Lemon Battery Connected in Series Charging a 2.2 Farad Capacitor SECTION #1 - The experimental setup 1. The goal of this experiment is to see if I can connect 2, 3 or 4 lemons together in a series configuration to make one battery which will charge up a 2.2 Farad Super Capacitor. 2. Diagrams of the battery configurations are presented in Figure 1 and the experiments are labeled as: Experiment #1: 2 lemons in series, 2.2F Cap Experiment #2: 3 lemons in series, 2.2F Cap Experiment #3: 4 lemons in series, 2.2F Cap 3. The lemons were frozen over night and then allowed to thaw for several hours before the experiment. 4. The negative electrodes are Zn plated iron nails. Positive electrodes are Canadian Silver Dollars (80%Ag / 20%Cu), years - 1958, 1958, 1958 & 1965. 5. I connected one 2.2 Farad electrolytic Super Capacitor in series with my lemon battery configurations, which theoretically should be able to store 2.2 Farads of charge (or 2,200,000 μf or 2.2 million micro Farads). See Section 8 at the end of the document for the equivalent circuit. The device number is: 2.5DMB2R2M8X16 CAP, ALU ELEC. Manufacturer specifications: 2.2F @ 2.5 Volts. Also, these devices have a polarity and MUST be connected into the circuit in the correct orientation! 6. The experiment used the following load resistor: 4700Ω. 7. Experiments were run continuously, i.e., once connected to the battery, the experiment was allowed to run for the indicated time period before the load resistor was disconnected from the battery. 8. My design also uses a germanium diode at the negative side of my battery configurations. Strictly speaking, I do not need a diode since this is a DC circuit. Diodes are generally used in AC (alternating current) circuits. But, having said that, many "Energy Harvesting" applications, such as solar panels which are used to charge up Super Caps, use a diode to ensure that the current will flow in only one direction, i.e., we do not want a flow reversal - current flowing in the opposite direction of that desired. 9. Since a diode will consume power from the battery (Power = Volts x Amps), I chose to use a Germanium diode, which has a low "forward" voltage (V F ). The device number is 1N34A. I measured the V F and found it to be 0.361 Volts, which is considerably lower than the Silicon diodes that I previously used (V F = 0.629V). Thus, we are saving some power by using the Germanium diode. 10. As well, diodes have polarity, thus, MUST be connected in the correct orientation to work properly. The 1N34A device has a RED band to mark positive and a GRAY band to mark negative. 11. A cautionary note regarding Germanium devices (e.g., diodes and transistors) - they are "static electricity" sensitive, i.e., static electricity is sufficient to destroy the device. Thus, one must wear a Anti-Static Wrist strap while handling the device, to drain off any static electricity that might have built up on your body and the devices should be stored in anti-stat bags when not in use. 12. Electrical measurements were taken at several points: (i) Circuit current at point "a" (I circuit ). (ii) Voltage across the capacitor, (V cap ). (iii) Charging current (I charging ) at point "b". The reason for taking measurements at these points is to ensure that I do not exceed the manufacturer's specifications while using the Super Capacitor. As well, I have never used Super Caps before, so I want to learn as much as I can regarding the operation of this device. 13. Additionally, when the current at point "b" is equal to the current at point "a", I will assume that the capacitor is fully charged, or at least very close to fully charged. 14. The charging current is monitored as a percentage of the maximum circuit current in order to determine when the two currents are equal (I charging reaches 100% of the maximum I circuit ), i.e., the capacitors are fully charged, using the following equation: % charging = (I charging / I circuit, max ) x 100 15. At the end of each experiment, the charge was drained from the capacitor by disconnecting the Positive terminal and by using an additional resistor of lower value, in parallel with the load resistor, in order to SAFELY drain off the stored up charge. I use the symbol " " to mean "in parallel with".
SECTION #2 - Figure 1: The experimental diagram {also see equivalent circuit at the end of the document, Section #8} V R 4700 Ω Resistor A = Ammeter, to measure the current, point "a" Negative rail of battery (-) terminal A 2.2 Farad Super Capacitor (rated @2.5V) A = Ammeter, point "b" - + {2.5DMB2R2M8X16, ALU, ELEC} Germanium diode, V F = 0.361V Positive rail of battery (+) terminal V d A S #2 S #4 V cell1 Cell #1 V cell2 Cell #2 S #1 V cell3 V cell4 Cell #3 Cell #4 S #3 (-) (+) (-) (+) (-) (+) (-) (+) Expt #1, two lemons Experiment #2, three lemons Experiment #3, four lemons
SECTION #3: Data for Experiment #1 Table 1a - Experiment #1a: Two Lemon Cells in Series, 4700 Ohm Resistance, 2.2 Farad Capacitor. Charge Time Charge Current Capacitor Circuit Current (I (minutes) (I charging ), "b"; ma charging / I circuit, max ) x 100 Voltage (V cap ) (I circuit ) "a"; ma Instantaneous 1.5 2055% ---- ---- 0.5 0.707 968% 0.018 0.004 1 0.558 764% 0.030 0.006 3 0.412 564% 0.062 0.013 5 0.364 499% 0.088 0.018 7 0.329 451% 0.115 0.024 10 0.303 415% 0.141 0.029 15 0.282 386% 0.188 0.039 20 0.255 349% 0.222 0.046 25 0.241 330% 0.255 0.053 30 0.229 314% 0.284 0.059 35 0.223 305% 0.313 0.065 40 0.218 299% 0.334 0.069 45 0.226 310% 0.338 0.070 48 0.200 274% 0.350 0.073 Voltage of each cell start: V cell 1 = 1.0V, V cell 2 = 1.0V; end of experiment: V cell 1 = 0.95V; V cell 2 = 0.95V Voltage (V d ) across 1N34A diode @ 2.5 min = 0.23V and @ 45 min = 0.20V {Analog meter} Estimated # s & Farads: Charge = 0.618 s, V = 0.350 Volts, Capacitance = 1.76 Farads Table 1b - Experiment #1b: Discharge of 2.2 Farad Capacitor through 4700 Ohm Load Discharge Time (minutes) 0 0 0.350 0.073 25.6 1 0 0.349 0.073 25.5 3 0 0.342 0.071 24 5 0 0.337 0.070 23.6 10 0 0.323 0.067 22 15 0 0.312 0.065 20 20 0 0.300 0.062 19 30 0 0.279 0.058 16 Current @ "b"; ma V cap (Volts) I circuit, "a"; ma Power (μw) Table 1c - Experiment #1c: Discharge of 2.2 Farad Capacitor through 4700 Ohm 1000 Ohm, ( 825Ω) Discharge Time (minutes) 1 0 0.265 0.290 77 2 0 0.252 0.276 70 5 0 0.229 0.250 57 10 0 0.189 0.207 39 15 0 0.157 0.172 27 Current @ "b"; ma V cap (Volts) I circuit, "a"; ma Power (μw)
Table 1d - Experiment #1d: Discharge of 2.2 Farad Capacitor through 4700 Ohm 300 Ohm, ( 282Ω) Discharge Time (minutes) 1 0 0.155 0.410 64 5 0 0.096 0.250 24 10 0 0.063 0.165 10 Current @ "b"; ma V cap (Volts) I circuit, "a"; ma Power (μw) SECTION #3: Discussion of Data for Experiment #1, Two Lemon Battery (Series) Discussion of the Data for Experiment #1: 1. A set of switches (S #x) were put in place so that I could connect the lemons in series for each experiment: Expt #1: S1 is up, S2 is open (thus, two cell battery). Expt #2: S1 is down, S2 is down, S3 is up, S4 is open (thus, three cell battery). Expt #3: S1 is down, S2 is down, S3 is down, S4 is down (thus, four cell battery). 2. The instantaneous charging current generated when the positive terminal is connected, is 1.5mA, which drops to 0.707mA within 30 seconds. Note at 30 seconds that the circuit current is only 0.004mA. Thus, the majority of the current leaving the battery is going directly to charging the Super Cap. 3. Similar to a previous post, as the charging current decreases, the voltage across the capacitor and circuit current increases. 4. The goal was to let the experiment run until the charging current was equal to the circuit current, but I terminated the experiment early due to a problem with lemon #2 (cell#2). At the end of experiment, I charging = 0.200mA and I circuit = 0.073mA, thus, not equal. 5. Thus, the I charging / I circuit drops to only 274% and never reaches the goal of going down to 100%. 6. The voltage across the Super Cap at 48 minutes is 0.350V. But, considering that the diode is dropping between 0.20-0.23V, the total voltage is: V cap + V d = 0.350V + 0.215V = 0.565V 7. Thus, this two cell battery is delivering 0.565V. 8. The initial cell voltage before the experiment for each cell was 1.0V and at the end of the experiment, the voltages decreased to 0.95V for each cell. Thus, the battery has lost some of its potential voltage. 9. Calculations for the number of ombs of charge and Farads is reported at the very end of this document. The calculations were done in Excel and I subtracted the circuit current from the charging current in order to get a better estimate of the current going to charge the Super Cap. 10. The results are under estimated, because I used the low current value at each time point (rather than an average). 11. The charge transferred in the experiment is 0.618. I have been able to achieve this in previous posts. 12. The number of electrons transferred is: 0.681 /1.602 x 10-19 /e - = 4.3 x 10 18 electrons. 13. The Capacitance, based on the final voltage on the capacitor of 0.350V, is 1.76 Farads. This is very close to the theoretical value of 2.2F. Thus, the capacitor was not fully charged to its maximum potential. 14. Three discharge experiments were done, i.e., experiments 1b, 1c and 1d. In experiment 1b, I simply disconnected the positive terminal of the battery and let the capacitor discharge through the 4700Ω load. By 30 minutes, the voltage had dropped by only 0.071V (0.350V-0.279V = 0.071V) and was still producing 0.058 ma. Thus, there was still a significant amount of charge on the Super Cap. 15. In experiment 1c, I connected a 1000Ω resistor in parallel, thus giving me an equivalent resistance of 825Ω. At 15 minutes, there was still 0.157V & 0.172mA, thus not fully discharged. 16. Experiment 1d, I used an equivalent resistance of 282Ω for discharging the Super Cap and even after 10 minutes, there was still Voltage left on the capacitor, 0.063V, thus not fully discharged. 17. The amount of effort that it took to discharge the capacitor supports the fact that the capacitor stored a significant amount of charge, which is consistent with the calculation of 1.76 Farads. 18. Calculations of the Power dissipated during the discharge experiment also support the fact that the Super Cap stored a significant amount of charge, i.e., from 10-77 μw of Power. 19. The data suggest that the two cell battery, in series, is capable of charging a 2.2 Farad Super Capacitor.
SECTION #4: Data for Experiment #2 Table 2a - Experiment #2a: Three Lemon Cells in Series, 4700 Ohm Resistance, 2.2 Farad Capacitor. Charge Time Charge Current Capacitor Circuit Current (I (minutes) (I charging ), "b"; ma charging / I circuit, max ) x 100 Voltage (V cap ) (I circuit ) "a"; ma Instantaneous OL on 2mA scale 1471% ---- ---- 0.5 0.713 524% 0.028 0.006 1 0.632 465% 0.040 0.009 3 0.566 416% 0.080 0.017 8 0.509 374% 0.178 0.037 10 0.504 370% 0.202 0.042 15 0.463 340% 0.276 0.057 20 0.438 322% 0.336 0.070 25 0.413 304% 0.392 0.081 30 0.395 290% 0.430 0.091 35 0.370 272% 0.484 0.100 40 0.349 257% 0.521 0.108 45 0.330 243% 0.544 0.113 50 0.313 230% 0.573 0.119 55 0.289 212% 0.596 0.124 60 0.251 185% 0.615 0.127 70 0.212 156% 0.638 0.132 75 0.196 144% 0.646 0.134 77 0.191 140% 0.648 0.135 80 0.185 136% 0.652 0.135 82 0.182 134% 0.653 0.135 83 0.180 132% 0.654 0.136 85 0.176 129% 0.656 0.136 Voltage at start: V cell 1 = 0.95V, V cell 2 = 0.95V, V cell 3 = 0.97V; end of expt = not measured Voltage (V d ) across 1N34A diode @ 17 min = 0.25V and @ 30 min = 0.23V {Analog meter} Estimated # s & Farads: Charge = 1.185 s, V = 0.656 Volts, Capacitance = 1.81 Farads Table 2b - Experiment #2b: Discharge of 2.2 Farad Capacitor through 4700 Ohm 150 Ohm, ( 145Ω) Discharge Time (minutes) Current @ "b"; ma Instantaneous 0 --- 10 ----- 1 0 0.528 3.41 1800 3 0 0.341 2.20 750 5 0 0.241 0.986 238 7 0 0.185 0.758 140 10 0 0.125 0.515 64 15 0 0.066 0.272 18 17 0 0.053 0.220 12 20 0 0.038 0.154 6 Dead short @ 2 & 4 0 0.017 & 0.010 0.169 & 0.098 ----- V cap (Volts) I circuit, "a"; ma Power (μw)
SECTION #4: Discussion of Data for Experiment #2, Three Lemon Battery (Series) Discussion of the Data for Experiment #2: 1. This experiment gave similar results to experiment #1, i.e., as the charging current decreases, the voltage across the capacitor and circuit current increases. 2. The instantaneous current at connection of the positive electrode exceeded the meter setting (OL - Over Load), thus, the instantaneous charging current was greater than 2mA. 3. After 30 seconds, the charging current dropped to 0.713mA, which was very similar to 0.707mA observed at 30 seconds in experiment #1a. 4. The goal was to let the experiment run until the charging current is equal to the circuit current, thus (I charging / I circuit, max ) x 100 = 100%. At 85 minutes, the percentage is 129%, thus, the capacitor can be considered to be very near completely charged at this point. 5. Since the charging time is 85 minutes, the series battery configuration (3 cells) is having a very difficult time charging the Super Capacitor. A same conclusion can be made for experiment 1a (two cells in series). 6. These conclusions make sense because in order to charge up a capacitor, the circuit needs current and batteries connected in series will always "ADD" their voltages and have the same current. Thus, a better battery configuration would be to connect the cells in parallel, because the cells "ADD" their currents and be better able to supply current for charging up the capacitor. 7. The final voltage across the Super Cap is 0.656V. In addition, the diode is dropping 0.23-0.25V (average of 0.24V). Thus the three cell battery is delivering: V cap + V d = 0.656V + 0.24V = 0.896V. 8. The charge and capacitance were calculated (reported at end of document) and found to be: Charge = 1.185 ombs Capacitance = 1.81 Farads 9. The charge transferred in this experiment is about 1.185/0.618 = 1.9 greater than in experiment 1a. 10. The number of electrons transferred is: 1.185 /1.602 x 10-19 /e - = 7.4 x 10 18 electrons. 11. Based on the final voltage drop of the capacitor, the Capacitance is 1.81 Farads. This is very similar to the 1.76 Farads of Capacitance determined in experiment 1a. This makes sense because even though there was about twice as much charge transferred in experiment 2a compared to experiment 1a, the voltage drop in experiment 2a is: 0.656V/0.350V = 1.9 times greater. 12. Capacitance has two variables, i.e., charge and voltage and since both variables change at the same rate between the two experiments, the capacitance will be similar. 13. Thus, both experiments charge up to the same (or similar) capacitance, but, experiment 2a stores about twice the charge. 14. In experiment 2b, I used an equivalent resistance of 145Ω to discharge the Super Capacitor, thus a considerably lower resistance than in experiment 1b, 1c or 1d. Thus, it is not surprising to see significantly larger currents in experiment 2b than in experiment 1b, 1c or 1d. 15. It should be noted that even though the battery configuration is supplying about 0.896V (point #7 above), when the positive lead is disconnected from the battery, one only gets the voltage (or potential) that is stored on the capacitor, i.e., at 1 minute of discharge, the voltage across the capacitor is 0.528V (Table 2b), which is comparable to 0.656V (@ 85 minutes, Table 2a). 16. From Table 2b at 1 minute, the power dissipated is about 1800μW (1.8mW), which is a considerable amount of energy. 17. From Table 2b at the 5 minute time point, the capacitor is still delivering about 1mA (0.986mA) of current, further demonstrating that the Super Capacitor has stored a significant amount of charge. 18. At 20 minutes (Table 2b), there still is charge left on the capacitor (i.e., 38mV & 154μA) and even a dead short across the capacitor for 4 minutes still did not fully discharge the capacitor. 19. Both experiments 1a and 2a demonstrate that the two or three lemon battery configuration is capable of supplying a significant amount of energy, which can be stored by the capacitor. 20. Finally, it is clear that the series configuration is experiencing difficulty in charging up the Super Capacitor, based on the fact that the charging time is 1-1.5 hours.
SECTION #5: Data for Experiment #3 Table 3a - Experiment #3a: Four Lemon Cells in Series, 4700 Ohm Resistance, 2.2 Farad Capacitor. Charge Time Charge Current Capacitor Circuit Current (I (minutes) (I charging ), "b"; ma charging / I circuit, max ) x 100 Voltage (V cap ) (I circuit ) "a"; ma Instantaneous 1.3 ma 949% ---- ---- 0.5 0.618 451% 0.026 0.006 1 0.548 400% 0.037 0.008 3 0.491 358% 0.056 0.012 5 0.436 318% 0.102 0.022 7 0.418 305% 0.129 0.027 10 0.402 293% 0.168 0.035 15 0.376 274% 0.226 0.044 20 0.359 262% 0.278 0.058 25 0.346 252% 0.325 0.067 30 0.344 251% 0.368 0.076 35 0.313 228% 0.406 0.084 40 0.298 218% 0.439 0.091 45 0.351 256% 0.473 0.098 50 0.331 242% 0.509 0.106 55 0.289 211% 0.539 0.112 60 0.250 182% 0.561 0.116 65 0.244 178% 0.578 0.120 70 0.240 175% 0.595 0.123 80 0.230 168% 0.625 0.129 90 0.224 164% 0.648 0.134 95 0.222 162% 0.660 0.137 Voltage at start: V cell 1 = 0.57V, V cell 2 = 0.93V, V cell 3 = 0.90V, V cell 4 = 1.07V; end of expt = not measured Voltage (V d ) diode @ 16 min = 0.23V; V battery, 42 min = 0.70V{V cap + V d = 0.454V + 0.23V = 0.684V} Estimated # s & Farads: Charge = 1.189 s, V = 0.660 Volts, Capacitance = 1.80 Farads Table 3b - Experiment #3b: Discharge of 2.2 Farad Capacitor through 4700 Ohm 68 Ohm, ( 67Ω) Discharge Time (minutes) Current @ "b"; ma 0 0 --- 9 --- 0.5 0 0.512 6.59 3374 1 0 0.418 5.35 2236 3 0 0.160 2.05 328 5 0 0.092 0.545 50 7 0 0.066 0.400 26 9 0 0.048 0.290 14 15 0 0.021 0.123 2.6 20 0 0.011 0.064 0.7 V cap (Volts) I circuit, "a"; ma Power (μw) ----------------------------------------------- Continued below ------------------------------------------------------
SECTION #5: Discussion of Data for Experiment #3 Discussion of the Data for Experiment #3: 1. From Table 3a, it can be seen that the charging current very closely approaches the circuit current, i.e., charging is 162% of circuit current. 2. The results for experiments 3a, i.e., a four cell battery, performed no better than the three cell battery, i.e., the final voltage of 0.660V is very similar to 0.656V (Table 2a). The ombs (1.189C & 1.185C) and Farads (1.81 & 1.80) are also very similar. 3. The reason for this result can be seen from the data obtained from the measurement of the individual cell voltages before experiment 3a: V cell 1 = 0.57V; V cell 2 = 0.93V; V cell 3 = 0.90V; V cell 4 = 1.07V. This data suggests that cell #1 was very worn down, i.e., it could only produce 0.57V versus the 0.90-1.07 volts for the other 3 cells. 4. Examination of the Zn plated iron nail of cell #1, after all of the experiments, showed that all of the Zn plating had been removed (dissolved) and the nail was black and corroded. In fact, it appeared as though the iron portion of the nail was starting to dissolve (appeared very pitted and rough). 5. Thus, this suggests that cell #1 was "depleted" or "drained" and the additional cell provided no advantage to the charging of the capacitor, i.e., the charging current must come from cell #1 and since it was "depleted" (Zn was literally stripped from the nail), adding an additional cell is series had no benefit. 6. This suggests that a battery configuration that has some combination of cells in parallel would be a better option (because there would be more than 1 cell that could provide charging current). 7. The voltage drop across the diode is 0.23V (Table 3a), as found in the first two experiments. 8. Considering the voltage dropped by the diode and the capacitor, the four cell battery was developing a voltage of 0.684V (measured = 0.70V), Table 3a, which is considerably less than the 0.896V(point #7 in discussion of experiment 2 data) produced by the three cell battery. This further suggests that cell #1 was unable to provide the required current for charging. 9. The number of electrons transferred is: 1.189 /1.602 x 10-19 /e - = 7.4 x 10 18 electrons. 10. Table 3b reports the discharge experiment, which contains an equivalent resistance of 67Ω, and the instantaneous current measured is 9mA, which is very similar to experiment 2b (10mA). 11. The difference being that in experiment 3b, the power dissipated at 30 seconds is 3374μW (3.37mW), which is almost double that in experiment 2b at 30 seconds (1800μW). This makes sense because the resistance in experiment 3b is almost 1/2 of that in 2b. 12. It is amazing to see that at 20 minutes (Table 3b), the capacitor is still dissipating about 1μW (0.7 μw) of power. This discharge experiment suggests that a lot of energy has been released or dissipated. 13. If one adds up all of the s from the 3 different experiments, the total is 0.681 + 1.185 + 1.189 = 3.055 s. This is equivalent of 1.9 x 10 19 electrons! 14. If one were able to move 3.055 s of charge and develop a voltage of 0.660V (as in experiment 3a), this would equal 4.6 Farads of capacitance. That is my next experiment. I will connect two 2.2F Super Caps in parallel (equivalent of 4.4F) and try to charge it with the four lemon battery as developed in experiment #8 (posted earlier), which is the series/parallel configuration. SECTION #6: Lessons Learned 1. The current battery designs can charge up a 2.2F Super Capacitor. 2. The number of s of charge stored will be 0.618C @ 0.350V or 1.189C @ 0.660V. 3. The total number of s of charge transferred in all 3 experiments is 3.055 s. 4. The four lemon battery potentially has the amount of energy required to charge up a 4.4F Super Cap, given the correct configuration of cells, preferably a series/parallel configuration. 5. The total number of electrons transferred in all 3 experiments is 1.9 x 10 19 electrons! 6. Discharge experiments indicate that, for short periods of time, the Super Caps can dissipate on the order of 10's to 100's and even 1000's of μw of Power.
SECTION #7: Calculations of ombs and Farads Expt #1 time ma @ b ma @ a diff in Curr Curr in time, Amp sec 0.5 0.707 0.004 0.703 0.000703 30 0.02109 1 0.558 0.006 0.552 0.000552 30 0.01656 3 0.412 0.013 0.399 0.000399 120 0.04788 5 0.364 0.018 0.346 0.000346 120 0.04152 7 0.329 0.024 0.305 0.000305 120 0.0366 10 0.303 0.029 0.274 0.000274 180 0.04932 15 0.282 0.039 0.243 0.000243 300 0.0729 20 0.255 0.046 0.209 0.000209 300 0.0627 25 0.241 0.053 0.188 0.000188 300 0.0564 30 0.229 0.059 0.17 0.00017 300 0.051 35 0.223 0.065 0.158 0.000158 300 0.0474 40 0.218 0.069 0.149 0.000149 300 0.0447 45 0.226 0.07 0.156 0.000156 300 0.0468 48 0.2 0.073 0.127 0.000127 180 0.02286 Total 0.618 Volts 0.35 Farads 1.76 Expt #2 time ma @ b ma @ a diff in Curr Curr in time, Amp sec 0.5 0.713 0.006 0.707 0.000707 30 0.02121 1 0.632 0.009 0.623 0.000623 30 0.01869 3 0.566 0.017 0.549 0.000549 120 0.06588 8 0.509 0.037 0.472 0.000472 300 0.1416 10 0.504 0.042 0.462 0.000462 120 0.05544 15 0.463 0.057 0.406 0.000406 300 0.1218 20 0.438 0.07 0.368 0.000368 300 0.1104 25 0.413 0.081 0.332 0.000332 300 0.0996 30 0.395 0.091 0.304 0.000304 300 0.0912 35 0.37 0.1 0.27 0.00027 300 0.081 40 0.349 0.108 0.241 0.000241 300 0.0723 45 0.33 0.113 0.217 0.000217 300 0.0651 50 0.313 0.119 0.194 0.000194 300 0.0582 55 0.289 0.124 0.165 0.000165 300 0.0495 60 0.251 0.127 0.124 0.000124 300 0.0372 70 0.212 0.132 0.08 0.00008 600 0.048 75 0.196 0.134 0.062 0.000062 300 0.0186 77 0.191 0.135 0.056 0.000056 120 0.00672 80 0.185 0.135 0.05 0.00005 180 0.009 82 0.182 0.135 0.047 0.000047 120 0.00564 83 0.18 0.136 0.044 0.000044 60 0.00264 85 0.176 0.136 0.04 0.00004 120 0.0048 Total 1.185 Volts 0.656 Farads 1.81
SECTION #7: Calculations of ombs and Farads Curr in time, Expt #3 time ma @ b ma @ a diff in Curr Amp sec 0.5 0.618 0.006 0.612 0.000612 30 0.01836 1 0.548 0.008 0.54 0.00054 30 0.0162 3 0.491 0.012 0.479 0.000479 120 0.05748 5 0.436 0.022 0.414 0.000414 120 0.04968 7 0.418 0.027 0.391 0.000391 120 0.04692 10 0.402 0.035 0.367 0.000367 180 0.06606 15 0.376 0.044 0.332 0.000332 300 0.0996 20 0.359 0.058 0.301 0.000301 300 0.0903 25 0.346 0.067 0.279 0.000279 300 0.0837 30 0.344 0.076 0.268 0.000268 300 0.0804 35 0.313 0.084 0.229 0.000229 300 0.0687 40 0.298 0.091 0.207 0.000207 300 0.0621 45 0.351 0.098 0.253 0.000253 300 0.0759 50 0.331 0.106 0.225 0.000225 300 0.0675 55 0.289 0.112 0.177 0.000177 300 0.0531 60 0.25 0.116 0.134 0.000134 300 0.0402 65 0.244 0.12 0.124 0.000124 300 0.0372 70 0.24 0.123 0.117 0.000117 300 0.0351 80 0.23 0.129 0.101 0.000101 600 0.0606 90 0.224 0.134 0.09 0.00009 600 0.054 95 0.222 0.137 0.085 0.000085 300 0.0255 Total 1.189 Volts 0.66 Farads 1.80 SECTION #8: Equivalent Circuit A = Ammeter, point "b" A A = Ammeter, point "a" A - + - +