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: **
4.0,70
4.0,0
3.5,3.125
3.0,7.25
2.5,11.8
2.0,16.9
1.5,22.4
1.0,28.5
.75,33.0
.5,39.1
.25,41.0
These results show how the voltage in the capacitor discharges through a resistor over time.
** 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. **
19.2
15.9
13
7.6
My graph shows the voltage decreasing at a decreasing rate. After drawing the line of the curve, I drew lines to estimate each corresponding time and voltage above and subtracted each pair to get the difference.
** Table of current vs. clock time using same resistor as before, again starting with 4 volts +- .02 volts. **
100,0
75,14.7
50,33.9
25,50.8
12,61.3
6,71.9
3,84.5
These results show how the current in the capacitor diminishes through a resistor over time.
** Times to fall from initial current to half; 75% to half this; 50% to half this; 25% to half this, based on graph. **
33.9
41.7
16.9
10.5
My graph shows the current decreasing at a decreasing rate. After drawing the line of the curve, I drew lines to estimate each corresponding time and current above and subtracted each pair to get the difference.
** 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? **
These two patterns show a similar trend of decreasing current and voltage at a decreasing rate over time but there is quite a bit of difference in the actual values. I suspect this is due to my measurement of an initial current that is in fact much higher than 100mA, but the speed of the current drop in the first second made a more accurate measurement difficult.
** Table of voltage, current and resistance vs. clock time: **
9.6,2.73,80,34
28.1,1.07,60,18
40.5,.34,40,9
52.6,0,20,0
66.1,0,10,0
Each proportion was multiplied by the initial current and the graph was used to determine the corresponding clock time. These clock times were used in the V v. t graph to determine the corresponding voltages. Finally, R=V/I
** Slope and vertical intercept of R vs. I graph; units of your slope and vertical intercept; equation of your straight line. **
.625,-17
ohm/mA
R=.625I-17
My graph shows a direct proportion between resistance and current with a slope that decreases at a decreasing rate, showing how both of these decrease together in the circuit.
** Report for the 'other' resistor:; Resistance; half-life; explanation of half-life; equation of R vs. I; complete report. **
42
15.453125+-.5sec
Based on the speed at which voltage decreased, I felt confident that my click accurately recorded the time to within .5sec.
R=5.09I-221
voltage vs time
4.0,0
3.5,3.59375
3.0,6.828125
2.5,11.015625
2.0,15.453125
1.5,21.3125
1.0,28.890625
.75,34.4375
.5,39.953125
.25,47.21875
current vs time
108,0
100,11.17188
75,24.25001
50,41.18751
25,49.796885
12,57.75001
6,67.34376
current vs resistance
108,660
100,24
75,17
50,9
25,5
12,0
6,0
As with the previous system, these results show a similar trend of decreasing current and voltage with time.
** Number of times you had to reverse the cranking before you first saw a negative voltage, with 6.3 V .15 A bulb; descriptions. **
12
I counted out loud: 1,2,3,4,5,ONE! etc...so it was accurate.
Around 40 cranks, the bulb went dim. By the time I got to 100 cranks, the voltage was steady at 3.3V. As soon as I reversed, the bulb glowed like the sun and I still have a blind spot from looking directly at it. When I switched again, there was no glow. Reversed again, bright glow. Eventually, cranking forward made it glow again too, once the voltage had dropped sufficiently, but the brightest glow was always from cranking in reverse. This is because cranking backward discharges the capacitor through the bulb whereas cranking forward attempts to add more charge to an already charged capacitor, at least until near the end of the experiment when the capacitor is nearly discharged again.
** When the voltage was changing most quickly, was the bulb at it brightest, at its dimmest, or somewhere in between? **
The bulb is brightest when the voltage is changing most quickly because voltage is proportional to current and resistance, which are the factors responsible for making the bulb glow.
** Number of times you had to reverse the cranking before you first saw a negative voltage, with 33 ohm resistor; descriptions. **
5
Accurate using the same method as above.
The pattern here was identical to the trial above except that it took fewer crank reversals to get to negative voltage. This was expected because each time I reversed, I cranked 15x instead of just 5x,
1.4Hz max voltage 3.9. 42ohm 42s 118beeps in 84s. 15beeps in 10.5s.
** 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. **
26beeps, 18.6s
The voltage changed more quickly as I approached 0 voltage.
3.1V.
** Voltage at 1.5 cranks per second. **
approx 3.2V
** Values of t / (RC), e^(-; t / (RC) ), 1 - e^(- t / (RC)) and V_source * (1 - e^(- t / (RC) ). **
-2, .135, -6.39, 3.46
-84/(42*1)=-2, e^2=.135, 1-.135=.865, 4*.865=3.46
** 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): **
3.46, 3.9
88%
** According to the function V(t) = V_source * (1 - e^(- t / (RC) ), what should be the voltages after 25, 50 and 75 'beeps'? **
1.38, 2.29, 2.88
** Values of reversed voltage, V_previous and V1_0, t; value of V1(t). **
3.9, -7.9, 18.6, 1.07
3.9+(-4-3.9)(1-e^(-18.6/(42*1))=1.07. This equation gives the voltage at a time t after the voltage is reversed.
** How many Coulombs does the capacitor store at 4 volts? **
4C b/c 4*1=4
** 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
3.5*1=3.5, 4-3.5=.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? **
3.59375s, .139C/s
.5/3.59375=.139
** 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? **
103mA
As an average current of 103mA is flowing through the circuit during the first 3.59375 seconds, the capacitor loses .5 Coulombs of capacitance. I expect this relationship to be directly proportional because I know that the capacitor has lost all its capacitance once the capacitor is fully charged and no more current is flowing through the circuit.
It still has its capacitance, which is a measure of how much charge increases for a given increase in voltage; however when the voltage on the capacitor is equal to the voltage of the source (or very nearly so) there is no voltage across the resistance and no current will flow.
&#The current progressively diminishes; the graph decreasing at a decreasing rate toward a limiting current of zero.
When constant voltage is applied, as is the case here if you crank the generator at a constant rate, the charge on the capacitor approaches C * V, where C is the capacitance and V is the voltage.The current in this case decreases, as evidenced by the decrease in force. As the capacitor approaches its terminal charge, its voltage approaches that of the source, and opposes that source. As a result charge flows less and less quickly onto the capacitor (i.e., current decreases).
&#
** How long did it take you to complete the experiment? **
9hrs
** **
Excellent work. Let me know if anything is unclear.