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: **
I wasn't provided with a 33 ohms or 100 ohms resistor. my resistors were 15 ohms and 100 ohms; I believe that this caused some of results to be erroneous.
** Initial voltage and resistance, table of voltage vs. clock time: **
4, 15 Ohms
3.5, 1.32
3.0, 2.20
2.5, 5.00
2.0, 9.23
1.5, 13.63
1.0, 18.13
.75, 23.07
.50, 26.92,
.25, 31.36
After charging the voltmeter to 4.00 V, I then recorded the time elapsed between each interval. My results show that voltage decreases at an increasing rate.
** 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. **
9
11
9
9
My graph is downward sloping, quadratic graph. Each point corresponds to an x and y value; x is the clock time and y is the voltage. I estimated the times as stated above. They are almost all consistent at 9 second intervals.
** Table of current vs. clock time using same resistor as before, again starting with 4 volts +- .02 volts. **
0, 0.00125
13.57, 0.000938
244.86, 0.000625
878.32,0.000313
2056.28,0
** Times to fall from initial current to half; 75% to half this; 50% to half this; 25% to half this, based on graph. **
244.86,231.29,633.46,1178.26
he current vs. clock time graph is downward sloped and is represented by a quadratic equation. The graph shows that over time current decreases. I used the Timer program and the voltmeter to determine the values. It was difficult to measure the current because I had to switch from volts after charging the capacitor and quickly re-attach the leads to measure the current. I may have charged the capacitor too much, which could have resulted in the lengthy clock times for measuring the decreasing current.
The long time intervals are the result of using a resistor with greater resistance than that you read, or possibly connecting the voltmeter in series rather than parallel. Either way the resistance in the circuit was well into the thousands of ohms. However, except for the extra time it took you to get your readings, everything looks good.
** 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 times reported are not the same. They are not the same times I reported for the voltage drops; however, there is a pattern. The values for both voltage and current both decrease over time; however, voltage decreases at a much faster, more consistent rate than the current.
** Table of voltage, current and resistance vs. clock time: **
** Slope and vertical intercept of R vs. I graph; units of your slope and vertical intercept; equation of your straight line. **
I could not sketch a graph based on my data; however, I know that resistance is inversely proportional to current. The graph would have been downward sloping.
Please note that I was not able to complete this part of the experiment. My clock times when measuring voltage were much shorter than the clock times I measured using current. With my data I could not calculate the voltage for the current at specific clock times. I measured everything accurately; however, I believe that without 2 multimeters I could not complete the second part of the experiment as accurately as I should have. Also the resistors that were provided to me in the lab kit were not 33 and 100 ohms; they were 15 and 58 ohms. This could have thrown my experiment completely off.
** Report for the 'other' resistor:; Resistance; half-life; explanation of half-life; equation of R vs. I; complete report. **
55 Ohms
11.20 +- 1.00 seconds
I used the TIMER to measure t; my uncertainty came from the previous part of the experiment when I was asked to measure uncertainty.
Resistance is inversely proportional to resistance. If voltage increases, current increases and then resistance decreases. Resistance also depends on the resisitivity, length and area of the material in the experiment.
** Number of times you had to reverse the cranking before you first saw a negative voltage, with 6.3 V .15 A bulb; descriptions. **
About three cranks
I think my estimate was somewhat accurate; it might have been closer between the first and second crank when the voltage went negative.
The bulb’s brightness stayed consistent when I was turning the generator forward. As soon as I reversed the crank the bulb went dim.
** When the voltage was changing most quickly, was the bulb at it brightest, at its dimmest, or somewhere in between? **
The voltage changed most quickly at first and the bulb was at its brightest. The faster the voltage changes the faster the bulb will die out. the meter reads a decreasing voltage when the charge decreases, voltage decreases, and current decreases.
** Number of times you had to reverse the cranking before you first saw a negative voltage, with 33 ohm resistor; descriptions. **
I had to use the 15 ohm resistor; I do not have a 33 ohm resistor
I had to reverse the crank 1.5 times before I got a negative voltage reading.
I think the estimate was accurate because I was cranking at a relatively slow rate of .25 beeps per second.
I cranked the generator handle in the forward direction at 7.5 cranks per 30 seconds. I then reversed the crank at a rate of 1 turn in 3.75 seconds. The voltage after 1 crank in 3.75 seconds was almost negative. I continued slightly in the reverse direction and the voltage became negative.
** 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. **
2 beeps, .50 seconds
The voltage was changing more quickly as I approached 0.
My peak voltage was about 9V.
** Voltage at 1.5 cranks per second. **
17 volts
** Values of t / (RC), e^(-; t / (RC) ), 1 - e^(- t / (RC)) and V_source * (1 - e^(- t / (RC) ). **
.45, .639, .361, 6.137
I first calculated t using my .25 beeps per second rate. I found that t = 6.7. I then put t/ RC to find this value. I plugged in this value of .45 into the equation e^(-t/RC) to calculate .639. I then found the value of 1 minus this value; this was .361. My max. voltage was 17 V for the source. I then multiplied 17 x .361 to get 6.137V.
** 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): **
6.137, 9
2.863, 68.8%
** According to the function V(t) = V_source * (1 - e^(- t / (RC) ), what should be the voltages after 25, 50 and 75 'beeps'? **
1.78, 3.38,4.81
** Values of reversed voltage, V_previous and V1_0, t; value of V1(t). **
-8, 5, 3, 2
7.626 V
I used the rate of .25 cranks per second to find out t. In two seconds the voltage dropped to 0. I then plugged in my values for V previous and V1_0 to calculate V(t).
** How many Coulombs does the capacitor store at 4 volts? **
The capacitor is in 1 Farad. At 4 volts, the capacitor can store 4 Columbs. I used the equation above:
1F = Coulomb/ 4V
Coulomb = 1 F x 4 V
= 4 Coulomb
** 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, .5
I used the equation F = Coulombs/ V to find that the capacitor with 3.5 Volts contains 3.5 coulombs. The difference in coulombs is 4 – 3.5 V = .5 V. 1 F x .5 V = .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? **
42.49, .012
I solved for t:
17V = .5 V (1- e^(-t/15)
ln (17) = ln (.5) – [ ln(.5) + ln (e^-t/15)]
= ln (.5) – ln (.5) – (-t/15)
Ln (17) = t/15
T = 42.49
Then I set up a ratio to solve for the average number of Coulombs flowing per second when the capacitor is discharged
.5 V/ 42.49 = x/ 1
X = .012 Coulombs
** 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? **
When the voltage dropped from 4V to 3.5 V, the average current during this time was .001 A.
Coulomb by definition is the unit of electrical charge equal to the amount of charge transferred by a current of 1 amp in 1 second. The change in amp that corresponds to the .5 change in voltage over a time interval should be equivalent to the Coulomb of electrical charge transferred in that period.
** How long did it take you to complete the experiment? **
4 hours
** **
Your work on this experiment is excellent.