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.
** #$&* Your comment or question: **
** #$&* Initial voltage and resistance, table of voltage vs. clock time: **
4V, 42 ohms
4, 317.1719
3.5, 321.9375
3, 327.3887
2.5, 334.8281
2, 343.7637
1.5, 355.375
1, 373.0625
0.75, 387.8438
0.50, 403.8281
0.25, 427.9844
The first number is the voltage that I charged to the capacitor and the second number is the resistance of the resistor. The next set of numbers include the voltage and corresponding times.
** #$&* Times to fall from 4 v to 2 v; 3 v to 1.5 v; 2 v to 1 v; 1 v to .5 v, based on graph. **
26 s
28 s
30 s
30 s
As time increased, voltage decreased, creating a graph that decreased at a decreasing rate. I was able to calculate the time at 4 volts and then subtract the time that occured at 2 volts, and so on, to determine each set of times.
** #$&* Table of current vs. clock time using same resistor as before, again starting with 4 volts +- .02 volts. **
75mA, 804.4824
50mA, 815.1543
25mA, 849.2012
15mA, 881.8418
10mA, 914.2793
These numbers represent the current and corresponding clock times.
** #$&* Times to fall from initial current to half; 75% to half this; 50% to half this; 25% to half this, based on graph. **
25 s
28 s
38 s
50 s
The graph represents the current v. clock time and decreases at a decreasing rate.
** #$&* 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 started out close to one another, but the current times ended up being longer. The pattern is the increasing time that it takes for the current and voltage to decrease.
** #$&* Table of voltage, current and resistance vs. clock time: **
5, 3.4, 60, 56.67
15, 27, 45, 60
33, 1.7, 30, 56.67
77, 0.65, 15, 43
123, 0.15, 7.5, 20
I multiplied the initial current by the given decimal numbers and obtained a number which I used to look up the corresponding clock time by using the graph. I then used the graph to determine the voltage at each of these clock times. Lastly I divided the voltage by the current to obtain the resistance.
** #$&* Slope and vertical intercept of R vs. I graph; units of your slope and vertical intercept; equation of your straight line. **
0.76, 14.4
ohms, mA
60 = 0.76 (60) + b
This graph has the beginnings of a bell-shaped curve giving us the resistance v. current. I got the slope by dividing rise by run. I used that number to plug in to the above equation to obtain the vertical intercept.
** #$&* Report for the 'other' resistor:; Resistance; half-life; explanation of half-life; equation of R vs. I; complete report. **
150 ohms
+-71 s
I subtracted the beginning clock time from the time at 2 volts. t is the time occuring at 2 volts and change in time would be 1448.232 - 1377.154.
** #$&* Number of times you had to reverse the cranking before you first saw a negative voltage, with 6.3 V .15 A bulb; descriptions. **
20
Pretty close, +- 2 cranks
The bulb lit up with the initial cranking, then went out after a period of time, the immediately lit up brightly with reversing the direction. The capacitor voltage would decrease, then with reversing directions it would increase again, and so on. The bulb became brighter with reverse cranking with reverse cranking when the voltage in the capacitor decreased.
** #$&* When the voltage was changing most quickly, was the bulb at it brightest, at its dimmest, or somewhere in between? **
With the initial reversal it was brightest. The forward direction was a little dimmer. As capacitor voltage decreased the bulb became brighter due to increased current flowing from the capacitor to the bulb.
** #$&* Number of times you had to reverse the cranking before you first saw a negative voltage, with 33 ohm resistor; descriptions. **
8
within 1 crank, so this is pretty accurate.
It decreased more with each reversal.
** #$&* How many 'beeps', and how many seconds, were required to return to 0 voltage after reversal;; was voltage changing more quickly as you approached the 'peak' voltage or as you approached 0 voltage; 'peak' voltage. **
20, 15 s
more quickly as approaching peak voltage
3.5 V
** #$&* Voltage at 1.5 cranks per second. **
6 V
** #$&* Values of t / (RC), e^(-; t / (RC) ), 1 - e^(- t / (RC)) and V_source * (1 - e^(- t / (RC) ). **
0.9756, 0.3769, 0.6231, 3.7
** #$&* 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): **
3.7, 3.5
0.2
** #$&* According to the function V(t) = V_source * (1 - e^(- t / (RC) ), what should be the voltages after 25, 50 and 75 'beeps'? **
1.296, 2.3, 3.1
** #$&* Values of reversed voltage, V_previous and V1_0, t; value of V1(t). **
-3.5, 3.5, -7, 15 s
1.358
3.5 + (-7) * 0.306
** #$&* How many Coulombs does the capacitor store at 4 volts? **
C = F V = 1 F * 4 V = 4 C
** #$&* 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, 0.5
The capacitor contains 4 C @ 4 V and 3.5 C @ 3.5 V; therefore, with each decrease of 0.5 V, you will have a decrease of 0.5C
** #$&* According to your data, how long did it take for this to occur when the flow was through a 33-ohm resistor?; On the average how many Coulombs therefore flowed per second as the capacitor discharged from 4 V to 3.5 V? **
15 s, 0.27
** #$&* 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? **
75 mA
** #$&* How long did it take you to complete the experiment? **
4-5 hours
** **
&#This looks very good. Let me know if you have any questions. &#