this experiment, use the voltmeter probes to connect each piece of equipment as shown in Figure 4, which illustrates the t’ position. Move the switch down so it makes contact, and then set the power supply to 10 volts. Open the switch by placing it in a neutral zone with no contact and record the time it takes the voltage to drop to 3.68 V. Record the time for three separate trials and take the average. Rearrange the voltmeter probes to match t” (Figure 5), t2 (Figure 6), ts (Figure 7), and tp (figure 8), and repeat the entire process for each. Once the average for t’, t”, t2, ts, and tp are determined, Rm can be calculated from equation 6. C1 and C2 can then be calculated from equation 5, and Cs and Cp can finally be calculated from equations 1 and 2.

Figure 4: Setup of Capacitor C1 (t’)

Figure 5: Setup of Capacitor C1 and Rx (t")

Figure 6: Setup of Capacitor C2 (t2)

Figure 7: Setup of Capacitors C1 and C2 in Series (ts)

Figure 8: Setup of Capacitors C1 and C2 in Parallel (tp)

For the second part of the experiment, only C1 would be used. Repeat the setup shown in Figure 4, and make sure the power supply and voltmeter still read 10 volts. Lift the switch and record the voltage every 10 seconds until the voltage drops to about 0.3 volts. Results and Analysis

The data from the first part of the experiment can be found in Table 1. Rm was calculated to be about 9.759 x 106 Ω, which resulted in C1 equaling 10.69 µF. Using the same method, C2 was calculated to be 4.48 µF. Cs was 3.17 µF, and finally CP was 15.17 µF. Table 1: Measurement of Capacitance

Trial

t' (s)

t'' (s)

t2 (s)

ts (s)

tp (s)

1 2 3 Average

104.23 103.57 105.42 104.41

52.57 52.83 53.11 52.84

42.56 43.80 44.70 43.69

33.40 34.03 34.45 33.96

138.85 145.52 148.35 144.24

The results from the second half of the experiment are shown in Table 2. The graph produced by this table was a curve that showed the voltage decreasing as time went on, but the curve never reached zero volts because the calculations have the exponential function, e. The lowest voltage it reached was 0.36 V at 490 seconds, but proceeded to rise back up as time continued. The rate of discharge decreases while the voltage decreases, so theoretically, the voltage would get very close to zero, but never actually reach it. Table 2: Measurement of Exponential RC Decay (with C1)

t (sec) 0 10 20 30 40 50 60 70 80 90 100

V (volts) 10.00 9.04 8.19 7.43 6.70 6.10 5.55 5.04 4.60 4.19 3.82

t (sec) 110 120 130 140 150 160 170 180 190 200 210

V (volts) 3.52 3.23 3.01 2.78 2.58 2.40 2.24 2.09 1.94 1.81 1.70

t (sec) 220 230 240 250 260 270 280 290 300 310 320

V (volts) 1.60 1.49 1.40 1.31 1.23 1.16 1.09 1.03 0.97 0.91 0.86

t (sec) 330 340 350 360 370 380 390 400 410 420 430

V (volts) 0.81 0.77 0.73 0.69 0.65 0.62 0.58 0.55 0.52 0.50 0.47

t (sec) 440 450 460 470 480 490 500

V (volts) 0.45 0.43 0.41 0.39 0.37 0.36 0.37

Table 3 demonstrates the value change of the voltage when the exponential function, e, gets removed by taking the natural logarithm of both sides. This caused the graph to be linear, rather than a curve. Table 3: Measurement of Exponential RC Decay (with C1)

t (sec) 0 10 20 30 40 50 60 70 80 90 100

lnV (volts) 2.30 2.20 2.10 2.01 1.90 1.81 1.71 1.62 1.53 1.43 1.34

t (sec) 110 120 130 140 150 160 170 180 190 200 210

lnV (volts) 1.26 1.17 1.10 1.02 0.95 0.88 0.81 0.74 0.66 0.59 0.53

t (sec) 220 230 240 250 260 270 280 290 300 310 320

lnV (volts) 0.47 0.40 0.34 0.27 0.21 0.15 0.09 0.03 -0.03 -0.09 -0.15

t (sec) 330 340 350 360 370 380 390 400 410 420 430

lnV (volts) -0.21 -0.26 -0.31 -0.37 -0.43 -0.48 -0.54 -0.60 -0.65 -0.69 -0.76

t (sec) 440 450 460 470 480 490 500

lnV (volts -0.80 -0.84 -0.89 -0.94 -0.99 -1.02 -0.99

When n is 1, Vc equals 0.368V0 and t was calculated to be 104.3 seconds. When n equals 2, Vc is equivalent to 0.135V0, which makes the time 208.6 seconds. Finally, when n is 3, Vc equals 0.050V0, and the time equals 313.0 seconds. Discussion While analyzing the Volts vs. Time graph, the time when Vc equals 3.68V was at roughly 105 seconds, which directly compares to the theoretical value of 104.3 seconds. However, the results for 1.35V and 0.50V turned out higher than they were originally anticipated. The graph shows the time at 1.35V being 245 seconds, while 0.50V shows a time of 400 seconds. The greatest explanation for this relates to the accuracy of the calculations for the capacitance of the first and second capacitor. The times and voltages may not have been recorded precisely as they appeared, which could ultimately alter the results over time.

Conclusion This experiment clearly showed how the laws that govern the rate of charge and discharge effect the measurement of capacitance. It showed the different between parallel and series capacitors when they are charged and discharged. Questions 1. R = 20 MΩ = 20x106 Ω; E = 10V; t = RC = 100s t

100

C = = 20x106 = 5x10-6 F R

2. A) V = 0.5V0 −t

0.5 = e (1x10−6)(1x106 ) = 0.693 seconds B) V = 0.1V0 −t

0.1 = e (1x10−6)(1x106 ) = 2.303 seconds C) V = 0.02V0 −t

0.02 = e (1x10−6)(1x106 ) = 3.912 seconds 3. Q = CpV Q

Cp = V =

1 200

= 5x10-3 F

Cp = nC C = 10-6 n=

Cp C

=

5x10−3 10−6

= 5000...