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: **
2.3, 3.4
3.5, 1.10, 3.0, 1.25, 2.5, 1.47, 2.0, 1.69, 1.5, 2.07, 1.0, 2.31, .75, 2.59, .50, 2.98, .25, 3.10
I used the TIMER program to best estimate where the specified voltages were
** 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. **
1.69
.82
.62
.28
I measured the differnece on the y-axis (time) between points on the x-axis (voltage)
** Table of current vs. clock time using same resistor as before, again starting with 4 volts +- .02 volts. **
150, 2.4, 100, 3.2, 75, 4.3, 50, 4.9, 25, 5.3
** Times to fall from initial current to half; 75% to half this; 50% to half this; 25% to half this, based on graph. **
3.2
4.9
4.9
5.3
I graphed the current on the x-axis and time on the y-axis so that I could find the difference between points by simply subtracting the values on the x-axis.
** 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? **
They are similar. I think they are similar because current, resistance, and voltage are all related. If the current goes up and resistance stays the same, voltage must go up as well.
** Table of voltage, current and resistance vs. clock time: **
.8, 2.10, .6, 2.93, .4, 3.32, .2, 4.20, .1, 5.11
** Slope and vertical intercept of R vs. I graph; units of your slope and vertical intercept; equation of your straight line. **
Ohms, amps
y=-.698x+1.256E-4
My graph looks somewhat straight with a negative slope. It means there is an inverse relationship between resistance and current.
Terminology note: 'inverse' means reciprocal, and an inverse relationship wouldn't be linear. Your graph indicates resistance decreasing with increasing current. The phrase 'decreasing linear' would be one possible choice of terminology.
It is in any case clear what you are mean.
** Report for the 'other' resistor:; Resistance; half-life; explanation of half-life; equation of R vs. I; complete report. **
36.2
5.1 +/- 1.0
I measured the initial value as 0 and found half of the initial value. I am not exactly sure about 'dt but I tried to take the derivative of my graph and insert the time where it was 1/2 of it's initial value.
y=-1.56437+.773839
** Number of times you had to reverse the cranking before you first saw a negative voltage, with 6.3 V .15 A bulb; descriptions. **
12
I doubt this is accurate, but it is probably in the neighborhood of +/- 3
The bulb changed with varying reverses of the generator. At times, it seemed to be brighter than other times. Overall, I would say it maintained a pretty consistent glow, though, without too much fluctuation. The direction of the cranking is hard to say because each direction caused the bulb to increase and decrease brightness.
** When the voltage was changing most quickly, was the bulb at it brightest, at its dimmest, or somewhere in between? **
It is hard to tell, but I seemed like it was leaning towards brightest. This is probably because voltage is being sent and then changed so it has voltage at one point and then lacks that amount shortly thereafter.
** Number of times you had to reverse the cranking before you first saw a negative voltage, with 33 ohm resistor; descriptions. **
16
It is fairly accurate since I paid attention after the last time.
The capacitor voltage undoubtedly changed with time since the generator was being cranked in the reverse direction. This is not surprising since we continue to adjust the output from the generator. I am thinking the voltage decreased over time.
** 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. **
27, 12.1
No
5.4
** Voltage at 1.5 cranks per second. **
About 2.0 V
** Values of t / (RC), e^(-; t / (RC) ), 1 - e^(- t / (RC)) and V_source * (1 - e^(- t / (RC) ). **
.012, 3.2, 5.27E-6, 3.5E-4
** 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): **
I obtained.jk;lkwjr234@@#_)!)!
** According to the function V(t) = V_source * (1 - e^(- t / (RC) ), what should be the voltages after 25, 50 and 75 'beeps'? **
.jk;lkwjr234@@#_)!)!****DOS ERROR*****a2##!_)@)#@(((#))#_ADC,,s'[fix1]
** Values of reversed voltage, V_previous and V1_0, t; value of V1(t). **
-3.2, -2.6, 3.9, 5.2
5.2
I seriously doubt these findings because I'm not exactly sure what you are looking for.
** How many Coulombs does the capacitor store at 4 volts? **
2.3E-4
I multiplied to get Q, although this seems quite low.
** 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?; **
5.4E-5, 4.3E-4
** 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? **
3.9
3.4E-4
** 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? **
2.9
It is somewhat similar which should be expected since I=n*A*v*Q, so current is directly for velocity and Coulombs.
** How long did it take you to complete the experiment? **
2 hours
** **
Is there a way we can check our grades?
I can always give you an update, but until you've completed a test there's no good indication of your grade trend.
Your labs are good, at least at B level. I haven't seen enough of your work on the assigned problems to make an accurate grade estimate, but that work appears to be consistent with the level of your labs.
At this point an A on the test would give you an A average; a midrange C or a B on the test better would give you a B average.