Phy 202
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: **
25, 4.0
0, 1.03, 2.66, 5.52, 9.02, 15.31, 19.92, 24.77, 32.22, 39.08
** 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. **
9 sec
12 sec
11 sec
12 sec
My graph seems to show some curvature. As the voltage decreases it takes more time to decrease (it is decreasing at a
decreasing rate). I determined these numbers by drawing a best fit line, then finding the numbers that corresponded to
each of these voltages on my line. Then i found the differences.
** Table of current vs. clock time using same resistor as before, again starting with 4 volts +- .02 volts.
**
35, 0
30, 2.867
25, 8.68
20, 14.61
15, 22.7
10, 28.45
5, 35.20
0, 50.09
** Times to fall from initial current to half; 75% to half this; 50% to half this; 25% to half this, based on graph.
**
17 sec
18 sec
14.5 sec
4 sec
This graph is more linear. As time goes by the current seems to decrease at about a constant rate. I also have my voltage
curve i drew above on this graph.
** 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? **
No the current vs time graph is decreasing at a constant rate and the voltage vs time graph seems to be decreasing at a
decreasing rate.
** Table of voltage, current and resistance vs. clock time: **
6 sec, 2.5V, 28 mamp, 2.4mamp/V
15 sec, 1.5V, 21 mamp, 10mamp/V
24 sec, 1.0V, 14mamp, 24mamp/V
33 sec, .75V, 7mamp, 44mamp/V
38 sec, .25V, 3.5mamp, 152mamp/V
current = voltage / resistance, so resistance = voltage / current. Your units don't appear to be correct, and it's hard to tell how your resistances are connected to the voltages and currents you report.
** Slope and vertical intercept of R vs. I graph; units of your slope and vertical intercept; equation of your
straight line. **
m=-0.28, y int= 27mamp
y int= mamp m=mamp^2/V
I=-0.28R+27mamp
I got my slope by taking two points on my best fit line and plugging the x and y components in the equations y2-y1/x2-x1.
I found the y intercept by taking this slope and one point on my line and solving for b in the equation y=mx+b. My graph
is more linear then the previous graphs, but it does level off at the end which could suggest concavity. Overall it suggests
that as current decreases, resistance increases.
** Report for the 'other' resistor:; Resistance; half-life; explanation of half-life; equation of R vs. I; complete
report. **
100 ohm resistor
I charged my capcitor to 4 Volts, however the voltage did not reduce at all when the circuit with the 100 ohm resistor
was attached. I would assume that this means that the resistor is so strong that the voltage does not reduce. This is much
like when we put the voltmeter in a series circuit and saw no change in voltage.
Would the circuit have been complete had you removed the voltmeter? What you describe would indicate that you were attempting to discharge the circuit through the voltmeter, rather than in parallel with it.
Another possibility is that you misread the resistor, and in fact have a very high resistance. If the resistance is very high, the voltage will change very slowly.
** 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 generator for 20 turns
I think my estimate is pretty accurate.
The bulb didnt light up until I was reverse cranking. This may be because the capacitor was storing the voltage and when
i reverse cranked, the voltage was released and the bulb was exposed to some allowing it to light up.
** 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 the most quickly. This my be because the capacitor was loosing
voltage faster which provided more volts for the bulb.
** Number of times you had to reverse the cranking before you first saw a negative voltage, with 33 ohm
resistor; descriptions. **
I had to reverse crank about 40 times
I am not sure if this is accurate, but i did the procedure as indicated.
My capacitance was 1 ferad so my time constant was 33 sec. 2*33sec is 66 seconds. I set my beeps to 2 cranks/sec.
My count was 66. The voltage increased as i turned it and decreased when i reverse turned it. The resistor may have
resisted the decrease in voltage which is why i had to reverse crank a bit more then the previous example.
** 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. **
18 beeps which was 9 sec.
changed more quickly as i approached 0
peak was about 4.0V
** Voltage at 1.5 cranks per second. **
4.5V
** Values of t / (RC), e^(-; t / (RC) ), 1 - e^(- t / (RC)) and V_source * (1 - e^(- t / (RC) ). **
2, .14, .86, 3.89
I found two by taking my clock time which was about 66sec/33sec. Then i solved the rest of the equations my plugging
my information into the given lay outs.
** 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): **
3.89, 4.0
2%
** According to the function V(t) = V_source * (1 - e^(- t / (RC) ), what should be the voltages after 25,
50 and 75 'beeps'? **
2.3, 2.84, 3.49
** Values of reversed voltage, V_previous and V1_0, t; value of V1(t). **
-4.0, 4.0V, -4V ,5 sec
4.65
** How many Coulombs does the capacitor store at 4 volts? **
132 Coulombs
33F=C/4V, then i solved for C.
It's a 1 Farad capacitor; 33 is the resistance of the resistor, in ohms, and it has nothing to do with the charge and voltage of the capacitor.
A Farad is a coulomb / volt; so this capacitor stores 4 coulombs 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?; **
115.5, 16.5
33=c/3.5V, to solve fore coulombs at 3.5V, then i subtracted this from the coulombs at 4 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? **
1.03sec, .5
It took 1.03 sec for the voltage of the 33ohm resistor to fall from 4.0 to 3.5 so i took this time and divided change in
voltage by this 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? **
The current was about 30 at this clock time.
This doesnt really compare. It should be similar, but they arent. I am not sure exactally why.
** How long did it take you to complete the experiment? **
2 hours
Good work, overall. See my notes and let me know if you have questions.