#$&*
Phy 232
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.
** The RC Circuit_labelMessages **
7/22/12 submitted around 2:19 PM
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4.02 volts, resistance : 100 ohms
Voltage vs. Clock time table
Time (sec), Voltage(V)
11.5, 3.5
26, 3.0
44.5, 2.5
68, 2.0
99, 1.5
144.5, 1.0
176.5, .75
304, .25
@&
Good. It appears that, as expected, the time required for the voltage to drop to half is always in the neighborhood of 75 seconds, give or take 5 seconds.
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68 sec
73 sec
76 sec
79 sec
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Table of Current vs. Clock time
Time (sec), Current (mA)
11.5, 30
26, 28
44.5, 22
68, 17
99, 17
144.5, 9
176.5, 7
304, 3
@&
Again these are consistent with a 'half-life' of about 75 seconds (e.g., from 28 mA to 7 mA is two half-lives, from 28 to 14 then from 14 to 7 mA, and took just over 150 seconds).
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The results mean that the current decreased over the time and they were obtained by reading the meter that was parallel to the series circuit
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65sec
68sec
65sec
69sec
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The times do not seem to be the same. The numbers above are about 10 or so off from the other numbers for the voltages. I think there is a pattern since the numbers above seem to be about 10 or so off from the numbers for the voltages, which seemed kind of strange.
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Time (sec), Current ( mA), Voltage (volts), Resistance (ohms)
18, 28, 3.26, .116
46, 21, 2.47, .1176
89.5, 14, 1.64, .1171
155, 7, .9, .128
217.5, 3.5, .52, .1485
@&
The resistance would be the voltage divided by the current. Your resistances would all be in the range from about 110 ohms to about 130 ohms, which is reasonably consistent with the 100 ohm resistance you reported originally.
This is also consistent with the time constant, which is equal to the voltage multipled by the capacitance. The time constant is the time required for the voltage to fall to 1 / e times its original value (about 37% of the original value). With 1 Farad capacitors the time constant would be between 110 and 130 seconds. These results are consistent with your data, which indicate half-lives (time to fall to 50%) between about 70 and 80 seconds.
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I obtained these values by multiplying the initial current by the values and then look for the values on the graphs.
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y=-.0011x+.1405
The vertical intercept was .1405
I got the slope by using excel
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33ohm resistor
Voltage vs. clock time
Time (sec), Voltage (volts)
3.5,3.5
8.7, 3.0
15.5, 2.5
23.5,2.0
34.2,1.5
49.7,1.0
61,.75
103,.25
23.5sec +-1sec
25.5sec+-1sec
26.2sec+-1sec
26.8sec+-1sec
I estimated theses numbers from looking off the graph and the graph looked like a downward sloping curve.
Current vs. clock time table
Time (sec), Current (mA)
3.5, 82
8.5, 71
15.5, 58
23.5, 48
34.2,35
49.7, 35
61, 18
76.5,13
104, 6
24.5sec
26sec
26sec
27.5sec
The graph was a little noisy but I used it to come up with these numbers
The numbers are almost the same as the ones above and they seem to be very close to the voltage drop values. I think that there is a pattern.
Time (sec), Current (mA), Voltage (volts), Resistance (ohms)
8, 73.6, 3.08, .0418478216
18, 55.2, 2.32, .0420289855
33.5, 36.8, 1.53, .041576087
61.5, 18.4, .74, .0402173913
89, 9.2, .36, .0391304348
@&
Dividing voltage by current the results would all be close to 40 ohms, not .04 ohms. You divided voltage in volts by current in milliamps, which would give you units of 1000 ohms, not ohms.
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I obtained these values from the graph and I solved the resistance by using R=V/I
Graphing resistance vs. Current
My best fit line doesn’t fit very well but my line equation is y=4E-05x+.03913 and the vertical intercept is .03913
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25 reverses
I think that my estimate was pretty accurate
Over time when I was going forward it got dimmer and dimmer and then as soon as I reversed it glowed brighter but over the course of many reverses it grew dimmer. I think that the capacitor voltage would alternate depending on crank direction. The brightness of the bulb was affected by the current that alternated by the cranking direction. The voltage across the capacitor depended on the cranking direction as well.
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It is at its dimmest
The relationship between the brightness of the bulb and the rate at which capacitor voltage changes is since the bulb was getting dimmer over the amount of times I was going forward and reverse. The relationship would be directly proportional.
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10 times
It was very accurate
I cranked the generator forward for 99 cranks and then backward for 12 and continued cranking forward and backwards. The capacitor voltage decreased when cranking backwards and increased when cranking forwards.
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33 beeps, 22sec,
The voltage was changing more quickly as you approached the peak voltage
3.05volts
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3.5 volts
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22, 3.5 volts, 2.7894E-10, .99999, 3.5volts
I used the formulas above and filled in the numbers
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3.5volts,3.2 volts
8%
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1.38,1.27, .769
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V_previous: 3.2volts, reversed voltage:2.98volts, V1_0=.22, t=22sec, V1(t)=3.09volts
I used the formulas above to gather the variables to solve for V1(t) and the results mean that that is how many volts were at V1
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First I combined the formula: Q=coulomb/volt *volt. The volts cancel leaving Q=coulomb. So the answer should be Q=4 Coulombs.
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3.5, .5
I calculated the using the same formula above: Q=coulomb so the answer should be Q=3.5 so the difference would be .5 coulombs
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3.5sec, .142
To obtain this answer I took the time between 4 V and 3.5 V and then I divided .5 by 3.5
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87.33mA
It is smaller than the one above
Because of the additional resistance of the volt meter
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5 hours
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