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: **
4volts, 2500
4.76, 3.5
5.84, 3.0
6.52, 2.5
7.10, 2.0
9.88, 1.5
12.33, 1.0
15.06, .75
20.92, .50
28.19, .25
overtime as time increases, voltage decreases, I used the timer program and once the volt meter read the voltages, I clicked the timer to obtain times of when the voltages were reached.
** 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. **
7.10
4.04
5.23
8.59
I took the inital voltage and subtracted it from the falling voltage
** Table of current vs. clock time using same resistor as before, again starting with 4 volts +- .02 volts. **
.11amps, 4.76sec
.46amps, 1.08sec
.74amps, .68sec
.86amps, .58sec
within limits, as clock time decreased amps increased
** Times to fall from initial current to half; 75% to half this; 50% to half this; 25% to half this, based on graph. **
4.18sec
4.08sec
.5sec
.10sec
graph represents amps vs clock time and with in limits as clock time decreased, current increased.
** 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 are same
Yes there is a pattern
** Table of voltage, current and resistance vs. clock time: **
.52volts, .13ohms
.52volts, .54ohms
.527volts, .55ohms
1.04votls, 1.09ohms
I took the initial current time and multiplied the additional currents to conclude the time, then for voltage I used the formula.
** Slope and vertical intercept of R vs. I graph; units of your slope and vertical intercept; equation of your straight line. **
20.45, .2
volts, ohms
y= 20.45x + .2
I used the formula y2-y1 / x2-x1 to get the slope of the line, and then I graphed my points to obtain my vertical intercept. The graph displays as voltage increases, ohms increased as well.
** Report for the 'other' resistor:; Resistance; half-life; explanation of half-life; equation of R vs. I; complete report. **
3.98
3.78 +- .20
I took the differences from multiple trials and there was an average of .20 difference among them.
** Number of times you had to reverse the cranking before you first saw a negative voltage, with 6.3 V .15 A bulb; descriptions. **
25
close within 5 cranks
brightness was dimmer, and the voltage was less but quick when cranking was reversed
** When the voltage was changing most quickly, was the bulb at it brightest, at its dimmest, or somewhere in between? **
brighter
brightness depends on rate at which voltage changes, when boltage is less, the bulb is dimmer, when voltage is more the bulb is brighter.
** Number of times you had to reverse the cranking before you first saw a negative voltage, with 33 ohm resistor; descriptions. **
30
partially accurate because I could not remember the number of cranks +- 10
voltage varied quickly overtime and then decreased slowly to 0
** 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 20sec
more quickly when approached peak voltage
3.80, 2.70
** Voltage at 1.5 cranks per second. **
225
** Values of t / (RC), e^(-; t / (RC) ), 1 - e^(- t / (RC)) and V_source * (1 - e^(- t / (RC) ). **
.72, .486,.514,115.6
I substituted R=33ohms, C=1Farad,T=23.8sec,V_Source=225 accordingly to the formulas provided
** 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): **
115.6, 105
10.6
** According to the function V(t) = V_source * (1 - e^(- t / (RC) ), what should be the voltages after 25, 50 and 75 'beeps'? **
48.64, 97.27,145.9
** Values of reversed voltage, V_previous and V1_0, t; value of V1(t). **
-2.3, 3.7, -2.3 , 37.62sec
.4466
I substituted the numbers V_previous=3.7, V1_0=-2.7, R=33ohms, C=1Farad accordingly to formula provided.
** How many Coulombs does the capacitor store at 4 volts? **
4Coulomb
1farad= Coulomb / 4volts, multiply both sides by 4 volts which gave me 4Coulomb
** 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
I took 1farad= Coulomb / 3.5volts, multiply both sides by 3.5 volts which gave me 3.5 Coulomb and then I took 4coulomb-3.5coulomb= .5coulomb
** 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? **
It took 3.6sec, .14C
I used the timer program which gave me 3.6 sec and then I used the formula to get .14C
** 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 average current= 3.6
Compares with the one above since my timing was correct.
** How long did it take you to complete the experiment? **
3hours
** **
This was a very challenging lab!
You did very well here. Good data, good analysis, good insight.