#$&*
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 **
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4V, 45ohms
0,4
3.8, 3.5
6.4, 3
10.7, 2.5
17.6, 2
25.4, 1.5
39.7, 1.0
45.3, 0.75
59.9, 0.5
87.1, 0.25
The first line is the initial voltage, 4V, and the initial resistance, 45ohms. The 2nd through the 11th lines are the voltages as the capacitor discharges over time. The times are listed first in each line in seconds then the voltages are listed after each comma.
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17.82
20.02
21.13
22.23
The graph is decreasing at a decreasing rate and the line of best fit for the graph was y=3.6*0.9^x since it was exponential. I found the time required for these voltage drops by looking directly at my graph and subtracting the differences in the times required to reach these voltages.
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Very good. But for further information:
A bit of terminology:
In your model V = 3.6 * .9^t, the number .9 is called the 'growth factor'. Being less than 1, this growth factor implies a decreasing, as opposed to a growing quantity.
Very good, but you need more significant figures to get a good model. Your model posits a growth factor of .9. The half-life of this model would be about 6 seconds, whereas the halflife indicated by your data is closer to 20 seconds.
This model is very good for the level of this course, but you are clearly capable at a higher level so the following might be useful:
FYI:
It takes about 20 seconds for the voltage to drop to half its original value, and this interval is pretty consistent throughout your graph (voltage 4 V at t = 0, 2 V at t = 17.6 s, 1 V at t = 29.7 s, .5 V at t = 59.9 s).
If this is the case the voltage after t seconds will be
V(t) = V_0 (1/2)^(t / 20).
You can verify that (1/2)^(t/20) is 1/2 when t = 20, half of that when t = 40, and half of that when t = 60.
Now it's easily verified that the factor .9^t drops to .729 when t = 3. After another interval of 3 the value is .729^2, which is easily seen to be close to 1/2 (e.g., we know that .7^2 = .49). So your model V = 3.6 * .9^t would imply a voltage that drops to half in just about 6 seconds.
V = 3.6 * .97^t would be closer; to 3 significant figures your growth factor would be .966.
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4.01V, 40ohms
0, 100
5.2, 90
10.6, 80
15.1, 70
20.3, 60
30.7, 50
39.4, 40
50.1, 30
71.5, 20
120.2, 10
the first line shows the initial voltage, 4.01V and the initial resistance, 40ohms. The 2nd through the 11th lines show the time in seconds followed by the current in mA.
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The half-life here appears to be about 25 seconds. This is significantly but not drastically different that before.
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30.7
29.2
29.2
52.3
These differences are in seconds. The graph is decreasing at a decreasing rate. The exponential line of best fit is y=92.9*0.98^x. I got these differences by looking directly at my graph and subtracting the times it took to reach these points in the decreasing current.
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My times are not the same. They are larger overall than the voltage times. There is a slight pattern of the time differences increasing as the voltage or current drops get closer to 0.
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10.6, 2.5, 80, 31.25
20.3, 1.9, 60, 31.67
39.4, 0.99, 40, 24.75
71.5, 0.42, 20, 21
120.2, 0.18, 10, 18
Each line has a clock time in seconds, a voltage in volts, a current in mA and a resistance in ohms. First I multiplied 0.8, 0.6, 0.4, 0.2, and 0.1 by my initial current, 100mA, to get each of the currents I would be focusing on. Then using the current vs. time graph I found the times corresponding to these currents. Then using the time and the voltage vs time graph I found the corresponding voltages to these times. Lastly by using ohms law, R=V/I, I used each voltage and current to find each corresponding resistance.
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0.204, 16.8
The slope in in Amps for the current and the y-intercept is in ohms for resistance.
R=0.204I+16.8
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If the slope was in amps the, for example a current of 80 milliamps would imply a resistance of
R = .204 amps * .080 amps + 16.8 ohms = .016 amps^2 + 16.8 ohms.
This doesn't make sense. You can't add amps^2 to ohms.
Clearly your slope is not in amps.
I do think you have a valid slope, but what are its units?
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My graph has a positive slope that has a small curve at the very top right portion of the graph. This shows a positive relationship between current and resistance. Other than the small curve at the top of the graph, the linear function best fits this data. I found the equation using the linReg(ax+b) function on my calculator.
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145ohms
90.3 +-4.2s
I found the average time of the differences from the initial voltage to half, 75% to the half of this, 50% to the half of this, and 25% to the half of this value. Then I found the standard deviation of these as well.
R=1.01I+129.5
25.3, 4.7, 30, 156.7
41.7, 3.1, 20, 155
150.2, 1.4, 10, 140
257.8, 0.67, 5, 134
315.9, 0.46, 3.5, 131.4
First I set up the circuit like before in parallel to first measure voltage. Using this data I created a voltage vs time graph. Then I set up the circuit in series to get measurements for current for the current vs time chart. Then using the current vs time graph I found the times corresponding to the current at 30, 20, 10, 5, and 3.5mA. Using these times I found the corresponding voltages on the voltage vs time graph. Then using ohms law, R=V/I, I found the resistance for each voltage and current set. The I used the values for the current and resistance to make a graph and from which I found the line of best fit, R=1.01I+129.5. These findings show that as time passes, the voltage and the current and the resistance all decrease as the capacitor discharges. This also shows that as current is increased then resistance also increases. All this data is written in the table above in each line as time in seconds, followed by the voltage, then the current in mA, and lastly the resistance in ohms.
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15 crank reversals
I think my estimate is fairly accurate because I counted out loud each time.
when I cranked the generator to 2 volts the bulb glowed and then faded out. When I reversed the crank the bulb glowed again and then went back out again. This kept happening but eventually the bulb didn't glow as bright and would not dim as completely as before. The voltage each time increased a little but decreased more than it increased each time the crank was reversed. Eventually, because the voltage was decreasing faster than it could increase, the voltage ended up at 0.
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When the voltage changed the most quickly, the bulb was at its brightest. I think that as the voltage changes less rapidly the bulb will be dimmer and dimmer. This might be because since the capacitor stores voltage when the generator is cranked then when the crank is reversed the capacitor releases this energy in the form of lighting the bulb. If the voltage doesn't have enough time to build up between crank reversals then the capacitor will not build up enough energy to make the bulb glow brighter than before.
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10 crank reversals
I counted out loud each time so I think this is fairly accurate.
I followed the above procedures and after my initial 100 cranks I reversed the direction for 14 cranks and continued to reverse the direction every 14 cranks until I got a negative voltage. Each time the direction was reversed the capacitor lost more voltage than it could build up in the same amount of time causing the voltage to continuously decrease over time.
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45 beeps, 30 seconds
The voltage changed more rapidly as it approached 0 voltage.
3.3 V
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3.3V
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1, 0.368, 0.632, 2.09
The first value is the t/RC which is 30s/30ohms*1 farad. The second value is e^(-t/RC) so I did e^(-1) to find 0.368. The third value is 1-e^(-t/RC) so I did 1-0.368 to get 0.632. The last value is V_source * (1-e^(-t/RC)) = 3.3*0.632 = 2.09.
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2.09V, 3.3V
difference: 1.21, 63%
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1.32, 2.17, 2.81
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-3.3V, 3.3V, -6.6V, 30s
0.77
The first value is the voltage at reversal. The second is V_previous, the third is V1_0 and the fourth value is the time. The second row is the the answer to the equation for V1(t). V1(t)= V_prev+V1_0*(1-e^(-t/RC)) = 3.3V+-6.6V(1-e^(-30s/30ohms*1farad)) = 0.77
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Q=CV
Q=1 farad * 4 volts = 4 coulombs
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3.5 C, 0.5C
Q=CV
Q = 1 farad *3.5V = 3.5C
Q= 1 farad* 4V = 4C
4C - 3.5 C = 0.5C
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20s, 0.025C/s
0.5C / 20s = 0.025C/s
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0.0245A
My answer is very close and is only off by approximately 0.0005A.
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3 hrs
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@&
Very good work.
I've given you a fairly extensive note on the mathematical model for the first data. This is mainly for your information, since you appear to have the background to understand it. No revision is necessary but if you have questions I'll be happy to answer them.
I do want one revision. The units on your model of resistance vs. current do not hold up, as will be clear from my note. I don't think you'll have any trouble coming up with a revision.
&#Please see my notes and submit a copy of this document with revisions, comments and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).
Be sure to include the entire document, including my notes. &#
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