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:
4.02V, 32.1ohms
3.5, 2.88
3.0, 7.60
2.5, 11.02
2.0, 17.05
1.5, 25.43
1.0, 36.03
0.75, 44.37
0.50, 55.89
0.25, 76.91
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.
17s
17.86s
19s
19.86s
Table of current vs. clock time using same resistor as before, again starting with 4 volts +- .02 volts.
115mA
100, 3.05
90, 6.88
80, 13.48
70, 17.54
60, 22.42
50, 28.34
40, 35.84
30, 46.38
20, 65.15
10, 83.28
Times to fall from initial current to half; 75% to half this; 50% to half this; 25% to half this, based on graph.
22s
22s
24s
22s
The graph was concave sloping. I used the graph by going to the initial point requested and determining the difference between this tine and the original time.
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 time intervals can be considered the same or close to the same with some error of uncertainty.
Table of voltage, current and resistance vs. clock time:
6.5, 3.2, 0.092, 34.75
17.5, 2.0, 0.069, 28.99
32, 1.4, 0.046, 30.43
65.5, 0.6, 0.023, 26.9
82, 0.4, 0.0115, 34.7
I first found the corresponding current for .8, .6, .4, .2, and .1 of the original current value. Then I used that time to determine with the volts vs time graph the voltage at the same time. Then I divided the voltage by the current. It was fairly close until I got to the last two intervals; the voltage is half of what it should be so therefore so is the resistance value calculated. This could be do to error in reading the graph or the error in clicking the timer program at the proper voltage reading. In order to do the next step I will enter the corrected values.
Slope and vertical intercept of R vs. I graph; units of your slope and vertical intercept; equation of your straight line.
0.021, 30.139
mA, ohms
R = 0.021(I) + 30.139
The graph should show that the resistance is almost stable regardless of the amount of current flow. However some thermal drift may occur if the quality of the resistor is poor. This is where the tolerance band comes into factor with low and high quality resistors. If you application requires a very stable resistance then you would pay more for a precision resistor that has very little drift.
Report for the 'other' resistor:; Resistance; half-life; explanation of half-life; equation of R vs. I; complete report.
148ohms, 4.02V
90.39s+/- 0.045s
I used the graph for voltage vs time to find the value t.
R= 0.0061(I)=151.93
Number of times you had to reverse the cranking before you first saw a negative voltage, with 6.3 V .15 A bulb; descriptions.
8
Fairly accurate
The bulbs intensity was greatest when I reversed direction of rotation. I think this is because the capacitor was discharging and adding to the voltage of the generator. The capacitor voltage began to decrease more and more as time progressed and I reversed the cranking direction until finally it began to show a negative voltage potential.
When the voltage was changing most quickly, was the bulb at it brightest, at its dimmest, or somewhere in between?
The bulbs intensity was greatest when I reversed direction of rotation. I think this is because the capacitor was discharging and adding to the voltage of the generator. The capacitor voltage rate of change did affect the brightness of the bulb; This would be related to the frequency of the voltage change, in a normal 60Hz household this can not be seen by the naked eye.
Number of times you had to reverse the cranking before you first saw a negative voltage, with 33 ohm resistor; descriptions.
I think I had about 12
But I don’t think I was very accurate
The capacitor began charging when I was cranking at 2xRC, but when I reversed it began to discharge, and with every cycle of (fwd/Rvs) for ¼ of RC it charged to a less and less value until finally it went negative.
This scenario is how a capacitor would react to AC voltage vs. DC. Capacitors block DC (unless it’s a changing DC) and pass AC. That’s why they are used as filters in AC circuits to filter out certain frequencies. In amplifiers they are used to isolate the DC Biasing voltage from stage to stage.
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.
200beeps, 100s
The voltage was changing more rapidly as I was approaching the peak voltage.
3.4v
Voltage at 1.5 cranks per second.
Approximately 4.2V
Values of t / (RC), e^(-; t / (RC) ), 1 - e^(- t / (RC)) and V_source * (1 - e^(- t / (RC) ).
4.2
-2
0.8646647168
3.63
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):
3.63
87
According to the function V(t) = V_source * (1 - e^(- t / (RC) ), what should be the voltages after 25, 50 and 75 'beeps'?
1.65
2.65
3.27
Values of reversed voltage, V_previous and V1_0, t; value of V1(t).
66seconds
How many Coulombs does the capacitor store at 4 volts?
4
Q=(1.0F)*(4V)=4C
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, 0.5
The capacitor charges to 3.5C with 3.5V across it because Q=C*V and C=1.0F.
So between 4 and 3.5V the capacitor would lose 4C-3.5C=0.5C
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?
2.88, 1.44
If it 0.5C in 2.88seconds then the flow rate would be 2.88*(0.5)=1.44C/s
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?
Not sure what we’re looking for. But I do understand that resistance in series with a capacitor creates an RC time constant which corresponds to the time required for the capacitor to charge and discharge. When t-RC the capacitor will have charged to 63% of the applied voltage, or discharged to 37%. It is said that the capacitor is assumed fully charged at 5RC.
How long did it take you to complete the experiment?
Way too long, over 5 hours.
Very good report. You obviously have an excellent understanding of the behavior of capacitors and AC circuits in general, and I'm glad you were able to so clearly relate that behavior to this system.