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 V, 33 ohms
3.5, 1.659
3, 4.995
2.5, 10.721
2, 20.237
1.5, 33.018
1, 52.659
.75, 74.163
.5, 88.913
.25, 101.221
My results show that as the voltage decreases the resistance causes the interval to discharge half a volt to increase. I found my results by adding each interval to the sum before it.
** 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. **
From 4 to 2 V = 16 seconds
From 3 to 1.5 V = 22 seconds
From 2 to 1 V = 26 seconds
From 1 to .5 V = 36 seconds
My graph has a slope that is decreasing at a decreasing rate. It crosses the y axis at 4.
I found my times by drawing a line from the voltages that for each range to the point they crossed my line then I drew lines from the line of best fit to the x axis and then counting the seconds that lay between the range limits.
** Table of current vs. clock time using same resistor as before, again starting with 4 volts +- .02 volts. **
.1, 0
.09, 1.164
.08, 3.936
.07, 8.784
.06, 15.928
.05, 23.971
.04, 33.365
.03, 46.595
.02, 61.778
.01, 89.055
This shows that as the current decreases over time. Not shown with the running sum above it also shows that the interval of time from one whole unit of current to the next also increases as the current decreases over time. I obtained my results by taking a running sum of the times given by the timer program.
** Times to fall from initial current to half; 75% to half this; 50% to half this; 25% to half this, based on graph. **
20 seconds
15 seconds
36 seconds
21 seconds
The slope of the line is decreasing at a decreasing rate. The graph represents the rate of change of the current. I used it as I used the other graph. I found where the current and corresponding time intersected my line.
** 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? **
None are the same, but some values are reasonably close. They are close to the times for the listed drop in voltage. There is a pattern except for a noticable switch in the first two values.
** Table of voltage, current and resistance vs. clock time: **
4s, 3.1 V, .08 A, 38.75 ohms
13s, 2.2 V, .06 A, 36.6 ohms
28s, 1.4 V, .04 A, 35 ohms
65s, .75 V, .02 A, 37.5 ohms
87s, .4 V, .01 A, 40 ohms
** Slope and vertical intercept of R vs. I graph; units of your slope and vertical intercept; equation of your straight line. **
m = -.056, b = 3.18
slope units are amps/ohms. vertical intercept units are amps
y = -.056x + 3.18 y = ((I/R)x) +
My graph is nearly a vertical line in the middle of the data points that is decreasing. I got the slope by taking dividing the change in y by the change in x. I found the vertical intercept by using the slope I found and pluging in values of (x,y) and solving for b. I found the equation of the line by inputing the values I found. I think knowing that the y of my graph was current and the x was resistance changed the slope in my equation to I/R.
** Report for the 'other' resistor:; Resistance; half-life; explanation of half-life; equation of R vs. I; complete report. **
100
17 +- .5s
I determined dt by the line of best fit. I used .5s because it was half the unit I used on my scale that measured time.
y = .01 x -.11 y = ((R/I)x) -.11
15s, 1.5 V, .08 amps, 18.7 ohms
20.7s, .9 V, .06 amps, 15 ohms
26s, .5 V, .04 amps, 12.5 ohms
30.8s, .26 V, .02 amps, 13 ohms
33 s, .12 V, .01 amps, 12 ohms
I followed the procedure outlined for the other resistor, sketched the graphs, and estimated the values from the graphs.
** Number of times you had to reverse the cranking before you first saw a negative voltage, with 6.3 V .15 A bulb; descriptions. **
6 periods reverse cranks before neg voltage
Fairly accurate but mostly like +-3 cranks
The bulb was glow for a short period time then dimmed until it no longer glowed for the majority of the time I was charging the capacitor. The voltage increased, peaked and stayed around 5 V for the charging. When I began the reverse cranking the bulb began to glow very brightly. The voltage began to drop quickly when I began to reverse the crank. I think the voltage was pulled out of the capacitor and back the through the bulb very quickly by the generator so thats why the bulb was so bright and the voltage fells so fast.
** When the voltage was changing most quickly, was the bulb at it brightest, at its dimmest, or somewhere in between? **
The voltage was changing the most quickly when I reversed the crank.
The bulb was at its brightest when I reversed the crank.
It was at is dimmest when I was charging the capacitor probably after 20-30 cranks it not glowed at all.
The relationship is that as the capacitor stores energy the bulb does not glow and as it expells energy the bulb glose brightly. The voltage as increases more slowly than it falls. The reason is most likely that the generator is pulling the energy from the capacitor back through the bulb so it is glowing its brightest the the energy is falling its fastest.
** Number of times you had to reverse the cranking before you first saw a negative voltage, with 33 ohm resistor; descriptions. **
4 times reversing
Fairly accurate +- 5 cranks
It rose at a steady rate and then fell more rapidly when the reverse cranking started.
** 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. **
** Voltage at 1.5 cranks per second. **
** Values of t / (RC), e^(-; t / (RC) ), 1 - e^(- t / (RC)) and V_source * (1 - e^(- t / (RC) ). **
** 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): **
** According to the function V(t) = V_source * (1 - e^(- t / (RC) ), what should be the voltages after 25, 50 and 75 'beeps'? **
** Values of reversed voltage, V_previous and V1_0, t; value of V1(t). **
** How many Coulombs does the capacitor store at 4 volts? **
** 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?; **
** 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? **
** 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? **
** How long did it take you to complete the experiment? **
4 hours
** **
I think I burnt out the fuse in my voltmeter. I will submit the rest as soon as it is repaired/replaced.
Good work up to the point where your meter went out. That is usually the result of putting the meter, set to measure amps, in parallel with a circuit element.