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: **
4.0 V, 33 ohm ??
4.0, 0.0
3.5, 20.625
3.0, 44.26563
2.5, 73.21876
2.0, 109.21876
1.5, 155.70314
1.0, 185.34377
.75, 221.93752
.50, 342.25002
.25, 477.67192
Evertime the multimeter would get to one of the measurement marks, I would push the timer button to capture the time the marker passed. Then I had a list of the times at which certain voltages occurred. The rate between measured voltages is decreasing as time goes on.
** 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. **
109
111
76
157
My graph is decreasing at a decreasing rate from left to right. To get my estimates I figured out the time at the two voltages and then subtracted them from eachother to get the time between each group of voltages.
** Table of current vs. clock time using same resistor as before, again starting with 4 volts +- .02 volts. **
100 mA
75 mA
50 mA
25 mA
I could only get 4 data points because I charged the capacitor up to about 6 Volts, and pulled off the generator clamps, and waited for the multimeter to read 4 Volts before changing over to the current reader. Once changed over to the current readings it was 100 mA and going down fast, so I did my best to get as many points as possible. I don't like depending on a graph with few data points, but in this case I feel I have to.
** Times to fall from initial current to half; 75% to half this; 50% to half this; 25% to half this, based on graph. **
38 sec
52 sec
27 sec
80 sec
** 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. They are not the same as the times for voltage. But the graphs do look similar. They are both decreasing at a decreasing rate.
** Table of voltage, current and resistance vs. clock time: **
14 sec, 4.7 V, .08 A, 58.75 Ohms
28 sec, 3.3 V, .06 A, 55 Ohms
47 sec, 3.0 V, .04 A, 75 Ohms
70 sec, 2.7 V, .02 A, 135 Ohms
I found the coordinating points on each graph and drew straight lines and made an estimation to get the point values, then just used the forlmula, Resistance = Volts/ Amps to find the resistance in Ohms.
** Slope and vertical intercept of R vs. I graph; units of your slope and vertical intercept; equation of your straight line. **
-1395, 150
Ohms/Amp, Ohms
Resistance = -1395 * Current + 150
I drew a best fit line through the points on the graph, which were decreasing at a decreasing rate and then beginning to increase a little at the very end. (that may have been just a miscalulation or something). To Find the y-intercept, I extended the best fit line all the way to the y-axis and it hit around 150 Ohms. To get the slope I chose two points along the best fit line and used the slope formula m = (y1 - y2)/(x1-x2).
** Report for the 'other' resistor:; Resistance; half-life; explanation of half-life; equation of R vs. I; complete report. **
** Number of times you had to reverse the cranking before you first saw a negative voltage, with 6.3 V .15 A bulb; descriptions. **
Beep Program refuses to cooperate.
** When the voltage was changing most quickly, was the bulb at it brightest, at its dimmest, or somewhere in between? **
N/A
** Number of times you had to reverse the cranking before you first saw a negative voltage, with 33 ohm resistor; descriptions. **
N/A
** 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. **
N/A
** Voltage at 1.5 cranks per second. **
N/A
** Values of t / (RC), e^(-; t / (RC) ), 1 - e^(- t / (RC)) and V_source * (1 - e^(- t / (RC) ). **
N/A
** 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): **
N/A
** According to the function V(t) = V_source * (1 - e^(- t / (RC) ), what should be the voltages after 25, 50 and 75 'beeps'? **
N/A
** Values of reversed voltage, V_previous and V1_0, t; value of V1(t). **
N/A
** How many Coulombs does the capacitor store at 4 volts? **
On the capacitor there is writing that says 5.0 V and 1.0 F. Sooooo, I am assuming that the capacitor holds 1 Farad or 5 Volts. That would make the number of Coulombs also 5. So to get the answer, you would just have to divide the number of volts by the number of Coulombs to get the total, which would be 4 Volts/5 Coulombs = 0.8 Farad.
** 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?; **
2.45 Coulombs, 2.55 Coulombs
I divided 3.5 V (current capacity) by 5 V (total max capacity) to get the percentage that was still left. Then multiplied that percentage by 1 F (max Farad of the capacitor). Then I filled these numbers into the formula to get the number of Coulombs that capacitor contains at 3.5 V. F = C/V .7 = C/3.5 C = 2.45
Then we know that the number of Coulombs at 4 Volts is 3.2 so we just subtract the 2.45 from the 3.2 and get the difference, which is .75.
** 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? **
N/A
** 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? **
N/A
** How long did it take you to complete the experiment? **
1.5 hours
** **
&#Your work looks very good. Let me know if you have any questions. &#