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: **
I have three resistors. All three have only four bands. I went to a website(http://brunningsoftware.co.uk/ResVal.htm) that had the readings for a four band resistor but I still am not 100% sure which is the top or bottom of the resistor--so I may have an incorrect reading right off the bat. By my calculations, none of the resistors I have are over 0.27 ohms. I have one that is red, violet, brown, and gold (27 * 10^1 = 270 or 0.27 ohms). A second that is orange, orange, black, and gold (33 * 10^0 = 0.03 ohms). A third that is brown, black, black, and gold (10 * 10^0 = 0.01 ohms). Is this correct?
** Initial voltage and resistance, table of voltage vs. clock time: **
4.00 Volts, 0.27 Ohms
3.5V, 28s
3.0V, 34s
2.5V, 43s
2.0V, 55s
1.5V, 73s
1.0V, 104s
0.75V, 75s
0.50V, 109s
0.25V, 198s
My results were obtained by patiently watching my voltmeter and timing when the requested values were reached. It appears as the capacitor drained it took longer for the volts to reduce. This was with the red, violet, brown, and gold lined resisitor.
Voltage vs. clock time would have clock time in the first column.
Clock times can't increase and decrease--I suspect you reported time intervals rather than clock times.
Clock time is the time shown by a running clock which starts at the beginning and doesn't stop until the end. Your last clock time would be several hundred seconds.
** 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. **
160s
170s
120s
180s
I just referenced my graph and extrapolated the times it took for the values to cross.
** Table of current vs. clock time using same resistor as before, again starting with 4 volts +- .02 volts. **
15.2mA, 0s
13.3mA, 28s
11.7mA, 34s
9.7mA, 43s
7.6mA, 55s
5.8mA, 73s
4.0mA, 104s
2.8mA, 75s
2.0mA, 109s
1.0mA, 198s
This table appears to show time interval vs. current, not current vs. clock time.
** Times to fall from initial current to half; 75% to half this; 50% to half this; 25% to half this, based on graph. **
160s for initial to 1/2.
190s for 75% of initial to 1/2 of that.
150s for 1/2 initial to 1/2 of that.
170s for 1/4 the initial to 1/2 of this.
My graph has time in the x axis and current in the y axis. Similar to voltage, it shallows out at the bottom of the graph.
These times are consistent, with relatively small differences that are probably due to the inevitable limitations of your meter.
** 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 current times are the same as the timer program voltage times--within experimental uncertainty (they are as good as my $15.00 Timex stopwatch could provide.) There is a definite pattern, with voltage and current behaving very similarly in terms of percentage of rate of dissipation.
** Table of voltage, current and resistance vs. clock time: **
35s, 3.2V, 12.16mA, 26.3ohm
100s, 2.4V, 9.16mA, 26.2ohm
175s, 1.9V, 6.08mA, 31.3ohm
352s, 0.75V, 3.04mA, 22.1ohm
621s, 0.30V, 1.52mA, 19.7ohm
These results we obtained by extrapolating the data from the different graphs that correlated with the appropriate times. The mA is the most precise value, because I multiplied that off the original current of 15.2. The voltages are much times elapsed and the voltages are more of a best guess of my graph. The resistance is found using the precise amperage and imprecise voltage and would reflect that inherent inaccuracy.
** Slope and vertical intercept of R vs. I graph; units of your slope and vertical intercept; equation of your straight line. **
Slope= 1.16 , Vertical Intercept= -21.2
Unsure of the units (vertical intercept is current?)
On a graph of R vs. I, the resistance R would be on the vertical axis, current I on the horizontal. In this case the vertical intercept would be the resistance corresponding to 0 current.
I= 1.16 (R) -21.23
My line is a little off because I have a value of 31.3ohms at my 6.08mA reading. It has a negative I (Y) intercept (current) and was found by m=(y2-y1)/(x2-x1) and then plugging in y=mx+B using two known quantities for x and y (R and I). Please let me know how far I have strayed from reality with this.
** Report for the 'other' resistor:; Resistance; half-life; explanation of half-life; equation of R vs. I; complete report. **
Line 1: Resistor resistance rating is 0.03 ohms (orange, orange, black). If this is incorrect, please advise once more.
Line 2: 19s +/-1s
Line 3: time was calculated by directly observing my digital voltmeter and timing with a stopwatch exactly how long it took to drop from 4 to 2 volts. I am assigning myself a personal margin of error of approx one second.
Line4: I=5.09 (R) - 148.5
I pretty much set this up as I did the previous experiment. To find the equation I also used the same procedure as before...I found the slope and then found the y intercept. Interesting that this resistor allowed the capacitor to drain in a much quicker fashion, and also that it afforded current readings that were much higher. This resistor allowed more current to flow and voltage to drop than the previous resistor.
You need to redo these graphs and get the correct vertical intercept.
The line you report would intercept the R axis at about 30 ohms; except for the power of 10 it appears you read the color code correctly.
** Number of times you had to reverse the cranking before you first saw a negative voltage, with 6.3 V .15 A bulb; descriptions. **
14 times to reverse.
My estimate was fairly accurate (+/- 1 revolution).
This was very interesting. During about the first 50 of the 100 forward revolutions the bulb was glowing initially, however it progressively grew dimmer in a linear fashion as I neared the 50 crank mark. For the second 50 forward revolutions (and after voltmeter read approx 3.5V) it was extinguished. I reached a maximum of 5 volts on the voltmeter with 100 forward revolutions. As I started the crank reversal, the light obviously grew its brightest as the light was getting the full force of the current as it was flowing to the bulb first. Initially I would lose almost 25% of my voltage (1 full volt) with a course reversal and then gain roughly half a volt back again as I went to my forward 5 cranks. As I came to the lower voltage I would lose never less that approx. 0.70V). Consistently, while I was always getting the full brightness of the bulb on my 5 reverse cranks, I did not get any illumination on the 5 forward cranks until the capacitor voltage dropped to roughly 3.5V and then it was in reverse linear fashion as mentioned above with the brighter light on the 5 forward cranks at the lower capacitor voltage. Wow, this was a very illuminating (forgive the pu... ahh, never mind) portion of what so far has proved to be a fairly rigorous and challenging experiment.
** When the voltage was changing most quickly, was the bulb at it brightest, at its dimmest, or somewhere in between? **
When the voltage was dropping most quickly (for the most part, I noticed a roughly constant PERCENTAGE of voltage drop on the 5 reverse cranks, obviously resulting in a higher value of change at the higher voltage levels but never diminishing below about a 0.7V drop even at the lower voltage values. The gain I would receive from the subsequent 5 forward cranks always seemed more constant and would consistently be in the 0.5V range. The largest value change came at the 5.00V to 4.00V level as the first reversal dropped it almost 1 full volt and the subsequent 5 forward cranks gained back slightly more than half of that. The course reversal seemed to result in constant bright illumination of the bulb at any voltage range, while the following forward cranks did not result in any illumination until we reached below the 3.5V range and then the brightness was inversely proportional to the volts from 3.5 V until less than zero (higher volts/dimmer light, lower volts, brighter light). To answer your question When the voltage was changing most quickly, was the bulb at it brightest, at its dimmest, or somewhere in between specifically, I would say on the crank reversal it was at it's brightest, but at the forward cranks when it was at the higher voltages it was at it's dimmest, and at the lower voltages it was at its brightest with respect to forward cranking, but never as bright as when I was reversing cranks. The second question is a little harder for me to wrap my brain around. During the charge up to 5.0 volts I would get illumination of the bulb although it would grow progressively dimmer until approx 3.5V, and then was extinguished from 3.5V to 5.0V. This was the same on the way down (dark from 5.0V-3.5V, dimmest at 3.4V and progessively brighter down to 0V). I have duplicated this multiple times to make sure I was observing it correctly. At the higher voltage ranges the capacitor is doing more to drive my generator (even though I am maintaining a constant crank rate my force applied is dropping on the forward cranks as the capacitor is helping me more and more, and inversely, as I flow current against the capacitor at the higher voltages my cranking force seem to be slightly higher as I work against the higher power (resistant force) of the more highly charged capacitor. Also, during the initial 100 crank power up, at the higher voltages the gain I am receiving on the capacitor seems to slow for a given number of cranks (i.e 10 forward cranks in the beginning of the 100 seems to result in a higher gain than the 10 forward cranks at the end.) In the end, the current is bypassing the light and going to my generator and acting like a motor powering my generator, and this is why I believe the voltage is not gaining as fast, because there is less load (perhaps more load? This is part of my confusion...). In the beginning of the 100 cranks, it the voltage is gaining faster, and able to light the light bulb as well because there is more load (perhaps less? Again, I hope I don't have this reversed but I am sure you will let me know) and the current is not yet powerful enough to bypass the light and help to power the cranking of the generator. This I know to be fact though: it is always easier to push a charge on a uncharged capacitor as there is little charge to repel it. As the charge stored increases, the repulsion is greater and more work must be accomplished to push the next charge on.
** Number of times you had to reverse the cranking before you first saw a negative voltage, with 33 ohm resistor; descriptions. **
Reversed cranking 10 times
Again within =/- 1 reversal.
The capacitor voltage behavior was similar to the light bulb experiment. I was cranking at 2 cranks per second for 132 cranks total (33 ohms * 1.0F capacitor was 66 cranks * 2 = double the time constant of 132 cranks) and this resulted in a total capacitor voltage of 4.51 volts before the first 16 crank (at two cranks per second) reversal at the quarter time constant. The capacitor voltage always dropped slightly more during the crank reversal than it gained during the following forward quarter time crank, allowing it to reach negative status on the 10th reversal.
Special Note: As embarrassing as this is I am still not convinced I am rating the resistors correctly. For the purposes of this experiment, I used the orange, orange, black, gold banded resistor, that I can only assume I have rated incorrectly at 0.033 ohms when in reality it is 33 ohms (at least I feel confident it has two threes in it--no other resistor I have been provided with can make that claim). So for the purpose of this part of the experiment I will assume it is the 33 ohm resistor. Also, under the first bullet point above, it states Set the 'beeps' program to beep at the rate which will generate 4 volts. If you aren't sure what rate is required, hook up the meter and the generator and find out. My question: Does this mean the rate at which 100 beeps will result in four volts? I was unsure of this meaning. Based upon the second bullet point below, I have to assume the answer is yes.
I do believe you misread the power of 10 on the resistor--easy to fix, since you apparently read the other details correctly.
It would take around 3 cranks per second to give you 4 volts (see your previous information on voltage vs. cranking rate); whatever gives you 4 volts, that's where you should set the Beeps program.
** 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. **
20 seconds, 40 beeps.
Voltage seemed to drop at a rather constant rate throughout.
Peak voltage was 4.9V.
** Voltage at 1.5 cranks per second. **
For 100 beeps at 1.5 cranks per second I generate a total of 4.05V.
** Values of t / (RC), e^(-; t / (RC) ), 1 - e^(- t / (RC)) and V_source * (1 - e^(- t / (RC) ). **
1.5, 0.22, 0.78, 3.8V
t/(RC)= 50/33 = 1.5. Next, e^-1.5 = 0.22. Then 1-0.22=0.78. Finally 4.9 * 0.78 = 3.8V. (The fourth part confuses me because in the 6th line above the experiment states: V_source is the voltage of the source, which you reported in the preceding box. Well, in the immediately preceding box I have 4.05V which would make my calculation even farther off the mark. So I went with 4.9V as this seems to make more sense. Again, I look forward to your clarification of the possible errors in my ways. As an aside, I do find this fascinating and it makes me wish I had a more natural talent for this field. I am in awe of the people who do--I seem to have a more right-brained disposition that does not lend itself to efficiently working these problems out.
there's a slight discrepancy here but nothing to worry about
** 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): **
3.8V, and 100 cranks this time (2 cranks per second) is 4.1V.
Difference is 0.3V (92%).
** According to the function V(t) = V_source * (1 - e^(- t / (RC) ), what should be the voltages after 25, 50 and 75 'beeps'? **
25 beeps: 13/33=0.39, e^-0.39=0.67, 1-0.67=0.33, 4.9*0.33=1.6 Volts
50 beeps: 25/33=0.76, e^-0.76=0.47, 1-0.47=0.53, 4.9*0.53=2.6 Volts
75 beeps: 35/33=1.06, e^-1.06=0.35, 1-0.35=0.65, 4.9*0.65=3.2 Volts
** Values of reversed voltage, V_previous and V1_0, t; value of V1(t). **
-4.9V, V_previous = 4.9, V1_0 = -4.9, t=20s.
Already something is fundamentally wrong here with my values. Plugging these into the formula will always create an answer of 0 because V_previous (4.9V) plus V1_0 (-4.9V) will be 0. Please let me know where I am in error and I will resubmit. My first quantity comes from the experiment statement: After the first reversal what is the reversed voltage? Write this quantity down. (e.g., if the generator voltage was +2.5 volts, the reversed voltage would be -2.5 volts). I hit 4.9 Volts, so I am using a value of -4.9V. My second and third quantity comes from the experiment statement What are the values of V_previous and V1_0? Write these quantities down. Well, V_previous was defined as the voltage at the instant of reversal which is 4.9V. V1_0 is defined as reversed voltage (-4.9V) - V_previous (4.9) which = -9.8. Aha, I just found my error. Okay, to start again:
-4.9V, 4.9V, -9.8V, 20s
-4.5V
20/33=0.61, e^-0.61=0.54, 1-0.54=0.46, -9.8*0.46=-4.5V. I left all my thinking above so you can see hopefully more clearly my method of thinking for this problem. To me the result proves voltage is dissipate faster than it is created for the same cranking rate. Dissipates almost twice as fast as it accumulates.
** How many Coulombs does the capacitor store at 4 volts? **
4 Coulombs.
It is a 1 farad capacitor. Therefore, there is 1 Coulomb per volt.
1.0F capacitor is charged to 4.0V. Q = 4*4 = 16J?
** 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 Coulombs, 0.5 Coulombs.
This seems to have been made fairly simple by the fact that you provided me with a 1.0F capacitor.
** 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? **
4 seconds, 0.125 Coulomb's flowed per second.
I would lose 1 volt (1 Coulomb) in 8 seconds (16 cranks at 2 cranks per second) therefore you can extrapolate that I lost 0.5 Coulomb's when I dropped 0.5V, this would take 1/2 of 8 seconds (4 seconds) and 1/4th of 0.5 Coulomb's is 0.125C.
** 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? **
Average current is 13.5mA.
With voltage and current being more or less joined at the hip here, and in this case, I have (hopefully correctly) joined Coulomb's and Volts at the hip they have all come together. I dropped from a measured 15.00mA at 4.0V and registered 12.00mA at 3.2V so therefore I will say average amperage from 4.0V dropping to 3.5V will be 13.5mA.
** How long did it take you to complete the experiment? **
A stunningly long time. Quite a bit of time yesterday and 3 hours today. I am going to go reintorduce myself to my family now.
** **
This was a good experiment. There seemed to be many opportunities for me to take a wrong turn and I regret that some of them showed up early and may of invalidated many of my results. I tried to make my processes clear though so that if I did have a mistake early I could at least see how it affected my later calculations and further my knowledge of the lesson.
Very good work on this experiment. See my notes and let me know if you have additional questions. Enjoy getting reacquainted with your family.