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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.
** The RC Circuit_labelMessages **
4/28 2:30
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About 6.5 hours
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The RC Circuit
As on all forms, be sure you have your data backed up in another document, and in your lab notebook.
Your course (e.g., Mth 151, Mth 173, Phy 121, Phy 232, etc. ):
Measure capacitor voltage vs. t when you discharge the capacitor through the resistor
In the preceding experiment you discharged a capacitor through a bulb, and from your graphs of current vs. clock time and voltage vs. clock time you concluded that the resistance of the bulb varies very significantly with the current.
In this experiment you will measure a similar system, but will use resistors rather than a bulb. You will confirm that the resistor has a very nearly constant resistance, in contrast with the bulb.
If your kit doesn't contain a resistor, you should have noted this when you checked the contents of the kit and requested a set of resistors. However, if you do not have a resistor, you may substitute a bulb (different from the one you used previously).
You should determine the resistance of the resistor. Each resistor has a series of colored rings; the colors tell you the resistance.
To determine the resistance:
Note the color of the first ring and write down the corresponding digit, according to the table given below.
Write down the the digit corresponding to the color of the second ring.
Your two digits give you a two-digit number.
Raise 10 to the power of the number corresponding to the third ring.
Multiply the decimal number by the power of 10 and you have the resistance.
Black Brown Red Orange Yellow Green Blue Violet Gray White
0 1 2 3 4 5 6 7 8 9
For example the resistor shown above has a blue first ring, corresponding to the number 6. It has a red second ring, corresponding to the number 2. These two digits give you the decimal number 62. The third ring is red, indicating digit 2, so raise 10 to the power 2 and multiply by your two-digit number to get 62 * 10^2 = 62 * 100 = 6200.
To find out more search for 'reading resistors' and look for a page with a color picture of the rings around the resistor, accompanied by an explanation. There are hundreds of good sites and a search will quickly lead you to some good ones.
To begin the experiment, you will charge the capacitor to 4.00 volts +- .02 volts, then set up a series circuit consisting of the capacitor and a resistor whose resistance is between 25 ohms and 150 ohms. The voltmeter should be connected in parallel with the capacitor.
Use the TIMER program to record the clock times at which the voltage reaches 3.5 volts, 3.0 volts, 2.5 volts, 2.0 volts, 1.5 volts, 1.0 volt, .75 volt, .50 volt and .25 volt.
Give your initial voltage and the resistance in the first line; then starting in the second line give your voltage vs. clock time table, with one clock time and one voltage per line, in comma-delimited format. Starting in the line following your table give a short synopsis of what your results mean and how they were obtained.
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4.0, Did not have a resistor so I used the 14 Volt bulb
3.40, 3.5
7.80, 3.0
12.84, 2.5
17.70, 2.0
25.50, 1.5
33.79, 1.0
39.73, 0 .75
45.82, 0.5
53.70, .025
Table is clock time in seconds, voltage in Volts.
I used the TIMER program and then did a running sum of all the time intervals.
NOTE: I did not have a set of resistors. Unfortunately I did not do a thorough inventory inspection of my kit when I received it. I will be using the 14V bulb instead.
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Sketch a good graph of voltage vs. clock time, and sketch the curve you think best represents the voltage of the system vs. clock time.
Using your graph, estimate as closely as you can:
The time required for the voltage to fall from 4 volts to 2 volts.
The time required for the voltage to fall from 3 volts to 1.5 volts.
The time required for the voltage to fall from 2 volts to 1 volt.
The time required for the voltage to fall from 1 volts to .5 volts.
Give your estimated times, one time per line. Starting in the first line after your table, describe your graph and explain how you used it to determine these times.
****
17 sec
18 sec
16 sec
13 Sec
These are the times in seconds for the voltages at various stages to decrease by half. These are based on a visual estimation using the graph not calculations based on the data points. The graph is a smooth curve with voltage decreasing as clock time increases. It appears to be decreasing at a decreasing rate.
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For a constant resistance the curve would be exponential.
The half-life is fairly consistent. It would be consistent if and only if the voltage was an exponential function of time, which would happen with constent resistance. So it appears that the variation in the voltage isn't too drastic.
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Measure current vs. t when you discharge the capacitor through the resistor
You have observed and graphed the voltage across the capacitor as a function of time.
Now repeat the same procedure, but this time measure the current. In the process be sure to verify that the capacitor is in fact at 4.00 volts +- .02 volts.
Remember to be very sure the meter is connected in series rather than in parallel. The safest way to do this is to first put the meter in series with the bulb, the capacitor and an open switch, set to read volts, then close the switch and watch for a few seconds to be sure the reading on the meter doesn't change. If you don't have a switch, you can simply unclip one of the leads leading to capacitor, bulb and meter in order to simulate an open switch, then connect it to simulate a closed switch.
The reason this works is that the voltmeter has a very high resistance and will not permit significant current to flow. However if the meter is in parallel with the bulb and capacitor, the capacitor will discharge, resulting in a change of voltage.
If the voltmeter shows unchanging voltage, then it's probably safe to open the switch, then turn the dial to the 200 mA setting.
The ammeter has a very low resistance and if connected in parallel, a very large current will flow, with the potential to damage the meter.
After setting up the circuit go ahead and close the switch (or connect the lead) and observe current at a function of clock time, then give your table of current vs. clock time in the box below, using the same conventions you used above in reporting voltage vs. clock time. Starting in the following line give a short synopsis of what your results mean and how they were obtained.
****
0, 100
3.93, 90
10.02, 80
17.80, 70
26.94, 60
35.38, 50
45.74, 40
56.87, 30
67.69, 20
77.21, 10
Results are clock time in seconds, current in mA
I used the TIMER program to find the intervals between current levels and then did a running sum on Excel.
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Sketch a good graph of current vs. clock time, and sketch the curve you think best represents the voltage of the system vs. clock time.
Using your graph, estimate as closely as you can:
The time required for the current to fall from the initial current to half the initial current.
The time required for the current to fall from 75% of the initial current to half of this value.
The time required for the current to fall from half the initial current to half of this value.
The time required for the current to fall from 1/4 the initial current to half of this value.
Give your estimated times, one time per line. Starting in the first line after your table, describe your graph, explain what the graph represents and explain how you used it to determine these times.
****
32 seconds
30 Seconds
22 Seconds
8 Seconds
Graph is a line showing current decreasing as time increases. With the exception of the very last data point for 1 mA, it seems to be a linear decrease.
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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?
****
This does not follow the same pattern as the pattern seen with voltage.
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I would expect changes in resistance, so the pattern would differ somewhat.
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Determine voltage vs. current; conclude resistance vs. current for resistor
Using your graph of current vs. clock time determine the clock times at which the current is .8, .6, .4, .2 and .1 times the initial current.
Then using the graph of voltage vs. clock time determine the voltage at each of these clock times.
Using voltage and current at each clock time, determine the resistance at each clock time.
Report a table of voltage, current and resistance vs. clock time, giving in each line in comma-delimited format a clock time, the corresponding voltage and current, and the resistance. Starting in the first line following the table, explain how you obtained your values.
****
9, 2.8, 72, 39
24, 1.6, 54, 30
40, .74, 36, 21
59, .2, 18, 11
64, .1, 9, 11
Each is clock time in seconds, voltage in Volts, current in miliamps, resistance in Ohms
Current and voltage were determined by visual estimation based on graphs. Ohms were determined by divided voltage by current. Note that current was converted from miliamps to amps prior to the division.
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Sketch a graph of resistance vs. current. Fit the best possible straight line to your graph. Give in the first line the slope and vertical intercept of your graph. In the second line give the units of your slope and vertical intercept. In the third line give the equation of your straight line, using R for resistance and I for current. Beginning in the fourth line describe your graph, explain what you think it means and explain how you got the slope, the vertical intercept and the equation of the line.
****
516, 1
Ohms/amp, Ohms
R=561 Ohms/Amp I + 1 Ohm
One data point seemed erroneous, but other than that one, the line was a rather straight line. Ohms increased linearly with Amps. Slope was determined by taking two points far apart on the best fit line and determining the difference in y (rise) /divided by difference in x (run). Y intercept was determined visually from the graph.
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Resistance is expected to rise with temperature, and filament temperature will rise with current.
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Repeat for the 'other' resistor
Repeat all the above measurements for the 100 ohm resistor (assuming you started with the 33-ohm resistor; if you started with the 100-ohm resistor, then use the 33-ohm resistor here).
Give in the first line the resistance in the first line.
In the second line give the time required for current or voltage to fall to half its original value. Give in the form t +- `dt, where t is the single time you think you best answers the questions and `dt is the uncertainty in the time. In the third line explain how you determined t and `dt.
In the fourth line give your equation of R vs. I for this resistor.
Starting in the fourth line, in any order you deem appropriate to the needs of the reader, report your procedure, data, analysis and interpretation of results:
****
I used a 6.3V .25A bulb not a resistor
9.045 sec +-.8419
I determined the time for voltage at four different data points to fall to half voltage. I then found the mean and standard deviation which are the values I included in line two. For voltage the the halving seemed to follow a predictable pattern. Like the last time, however, current did not follow this predictable pattern. I think this may be a factor of using a bulb instead of an actual resistor.
R= 712 Ohms/Amps - 40 Ohms
I used same procedures with the 6.3V .25A bulb as I had used with the 14V bulb. I used corresponding points on graphs of current vs clock time and voltage vs clock time to find corresponding currents and voltages. Then I divided Voltage by current to get resistance. The final graph of resitance vs. current revealed again the linear relationship between resistance and current, although the equations were rather different.
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Charge capacitor through bulb then use 'square wave' pattern until voltage reaches 0
Set up a circuit using the 'switch', the 6.3 volt .25 amp bulb, and the capacitor, in series with the generator. Put the voltmeter in parallel with the capacitor. Set the 'beeps' program to 1.5 beeps per second.
Discharge the capacitor and close the switch.
Crank the generator at this rate for 100 'beeps'. Keep an eye on the voltmeter and the bulb.
Immediately reverse the generator for 5 'beeps'; don't miss a beat. Keep an eye on the voltmeter and the bulb.
Again, immediately reverse the generator for 5 'beeps'. Keep an eye on the voltmeter and the bulb.
Continue reversing the generator every 5 'beeps' until you first register a negative voltage.
Report in the first line about how many times you had to reverse the cranking before you first saw a negative voltage. You probably won't remember the exact number; just give your best estimate. In the second line report whether you think your estimate was accurate, and if not how close you think you probably were. Starting in the third line describe the behavior of the bulb, your best explanation for this behavior, your impressions of how the capacitor voltage changed with time, and how you think the brightness of the bulb was affected by the direction of cranking and the voltage across the capacitor.
****
20
The number was farily accurate, within +-1.
During the intitial 100 beeps, the bulb glowed at first and then grew dimmer as would be expected from previous labs. Upon reversing directions, the bulb glowed very brightly, when cranking reversed again to what was the original direction the bulb got very dim only to shine brightly on the next reversal. The Voltage which had been hovering around 3.5 V for most of the 100 beeps phase started to reduce. The voltage would reduce more on the cranking counter to the origingal direction than it would gain on the cranking in the original direction, thus it was able to eventually reduce to zero and a negative number. Of note, is that if the reversals were done 20 times with 5 cranks each, this would equal a total of 100 cranks which is the number we did in the initial phase. It would seem that the bulb glowed brightly when the direction was reversed because voltage was released from the capacitor. The bulb buring brigtly coincided directly with the voltage across the capacitor dropping.
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Good. As the capacitor is charged its voltage approaches that of the source, but in the opposite direction. Hence the dimming of the bulb.
When cranking direction is reversed, the capacitor voltage is in the same direction as the generator voltage, resulting in the enhancement you observed.
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When the voltage was changing most quickly, was the bulb at it brightest, at its dimmest, or somewhere in between? What do you think is the relationship between the brightness of the bulb and the rate at which capacitor voltage changes, and what might be the reason for this relationship?
****
The voltage was changing most quickly when the bulb was at it brightest. This was when the cranking was done opposite to the original direction of cranking. The voltage decreased faster when the cranking was oppposite than it increased when the cranking was in the same as the original direction. The faster the voltage changed, the brighter the bulb was. I assume that when the voltage drops that voltage is leaving the capacitor and thus powering the bulb.
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Charge capacitor through resistor then use 'square wave' pattern until voltage reaches 0
Now set up the circuit using the 'switch', 33 ohm resistor, and the capacitor, in series with the generator. Put the voltmeter in parallel with the capacitor. The circuit will be the same as before but with the resistor in place of the bulb.
Using the 'beeps' program, you will do the following. Read through the complete set of instructions before you begin:
Set the 'beeps' program to beep at the rate which will generate 4 volts. If you aren't sure what rate is required, hook up the meter and the generator and find out.
Multiply the resistance in ohms by the capacitance in Farads (the capacitance is marked on the capacitor; it is probably either 1 Farad or .47 Farad). This will give you a quantity called the 'time constant'. Its units are seconds.
Figure out, to the nearest whole number, how many beeps there are in a time equal to double the time constant, and also in a time equal to 1/4 of a time constant.
For example if your time constant was 25 seconds then double the time constant would be 50 second, a quarter of the time constant would be 6.25 seconds. If your beeping rate is 2.5 cranks per second, then you would need 2.5 * 50 = 125 cranks to get double the time constant, and 2.5 * 6.25 = 16 cranks, approximately, for 1/4 of a time constant.
Discharge the capacitor and close the switch.
Crank the generator at this rate for a time equal to double the time constant, and keep an eye on the voltmeter. Just count cranks, according to your calculation above.
As soon as you get to this count, reverse the crank and crank for 1/4 of a time constant (e.g., 16 cranks using the above calculation). Keep an eye on the voltmeter.
Immediately reverse the cranking and continue for another 1/4 of a time constant.
Continue the process, reversing the cranking every 1/4 of a time constant.
Continue the process until the voltage first becomes negative.
Report in the first line about how many times you had to reverse the cranking before you first saw a negative voltage. You might not remember the exact number; just give your best estimate. In the second line report whether you think your estimate was accurate, and if not how close you think you probably were. Starting in the third line describe what you did here and give your impressions of how the capacitor voltage changed with time.
****
19 times
Accurate within +-1
I used the 6.3 V .15 A bulb and used an esimation of 15 Ohms for the resistance. My capacitor was 1.0 F so my time constant was 15 seconds. Double was 30 seconds, and quarter was 3.75 Seconds. My cranking rate was 1.5 beeps per second so for Double I cranked for 45 beeps and for quarter I cranked for 6 beeps (rounded up from 5.625). As before, I noticed that the voltage decreased more on the reverse crank than it did on the subsequent forward crank. I did find it difficult however to focus on cranking rate, counting cranks, counting reversals and being able to closely pay attention to numbers on the meter.
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I would agree that this is at or even a little beyond the limit of what could reasonably be asked on a lab to be done by a single person.
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Charge capacitor through resistor then reverse until voltage reaches 0:
You will repeat the preceding, except you will only reverse the cranking once:
Discharge the capacitor and close the switch.
Crank for 100 'beeps', then reverse. Keep an eye on how quickly or slowly the voltage changes. Note the voltage at the instant you reverse the cranking, and try to remember it.
Count the 'beeps' in reverse, until the voltage reaches 0. Keep an eye on how quickly the voltage changes.
Sketch the graph you think best represents voltage vs. clock time for the duration of this trial. Label the clock times at which you first reverse the system, and at which the system first reaches 0.
Report in the first comma-delimited line how many 'beeps', and how many seconds, were required to return to 0 voltage after you reversed the cranking'. In the second line report whether the voltage was changing more quickly as you approached the 'peak' voltage or as you approached 0 voltage. In the third line report your 'peak' voltage, the voltage at the instant you reversed the cranking.
****
32 beeps, 48 seconds
Voltage changed more quickly as voltage approached zero.
3.5 Volts
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What voltage is produced by your generator at 1.5 cranks per second? If you aren't sure, hook the generator up to the voltmeter and find out.
****
4.0 Volts (this is very much an appoximation of a voltage that varies greatly - I would not be a good drummer)
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When charging the initially uncharged capacitor the voltage should be given by the function
V(t) = V_source * (1 - e^(-t / (RC) ) ).
where R and C are the resistance and capacitance. If R is in ohms and C in farads, RC is in seconds.
V_source is the voltage of the source, which you reported in the preceding box.
What is the value of t / (R C), where t is the t at which you reversed voltage? Write down this number.
Plug into the expression e^(-t / (RC) ) the values of R, C and the clock time t at which you reversed voltage, and write down the result.
Note on evaluating the exponential function: e^x would be evaluated using the value of x and the e^x button on your calculator. Alternatively it can be evaluated using most spreadsheets by entering into the cell the expression = exp(x), where x is a number. In the case of this expression, x would be the value of - t / (R C).
What then is the value of 1 - e^(-t / (RC))? Write down this number.
What therefore is the value of V_source * ( 1 - e^(-t / (RC) ) )? Write down this number.
Report the four numbers you have written down, in order, in comma-delimited format in the first line below. Starting in the second line explain how you evaluated your results.
****
-10, 4.53999* 10^-5, .9999546, 3.9998184
I used more significant figures than probably was necessary, but I could see that (1 - e^(- t / (RC) ), was heading towards 1 and I wanted to see if the extra digits would give me some variation. Upon reaching the end, it would seem that the number 1 was exactly where it perhaps was supposed to head.
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The final value reported above, that of V(t) = V_source * (1 - e^(- t / (RC) ), should theoretically be equal to the voltage you observed at the time of reversal. However the resistor you used is accurate only to within +-2%, and the meter isn't completely accurate either.
If necessary in order to get an accurate result, you may repeat the charging process, first discharging the capacitor then charging through 100 'beeps'. If you obtained an accurate reading before, you may use that reading.
Report in the first line your reported value of V(t) = V_source * (1 - e^(- t / (RC) ) and of the voltage observed after 100 'cranks'. In the second line give the difference between your observations and the value of V(t) as a percent of the value of V(t):
****
4.00, 3.5
12.5%
Part of the error here is like due to my erratic cranking rate - I can't always hit the beat and a find the the location of the crank on the beat tends to migrate forward. Also, when I can hit the beat at the appropriate time, I'm usually having to adjust speed mid crank because I can tell that I'm either ahead or behind - which makes the meter go crazy. Underlying all of this is that 1.5 beeps per second will give me a Voltage that could easily be 4.0 or just as easily by 3.5, the digits on the meter flucuating as rapidly as they do.
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According to the function V(t) = V_source * (1 - e^(- t / (RC) ), what should be the voltages after 25, 50 and 75 'beeps'?
****
3.67, 3.97, 3.998
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Now, when you reverse the voltage the function becomes
V1(t) = V_previous + V1_0 * (1 - e^(- t / (RC) ),
where V_previous is the voltage at the instant of reversal, V1_0 = reversed voltage - V_previous and t is the time since the instant of reversal.
After the first reversal what is the reversed voltage? Write this quantity down. (e.g., if the generator voltage was +2.5 volts, the reversed voltage would be -2.5 volts)
What are the values of V_previous and V1_0? Write these quantities down.
How long after the reversal did the voltage reach 0? This is the value of t. Write it down.
What do you get when you use these values to find V1(t)?
Report the four numbers you have written down in the first line, in comma-delimited format. Report the value of V1(t) in the second line. Starting in the third line explain how you did your calculations, and what the result means.
****
-4.0 volts, 3.5 volts, -7.5 volts
-3.69V
-4.0 was the reverse voltage, 3.5 Volts was the previous voltage, -7.5 was reverse voltage - previous voltage. The result would seem to mean that the voltage 48 seconds after reveral was -3.69 which was rather far from the actualy observed Voltage of 0.0 volts
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When its voltage is V the charge on the capacitor is Q = C * V. A Farad is a Coulomb per volt: Farad = Coulomb / Volt.
How many Coulombs does your capacitor store at 4 volts? Explain how you got your result.
****
4 Coulombs.
The capacitor is rated at 1.0 Farad or 1.0 Coloumb/V. 1.0 Coulomb/V * 4 Volts = 4 Coloumb.
That seems odd to me, but we shall see.
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That is correct.
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The negative charge on the capacitor is repelled by other negative charges and attracted by positive charges. The capacitor keeps the negative and positive charges separate. The positive charges are in contact with one of the posts, the negative charges with the other.
If you connect a lead between the two posts, the negative charge carriers (the electrons) in the wire are repelled from the negative post and attracted to the positive post, so that electrons migrate from the negative to the positive. As electrons flow from the wire onto the positive 'plate' of the capacitor, they are replaced by electrons from the negative 'plate'. This continues until both the positive and negative charges on the capacitor are neutralized. If only a lead is connected, there is no significant resistance to the flow of current and the exchange takes place quickly. The capacitor in your kit actually releases its charge very slowly compared to the capacitors used in most applications.
If a circuit containing resistance is connected across the posts of the capacitor, then the flow of current is slowed and it takes longer for the exchange to occur. This is the case for the circuits you have observed.
As the charge on the capacitor decreases, its voltage decreases.
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? Reply with two numbers, delimited by commas, in the first line, an explanation starting in the second.
****
3.5 Coulomb, .5C Coulomb
If the capacitor has a Capacitance of 1 Coloumb/Volt at 3.5 Volts it would have 3.5 Coulombs. Thus going from 4 volts to 3.5 volts would be going from 4 Coulomb to 3.5 Coulombs or a loss of .5 Coulombs
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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? Reply with two numbers, delimited by commas, in the first line, an explanation starting in the second.
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1.7sec, .29 Coulombs.sec
I did not have a resistor so this is based on the 6.3V .25A bulb used in the first part of this lab. The second value given is based on .5 Coulombs/1.7 sec = .29 Colmoumbs /sec or .29 Amps or 290mA
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You have graphs of voltage vs. clock time and current vs. clock time, and current vs. voltage. According to your data, what was the average current during the time the voltage dropped from 4 V to 3.5 V? Answer in the first line. In the second line, state how this compares with the result you reported in the previous box, how you think it should compare and why:
****
97.5 mA
This is about one third of the answer in the previous box. I would have thought they should be the same, but they are not even in the same ballpark.
If I compared this when using the 14V bulb I would get a Coloumb/sec drop of .147 Coulombs/sec or 147 mA. In that same time frame the average current was 87.5 mA (calculated by( inital current + final curent)/2. I cannot see a pattern between the results of these two different bulbs.
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The results should be consistent. Perhaps the capacitor's rated capacitance is in error.
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*#&!*#&!
&#Good responses on this lab exercise. See my notes and let me know if you have questions.
Revision isn't requested, but if you do choose to submit revisions, clarifications or questions, please insert them into a copy of this document, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).
Be sure to include the entire document, including my notes. &#