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: **
4.02, 100 ohms
0.00, 4.2
1.10, 3.5
2.41,3.0
3.90, 2.5
5.93, 2.0
8.90, 1.5
13.89. 1.0
16.97, .75
21.47, .50
29.71, .25
** 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. **
5.90 s
6.49
7.96
7.58
I sketched voltage on my y axis and clock times on my x axis. My voltage intervals start at .25, increments of .25 up to 4.00 volts. My clock times begin at 2, increments of 2 up tp 28 seconds. I plotted my data points from my chart and then used my best fit curve to estimate to estimate the times above. I also used the times from my initial data chart to help determine the range of my times on the best fit curve.
** Table of current vs. clock time using same resistor as before, again starting with 4 volts +- .02 volts. **
0.00 s, 190 mAmp
2.53, 180
3.29, 170
4.06, 160
4.83,150
5.88, 140
6.75,130
7.69,120
8.84, 110
10.11, 100
The above tables represents current in mAmp as a function of time in seconds. I obtained my results my setting up a series circuit with the resistor, voltimeter and capicitor and using the timer progam to time the mAmps in increments of 10.
** Times to fall from initial current to half; 75% to half this; 50% to half this; 25% to half this, based on graph. **
5.2
9.2
8.0
7.4
I graphed current(y) and clock times on x. Intervals on y start at 10, increase in increments of 10 to 190 mA. Intervals on x start at 1 and in increase by .5 up to 13.5 seconds. I used my best fit curve line to chart the above times. Some of the above time intervals fell out of the range of data points where I recorded my original data so I had to rely heavily on my curve to estimate the data.
** 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? **
I notice that the times for the voltage and current fall within the range of 5 -9 seconds, which I think is interesting, but I'm not sure if I should draw any conclusions yet about why that would be.
Theoretically the time to fall to half a given voltage, or to half a given current, should be the same. The capacitor and resistor should follow the theoretical model very closely, within 5% or so.
However most students report an increase in the times, and your results also seem to indicate an increase. I don't have an explanation for that, but we're checking it out in the lab.
Incidentally your results are consistent with a 10-ohm resistor rather than a 100-ohm resistor. The time to fall to half, for the given capacitor, should be equal in seconds to about 2/3 the resistance in ohms. If you have the resistor and the meter handy, you can check this out by setting the meter to read resistance and attaching the leads to the resistor. However that isn't necessary.
** Table of voltage, current and resistance vs. clock time: **
4.5s,2.25V, .152 Amp, 14.8 ohms
8.0,1.7, .144, 11.8
14.0, .9, .076, 11.8
20,.37, .038, 9.7
Could not determine .1 times the initial current from my graph of current vs time, because the best fit curve did not extend to a data point that low.
I used the formula, R=V/R to obtain my resistance values. To do this correctly, I had to convert my mAmps measured in the experiment to Amps.
** Slope and vertical intercept of R vs. I graph; units of your slope and vertical intercept; equation of your straight line. **
51.8, 6.1
ohms/amp
R=51.8(I)+6.1
I sketched the resistance on ohms(y) and current(x) in amps. The intervals on my y axis start at 2, increments of of 2, up to 16 ohms. My intervals on my x, start at .03, increments of .03 up tp .21 Amps.
I then graphed my data points and drew the best fit line throught the points. I choose two random points on each end of my best fit line and obtained the slope using the formula y2-y1/x2-x1. Then I used b=y-mx to obtain my y intercept, plugging in the value for one of my data points and the slope to solve for the y intercept. I thought my slope was high until I solved for the y intercept and it seemed to fit the coordinates of my graph. The slope means the ratio of the increase of the y axis with the increase the of the x axis. The y intercept is simply where my best fit line intercepts the y axis.
** Report for the 'other' resistor:; Resistance; half-life; explanation of half-life; equation of R vs. I; complete report. **
33 ohms
26.80 +/- 1.00
I looked at voltage vs time graph and my best fit curve and tried to determine to the best of my ability when the voltage would fall to half of it's original value.
R=(20.40)I+43.3
I ran all of the appropriate procedures and obtained three graphs, voltage vs time, current vs time. I used the data points from each of these graphs and the equation V/I=R, to find the appropriate resistance. I then drew another graph for resistance vs current and drew my best fit line. From two points placed at opposite ends of my best fit line, I used y2-y1/x2-x1 to find the slope of the line. I then plugged the slope of the line into the equation y=mx+b to get my y intercept. I noticed that my resistance values that i obtained from my current/voltage data were higher with this resister and I guess I expected them to be lower because the resistance of this resistor is lower. I will go further and see if the rest of the lab helps me to understand why this is so.
Your results are in the right range. Your results appear to be based on good data and your analysis is correct. However there is more variability in the resistance than would be expected from the claims of the manufacturer. We'll check out other resistors in this batch sometime this week.
** Number of times you had to reverse the cranking before you first saw a negative voltage, with 6.3 V .15 A bulb; descriptions. **
16
I lost count a few times so I think the accuracy of this estimate is fair.
As I expected when I started cranking the generator at 1.5 rev/second, the bulb was bright, then I as I kept cranking at a steady rate, the bulb grew dimmer and dimmer until it was not lit at all. I know from past experiences that to this is because I kept the cranking at a steady rate and the bulb required more voltage to stay lit. When I started to crank in the reverse direction the bulb was very bright. Every time I cranked in reverse the bulb was very bright and the voltage on the voltimeter would decrease significantly.
** When the voltage was changing most quickly, was the bulb at it brightest, at its dimmest, or somewhere in between? **
When the voltage was changing most quickly was when the bulb was its brightest; when I reversed the cranking of the generator. It seems like there is an indirect relationship between the brightness of the bulb and the rate at which the capacitor voltage changes. The brighter the bulb, the quicker the voltage decreased. Is it possible, that when we reverse the cranking of the generator, the capacitor becomes the source of the voltage, and the current reverses, causing the bulb to become brighter, yet because the capacitor is the source of the voltage, it decreases much quicker?
** Number of times you had to reverse the cranking before you first saw a negative voltage, with 33 ohm resistor; descriptions. **
11 times
I think my estimate was pretty accurate.
It seemed like the capicitor voltage changed faster with the passage of time, then it did when the I used the bulb in the circuit. I think this because it took roughly 16 reverse crankings with the bulb, and 11 reverse crankings with the resistor.
** 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. **
25 beeps, 25 s(1 beep per second)
It seemed like the rate at which the voltage changed was the same for when it was approaching the peak voltage and when it was approaching zero voltage. I really did not notice a significant difference.
3.03 V was my peak voltage.
** Voltage at 1.5 cranks per second. **
4.0 V
** Values of t / (RC), e^(-; t / (RC) ), 1 - e^(- t / (RC)) and V_source * (1 - e^(- t / (RC) ). **
.75, .472, .528, 2.11
First i took, 25 seconds from my data above and divided by 33 ohms to get t/RC. Then I took .75 and took the e^x on my calculator to arrive at .472. I then took 1-.472 to arrive at .528. Then I multiplied .528 by 4.0 volts to get 2.11.
** 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): **
2.11, 3.03
Percent difference=value 1-value2/average of values*1oo
3.03-2.11/2.57=.35, or 35 % difference.
** According to the function V(t) = V_source * (1 - e^(- t / (RC) ), what should be the voltages after 25, 50 and 75 'beeps'? **
2.1, 3.1, 3.6
** Values of reversed voltage, V_previous and V1_0, t; value of V1(t). **
3.03, -3.03, 25 s, 1.44
1.44
I understand from above that V previous is the voltage when I started to reverse the crank and v1_0 is the negative value of the v previous. I then found my value for 1-e^(-t/RC) from above and plugged into the equation with my value for v previous and v1_0.
** How many Coulombs does the capacitor store at 4 volts? **
1 C
My capacitance is 1F. If Farad=Coulomb/Volts, then I plugged in 1 for Farad, 4.00 for volts and solved for coulombs.
** 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, it loses .5 Coulombs,
I used the same formula 1F=C/V to calculate the coulombs in the capacitor at 3.5 volts. Since there are 4.00 coulombs at 4.0 volts and 3.5 Coulombs at 3.5 volts, we know that .5 Coloumbs are lost in the interval between 4 volts and 3.5 volts.
** 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? **
.12 Coulombs flowed per second
From my data, I can see that at a voltage of 3.03, and a capitance of 1F, there are 3.03 coulombs. It took 25 seconds to reach a voltage of zero, so .12 coulombs were lost per second.
** 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? **
90 mAmps, or .090 Amps
Amps is calculated as columbs/second.Thus .090*25 second=2.25 coloumbs which does not agree with my result above and I think the answer should be close if not the same as above. There may be some errors in my data that's preventing this answer from being more accurate, but I can't seem to figure out where the error has occured.
** How long did it take you to complete the experiment? **
5-6 hours
** **
Your data, analysis and results are all very good. See my notes and let me know if you have questions.