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: **
4.2 Volts, 33 ohms resistance
6 (seconds), 3.5 (Volts)
9, 3.0
12 2.5
15 2.0
22 1.5
32 1.0
39 .75
48 .50
1:07 .25
The results above are the times, in seconds, it took for the meter to reach the corresponding voltages. They were obtained by using the timer to observe the voltages.
** 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. **
about 12 seconds
about 13 seconds
about 17 seconds
about 16 seconds
The graph was clock time vs. voltage. As the time increased, the voltage decreased, with the clock time going from 6 seconds starting at 3.5 volts, going up to 1 minute and 7 seconds when the meter reached .25 volts.
** Table of current vs. clock time using same resistor as before, again starting with 4 volts +- .02 volts. **
6 (seconds) ,0 (mA)
9 0
12 0
22 0
32 0
39 0
48 0
1:07 0
The above results is set up as voltage vs. clock time. The same clock times were used from the previous procedure, except this time using the setting on 200 mA. All of the mA readings turned out to be 0 because the capacitor discharged very quickly. It discharged within just a couple of seconds. I repeated the procedure several times to see if I was doing something wrong, and came up with the same data each time.
If the capacitor discharged that quickly, it's probably because you had the ammeter in parallel with the capacitor, allowing the current to bypass the resistor or bulb. An ammeter is a very low-resistance device, and if it is connected in parallel it's likely to be damaged.
If the ammeter is connected in series with the capacitor and the resistor or bulb, the capacitor will discharge at about the same rate as in the first setup.
** Times to fall from initial current to half; 75% to half this; 50% to half this; 25% to half this, based on graph. **
The times for all of these turned out to be 0 because of my data that i observed.
** 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? **
Yes, the times I reported above are the same. They are all 0 but they are not the same times that I reported for voltages from the previous procedure.
** Table of voltage, current and resistance vs. clock time: **
The clock times are all 0 for each of the currents. This was because the capacitor discharged so quickly.
For voltage vs. clock time:
.8 times initial current about 8 seconds
.6 about 13 seconds
.4 about 18 seconds
.2 about 36 seconds
.1 about 58 seconds
** Slope and vertical intercept of R vs. I graph; units of your slope and vertical intercept; equation of your straight line. **
** Report for the 'other' resistor:; Resistance; half-life; explanation of half-life; equation of R vs. I; complete report. **
100 ohms resistance
about +-1 minute and 5 seconds
I used the same procedure as the first procedure, using the same giving voltages and observing how long it took to reach the voltages.
** Number of times you had to reverse the cranking before you first saw a negative voltage, with 6.3 V .15 A bulb; descriptions. **
I had to reverse the cranking time about 3 times before I saw mu first negative sign.
I think that my estimate was pretty accurate because I preformed the procedure twice and came up with the same observations both times.
The bulb looked like to would shine at its brightest when I should switch from cranking the generator from inverse to reverse. The bulb would become bright when I would first start cranking, then it would dim a little and stay at a constant shine. I think that the bulb was also brighter when I was cranking backwards.
** When the voltage was changing most quickly, was the bulb at it brightest, at its dimmest, or somewhere in between? **
When it was changing most quickly, the bulb was at its brightest.
I think that the relationship between the brightness of the bulb and the rate at which the capacitor voltage changes is that as the capacitor voltage changes at a faster rate (which was when the generator was reversed), the bulb becomes brighter.
** Number of times you had to reverse the cranking before you first saw a negative voltage, with 33 ohm resistor; descriptions. **
reversed about 3 times before I saw a negative voltage
I think my estimate is correct because I did the procedure twice and it took about 4 cranks in reverse both times before I saw negative voltage.
When I first cranked the generator at the double time constant rate of 165 cranks, I watched the voltmeter and the voltage gradually increased as I cranked the generator and by the time I got to 165 cranks, I was past 4 volts. When I then began to crank reverse for 1/4 of the time constant, the time then began to decrease at a steady pace, but quicker than it increased. I then reversed the generator in the opposite direction and the voltage began to go up again, I cranked for another 1/4 of a time constant. The third time I reversed the cranking, I then saw a negative voltage because the voltage started to decrease more rapidly.
** 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. **
about 21 beeps, about 22 seconds
The voltage was changing more quickly as it approached the 0 voltage
2.3 volts
** Voltage at 1.5 cranks per second. **
about 3.6 volts
** Values of t / (RC), e^(-; t / (RC) ), 1 - e^(- t / (RC)) and V_source * (1 - e^(- t / (RC) ). **
.66, .51, .49, 1.99
I evaluated the above results by reading the instructions above and using my values that I obtained to plug into the equations.
** 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): **
1.99, 2.3
.31
** 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). **
-2.3, 2.3, -4.6, 22 seconds
13.7
I did the calculations by using the directions above to obtain the numbers and then plugged the numbers into the first equation to get v1(t)
** How many Coulombs does the capacitor store at 4 volts? **
It would have 4 Coulombs. I got this result by using the equation Farad= Coulomb / Volt. I then used my farad, being 1.0, and multiplied it by the volt, being 4, to come up with the 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
** 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? **
6 seconds, .5
** 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? **
about 3 hours
** **
I found this lab a little difficult and confusing, especially towards the end
The lab gets a little more challenging toward the end.
See my notes and let me know if you have questions.