Your 'the rc circuit' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** Your comment or question: **
I accidently submitted this with no answers previously. I'm using the 6.3 V, .15 A light bulb AND 14 V, .2 A bulb as my resistors.
** Initial voltage and resistance, table of voltage vs. clock time: **
4.02
3.765625, 3.5
8.84375, 3
11.96875, 2.5
14.203125, 2
20.03125, 1.5
29.609375, 1
35.09375, 0.75
40.34375, 0.5
48.921875, 0.25
The table above represents the the time for a specified voltage. The time is in seconds in the first column, and the voltage is in volts in the second colums. This was performed by creating a circuit connecting a resistor in series with a capacitor, and a voltmeter in parallel with the capacitor. The capacitor was charged to 4 volts with a generator, and then a timer program was used to find the length of time it took for the voltage to decrease once the resistor was in place.
** Times to fall from 4 v to 2 v; 3 v to 1.5 v; 2 v to 1 v; 1 v to .5 v, based on graph. **
14.2
11.2
15.4
10.7
The graph of voltage vs. clock time shows that voltage decreses with clock time. I used the graph to determine the time required for the voltage to fall a specific number of volts by subtracting the one x value from the other. Ex: Time required for voltage to fall from 1 volt to .5 volts = 40.34375 sec - 29.6093 sec = 10.7 sec.
** Table of current vs. clock time using same resistor as before, again starting with 4 volts +- .02 volts. **
0, 90
9.0625, 75
28.7, 50
56, 25
The results above represent current through a circuit vs. clock time. The time in the first column is measured in seconds, while the current in the second column is measured in mA. These results were found by setting up a circuit in parallel including a resistor, capacitor and ammeter. The capacitor was charged to 4 V and then the time was noted as the current flowed through the circuit until the capictor became de-charged.
** Times to fall from initial current to half; 75% to half this; 50% to half this; 25% to half this, based on graph. **
33.3
25.7
23.7
15.3
The graph of current vs. time lapse reveals that current decreases with time at a decreasing rate. This graph was used to find the time required for the current to fall from one specified point to half of that point. This was found by adding a trendline to the graph and using the linear equation to plug in time current values to find time values.
** Within experimental uncertainty, are the times you reported above the same?; Are they the same as the times you reports for voltages to drop from 4 v to 2 v, 3 v to 1.5 v, etc?; Is there any pattern here? **
The results for current drop were higher than the results for voltage drop. However there is a pattern of decreasing values.
** Table of voltage, current and resistance vs. clock time: **
16.4, 1, 0.072, 13.9
26.3, 0.3, 0.054, 5.6
41.8, 0.57, 0.036, 15.8
62.9, 1.8, 0.018, 100
75.6, 2.8, 0.009, 311
The table above represents clock time in seconds in the first column, voltage in V in the second column, current in amps in the thrid column, and resistance in ohms in the fourth column. These results were found by creating a trend line on the current vs. time graph and using that linear equation to find the times when the current was .8, .6, .4, .2 and .1 times the initial current. The current was converted from milliamps to amps. Then these times were plugged into the linear equation for the voltage vs. time graph to find the corresponding voltages for each of these times. The resistance was found by dividing voltage by current.
** Slope and vertical intercept of R vs. I graph; units of your slope and vertical intercept; equation of your straight line. **
-3914.2, 237.22
ohms/amp, ohms
R = -3914.2I + 237.22
The graph is not consistent. At first, ohms decrease dramatically with current, and then the graph levels off and slightly curves back up. I think this has to do with using a light bulb rather than a regular resistor. I found the slope and vertical intercept by my using excel to tell me the linear equation. Slope is rise/run.
** Report for the 'other' resistor:; Resistance; half-life; explanation of half-life; equation of R vs. I; complete report. **
45.45 R
19.84 sec +- 0.1 for voltage, 34.7 sec +- 0.1 for current
I determined t+-`dt by using excel to find the linear equations of the graph of time vs. current and observing the time that passed for the current to drop to half that of the original. I had already found the time for half of the original voltage when I was collecting data.
R = 84.161I + 28.784
First I found the correlation between voltage and time as well as current and time. The voltage and time results were found by setting up a circuit in parallel including a resistor, capacitor and ammeter. The capacitor was charged to 4 V and then the time was noted as the current flowed through the circuit until the capictor became de-charged. Then the same procedure was used, with the exception that the circuit was in series, to find the current flowing through the circuit. Then I found resistance vs. current. These results were found by creating a trend line on the current vs. time graph and using that linear equation to find the times when the current was .8, .6, .4, .2 and .1 times the initial current. The current was converted from milliamps to amps. Then these times were plugged into the linear equation for the voltage vs. time graph to find the corresponding voltages for each of these times. The resistance was found by dividing voltage by current. This graph of resistance vs. current does not seem to be appropriate for the intended results which should show that the two quantities are indirectly proportional.
** Number of times you had to reverse the cranking before you first saw a negative voltage, with 6.3 V .15 A bulb; descriptions. **
37
I think this estimate was fairly accurate, within +-2 cranks.
The bulb glowed at first, then diminished as I approached 100 cranks. When I reversed directions the first time (now turning counterclockwise), the bulb began to glow brightly and the voltage decreased. When I reversed to the original direction (clockwise), the bulb glowed dimly and the voltage increased. Each time I reversed to the counter clockwise position, the bulb glowed a little less brightly. Each time I reversed to the clockwise position, the bulb glowed slightly more bright. This pattern continued until the end, when the bulb had nearly the same brightness regaurdless of the direction I was cranking. I think the reason for the bulb glowing brightly in the counterclockwise cranking direction was due to decreased voltage in the capacitator which allowed for decreased resistance so that more current would flow to the bulb. Reversing to the clockwise direction did the opposite of this.
** When the voltage was changing most quickly, was the bulb at it brightest, at its dimmest, or somewhere in between? **
I think the voltage was changing most quickly when the bulb was brightest. The brightness of the bulb is related to the rate of voltage change because of the relationship between voltage, resistance and current.
** Number of times you had to reverse the cranking before you first saw a negative voltage, with 33 ohm resistor; descriptions. **
5 times
This is an accurate measurement of the number of reversals required. However the measurement of the time constant may be off because I found resistance by the voltage and current listed on the light bulb, which may not be accurate.
I used the 14 V bulb in place of the resistor. I found the time constant for the bulb and generator by multiplying ohms by Farads to be 70 seconds. I cranked the generator at a rate of 1.6 cranks per second for 224 cranks (which was found by multiplying the time constant by 2 and then by 1.6 beeps/sec. I then reversed the direction of the generator to counter clockwise for 28 cranks (found by multiplying the time constant by .25 and then by 1.6 beeps/sec). I then reversed the direction of the every 28 cranks thereafter. Every time I cranked counterclockwise the voltage decresed. It increased at a slower rate when cranking clockwise. So overall the capacitor voltage was decreasing becuase it decreased at a faster rate than it increased.
** How many 'beeps', and how many seconds, were required to return to 0 voltage after reversal;; was voltage changing more quickly as you approached the 'peak' voltage or as you approached 0 voltage; 'peak' voltage. **
39, 62.4
The voltage seemed to change more quickly as I approaced peak voltage than 0 voltage, but I'm not sure because my mind can't think about so many things at one time.
3.45, 3.30
** Voltage at 1.5 cranks per second. **
3.3
** Values of t / (RC), e^(-; t / (RC) ), 1 - e^(- t / (RC)) and V_source * (1 - e^(- t / (RC) ). **
.89, .41, .59, 1.9
The number in the first column was found using the the fomula t / (R C). The second number was found by using the formula e^ -t / (R C). The third number was found by subracting the second number from 1. The last number was found by multiplying the third number by 3.3
** Your reported value of V(t) = V_source * (1 - e^(- t / (RC) ) and of the voltage observed after 100 'cranks'; difference between your observations and the value of V(t) as a percent of the value of V(t): **
1.9, 3.3
There is a difference of +1.4 V between these two values, which means something probably wasn't calculated correctly. But I got nearly the same answer twice. I think this is because resistace was caluclated from the figures written on the bulb.
** According to the function V(t) = V_source * (1 - e^(- t / (RC) ), what should be the voltages after 25, 50 and 75 'beeps'? **
2.6
2.1
1.69
** Values of reversed voltage, V_previous and V1_0, t; value of V1(t). **
-3.30
3.45, -3.30
62.4
-6.27
I'm lost as to what number you are looking for and whether they are supposed to be positive or negative. I also don't know what V1(t) stands for. What did you mean by after the first reversal? Does this mean after one crank of the first reversal, or after it has been reversed down to 0 V?
** How many Coulombs does the capacitor store at 4 volts? **
4
A capacitor of 1.0 farads stores 4 Coulombs at 4 volts. This was found using the equation: Coulomb = (1 Coulomb/Volt) (4 Volts) = 4 Coulombs
** How many Coulombs does the capacitor contain at 3.5 volts?; How many Coulombs does it therefore lose between 4 volts and 3.5 volts?; **
3.5, .5
The capacitor is charged with 4 Coulombs at 4 volts, and 3.5 Coulombs at 3.5 volts, so that it loses .5 Coulombs between the voltages of 4 and 3.5.
** According to your data, how long did it take for this to occur when the flow was through a 33-ohm resistor?; On the average how many Coulombs therefore flowed per second as the capacitor discharged from 4 V to 3.5 V? **
4.1, .12
The capacitor took 4.1 seconds to discharge from 4 to 3.5 V. Divide this by .5 Coulmbs to get .12 Coulombs per sec.
** According to your data, what was the average current as the voltage dropped from 4 V to 3.5 V?; How does this compare with the preceding result, how should it compare and why? **
76.6
This is the amount of current that correlates to the time in the previous box of 1.4 seconds
** How long did it take you to complete the experiment? **
5 hours
** **
Good work on this experiment.