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, 32
3.0, 60
2.5, 74
2.0, 98
1.5, 126
1.0, 178
0.75, 130
0.50, 172
0.25, 307
This table appears to be reversed, showing clock time vs. voltage.
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.
225
298
301
301
As time increases the volts decrease. The graph has a negative slope that is almost constant. I got these times by looking at the graph and adding together the intervals.
Table of current vs. clock time using same resistor as before, again starting with 4 volts +- .02 volts.
40 amps, 100 ohms
35, 0.60
30, 0.60
25, 0.60
20, 0.65
15, 0.65
10, 0.65
7.5, 0.82
5.0, 0.62
2.5, 0.64
Clock times ran from about 0 sec to about 300 sec in your preceding table; note that the clock times will be strictly increasing. This table doesn't show current vs. clock time; it doesn't have a column that could be interpreted as clock time.
Times to fall from initial current to half; 75% to half this; 50% to half this; 25% to half this, based on graph.
1.75
2.40
0.65
0.63
The graph is decreasing from high amps to low amps. I got these numbers like I did before, I looked at the graph at the intervals and added them together.
If your graph is based on the table you gave for current vs. clock time, your conclusions will be faulty.
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?
Because of the amps decreasing a much faster rate, it is harder to get it accurate reading. There are similarities between the voltage vs. time and the current vs. time because both of the 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 the percentages given and estimated the volts and time.
This table appears to be plausible, except that your currents are probably in milliamps so your resistances will be in the 8-10 ohm range rather than the .08 to .1 range.
Slope and vertical intercept of R vs. I graph; units of your slope and vertical intercept; equation of your straight line.
-0.33, 39
these results don't follow from your table. The resistances you report vary little, while current varies greatly.
mAmps/ohms
y=-0.33x+38
The graph looks like an incomplete parabola that is upside-down. It decreases at a faster rate as resistance increases. I used the slope equation with two points from the graph from the best fit line on the graph. The vertical intercept is where the line of best fit crosses 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
The time found at 4V was taken and then added the intervals from there until 2V.
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.
7 times
This is pretty accurate and only maybe off a little bit by a turn.
It depended on the direction of the cranking on the brightness of the bulb. This is because the generator in the starting direction was storing more energy, and then when in reverse,it was storing more energy as well as using the energy that was stored. 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?
The bulb 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.
28 times
I think it was accurate enough, at least within a rotation or two.
The voltages dropped a lot slower when it had the resistor in the set up. It also took a lot more turns to drain the capacitor.
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.
59 beeps, 29.4 sec
The voltage changed more quickly as it approached 0.
The peak voltage was 1.02
Voltage at 1.5 cranks per second.
3.9 volts are produced
Values of t / (RC), e^(-; t / (RC) ), 1 - e^(- t / (RC)) and V_source * (1 - e^(- t / (RC) ).
0.86, 0.42, 0.58, 0.87
I plugged the numbers 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
0.19 volts
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 used the equation above to get these numbers. These are the relationship between the starting point and the number of beeps per second.
How many Coulombs does the capacitor store at 4 volts?
4C/V because when I plugged in 1C*4 volts is 4C/V.
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?;
1C, it loses 0.5C/V
I used the equation that was given and plugged in the numbers
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
The first number is how long it took me to get to 4 volts and the second number is the difference between 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 the voltage is high, the current flows freely.
How long did it take you to complete the experiment?
4 hours
One of your tables didn't appear to report clock times; and it isn't clear how you obtained the straight-line approximation for R vs. I. See my notes. Otherwise your work looks very good.