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 volts, 100 ohms
3.5, 34
3.0, 59
2.5, 76
2.0, 97
1.5, 124
1.0, 176
0.75, 125
0.50, 175
0.25, 305
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.
226
297
300
300
The graph is increasing in time as the volts decrease. So it has a negative slope, that is pretty constant. I used the graph to get these numbers for the table above by looking at the intervals between the values I wanted to obtain and adding them together.
Table of current vs. clock time using same resistor as before, again starting with 4 volts +- .02 volts.
40 Mamps, 100ohms
35,0.58
30,0.58
25,0.58
20,0.64
15,0.63
10,0.63
7.5,0.81
5.0,0.58
2.5,0.63
The table you gave in response to a previous question showed clock time vs. voltage rather than voltage vs. clock time. Clock times appear to have ranged up through 305 seconds while voltages dropped through 3.5 volts, ..., 2 volts, ..., .25 volts. The convention is 'y vs. x', so clock times should have been in the first column, voltages in the second.
In this table I do not see clock times at all, and the readings in the first column seem to indicate the currents in milliamps (your first column would therefore make a good second column). However I don't see the clock times in either column. Also there doesn't seem to be any discernable trend in the second column, and I can't tell what that column represents.
Can you clarify?
Please respond by just copying this question, your responses and my comments into a text document and inserting your explanation. As my comments are set off by double asterisks **, set off your inserted comments using &&. Just submit a copy using the Submit Work form. You are of course welcome to also submit additional questions if you have them.
Times to fall from initial current to half; 75% to half this; 50% to half this; 25% to half this, based on graph.
1.74 sec
2.38 sec
0.64 sec
0.63 sec
The graph looks like an exponential graph that is decreasing, going from high amps to low amps, the intervals between them get longer. I used the graph to get these numbers for the table above by looking at the intervals between the values I wanted to obtain and adding them together.
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?
It was much harder to get the amps, they decreased at a much faster rate, they arent that accurate. Voltage is pretty accurate though. There are simularities between the voltage vs time and the current vs time because both of the grpahs intervals increase in time as they decrease in the amount of volts or amps.
Table of voltage, current and resistance vs. clock time:
47,3.2, 39,0.082
83,2.4, 28,0.086
108,1.6, 19,0.089
139,0.8, 9,0.089
161,0.4, 4,0.100
I looked at the graph at each of the percentages given above and estimated the amount of volts and the time at that point by looking at the line on the graph from before.
Slope and vertical intercept of R vs. I graph; units of your slope and vertical intercept; equation of your straight line.
-0.33, 38
mAmps/ohms, 38
y=-.033x+38
The graph is a half of a parabola turned upside-down. It is starting high and decreases at a faster as resistance increases. I got slope by drawing a best fit line on the graph, taking two points and using the slope equation. The vertical intercept I got was by figuring the line to where it would cross the y axis.
Report for the 'other' resistor:; Resistance; half-life; explanation of half-life; equation of R vs. I; complete report.
33 ohms
35.2 sec +/-0.5 sec
I took the voltage at 4 V and found that time and added the intervals from there till 2 volts.
y=0.15x+0
Number of times you had to reverse the cranking before you first saw a negative voltage, with 6.3 V .15 A bulb; descriptions.
9 times
This was an accurate reading of how many times I reversed the cranking.
The brightness of the bulb changed depending on the direction of cranking. When I cranked in the starting direction it was dimmer than when cranked in the reverse direction. This is because the generator in the starting direction was storing more energy and in the reverse direction it was not only putting in more energy, but also using the energy that was stored in the capacitor. The voltage decreased faster than what it gained.
When the voltage was changing most quickly, was the bulb at it brightest, at its dimmest, or somewhere in between?
It was at the brightest. They are negatively correlated, the capacitor loses voltage as the bulb grows brighter.
Number of times you had to reverse the cranking before you first saw a negative voltage, with 33 ohm resistor; descriptions.
27 times
I think it was pretty accurate within one or two rotation
It took a lot more turns to drain the capacitor. The voltages dropped a lot slower when it had the resistor in the set up. With the set up this way it only dropped about 0.05 volts for each change in direction where as the other one dropped much faster.
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.
57 beeps, 28.5 sec
The voltage changed more quickly as it approached 0.
Peak voltage 1.06
Voltage at 1.5 cranks per second.
3.9 volts
Values of t / (RC), e^(-; t / (RC) ), 1 - e^(- t / (RC)) and V_source * (1 - e^(- t / (RC) ).
.86, 0.42, 0.58, 0.87
I got these numbers by plugging them into the equation.
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):
0.87 volts, 1.06 volts
+0.19volts
According to the function V(t) = V_source * (1 - e^(- t / (RC) ), what should be the voltages after 25, 50 and 75 'beeps'?
0.69
1.70
3.17
Values of reversed voltage, V_previous and V1_0, t; value of V1(t).
1.45, 0.49, 1.20, 2.24
I got these results from using the equation above. These show the relationship between the starting point and the number of beeps per second.
How many Coulombs does the capacitor store at 4 volts?
The charge is 4 C/V. I plugged in 1 coulomb*4volts.
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?;
1 C, it loses 0.5C/V.
I found this by using the equation above.
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?
28.5, 0.5
This is how long it took me to get to 4 volts and the 0.5 is from the differece betweeen the 4 volts and the 3.5 volts.
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?
37 mamps
As voltage is high current flows freely.
How long did it take you to complete the experiment?
4 Hours
Good work overall; please respond as requested on my one question.