Your work on the rc circuit 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:
4.02V,33ohm resistor
3.5V,2.17s
3.0, 6.51s
2.5, 11.49
2.0,17.58
1.5,25.78
1.0,35.36
.75,47.15
.50,61.07
.25,86.03
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.
15.41s
19.27s
23.87s
25.63s
Table of current vs. clock time using same resistor as before, again starting with 4 volts +- .02 volts.
17.1s
30.57s
18.57s
23.26s
Times to fall from initial current to half; 75% to half this; 50% to half this; 25% to half this, based on graph.
100mA start
90, 1.64s
80, 2.70
70,3.55
60,4.16
50,5.05
40,6.14
30,7.33
20,10.20
10, 16.11
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?
not exactly the same, but very similar. The pattern is that they are both decending
Table of voltage, current and resistance vs. clock time:
3.1V,.8mA, 3.875, 4.34s
2.4V,.6mA,4,12.05s
1.7,.4mA,4.25,23.24
.85,.2,4.26,40.77s
.64,.1,6.4,56.88
I believe the range here would be 10 mA to 80 mA. This would change the decimal points on your resistances.
However resistance = voltage / current, so resistances here would be on the order of 40 ohms (e.g., 2.4 V / (.060 amps) = 40 ohms).
Slope and vertical intercept of R vs. I graph; units of your slope and vertical intercept; equation of your straight line.
slope = -2.6692, y-intercept = 5.6761
y=-2.6692x + 5.6761
I used the excel program, put in my data, plotted the points and fit the best line then inserted the equation of the line.
Report for the 'other' resistor:; Resistance; half-life; explanation of half-life; equation of R vs. I; complete report.
100 ohm resistor
64.32s, +-.5s
We determined t but adding all the time intervals together and we determined 'dt by the time it takes to get the sound to a persons ear and the time to react to that sound.
Note the instruction:
Starting in the fourth line, in any order you deem appropriate to the needs of the reader, report your procedure, data, analysis and interpretation of results:
Number of times you had to reverse the cranking before you first saw a negative voltage, with 6.3 V .15 A bulb; descriptions.
15 times before we saw the negative voltage
yes it was accurate
The bulb glowed the brightest when you crank backward. When you crank backward you are taking all the energy you put into the system back out and therefore using up the energy quicker.
When the voltage was changing most quickly, was the bulb at it brightest, at its dimmest, or somewhere in between?
The bulb was at its brightest when the voltage was changing most quickly. The brighter the bulb the more voltage was changing.
Number of times you had to reverse the cranking before you first saw a negative voltage, with 33 ohm resistor; descriptions.
4 times
yes the estimate was accurate
The voltage built up very slow then when reversed it decreased very quickly.
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.
15 beeps,15s
The number of beeps won't equal the number of seconds, unless you are cranking at 1 crank per second; 1 crank per second won't give you 4 volts.
The voltage changed more quickly as you appraoch the 0 voltage.
peak voltage = 3.19V
Voltage at 1.5 cranks per second.
3.5V
Values of t / (RC), e^(-; t / (RC) ), 1 - e^(- t / (RC)) and V_source * (1 - e^(- t / (RC) ).
3.03
.048
.952
3.332V
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):
3.332V,3.19V
.142 difference
According to the function V(t) = V_source * (1 - e^(- t / (RC) ), what should be the voltages after 25, 50 and 75 'beeps'?
1.14 at voltage 25
1.92 at voltage 50
2.43 at voltage 75
You should be reporting voltages at 25, 50 and 75 beeps, not at voltages 25, 50 and 75. You probably intended the correct result and just described in incorrectly.
Values of reversed voltage, V_previous and V1_0, t; value of V1(t).
-3.19V, +3.19V, -3.19V, 15s
1.9125
I used the equation to find the values of the above numbers.
How many Coulombs does the capacitor store at 4 volts?
4
Because Q = C*V = 1F * 4V = 4
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, .5
Using the same equation above gives me those 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?
Your graph of voltage vs. clock time will tell you how long it took for the voltage to drop from 4 V to 3.5 V.
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
Your graphs of current vs. clock time and voltage vs. clock time will give you a very good idea of the average current between the times when the voltage was 4 V and 3.5 V.
1.5hours
Please let me know if you have any questions related to this orientation assignment.