#$&*
PHY 122
Your 'the rc circuit' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** #$&* Your comment or question: **
11/13 6pm
** #$&* Initial voltage and resistance, table of voltage vs. clock time: **
4.2,98.3
10.5625,3.5
22.875,3
38.32813,2.5
58.14063,2
86.25001,1.5
122.40626,1
150.17189,0.75
189.85939,0.5
260.81252,0.25
This data represent the discharge voltag rate. This data represents how the capacitor discharges voltages over a period of time. These data points were determined using a digital multimeter and the timmer program provided for this course.
** #$&* 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. **
58
63
64
67.5
the graph starts on the left at the highest point (0,4.2) then decresases at a decreasing rate to the endpoint of (260.81,0.25). The numbers above we determined by finding the points where coresponding voltages and time meet and then subracting these data point from one another to find the differences in times.
** #$&* Table of current vs. clock time using same resistor as before, again starting with 4 volts +- .02 volts. **
0.04,98.3
30,3.28125
20,30.35937
10,74.21875
0,154.14062
This data shows how the current decreases with time. These measurements were made with a digital multimeter, the resolution of the meter would not allow for measurements be made of less than two decimal places. If the resolution of one more decimal place could have been made it would have made for a much better set of data points that would have shown much more of the true mannor in which the current behaved.
** #$&* Times to fall from initial current to half; 75% to half this; 50% to half this; 25% to half this, based on graph. **
123
47
44
44
The graph of current vs time begins at the min on the left of graph of (154,0) then decreases on a decreasing rate as the graph moves to the right and finaly ends at ((40,0). The times were determined as before by finding the data points that coinsided with the currents then the times were found by subracting the times.
** #$&* Within experimental uncertainty, are the times you reported above the same?; Are they the same as the times you reports for voltages to drop from 4 v to 2 v, 3 v to 1.5 v, etc?; Is there any pattern here? **
They are not the same as found for voltages, but the resolution of the meter for current measuring limited the measurments that could be made. From the graphs of both current and voltages both seemed to decrease on a decreasing rate so the times would very likley be the same for both current and voltages.
** #$&* Table of voltage, current and resistance vs. clock time: **
0.12,3.77,.032,117.8
14.90,3.31,.024,137.9
45.29,2.48,.016,155.0
91.12,1.47,.008,183.75
119.88,0.99,.004,247.5
Thes values were obtained by finding the corressponding currents due to the percentage of currents. Then the times for the currents were obtained by using the graph of current vs time to find out the corresponding times for each current. Then with the times know the voltage values were found on the voltage vs time graph for the corresponding times. Then the resitanc for each current-voltage-time was found using ohms law.
** #$&* Slope and vertical intercept of R vs. I graph; units of your slope and vertical intercept; equation of your straight line. **
-4.1614,235.8
ma,235.8
R=-4.1614*I+235.8
The graph starts on the left at point (32,117.8). The graph then decrases rapidly to point (21,137.9) then gradualy decreases through points (16,155),(8,183.75) and finaly ends at point ((4,247.5). This shows that the resistance stays fairly consistance over the current ranges shown. The slope and vertical intercept and equation of the line were found by entering the data points into an excel program and using the treadline capability of excel.
** #$&* Report for the 'other' resistor:; Resistance; half-life; explanation of half-life; equation of R vs. I; complete report. **
10.1
7.4+-.5
R=0.138*I+8.296
The results we obtained by using voltages and currents over a time period in order to obtain resistance values. These values were then entered into an excel spreedsheet and finding the best fit line using the treadline funtion avalible in order to obtain the equation of the line shown above.
** #$&* Number of times you had to reverse the cranking before you first saw a negative voltage, with 6.3 V .15 A bulb; descriptions. **
2
The number was accurate, on the second time i reversed the cycle of cranking I ssw a negative voltage. The bulb was fairly bright until i started cranking the handle in reverse and then the bulb dimed and final went out completly and then started to glow again until the five cycles were completed and the cranking was reversed again. everytime the cycle was reversed the bulb would begin to start glowing dimer and start the cycle over again. It would be assumed that the voltage across the capacitor would be working against the generator when the handle was reversed each time.
** #$&* When the voltage was changing most quickly, was the bulb at it brightest, at its dimmest, or somewhere in between? **
The bulb was somewhere inbetween at times when the handle was reversed and went completly dim after a few cranks. It seems the capacitor and generator are working against each other when the handle is reversed.
** #$&* Number of times you had to reverse the cranking before you first saw a negative voltage, with 33 ohm resistor; descriptions. **
2
I saw a negavite voltage on the second change in direction.
I continued reversing for several more cycles and saw that the more cycles occured the more the reversed cycles were negative. I also had to replace the 33 ohm resitor with the 10 ohm resitor due to the fact that the only two low value resistors I had was the 100 ohm and the 10 ohm.
** #$&* 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. **
6,15
zero voltage
3.35,3.15
** #$&* Voltage at 1.5 cranks per second. **
3.25
** #$&* Values of t / (RC), e^(-; t / (RC) ), 1 - e^(- t / (RC)) and V_source * (1 - e^(- t / (RC) ). **
24.75
Your half-life times were on the order of tens of seconds, and for the 1-ohm resistor and 1-farad capacitor RC would be 10 seconds.
17.78*10^-12
999.999*10^-3
3.24
** #$&* Your reported value of V(t) = V_source * (1 - e^(- t / (RC) ) and of the voltage observed after 100 'cranks'; difference between your observations and the value of V(t) as a percent of the value of V(t): **
3.24,3.15
2.77%
** #$&* According to the function V(t) = V_source * (1 - e^(- t / (RC) ), what should be the voltages after 25, 50 and 75 'beeps'? **
3.23
3.24
3.238
** #$&* Values of reversed voltage, V_previous and V1_0, t; value of V1(t). **
-3.14,3.35,0.210
2.639
** #$&* How many Coulombs does the capacitor store at 4 volts? **
1(C/v)*4V=4C
** #$&* How many Coulombs does the capacitor contain at 3.5 volts?; How many Coulombs does it therefore lose between 4 volts and 3.5 volts?; **
1(C/V)*3.5V=3.5C
.5C
** #$&* 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? **
2,0.25
By observation of the graph it took approxamatly 2 seconds for the cap to discharge for 4 to 3.5 volts. 0.5C/2(C/s)=0.025C
** #$&* According to your data, what was the average current as the voltage dropped from 4 V to 3.5 V?; How does this compare with the preceding result, how should it compare and why? **
.0325
The flow of current should be proportianal to the flow of Coulombs. The easy at which current is flowing should be proportional to the flow at which the Coulombs are flowing.
** #$&* How long did it take you to complete the experiment? **
8 hours
** **
&#Very good responses. Let me know if you have questions. &#