ENERGY AND TIME CONSTANTS IN RC CIRCUITS By: Iwana Loveu Student No. 416 614 5543 Lab Section: 0003 Date: February 8, 2004
Abstract: Two charged conductors consisting of equal and opposite charges forms the basis of a capacitor. The capacitor is a device used to store charge and energy. The capacitor can be charged with equal and opposite charges by connecting the conductors to the terminals of a power supply. Once the battery is disconnected, the charges remain in the conductor and can be discharged. The focus of this experiment is on the discharging and charging of a 1 farad capacitor. It is confirmed in this experiment that there is a direct relation between resistance and the time required (time constant) for a capacitor to reach a maximum or minimum voltage. The time constants calculated in this experiment were 28.57s and 47.62s for discharge to 30 Ω and 60 Ω respectively. Energy of the capacitor was also analyzed and found to be 10.799J and 10.789J. Theory: How much charge a capacitor can store depends on the voltage developed across the conductors of the capacitors. That is, the amount of charge stored in the capacitor is directly proportional to the voltage drop across the capacitor. Thus, by charging the capacitor, current (which is the flow of charge) moves from the power supply or the battery to the capacitor. The equal and opposite charge to each conductor develops a voltage drop across the capacitor. This voltage V across capacitor allows for electric potential energy U to exist in the capacitor because work is done to move charges from a source to the capacitor. This relation is given below in Equation 1. 1 CV 2 U = 2 [1] However, during the charging of the capacitor, the maximum voltage V is not achieved instantaneously, but requires some time. This is caused by the effect of electrical inertia
due to the fact that when charging begins, at time t = 0, the capacitor behaves almost like a wire with nearly zero resistance. Thus there is small time delay in reaching maximum potential V. During the charging of the capacitor, an initial voltage drop, v develops. This voltage v continues to increase over a period of time, t as more charge is transferred from the source. The voltage drop can therefore be described as function of time, given by Equation 2 below, where R is the total effective resistance of the circuit. The term R C is called the time constant and represents the amount time it takes for the potential difference across the capacitor to rise within 1/e of V. t = R' C v( t ) V 1 e [2] To now discharge the capacitor to a minimum potential difference v from maximum potential difference V also requires some time. This is given by Equation 3 where R is the total load resistance in the circuit. v( t ) = Ve... [3] The term RC again is the time constant, τ and is the time required for the voltage to drop from a maximum voltage V to a voltage 1/e of V (Equation 4). τ = RC [4] Apparatus: For the experiment, an MW 12V power supply connected to a circuit box is required. The circuit box is then connected to a MacOS computer with LabPro. The LoggerPro software is used to log voltage over a course of time. The circuit box used in this experiment is shown below in Figure 1. t RC
Figure 1. Circuit Box Procedure and Results: U of T PHYA21S3 Laboratory Manual After checking that all equipment was connected in the right place, the capacitor was discharged to make sure minimum or zero potential was obtained. LoggerPro was then started and was used to log the potential increase across the capacitor for 120 seconds. Immediately after switching S1 from right to left to begin charging, the lamp in the circuit box (Figure 2) was observed to be very bright. This was expected since initially at t = 0 the capacitor has nearly zero resistance. And, over period of time, as voltage increased across the capacitor the bulb inside the circuit box became gradually dim. This was because the amount of current reaching the lamp decreases as more and more current is delivered to the capacitor. So the voltage drop across the capacitor reaches a maximum of 5 volts, and the lamp goes completely off. This occurs because no current is able to travel to the lamp. From LoggerPro it was noticed that it takes approximately 120 seconds for the capacitor to reach the maximum 5 voltage. Therefore charging is not instantaneous. Furthermore the charge curve obtained from LoggerPro, was very similar to that of an exponential curve, so therefore maximum voltage is achieved exponentially. Once LoggerPro had stopped collecting, switch S1 was moved to the right to allow the capacitor to discharge with the 60Ω resistor. And LoggerPro was used once
again to log voltage against time. It was noticed that the capacitor does not discharge completely in 120 seconds. In addition, the capacitor seems to charge faster then it discharges. This is due to the different time constants during the charging and discharging of the capacitors. This time constant can be calculated directly from Equation 4, as shown. So theoretically, the time constant that should be expected for discharging the capacitor from a maximum voltage V to a lower voltage V/e should take 60 seconds. To verify if this is true, the time constant was calculated from the measurements of the voltage taken by LoggerPro. And the data obtained during discharging from LoggerPro appeared to be an exponential (decay). So the data was then fitted using a natural exponential function of the form: f ( x ) = Ae C x + inverse exponential function: B. Similarly the charging curve was also fitted, but using an f ( are illustrated below in Figure 3. τ = RC = 60Ω 1F = 60 sec C x x ) = A( 1 e ) + B. The data logged and fits acquired Figure 2. LoggerPro Data & Fits for Charging and Discharging (With 60 Ω )
For the data obtained during the discharge, the exact equation used by LoggerPro to fit the curve was: 4 652 0. 021 x f ( x ) =. e + 0. 416. From this equation, the constant C was found to be 0.021. Since this equation is almost the same as Equation 3, then C is equivalent to one over the time constant. The units for time constant were also determined from the theoretical equation. = s Therefore, the time constant obtained for discharging the capacitor was approximately 47.62 seconds. 1 C = τ 1 0. 021 = τ τ = 47. 62 sec Following this, the next step was to determine the amount of energy stored in the capacitor. Since the capacitance of the capacitor is 1 farad and the maximum voltage reached was 5V then the energy (Equation 1) expected to be stored in the capacitor is: 1 U = 1 CV 2 = 1. 00F 5. 00V = 12. 5J 2 2 To determine if this was true, LoggerPro was used to graph the power (which is V 2 /R, where R in this case is 60 Ω ) against time. Since power is the transfer of energy in a unit time, then finding the area under the curve for the graph of Power versus Time would result in total energy transferred. So LoggerPro was used once again to calculate the integral of the graphed function. The integral (energy) of the curve of Power versus Time was approximately 10.789 Joules. τ = RC unit for time constant = Ω F R has units Ω, C has units F V = A C = This entire process was repeated again by first charging the capacitor and then discharging it with the 30 Ω resistor (by moving switch S2 to the left). The theoretical C s C V = C A
time constant with the 30 Ω resistor was determined to be 30.0 seconds. The equation of the fit as shown below (Figure 4) was obtained for discharging: f ( x ) = 5. 246e 0. 035 x + 0. 070. Figure 2. LoggerPro Data & Fits for Charging and Discharging (With 60 Ω ) From this equation the time constant was determined experimentally to be 28.57seconds. Similarly, the same procedures were used as mentioned above to calculate the energy of the capacitor. Theoretically, the capacitor had gained 12.5 joules of energy again when charged to the maximum voltage of 5 volts. To verify this, the integral of the Power versus Time curve was taken again. This is the energy of the capacitor and was found to be approximately 10.799 Joules. Discussion of Error: The experimental time constant (47.62 sec) to discharge the capacitor to the 60 Ω resistor was off by 20.6% from the expected value of 60 seconds. This large error may have been caused because all of the current was not reaching one or both resistors. A
possibility existed that current was leaving the circuit, perhaps because there were ionized particles in the circuit box that may have allowed the electron to leave the circuit. However, for the discharging of the capacitor to 30 Ω resistor, the experimental time constant (28.57s) was very close to the theoretical time constant (30s). The difference was less then 4.76% of the theoretical value. So it was most likely that not all of the current was reaching the resistor, or the capacitor was not initially charged to a maximum voltage. This same reason may account for the large difference between the experimental and the theoretical values in the energy of the capacitor. When the capacitor was initially charged, the power supply theoretically delivered 12.5 joules of energy to the capacitor. However, from the discharging with the 30 Ω and 60 Ω resistor, the energy of the capacitor was found to be only 10.799J (differs by 13.6%) and 10.789J (differs by 13.7%) respectively. This was mot likely caused by some energy dissipation to the surrounding. For example, when the capacitor was being charged, the lamp was initially on, so some of the energy is dissipated as heat through the lamp into the air. Other sources of error may have accumulated from systematic errors such the LoggerPro, which may have given an unsuitable fit, which would give different time constant. Conclusion: In this experiment the time constants of 28.57s for 30 Ω resistor and 47.62s for 60 Ω were obtained with 4.76% and 20.3% difference from the theoretical value. Although the time constants obtained were not exactly as expected, this experiment shows that the time constant is proportional to resistance. So when resistance is
increased by a factor of 2, time constant should also increase by a factor of 2 (assuming that the capacitance is constant). This explains why discharging is slower then charging the capacitor. During charging, there is a small total resistance of the lamp plus the power supply and during discharging, there is a resistance of the load resistor (30 Ω or 60 Ω ). Since the resistance during charging is smaller then the resistance during discharging, a larger time constant is obtained for discharging. Because of the larger time constant during discharge, it takes longer for the voltage to decay. Another conclusion that can be made is that the energy of the capacitor does not seem to depend on resistance, but only on the voltage of the source. This is evident from the experiment because the energy of the capacitor was nearly the same: 10.799J and 10.789J. The two values only differ by 0.01, which can be counted as experimental error.