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Engineering
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Case Study
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Topic:

Discharging Of A Capacitor Through A Resistor Case (Case Study Sample)

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DISCHARGING OF A CAPACITOR THROUGH A RESISTOR; DID A RESEARCH ON THE TOPIC

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DISCHARGING OF A CAPACITOR THROUGH A RESISTOR
Introduction
The main aim of this experiment is to investigate the exponential discharging of a capacitor through a resistor. The activities of this lab focus on showing that an exponential function is followed when discharging a capacitor and RC is the time constant. In addition, the experiments will enable us to investigate how resistance and capacitance affects the capacitive time constant during discharging. This experiment features a simple RC circuit, in which two capacitors of large value such as 1000micro farads 1500micro Farads and a variable resistance are connected in series, a toggle switch, stopwatch and a voltmeter. The double pole switch will be used to either connect the capacitor to the power supply to charge the capacitor or it can be used to connect the capacitor and resistor to allow the discharging process to take place and be investigated. A variable resistor R is used so that we can be able to alter and set the resistance to a desired value at any moment. The time constant will be obtained graphically after plotting the graph from the data recorded. In the simple RC circuit as shown below, a short circuit is used in place of the battery in order to eliminate the power supply; as a result the capacitor will be discharging through the resistor.
ROR
Capacitor C
Switch at t =0
Apparatus
* Low voltage power supply
* Variable resistance
* Two capacitors of 1000 µF and 1500 µF
* Digital multi meter
* Stopwatch
* Double pole switch
* Connecting wires
THEORY
A capacitor is a passive device that comprises of two parallel plates separated from each other by a dielectric medium and it is capable of storing charges when it is charged. The ratio of charge on a plate of a capacitor to the potential difference between the two plates is called the capacitance, that is,
C = Q/V
And,Q = CV.
The capacitor is considered to be fully charged when the charge, Q and Potential difference, V have reached their maximum value. Considering a capacitor of capacitance C that is fully charged and having an initial charge of Qo being allowed to lose its charges through a resistor of resistance R. The initial voltage of the capacitor plates decays exponentially as time progresses in accordance to the equation shown below:
V (t) = V0e−t/R
Given that Vo is the initial voltage of a fully charged capacitor. The graph below of V vs t shows how voltage is decaying exponentially with time.
The time constant, Ƭ, for the circuit is the time taken for the voltage to decay exponentially to 37% of its initial value. Thus, it represents the response time associated with the circuit.
Ƭ = RC
Taking logarithm of both sides of this equation V (t) = V0e−t/R, we obtain
V (t)/V(o) = -t/RC
LnV = – t/RC + LnVo
Plotting a graph of ln V against t, a straight line that has a gradient equal to -1/RC = -1/ Ƭ as shown in the figure below.
V
lnV Slope = -1/RCt
PROCEDURE
Use the multi meter to measure the nominal value of the capacitor’s capacitance which in this case should be 1000µF and set the of value resistance of the variable resistor to 50kΩ. Set up the circuit as shown below. (In this case the differential voltage sensor is the multi meter that is connected to the connected by use of the black and red cords from the circuit)
If the capacitor is fully charged, toggle the switch to position B and observe the rate at which the volt meter reading is decreasing. Record the voltage values of the multi meter at intervals of 5 seconds. Repeat the steps above twice more and calculate the average voltage of the recorded values. Record this data in a table. Find the natural logarithms of the average values of the voltage. Plot a graph of lnV against t.
Repeat the experiment using a capacitor of 1500µF capacitance and setting the value of variable resistance to 50 kΩ and also using a 1000 µF and a resistor of 60kΩ. Compare the time taken to discharge the capacitor for the two experiments and the value of time constants.
Observations
The value of the voltage starts to decrease when the position of the switch is at B. The voltage decreases rapidly at the beginning but gradually decreases to a much slower rate. More time is taken to discharge the capacitance when the experiment is repeated with capacitor and resistor having larger values of capacitance and resistance compared to the ones used for the first experiment.
Discussion
During the discharging process, the capacitor loses its charges at a decreasing rate. The initial conditions before the capacitor begins to discharge are; t=0, I = 0, q = R and Vc = Vs, that is, the supply voltage equals the voltage across the capacitor. The value of the capacitor required is large in order for the time constant produced to be measured and to make it easier to track with a voltmeter and a stopwatch. The capacitor should be handled with care since it is an electrolytic type which is terminal sensitive. The experiment and recording of data was done twice and the averaged in order to reduce the error. The discharging time is faster at the initial stages because of high value of capacitance but the rate of discharging fades of at later stages as the quantity of charges is now small. The time constant in an RC Discharging Circuit has the similar value of 63% of its previous value that corresponds to 37%.
In this case, the time constant will be the time in seconds that an initial charged capacitor will take to discharge and fall to 37% of its initial value. The capacitor is fully discharged when the charge on the plates and potential difference between the plates is zero.
The current equation discharging: I = - Ioe-t/RC, where Io =∆Vo/ R. Io is the maximum current in the RC circuit when t= 0
The voltage discharging equation: ∆V = -∆Voe-t/RC
NOTE: I and ∆V are negative because the current will be flowing in the opposite direction during the discharging process as shown in the RC circuit above.
The capacitor starts to discharge when the switch is closed. The curve of discharging Capacitor is steeper at the first seconds of discharging because the rate of discharging is fastest at this stage but it tapers off as the capacitor discharges at a dawdling rate.
Analysis
The data recorded from both experiments are presented in a tabular form. The natural logarithm of the voltage remaining across the capacitor is also evaluated and recorded in the same table.
Note: The values of Vc recorded are the average values computed after repeating the procedure of the experiment twice.
Experiment 1: using a 1000µF capacitor and 50kΩ resistor.
R=49.7kΩ C=1000UF

Time(s)

Voltage

LnV

0

10.81

2.3805

10

8.95

2.1917

20

7.38

1.9988

30

6.09

1.8066

40

5.00

1.6094

50

4.15

1.4231

60

3.41

1.2267

70

2.82

1.0367

80

2.35

0.8544

90

1.94

0.6627

100

1.61

0.4762

110

1.33

0.2852

120

1.11

0.1044

130

0.91

-0.0943

140

0.76

-0.2744

150

0.64

-0.4463

160

0.54

-0.6162

170

0.45

-0.7985

180

0.38

-0.9676

A graph of Ln Vc against time (s)
From the graph, the gradient of the line is -0.0191 = -1/Ƭ
Ƭ = time constant.
Therefore, Ƭ = 1/0.0191 = 52.35 seconds
We also know that time constant is given by R*C = 1000µF * 49.7KΩ = 49.7 seconds
The percentage error is (2.65/49.7) * 100 = 5.3 %
Experiment 2: using a 1000µF capacitor and 60KΩ
R=60.1kΩ C=1000UF

Time(s)

Voltage

LnV

0

10.82

2.3814

10

9.15

2.2138

20

7.70

2.0412

30

6.61

1.8886

40

5.62

1.7263

50

4.79

1.5665

60

4.08

1.4061

70

3.50

1.2528

80

2.95

1.0818

90

2.49

0.9123

100

2.15

0.7655

110

1.84

0.6098

120

1.56

0.4447

130

1.32

0.2776

140

1.15

0.1398

150

0.96

-0.0408

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