8. THE CAPACITOR You will study several aspects of a capacitor, how the voltage across it changes with time as it is being charged and discharged and how it stores energy. The most well known device for storing electrical energy is no doubt the battery. But other devices have the ability to store energy the capacitor and inductor to name two. Since the capacitor is the easier of the two to study quantitatively, and the easier to understand in terms of the physics, it is the subject of study in this experiment. But capacitors are worthy of study for other reasons. Most consumers are aware that in many of the technological devices they buy there is a limitation placed on frequency or how quickly in the devices voltages and currents can change. The speakers in our audio system might respond to a maximum of 15 khz, or the CPU in our computer system might function at a maximum speed of Theory Why Study a Capacitor? 200 MHz. Serial or ethernet cables might not be able to sustain a certain bandwidth beyond a certain distance, and so on. These facts are manifestations of capacitance and inductance in the circuits. Capacitors and inductors have impedances (or effective resistances) that depend on frequency; the effects therefore also depend on frequency. This can be described as electrical inertia, a reluctance for the circuits to respond instantaneously to electrical changes. The object of this experiment is to study one example of this reluctance. For practical reasons we focus not on AC changes, but rather on a DC current and voltage that changes with time slowly enough to be easily observed. 1 Recall that in the experiment DC Circuits, you measured voltage and current with digital multimeters some time after you connected up the source of electrical energy (using a circuit as is shown in Figure 1a). In fact, you may have found it desireable to wait to enable the circuit to come to an electrical equilibrium. 2 In any case, this time was so short that for all practical purposes equilibrium was instantaneous (with a change in voltage as is shown in Figure 1b). If it were possible to use a high speed device to The Idea of Electrical Inertia measure the voltage across the resistor during the very short time just before and just after the connection, the voltage response might resemble Figure 1c. The voltage cannot go instantaneously from 0 to volts, but rather, requires some time to do so. This is an observation of electrical inertia. This effect is caused by very small capacitive and inductive components in what might otherwise be regarded as an ideal resistor. In this experiment this effect is deliberately emphasized for study by using a capacitor instead of a resistor. The Capacitor A capacitor is a device with the ability to store electrical energy (and therefore with a built-in electrical inertia). It is commonly made from two pieces of metal called plates separated by an insulating material called a dielectric. A capacitor s capacitance depends on the size of the plates, the distance between the plates and the ability of the dielectric to respond to an electric field. 3 It is found B8-1
8 The Capacitor experimentally that the voltage developed across a capacitor is proportional to the charge q (in coulombs, abbreviated C) stored in the capacitor. That is, q = Cv, [1] where the constant of proportionality, C, is the capacitance. C has the unit farad (abbreviated F). 1 F = 1 C. 1. If v in eq[1] is time dependent then q is also time dependent. 4 Though the capacitor used in this experiment is nominally 1 farad, a farad is a very large capacitance. More typical values are 0.1 µf (used at audio frequencies) and 0.01 µf (used at radio frequencies). (µ means micro, or 10 6.) The value of C depends on the construction of the capacitor, especially on the conductors and the kind of dielectric used. Details on the special construction of the capacitor used in this experiment are given in the Addendum. You can read this at your convenience. S = 5 volts R A (a) 5 5 t 0 time t t 0 time t (b) (c) Figure 1. When the switch S is closed at time t 0 in a typical DC circuit shown in (a) the voltage across the resistor, if observed on a multimeter, would be seen for all practical purposes to rise instantaneously from 0 to volts (b). However, if a high speed measuring device were used the voltage might be seen to rise in a curve as shown in (c). The curve is a manifestation of electrical inertia. The time axis of (c) is expanded with respect to (b). Charging/Discharging a Capacitor Experimental Details The capacitor is actually charged and discharged by means of the circuit shown in Figure 2. Figure 2 has certain features in common with Figure 1a, but the source of electrical energy here is a power supply not a battery. The capacitor is charged by connecting it to the power supply (switch S1 is thrown to the left). Subsequently the capacitor is discharged by connecting it to the load resistor R (switch S1 is thrown to the right). The charge stored in the capacitor flows through the resistor B8-2
The Capacitor 8 producing a voltage drop across the resistor. This voltage could be measured with a digital multimeter (as described above), but is here measured by means of a Serial Box Interface (SBI) and a computer. The use of an SBI avoids the problems associated with relatively slow responding digital devices like multimeters. And the computer also plots a graph of voltage versus time automatically (not to mention making possible subsequent data analysis). red lamp S1 30 Ω S2 to power supply 1.0 F 30 Ω black to SBI Figure 2. Schematic of the circuit box used in this experiment (within the dashed outline) which you can see through its transparent cover. Cables on the left plug into the power supply. Cables on the bottom coming from across the capacitor go to the SBI. Switch S1 controls charge/discharge, switch S2 provides a way of shorting one of the 30 Ω load resistors. The purpose of the lamp is to provide a visual indication of the current flowing through the capacitor as it is being charged. Charging/Discharging a Capacitor Mathematics Assumption In what follows we assume that when charging, the capacitor begins from a fully discharged state and when discharging begins from a fully charged state. Energy Required to Charge During charging the capacitor is brought from a begin state when the voltage across it is zero to an end state when the voltage is volts. The q and v in eq[1] change with time (which we examine in the next section), but eventually the voltage reaches some maximum volts and the charge a maximum Q coulombs. A little calculus is useful here. Charge is moved from the source to the capacitor in increments d q. For each increment the voltage is v. The element of work done d w by the supply in transferring a charge dq to the capacitor when at voltage v is given by dw = vdq = Cvdv, using eq[1]. Thus the total work W done is given by the integral 0 W = dw = Cvdv 0 B8-3
8 The Capacitor = 1 2 C 2, [2a] = 1 Q, [2b] 2 = 1 2 Q 2 C [2c] joules, by successive substitutions of eq[1]. The energy transferred to the capacitor can be calculated from any of these equations. oltage Change During Charging We now examine the physics as v is changing. Suppose at time t = 0, S1 is moved from the right to the left (Figure 2) to begin the charging process. By applying Kirchoff s voltage rule around the loop at some later clock time t when the capacitor s charge is q(t) and current i(t) is flowing we have q(t) C R' i(t) = 0, where R stands for the effective resistance in the charging circuit comprising lamp and power supply (not to be confused with the load resistor R in the discharging circuit). Substituting i = dq/dt and eq[1] and rearranging we obtain an equation in v(t): dv(t) dt + v(t) R ' C = R ' C. [3] This equation is a first order differential equation with solution v(t). You should be able to show that a solution is 5 v(t) = 1 e t RC. [4] The quantity R C (with units of s) is called the time constant. This is the time required for the voltage to rise to within 1/e of. A complementary treatment applies for the discharge of the capacitor. oltage Change During Discharge We suppose that at some clock time t = 0 seconds, S1 is thrown to the right (Figure 2) to begin the discharge process. By applying Kirchoff s voltage rule around the loop at some clock time t when the voltage across the capacitor is v C (t) and a current i(t) is flowing we have v C (t) Ri(t) = 0, [5] where now R is the resistance of the load resistor. Using eq[1] and the fact that i(t) = dq(t), dt (as current flows charge decreases) it can be shown that dv(t) dt + v(t) RC = 0. [6] This equation is a first order differential equation in v(t). You should be able to show that a solution is v(t) = e = e t RC, [7] where t C = RC. [8] Eq[7] is a relatively simple exponential function as is seen in Figure 3. t C is called the time constant and quantifies how sharply the voltage decreases. If R is in units of ohms and C in units of farads then RC has the unit of time in seconds. t C is the time in seconds for the voltage to fall from the initial voltage to /e = 0.368. The bigger the time constant the less is the rate of voltage decay. Objectives of this experiment are to test the validity of eqs[4] and [7] during charge and discharge and to examine the corresponding changes in energy. t t C B8-4
v C (t) The Capacitor 8 e =.368 t C = R C time t (s) Figure 3. The exponential decay of the voltage across a capacitor discharging through a resistor. The Experiment Exercise 0. Preparation Orientation Many of the technical aspects of this experiment you have seen before in other experiments. LoggerPro is used here too. You will use a Serial Box Interface (SBI) instead of a ULI board to detect the voltage across the capacitor but the functioning is similar. Identify the apparatus from Figure 4. You have a source of electrical energy (MW regulated 12 power supply), a circuit box (containing capacitor, switches and load resistors) connected to the output of the supply, lines running from the circuit box to the Serial Box Interface (SBI), and in turn to the computer. Confirm that the power supply is OFF. Using what you have learned in the experiment DC Circuits note the resistance of the two load resistors and their uncertainties. About the Circuit Box and the Capacitor For convenience you may wish to orient the circuit box so that it appears as in Figure 2 with switches S1 and S2 arrayed along the top. The capacitor is the green cylindrical object mounted upside down inside the circuit box. Follow what you can see of the circuit diagram through the transparent cover of the box and with the guidance of Figure 2. Confirm that the switch S1 is in the discharge position. Confirm that the switch S2 selects only one of the load resistors. Setting up the SBI Confirm that the power adapter for the SBI is plugged into the power bar and the SBI is ON, i.e., the light on the SBI is glowing green. Confirm that the lines from the circuit box are plugged into Port 1 of the SBI, and that the signal output cable from the SBI is plugged into the printer port of the computer. Opening LoggerPro 1 Boot the computer as you learned to do in the experiment "Linear Motion". ➁ Log into your account on the college network. Remember, if you can t log in you can always save your experiment in the Student Temp Save folder on the local hard drive. You can log in or out at any time. ➂ On the local hard drive "Macintosh HD" locate and open folders in this order: Physics >> PHYA10/A20 >> 8. The Capacitor. ➃ Inside 8. The Capacitor double click the icon The Capacitor Alias. LoggerPro and The B8-5
8 The Capacitor Capacitor setup should run. The Opening Screen Examine this screen and identify these aspects: The calibration is set to display voltage between 0 and 5 volts. The program once started will log data continuously for 2 minutes. At the end of that time logging will stop automatically. The calibration is set to collect two voltage measurements per second making for 240 datapoints. The voltage that is being measured at the present time is displayed at the bottom center of the screen. This is the voltage that is logged when you click the Collect button. Entering Formulas LoggerPro logs voltage; it must be deliberately instructed to calculate power. You have to enter the formula into LoggerPro s spreadsheet as you did in other experiments. If you have forgotten how to do this select Data >> New Column >> Formula. Figure 5 shows a fragment of the screen which then appears. Enter the formula for power ( potential ^2/R) including the load resistance you have selected via S2. Changing the Graph Displayed in an Area Recall that you can change the graph that is displayed in a graph area. To do this place the pointer over the graph label (down the left hand side of the graph), click the mouse and from the dialog box select the graph you wish displayed. Do this now to display the Potential (olts) graph. MW 12 power supply S1 S2 circuit box printer port SBI Figure 4. The equipment used in this experiment. The circuit box is shown in detail in Figure 3. Figure 5. A fragment of the screen that appears when New Column is chosen. B8-6
Exercise 1. Collecting the Data and Fitting a Function The Capacitor 8 The objective of this exercise is to collect data, obtain graphs for charging and discharging and to fit the graphs to obtain the time constants. Charging You may wish to begin with S2 in the position to short out one of the 30Ω load resistors. Turn the power supply ON, throw S1 to the charging position and click Collect. As the charging proceeds observe the brightness of the lamp and the voltage across the capacitor as printed at the bottom center of the screen. Here are some questions of a qualitative nature: Questions? Is the charging instantaneous?? Does the charge curve at least look exponential?? Roughly how long does it take for the voltage to reach maximum?? How bright is the lamp when the maximum voltage is reached? Why?? Explain what is happening as the lamp dims and then goes out. Saving the Data At the end of the 2 minutes you can save the data you have just collected. Select Data >> Save Latest Run. Discharging Now throw the switch to the discharge position and click Collect. After 2 minutes you should see a graph resembling Figure 6. Questions:? Does the capacitor completely discharge in 120 seconds?? Does the capacitor charge faster than it discharges? What is a likely explanation for this? Graph produced by LoggerPro, copied to the clipboard and pasted directly into Microsoft Word. Figure 6. An example of charge and discharge curves placed on the same graph. The little information windows can be placed anywhere on the graph by dragging with the mouse. B8-7
8 The Capacitor Finding the Time Constants You can conclude that the discharge is exponential only if the voltage function given by eq[7] really does describe your data. And if it does describe your data then the results of the fit should yield the time constant t C. Accordingly, apply what you have learned about fitting from other experiments to fit an exponential function to the decay curve. What value for the time constant does the fit return? Questions:? Can you estimate this value s uncertainty?? How well does this experimental value agree with the theoretical value predicted by eq[8]?? Assuming that R has the unit ohms and C has the unit farads, prove that t C has the unit seconds. Exercise 2. Aspects of Energy Energy In Based on eq[2a] how much energy did the power supply deliver when charging the capacitor to the maximum voltage? Did the power supply actually deliver more energy than this? Explain. Energy Out Display the power vs clock time graph. How is the energy stored in the capacitor related to the power graph? By diffentiation? Integration? Find the energy stored from the power vs clock time graph. For guidance an example is shown in Figure 7. How well does this number agree with the number obtained in the first paragraph of this exercise? B8-8 Graph produced by LoggerPro, copied to the clipboard and pasted directly into Microsoft Word. Figure 7. A typical output from LoggerPro showing the area under a power vs clock time curve. The area, given as 12.1 J differs by only about 3% from the expected value of 12.5 J.
Exercise 3. Repeat The Capacitor 8 Repeat the above with both load resistors in the circuit. Questions:?If R is increased by a factor of 2 what in theory is the effect on tc?? Does the answer to the previous question mean that in this particular experiment a discharging capacitor loses charge faster or slower than a charging one? Physics Demonstrations on LaserDisc from Chapter 46 Capacitance and RC Circuits Demo 18-19 Parallel Plate Capacitor Demo 18-28 RC Charging Curve Activities Using Maple E08The Capacitor If you choose you can input the data you collected in the experiment The Capacitor into this worksheet, plot it and fit it. This worksheet also attempts to shed light on the non-linear aspects of the voltage decay you have no doubt observed. As a tutorial, the differential equation describing the decay of the voltage across the capacitor is solved under assumptions of linear and non-linear decay. B8-9
8 The Capacitor Addendum. The Supercap 1 Farad Capacitor 6 The capacitor used in this experiment is the NEC Supercap 1 farad, developed to protect memory circuits during temporary power losses. The two materials serving as interface media in this capacitor are activated carbon and sulfuric acid. Activated carbon is ground to a powder and mixed with sulfuric acid to form a paste. The surface area of the activated carbon is approximately 1000 m 2 per gram. The combination of this very large surface area with the very small distance between the capacitor plates results in a capacitance of about 200 to 400 farads per gram of activated carbon. One problem with these materials is that if more than 1.2 volts is applied to this junction, the aqueous sulfuric acid solution breaks down. To remedy this, the Supercap incorporates a number of small cells, each consisting of a sandwich of the paste separated by a porous material. Each cell is encompassed by an impermeable gasket (to hold everything together) squeezed between two electrically conductive ends. The whole system is pressurized. Individual cells are stacked in series (end to end) and squeezed into a metal outer cover. By stacking them in series, the effective voltage of each cell is multiplied to useful proportions. Stuart Quick 97 EndNotes for The Capacitor 1 Here we regard a DC current as one that flows in only one direction through a circuit; its amplitude however may change with time. 2 Some waiting was required more due to the fact that the multimeters were digital devices and responded to changes relatively slowly. 3 This is quantified by the dielectric constant. All things being equal a capacitor s capacitance can be increased by choosing a dielectric with a larger dielectric constant. 4 Lower case letters are used in eq[1] to indicate quantities which may be variables. Eq[1] applies whether or not q and v are time dependent. When the voltage is a maximum then so also is the charge, ie., Q = C. 5 It is sufficient to show that eq[4] satisfies eq[3] by direct substitution. Take the first derivative of eq[4] and substitute it into eq[3] to show that the LHS = RHS. 6 Much of the material in this section is extracted from the pamphlet Supercap 1 Farad Capacitor P6-8012, Copyright 1990 by Arbor Scientific. B8-10