Phy 232
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: **
3.98 V, 68
3.98, 0
3.5, 7.52
3, 16.31
2.5, 27.79
2, 42.57
1.5, 62.01
1, 90.08
0.5, 139.29
** 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. **
42.57
45.70
47.51
49.21
The graph is exponential, time approaches infinity as the voltage approaches 0.
** Table of current vs. clock time using same resistor as before, again starting with 4 volts +- .02 volts. **
60, 0
50, 9.34
40, 20.00
30, 38.45
20, 60.97
10, 96.40
0, 200.10
After charging the capacitor to 4 V, I connected the meter, resistor, and capacitor in series. The graph of the current vs. resistance is similar to that of the voltage vs. resistance. Both have, approximately, an exponential trend. As time passes less and less current is absolved.
** Times to fall from initial current to half; 75% to half this; 50% to half this; 25% to half this, based on graph. **
38.45
42.38
32.35
38.98
Using my current vs. time graph, I visually interpolated the time values for currents for which I did not have precise values.
** 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 standard deviation of the numbers above is 4.174, but since there are not many numbers, this does not accurately describe their distribution. The numbers are very near to each other when taking into account that I eyeballed them from a hand-sketched graph. The drop in current seems to be very similar to the drop voltage which I observed earlier.
** Table of voltage, current and resistance vs. clock time: **
t, I, V, R
9.34, 48, 3.4, 70.8
33.95, 36, 2.3, 63.9
50.82, 24, 1.8, 75
92.40, 12, 0.9, 75
150.00, 6, 0.43, 71.7
I obtained the resistance values by dividing the voltage by the current for each time.
** Slope and vertical intercept of R vs. I graph; units of your slope and vertical intercept; equation of your straight line. **
-3.3, 281.6
current/resistance, current
I = -3.3*R + 281.6
Even though the equation above is linear, my data seemed a little scattered with no particular trend. I got the slope by using the first and last points (these two were in the middle of the others.) But, it seems to me that if V and I are decreasing, that R would stay relatively constant.
** Report for the 'other' resistor:; Resistance; half-life; explanation of half-life; equation of R vs. I; complete report. **
15
8.07+-0.05
t was determined by subtracting the time it took to go from 4V to 2V.
I = 105*R - 1288.5
It seems that the resistance in this situation stayed relatively constant also. I also noticed that my resistances are very close to what is labeled on the resistor. For instance, my resistance this time were 14.1, 15.9, 12.5, 12.5, and 12.5.
** 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 did about 20 reverse cranks.
This is fairly accurate, I cranked in sets of 5 until I got negative voltage.
When I reversed the direction of the cranking, the bulb got really bright. By reversing the direction of the cranking, I reversed the direction of the current. I think this caused the capacitor to release its current into the series, which increases the current and voltage in the system. This gave more power to the bulb.
** When the voltage was changing most quickly, was the bulb at it brightest, at its dimmest, or somewhere in between? **
The voltage seemed to change the fastest when the bulb was at its brightest. The higher the voltage drop across the circuit, the brighter the bulb is going to burn.
** Number of times you had to reverse the cranking before you first saw a negative voltage, with 33 ohm resistor; descriptions. **
19
I reversed the cranking for 1/4 the time constant consistently other than random errors it should be accurate.
When I cranked in a reverse manner, the voltage dropped very quickly. However as the total voltage became lower the voltage didnt drop at as fast of a rate. The capacitor voltage became less and less over time.
** 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. **
25, 12.5
Voltage seemed to drop more near peak voltage, but the difference didnt seem to be that great.
4.86 V
** Voltage at 1.5 cranks per second. **
3.3 V
** Values of t / (RC), e^(-; t / (RC) ), 1 - e^(- t / (RC)) and V_source * (1 - e^(- t / (RC) ). **
2, 0.135, 0.865, 2.85
I performed the calculations as instructed above. The time t at which I reversed the voltage was twice the time constant, which was R*C.
** 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): **
2.85, 4.86
171%
** According to the function V(t) = V_source * (1 - e^(- t / (RC) ), what should be the voltages after 25, 50 and 75 'beeps'? **
2.2, 2.94, 3.18
** Values of reversed voltage, V_previous and V1_0, t; value of V1(t). **
-3.3, 2.85, 0.45, 10
1.24
I think that the voltage at t = 10 should be 0. So V1_0 * (1 - e^(- t / (RC) ) should equal -V_previous.
** How many Coulombs does the capacitor store at 4 volts? **
4 C
The capacitance of my capacitor is 1 F, so 4V*1F would equal 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?; **
3.5 C, 0.5 C
This is the charge the capacitor loses as the voltage drops. Actually I was wondering how connecting one lead to the capacitor could cause it to discharge, but that makes a lot of sense.
** 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? **
1.21, 413 mA
The coulombs per second is the current that flowed from the capacitor during that time.
** 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? **
210 mA
The current I recorded is almost half the value of the current previously calculated.
** How long did it take you to complete the experiment? **
4 hrs.
** **
Very well done.