phy202
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: **
David,
I went ahead and pasted the info for this lab after the server came back up.
Thanks for the extra consideration on my labs. Now I can get down to really studying for the last exam on Problem set 4 on magnetism and Problem set on Modern Physics.
Rodger
** Initial voltage and resistance, table of voltage vs. clock time: **
4.02 vdc, 100 ohm
4.02, 1017.188 1017.188
3.50, 1028.781 11.59375
3.00, 1043.375 14.59375
2.50, 1064.422 21.04688
2.00, 1087.25 22.82813
1.50, 1121.297 34.04688
1.00, 1172.047 50.75
0.75, 1225.375 57.32813
0.50, 1278.328 65.95313
0.25, 1411.109 132.7813
The results mean that the capacitor is discharging at a specific rate depending on the capacitor voltage at that time. The results were obtained by charging a capacitor to 4 volts then starting a discharge process utilizing the timer program to determine specific clock times relevant to a specific voltage.
** 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. **
70 sec
78 sec
85 sec
106 sec
My graph I used 4 to represent 4 volts vertical and 4 to represent 400 sec horizontal. Math works out real good when the scaling process is easy to determine the actual points.
** Table of current vs. clock time using same resistor as before, again starting with 4 volts +- .02 volts. **
current vs clock time
40 mA, 4258.953 4258.953
35 mA, 4270.219 11.26563
30 mA, 4284.5 14.28125
25 mA, 4300.172 15.67188
20 mA, 4319.984 19.8125
15 mA, 4348.938 28.95313
10 mA, 4393.281 44.34375
7.5 mA, 4423.016 29.73438
5.0 mA, 4465.172 42.15625
2.5 mA, 4538.406 73.23438
1.2 mA, 4616.703 78.29688
This was a comparison of the discharge rate in current values versus the clock times for each current reading. Interesting though using a 100 ohm resistor while I had my backup multimeter reading the discharge voltage, the current value numbers were close to the same numbers but in a different scale. I guess it just goes to show you with the 100 ohm resistor and using the formula V = I * R can reveal some interest visual facts.
** Times to fall from initial current to half; 75% to half this; 50% to half this; 25% to half this, based on graph. **
58 sec
58 sec
68 sec
66 sec
My graph has a sharper discharge curve downward then as the capacitor discharges the slope of the graph becomes less steep. The graph's answer table is indicating various discharge times versus a different level in the curve. Ironically the times between the segments aren't off that much but it still indicates the capacitor discharges faster initially.
** 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? **
No not really. The slopes of both graphs look really similar. The total discharge times of 394 sec for the plotted voltage and 382 sec for the plotted current. My error may be on the beginning current value I used as 42 mA in my timer program.
** Table of voltage, current and resistance vs. clock time: **
3.4 vdc,33.6 mA, 101.2 ohms, 18 sec
2.72 vdc, 25.2 mA, 107.9 ohms, 41 sec
1.96 vdc, 16.8 mA, 116 ohms, 78 sec
1.1 vdc, 8.4 mA, 131 ohms, 148 sec
.6 vdc, 4.2 mA, 240 sec, 143 sec
I obtained my values by finding the clock times for the different percentages of my starting 42 mA of current. After I found my clock times, I used the clock times found to find my voltage in the voltage versus clock times graph. To find the resistance I used the formula V = I * R changed to R = V / I.
** Slope and vertical intercept of R vs. I graph; units of your slope and vertical intercept; equation of your straight line. **
slope = y/x: 3.01 for the initial circuit, y intercept = 144
ohms per milliamps,
y = 3.01 x + 144
The graph was an indication of what the resistance should be based on the voltage and current. The graph should have been a straighter line than I have drawn because the resistor's resistance shouldn't have changed. If the resistance was to change I believe electronic components would have a problem being consistent. The slope is the y value divided by the x value and the equation is y = (slope * x) + y-intercept. The y-intercept is where my best fit line crossed the y axis
** Report for the 'other' resistor:; Resistance; half-life; explanation of half-life; equation of R vs. I; complete report. **
voltage vs time for 47 ohm resistor
4.0 vdc 12148.05 12148.05
2.0 vdc 12174.75 26.70313
1.0 vdc 12207.77 33.01563
0.5 vdc 12245.45 37.6875
47 ohms is the resistor. I didn't have the 33 ohms out of the 4 supplied.
26.7 sec, +-2 sec
I determined the time from using the timer program starting at 4 volts and timing to 2 volts, 1 volt, and .5 volt.
Using the Ohms law formula: V = I * R, so the resistance is; R = V / I.
I replaced the 100 ohm resistor in my initial circuit where I had the components in series with one another except the voltmeter which was in parallel with my capacitor. I used two multimeters for the ease of determining the different values. I also utilized the computer timing program to find the different times of voltage and currents events happening. The results I could tell was that it took about half the time to discharge the capacitor which would make since the 2nd resistor was almost 1/2 the value of the 1st resistor. Using the 47 ohm resistor, the current went up which it should have upon starting the circuit through my home made switch.
** Number of times you had to reverse the cranking before you first saw a negative voltage, with 6.3 V .15 A bulb; descriptions. **
55 times.
I feel liked I counted fairly accurate.
The bulb's brightness was non existent until I reversed the generator then it brighten quite a bit but seemed liked it blink brightly which may have been from my inconsistent handle rotation speed.
** When the voltage was changing most quickly, was the bulb at it brightest, at its dimmest, or somewhere in between? **
The bulb was at its brightest when the generator cranking was reversed.
The bulb's brightness and the generator's negative half cycle had a definite relationship. When I cranked the generator in the charging direction, the bulb glow went out but when I cranked the generator in the opposite direction, the bulb glowed.
** Number of times you had to reverse the cranking before you first saw a negative voltage, with 33 ohm resistor; descriptions. **
8 reversals
The estimate is accurate just the cranking rate is suspect.
I cranked the generator 66 sec to charge the capacitor then I reversed the generator rotation for 24 cranks. As this process continued the starting voltage would drop while I reverse cranked the generator but would recharge the capacitor but not to the starting voltage. This continued until I stopped when the reversal voltage reached a negative value.
** 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. **
54 beeps, 26 sec
It looked like the voltage appeared close but it was hard to tell.
Peak voltage was 4.04 vdc, reversal voltage was jumped to 4.05 vdc, nothing real noticeable.
** Voltage at 1.5 cranks per second. **
Approx. 4.25 vdc
** Values of t / (RC), e^(-; t / (RC) ), 1 - e^(- t / (RC)) and V_source * (1 - e^(- t / (RC) ). **
I don't know how to do the e^(-t/RC) when you get e^-1.40-4.
** 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): **
I'm lost on this application
** According to the function V(t) = V_source * (1 - e^(- t / (RC) ), what should be the voltages after 25, 50 and 75 'beeps'? **
1.01 vdc
2.02 vdc
3.03 vdc
These are purely guesses
** Values of reversed voltage, V_previous and V1_0, t; value of V1(t). **
I'm lost on this application
** How many Coulombs does the capacitor store at 4 volts? **
Q = C * V; Q = 1 farad * 4 volt = 4 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 coulombs, .5 coulombs
The capacitor loses its charge when its voltage drops from 4.0 vdc to 3.5 vdc thus by use of Coulombs law the loss of charge is .5 coulombs
** 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? **
5.8 sec, .086 coulombs but with a 47 ohm resistor
The charge took 5.8 sec to lose from 4 vdc to 3.5 vdc from which the 4 vdc charge was 4 coulombs. In my graph it took 5.8 sec for the capacitor to discharge so I divided the time into .5 coulombs to determine that the circuit lost .5 coulombs to get the charge rate in coulombs 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? **
79 mA
With the resistance in the circuit, there is going to be a flow of current. As the circuit is closed through the switch, this flow is going to discharge the capacitor. Without a generator recharging the capacitor, the capacitor will lose its voltage which had a direct relationship with the charge of a capacitor which is measured in coulombs.
** How long did it take you to complete the experiment? **
6 hours and 45 minutes
** **
I started using two meters to keep from having to change from voltage to current to try to save a little bit of time. This lab was relatively time consuming versus the other four labs of this type.
You got very good results. For your course we don't worry too much about the exponential functions of form e^(-t / (RC) ).
In any case you've done outstanding work on your labs. I'll look forward to getting your last test.