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 couldn't figure out a way to do the last part of this experiment without the 33ohm resistor.
Initial voltage and resistance, table of voltage vs. clock time:
4,100ohms
3.5,9.41
3.0,21.49
2.5,35.85
2.0,54.24
1.5,78.49
1.0,113.17
.75,137.86
.50,174.19
.25,237.06
The capacitor discharged at a slower rate as the voltage decreased.
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.
54s
57s
60s
61s
Table of current vs. clock time using same resistor as before, again starting with 4 volts +- .02 volts.
35mA,8.42
30mA,20.14
25mA,34.87
20mA,53.80
15mA,78.28
10mA,114.11
7.5ma,140.05
5.0mA,176.80
2.5mA,244.70
The rate of discharge slowed as with time
Times to fall from initial current to half; 75% to half this; 50% to half this; 25% to half this, based on graph.
53sec
58sec
61sec
62sec
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 they are the same.
Yes they are when the resistance remains the same. The current and vlotage are directly prportional
Table of voltage, current and resistance vs. clock time:
18,3.2,32,100
33,2.4,24,100
78,1.6,16,100
120,.8,8,100
185,.4,4,100
I used ohms law to get the coresponding current to voltages.
I then used the graph to determine times corespondimg to the voltage
Slope and vertical intercept of R vs. I graph; units of your slope and vertical intercept; equation of your straight line.
1,0
V=I*R
The graph is a vertical line.
When the resistance is constant, the current and voltage are directly proportional
Report for the 'other' resistor:; Resistance; half-life; explanation of half-life; equation of R vs. I; complete report.
I don't have a 33ohm resitor,the only 3 resistors that are in my kit are the 100ohm, 10ohm, and a 460ohm resistor. I measured the time for the voltage to discharg, but I was unable to check the current because my meter's highest setting is 200mA. The curent from 4volts and a 10ohm resistance would exceed this. (400mA)
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
+- 2
As you cranked for the 100 beeps the bulb dimmed. When you reversed the crank the bulb got brighter. Reverse it again and it got dimmer.
The voltage of the cap dcreased wtih time. The brightness of the bulb changed because whren the direction was reversed the voltage was added to the voltage applied to the bulb from the discharging capacitor.
When the voltage was changing most quickly, was the bulb at it brightest, at its dimmest, or somewhere in between?
The bulb was brightest when the voltage was changing quickly
The faster the voltages changes the more voltage was applied to the bulb from the cap and the generator
Number of times you had to reverse the cranking before you first saw a negative voltage, with 33 ohm resistor; descriptions.
I tried this with both the 100ohm and the 10ohm resistor.
The100ohm had a time constant of 100seconds, That meant 300 beeps at the rate required to maintan 4V, for double the time constant. I couldn't keep track of that many beeps my data was all over the place.
I then tried with the 10 ohm resistor and the values where to small for me to react to.
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.
Voltage at 1.5 cranks per second.
about 4V
Values of t / (RC), e^(-; t / (RC) ), 1 - e^(- t / (RC)) and V_source * (1 - e^(- t / (RC) ).
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):
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).
How many Coulombs does the capacitor store at 4 volts?
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?;
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?
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?
4hrs.
I couldn't figure out how to finish the rest of this experiment with the choice of resistors that I have.
You did a good job on this experiment, and made a good, significant effort to substitute for the 33 ohm resistor. This is noted, to your credit.