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: **
** Initial voltage and resistance, table of voltage vs. clock time: **
4.01 Volts, I couldn't find the resistance on the resistor.
4.09375, 3.5
5.140625, 3.0
5.90625, 2.5
7.484375, 2.0
10.23438, 1.5
13.71875, 1.0
10.21875, 0.75
15.70313, 0.5
25.70313, 0.25
These results show that as time passes, the voltage across the circuit slows down. This means that resistance is building in the circuit and impeding the current.
Since your times as reported in the first column do not continue to increase (e.g., falling from about 13.7 to 10.2 between 1.0 and .75 volts) it seems likely that you have reported voltage vs. time interval rather than voltage vs. clock time. Clock time would represent the times showing on a clock which started at 0 and continued running, so clock time cannot decrease.
Before I can evaluate the rest of this document I need to know for sure if you are reporting clock time or time intervals, and I need to answer any question you might have on the difference.
Please respond with a note answering this question, and refer me to the date 070630 so I can easily find this document.
** 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. **
3.485 sec
5.094 sec
6.234 sec
2.0 sec
My graph of voltage vs clock time is decreasing at a decreasing rate. I drew a curve to try to fit most of the points on my graph and from there determined the approximate times that were associated with each voltage. As my times above indicate, it takes more and more time for the voltage to decrease. The voltage intervals here were 2V, 1.5V, 1V, and .5V.
** Table of current vs. clock time using same resistor as before, again starting with 4 volts +- .02 volts. **
6.5, 1.71875
6.0, 1.796875
5.5, 1.593755
5.0, 1.46875
4.5, 1.8125
4.0, 1.65625
3.5, 3.546875
3.0, 3.578125
2.5, 5.203125
2.0, 9.03125
1.5, 7.890625
1.0, 13.45313
.5, 27.3125
This graph looks very similar to the voltage vs time graph. It is decreasing at a decreasing rate. Therefore, the current flowing through the circuit is slowly decreasing.
** Times to fall from initial current to half; 75% to half this; 50% to half this; 25% to half this, based on graph. **
1.828125
3.734375
4.34375
5.562505
I drew a curved line to best fit the graph points and then found the times that corresponded to the currents. I then found the differences in times from one current to the other and reported them above. My graph shows that current decreases over time and that it takes longer for each subsequent decrease in current.
** 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 times are not the same, but they are fairly close. Time intervals increase with decreasing voltage and current.
** Table of voltage, current and resistance vs. clock time: **
1.5312525, 3.02, 5.2, 0.58
1.65625, 3.00, 3.9, .77
5.203125, 2.49, 2.6, .96
7.890625, 2.1, 1.3, 1.6
20.3828125, .29, .65, .46
I first found the time intervals corresponding to the desired current. I then used my graph to find the voltages which corresponded to those times. I then calculated the resistance by dividing this voltage by the current.
** Slope and vertical intercept of R vs. I graph; units of your slope and vertical intercept; equation of your straight line. **
-.0748, 1.0781
ohms/amp, ohms
R = -.0748I + 1.0781
My graph has a negative slope which indicates that the two variables are negatively correlated. The slope and intercept also indicate that as the resistance increases, the current decreases and as the current increases, the resistance decreases.
** Report for the 'other' resistor:; Resistance; half-life; explanation of half-life; equation of R vs. I; complete report. **
Again, my resistors were not labeled and therefore, I do not know the resistance. However, this resistor cause the voltage to decrease MUCH slower.
280.4844+-.0004
R = -73.184I+280
As for the actual slope and intercept, I am not very confident in my values. I tried to make this graph of several times and this equation seemed to be the best answer. It shows again the negative correlation between resistance and current. However, the difference from the last one is that the values for voltage and current were much different from each other. For example, in one case the voltage was 1.9V and the current was 45mA. Also, a big difference that I noticed was that the current decreased quicker with this reisistor, whereas the voltage decreased MUCH SLOWER with this resistor, compared to the previous. Therefore, it seems intuitive that the one used in these trials must have the higher resistance.
** Number of times you had to reverse the cranking before you first saw a negative voltage, with 6.3 V .15 A bulb; descriptions. **
I reversed the direction 17-19 times.
My estimate is definitely the correct range, I simply can't remember whether it was 17 or 19 reverses.
The bulb was glowing when I started cranking initially. Gradually the light faded and it became much easier to turn the crank. When I reversed the direction for the first time, the light shown VERY brightly (so bright I was afraid it would burn out) and the crank was hard to turn. When I reversed again, the light disappeared. It continued like this several times until the light began to glow evenly when I cranked in either direction. The voltage of the capacitor continued an OVERALL decrease. When the light was not glowing was when the capacitor had reached its limit of charge. When I reversed the direction, it seems that it pulled charge out of the capacitor, allowing the current to pass to the bulb. (I might have this idea backwards)
** When the voltage was changing most quickly, was the bulb at it brightest, at its dimmest, or somewhere in between? **
When the voltage was decreasing rapidly, the bulb glowed the brightest.
The faster the rate of change in capacitor voltage, the better the current flows through the circuit, and the brighter the bulb glows.
** Number of times you had to reverse the cranking before you first saw a negative voltage, with 33 ohm resistor; descriptions. **
12 reversals
This estimate is fairly accurate. I repeated this trial several times and the number of reverses was usually between 11-15.
The capacitor voltage change was similar to the previous trials with the light bulb. However, the voltage did not change as rapidly and it only decreases when I cranked backwards, instead of a steady decline in both directions. When I cranked forward, the voltage increased but just barely. When I cranked backward, the voltage decreased slowly, but still faster than it changed when I cranked forward.
** 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. **
22 beeps, 29.3 sec
The voltage was changing more quickly as it approached zero.
Peak voltage was 3.10 V.
** Voltage at 1.5 cranks per second. **
5.18 Volts
** Values of t / (RC), e^(-; t / (RC) ), 1 - e^(- t / (RC)) and V_source * (1 - e^(- t / (RC) ). **
.888, .411, .588, 3.05
From the calculations that I just made, the value of t/RC was .888, the value of e^(-t/RC) was .411, and the value of 1-e^(-t/RC) was .588. This gave me a theoretical voltage of 3.05V which is extremely close to the voltage that I observed at the time I reversed cranking.
** 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): **
3.05, 3.10
The difference between the theoretical voltage and the voltage that I observed is 0.05 Volts, which is 1.6% of the theoretical voltage.
** According to the function V(t) = V_source * (1 - e^(- t / (RC) ), what should be the voltages after 25, 50 and 75 'beeps'? **
3.29, 4.49, 4.93
I was a little confused in answering this question. To calculate the answers, I used the time I calculated with MY rate of beeps/sec.
** Values of reversed voltage, V_previous and V1_0, t; value of V1(t). **
-3.10, 3.10 and -6.20, 29.3, -.55
The reversal voltage is the opposite of my peak voltage, 3.10V. The V_previous is the peak voltage at the instant of reversal. V1_0 is the voltage which equals reversed voltage - v_previous, -3.10-(+3.10) = -6.2V. The time that was required for my voltage to return to zero was 29.3 seconds. It took 22 beeps to get back to zero at a rate of 0.75 beeps/sec, (22/.75 = 29.3 sec) The last number -0.55 is the voltage that I calculated from the equation V1(t) = V_previous + V1_0*(1-e^(-t/(RC)). I'm honestly not sure what this voltage represents. It seems from the formula that it is theoretical voltage that should have been reached in the time I cranked in reverse.
** How many Coulombs does the capacitor store at 4 volts? **
The capacitance on my capacitor is 1.0 Farad. Therefore the charge it holds is equal to Q = F*V = 1.0F * 4V = 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 Coulombs, 0.5 Coulombs
Since the charge on the capacitor is the product of its capacitance, 1.0F, and the voltage, 3.5, it is equal to 1.0F * 3.5V = 3.5 Coulombs. Therefore, the capacitor loses 0.5 Coulombs between 4V and 3.5V.
** 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? **
4.09375 sec, 0.122 C/sec
According to the first data I reported for this experiment, it took approximately 4.09375 seconds for the voltage to drop from 4.01V to 3.5V. Since the capacitance is 1.0F, the drop in voltage is the same value as the drop in charge. Therefore, the charge dropped 0.5C / 4.09375sec = .122 C/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? **
During the time that the voltage dropped 0.5V, the current should have been approximately 0.02 C/sec.
This data is less that my calculated value for current above. It makes sense that it is lower, however, because the resistance is accounted for on the graphs. During the experiments, the current faced a certain amount of resistance as it passed through the circuit. Another reason the two values are different is because of unavoidable discrepancies in obtaining the time intervals as the voltage quickly dropped.
** How long did it take you to complete the experiment? **
4 hours twice, 8 hours total
** **
You have good data, but before I go further with my commentary, please respond to the question in my note.