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 received a 10 ohm resistor. Its color code is brown-black-black with a gold stripe. This allowed the current to exceed my meter's limit, so I could not measure it. The more rapid drop in voltage compared to the 33-ohm resistor indicates that it has lower than 33 ohms of resistance.
** Initial voltage and resistance, table of voltage vs. clock time: **
4.0, 33
3.5, 3.43
3.0, 7.64
2.5, 13.13
2.0, 20.30
1.5, 29.84
1.0, 42.91
.75, 52.68
.5, 65.83
.25, 91.4
The table describes the voltage across the capacitor, with respect to the time it's been connected in series with a 33-ohm resistor. I charged the capacitor to 4.00V and used the timer program to measure the time at which the capacitor reached each voltage.
** 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. **
20.30
22.2
22.6
22.92
The graph decreases at a decreasing rate. I estimated about 21 seconds for each of the voltage changes. I got the table values by actually calculating the times for the voltage to decrease by one-half.
** Table of current vs. clock time using same resistor as before, again starting with 4 volts +- .02 volts. **
110, 0
100, 2.2
90, 5.0
80, 8.4
70, 11.9
60, 17.0
50, 22.6
40, 28.5
30, 38.3
20, 52.1
10, 72.6
8, 83.4
6, 93.3
4, 108.6
3, 121.5
The left-hand column of the table gives the current in milliamperes, and the right hand column gives the time in seconds after the capacitor begins to discharge through a 33-ohm resistor. I charged the capacitor to 4 volts, then connected the ammeter in series with the capacitor and resistor and used the timer program to record the elapsed time for the listed currents.
** Times to fall from initial current to half; 75% to half this; 50% to half this; 25% to half this, based on graph. **
21.67
22
21.5
The curve decreases at a decreasing rate, consistent with the voltage measured earlier and a constant resistance. As earlier, I estimated about 21 seconds for each of the current changes. I got the table values by averaging the intervals on either side of the target currents and estimating the time at each target current.
** 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? **
Within experimental uncertainty, these times can be considered the same. The time for current or voltage to decrease by one-half can be estimated at 22 seconds.
** Table of voltage, current and resistance vs. clock time: **
7.01, 3.02, 88.8, 34.0
13.59, 2.48, 66.6, 37.2
26.33, 1.69, 44.4, 38.1
51.02, 0.80, 22.2, 36.0
71.03, 0.44, 11.1, 39.6
I used MS Excel to graph the current and voltage vs. clock time and produce a regression equation for each. Using the equation for current, I found the clock times for each of the target currents. Using the equation for voltage, I found the voltage for each clock time. The resistance was calculated using the relationship R = V / I
** Slope and vertical intercept of R vs. I graph; units of your slope and vertical intercept; equation of your straight line. **
-.048, 39.235
The slope has no units,the intercept's units are ohms.
Resistance = -.048 * current + 39.2
The graph was scattered, with no clear trend. Again, I used Excel to calculate a best-fit line and generate the equation. Since resistance should not vary with current in this circuit, the equation is essentially demonstrating the experimental error. A horizontal line (slope = 0) with y-intercept of 33 would model a 33-ohm resistor.
** Report for the 'other' resistor:; Resistance; half-life; explanation of half-life; equation of R vs. I; complete report. **
10 ohms
5.7 s +/- .1s
I performed three trials of this measurement. The results were consistent and averaged 5.7 with .1s being the greatest deviation.
I was unable to measure the current through this resistor as the capacitor discharged from 4 V due to the limits of my ammeter. I expected the current to decrease at a decreasing rate, and a calculated resistance that was nearly constant (regression equation with slope less than .1 ).
** Number of times you had to reverse the cranking before you first saw a negative voltage, with 6.3 V .15 A bulb; descriptions. **
23
This is an exact number, I counted.
The bulb glowed initially as the capacitor was charged. It eventually stopped glowing as I finished the initial 100 cranks. When I reversed the cranking direction, the voltage of the generator was reversed. The capacitor was discharging, and the bulb glowed brightly. When I next reversed the cranking direction I was again charging the capacitor, and the bulb glowed-though not as brightly. The capacitor's voltage fluctuated after its initial charging. It discharged faster with the backward cranking than it charged with the forward cranking. The faster discharge corresponded with more current and brighter glowing of the bulb. The less voltage the capacitor had, the slower it discharged.
** When the voltage was changing most quickly, was the bulb at it brightest, at its dimmest, or somewhere in between? **
** Number of times you had to reverse the cranking before you first saw a negative voltage, with 33 ohm resistor; descriptions. **
9
The number of times I reversed the voltage was accurate, but my cranking is not always consistent.
The capacitor voltage definitely decreased with the backward cranking faster than it increased with the forward cranking.
** 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. **
32, 54.4
Voltage changed more quickly when approaching 0 voltage.
3.4V
** Voltage at 1.5 cranks per second. **
3.5V
** Values of t / (RC), e^(-; t / (RC) ), 1 - e^(- t / (RC)) and V_source * (1 - e^(- t / (RC) ). **
5.15, .00579, .99421, 3.977
I multiplied the number of cranks (100) by the cranking rate (1.7/sec), then divided that number by the product of the resistance (33 ohms) and the capacitance (1.0 F). I then raised e to the negative of that result. I subtracted this number from one, and multiplied it by the source voltage (4.0V).
** 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): **
3.977, 3.4
.577, 14% error
** According to the function V(t) = V_source * (1 - e^(- t / (RC) ), what should be the voltages after 25, 50 and 75 'beeps'? **
2.9, 3.7, 3.9
** Values of reversed voltage, V_previous and V1_0, t; value of V1(t). **
3.4, -7.4, 54.4, -2.6
-2.6
I used the observed value of 3.4V as V_previous and -4.0V as the reversed voltage. The difference between these is -7.4V, which I used as V1_0. The time for discharge was calculated previously. I used it to calculate the value of e^(- t / (RC)), then substituted the value (.19234) into the equation above. This should be the capacitor voltage 54.4 seconds after reversing the voltage. This seems like a considerable error, even taking my inconsistent cranking into account. Subsequent trials have given very similar results.
** How many Coulombs does the capacitor store at 4 volts? **
4 Coulombs, I multiplied the capacitance (1 Coulomb / Volt ) by the number of volts (4).
** 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
This capacitor stores one coulomb for every volt of potential difference between its leads.
** 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? **
3.43, .146
The time it took for the capacitor to discharge from 4.0V to 3.5V was 3.43 sec. This correlates to .5 C moving through the circuit during this time interval. .5 C/3.43 sec = .146 amperes, which is the average current. I measured a lower current at 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? **
99.65
The current calculated above is higher than the current I measured when discharging the capacitor. The current was limited by the resistor, so a lower value is to be expected.
** How long did it take you to complete the experiment? **
6 hours
** **
Very good work. You clearly understand this, but if you need clarification on anything please ask.