Discharging Of A Capacitor Through A Resistor Case - 1650 Words

Discharging Of A Capacitor Through A Resistor Case (Case Study Sample) Content: Student name:Course:Date:DISCHARGING OF A CAPACITOR THROUGH A RESISTORIntroductionThe 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.RORCapacitor CSwitch at t =0Apparatus * Low voltage power supply * Variable resistance * Two capacitors of 1000 F and 1500 F * Digital multi meter * Stopwatch * Double pole switch * Connecting wiresTHEORYA 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/VAnd,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) = V0et/RGiven 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, Æ ¬, 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.Æ ¬ = RCTaking logarithm of both sides of this equation V (t) = V0et/R, we obtainV (t)/V(o) = -t/RCLnV = t/RC + LnVoPlotting a graph of ln V against t, a straight line that has a gradient equal to -1/RC = -1/ Æ ¬ as shown in the figure below.VlnV Slope = -1/RCtPROCEDUREUse the multi meter to measure t he nominal value of the capacitors capacitance which in this case should be 1000F and set the of value resistance of the variable resistor to 50kâ„ ¦. 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 1500F capacitance and setting the value of variable resistance to 50 kâ„ ¦ and also using a 1000 F and a resistor of 60kâ„ ¦. Compare the time taken to discharge the capacitor for th e two experiments and the value of time constants.ObservationsThe 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.DiscussionDuring 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 experime nt 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= 0The voltage discharging equation: ∆V = -∆Voe-t/RCNOTE: 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.AnalysisThe 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 1000F capacitor and 50kâ„ ¦ resistor.R=49.7k C=1000UF Timeï ¼Ë†sï ¼â€° 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 -0.7 985 180 0.38 -0.9676 A graph of Ln Vc against time (s)From the graph, the gradient of the line is -0.0191 = -1/Æ ¬Ã† ¬ = time constant.Therefore, Æ ¬ = 1/0.0191 = 52.35 secondsWe also know that time constant is given by R*C = 1000F * 49.7Kâ„ ¦ = 49.7 secondsThe percentage error is (2.65/49.7) * 100 = 5.3 %Experiment 2: using a 1000F capacitor and 60Kâ„ ¦R=60.1k C=1000UF Timeï ¼Ë†sï ¼â€° 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