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course Phy 122
4/1 2151hrs. Total time of lab 5.5 hours. I ran into alot of problems in this lab. I know there is much wrong but after a while I had to push the I believe button.
Most student report an average time of 3-4 hours on this experiment. Some report considerably longer times, some as short as 1.5 hours.Using a basic multimeter the relationship between voltage and the cranking rate fora hand-held generator is quantified and modeled. Measurement of current vs. crankingrate indicates the internal resistance of the generator. Current and voltage relationships for various flashlight bulbs are quantified and resistances inferred. Current and voltage relationships for parallel and series circuits of flashlight bulbs arethen investigated.
Note video clip(s) associated with this experiments on the CD entitled EPS02. The links are Experiment 19: Batteries, Circuits and Measurement of Voltage and Current, Part I and Experiment 19: Batteries, Circuits and Measurement of Voltage and Current, Part II. The links will not work within this document; go to the CD, run the html file in the root folder which contains 'experiments' in the filename, and click on the link.
The 'beeps' program is located on the 'real' Physics II homepage under http://vhmthphy.vhcc.edu/ > Physics II > Simulations.
You will need the a basic multimeter, as mentioned under Sup Study ... > Course Information and specified at http://www.vhcc.edu/dsmith/genInfo/computer_interface_and_probes_cost_etc.htm. Watch the video clip before you attempt to use the meter; if you use the meter incorrectly you can burn out the fuse.
The figure below shows a meter connected in parallel with a capacitor and a bulb. The meter, the bulb and the capacitor are each connected across the terminals of the capacitor. The generator, with the 'silver-colored' leads, is not yet connected to anything.
If the generator leads are clipped to the two terminals of the capacitor, then the generator, the capacitor, the meter and the bulb will all be connected in parallel. The current flowing from the generator will flow into the circuit along one of the generator leads, where it will encounter a branch point from which some current will flow into the capacitor, some through the bulb and some through the meter.
If the meter is on a 'DC volts' setting, it will have a very high resistance to the flow of current, and very little current will flow through the meter. When measuring the voltage across a current element, the meter should always be connected in parallel with that element, giving the current a 'choice' of whether to flow through the meter or through the circuit element. Since the electrical resistance of the meter is so high, very little current will flow through the meter and there will be very little disruption of the circuit.
If the meter is on 'DC milliamps', 'DC mA' or 'current' setting its resistance is very low, and a great deal of current will flow into the meter. This will cause an overload, blow the fuse and possibly ruin the meter. The meter should therefore never be connected in parallel when it is set to read current (usually indicated by 'DC milliamps').
If the meter is set to a 'DC mA' scale and is connected in series with a circuit element, then the low resistance of the meter allows whatever current is already flowing in that element to flow with very little interference. As long as the current doesn't exceed the capacity of the meter, no harm will be done to the meter and you will get a good reading of the current. However, you still have to be careful. Most meters don't read much more that 250 mA, which is only 1/4 of an amp, and that's not a large current. Many devices operate at much higher currents than that, and your generator is easily capable of producing currents of several amps.
When using the meter to measure current, it is therefore important to start with the 'highest' current setting. For example the meter in the picture has settings for 200 mA, 20 mA and 2 mA. When measuring a current, you would want to start with the 200 mA scale; if the current reads less than 20 mA then it would be safe to switch to this scale; and if less than 2 mA it would be safe to switch to this scale.
Also, when using the generator to produce voltage and therefore current, start cranking slowly to be sure you don't overload the meter.
In the space below, specify the DC voltage and current scales on your meter (e.g., your meter might include scales like 15 V DC or 200 V DC, 1.5 V DC, 150 mA, 10 mA, 2 mA, etc.; give all these scales).
Your answer (start in the next line):
600V DC, 10 A fused
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voltage and current scales on meter,
You meter might be analog or digital. An analog meter will have a moving pointer, and the position of that pointer can be read on any of several scales, according to the scale you select. A digital meter will give you a digital readout of the current instead of a pointer. If you have both types of meter, you should take an extra few minutes to familiarize yourself with both. An analog meter is preferable for some of the experiments you will do in this course, but a digital meter will also suffice.
Do you have a digital meter, an analog meter or both?
Your answer (start in the next line):
Amprobe Digital
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Digital, analog or both
Summarize the rules given above for using these meters:
Your answer (start in the next line):
Dont use in parallel to read current, crank slowly and dont overload the meter
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summarize rules
Everything you do in this experiment will be with low voltage. However note that you can produce sufficient voltage to 'blow' a bulb, and though the bulbs are designed to be safe there is a chance that a bulb might burst and cause burns or eye damage. Normal precautions should be sufficient to protect you from injury, but you should wear basic safety glasses and should avoid handling hot bulbs. Please summarize these precautions in the space below:
Your answer (start in the next line):
Bulbs get hot, wear safety glasses.
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summarize precautions
You will later create parallel circuits such as the one shown above, but you will start with a series circuit.
The main rule to avoid burning out the fuse in the meter or otherwise damaging the meter is this:
Never, never, never connect the meter in parallel when it is set to measure current (the 150 milliamp setting).
If the meter is connected in parallel, double-check to be sure that it is NOT set to measure current (the 150 milliamp setting).
Other than this caution, you will not be working with voltages and currents capable of damaging the meter.
Also note that when the meter is not in use it should be turned to the OFF position to avoid running down the battery.
Summarize these rules below:
Your answer (start in the next line):
Dont measure current in parallel, turn multi meter off when not in use.
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summarize
Begin by observing how the cranking rate of the generator affects the voltage it produces:
If necessary, plug the probes into your meter. Generally the red plug goes into the + jack and the black plug in the - jack. In some meters the probes are permanently attached to the meter.
Turn the dial to an intermediate setting (e.g., the DC 15 volt setting; for each reading use the lowest DC voltage setting that exceeds the voltage being generated) and attach the leads of the generator to the probes coming from the meter, one lead to each probe.
Crank the meter in the most comfortable direction, not too fast, until the needle on the meter moves. If the needle moves in the correct direction, you may continue. Otherwise either reverse the direction of your cranking or unplug and reverse the plug which attaches the leads to your generator. You can damage an analog meter by allowing voltage or current to deflect the needle in the wrong direction.
If you are using a digital meter, the readout will indicate either a positive or a negative current. This is harmless to a digital meter. If necessary reverse the plug so the current will be positive.
Using the BEEPS program, determine the voltage obtained by cranking the generator at 1, 2, 3 and 4 complete cycles per second.
Plot a graph of voltage vs. the number of cycles per second and estimate the slope and vertical intercept of your best-fit straight line.
In the space below, give in the first line your voltages at 1, 2, 3 and 4 cranks per second. In the second line give the slope and vertical intercept of your best-fit straight line. In the third line give the equation of your straight line, using V for voltage and rate_crank for cranking.
Your answer (start in the next line):
2.4V, 5.9V, 9.75V, 13.4V/
voltages at 1, 2, 3, 4 cranks / sec:
slope and vertical intercept of best-fit line:
equation of straight line:
1, .2
y = 2.5 x + 0
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Now switch the meter to the 150 mA scale (or the highest scale that measures DC voltage) in order to measure the current flowing through the generator.
Crank the meter very slowly, and gradually speed up until the meter indicates that the current is 100 mA. Using a clock or another timing device, count the number of complete cycles of the generator crank in 10 seconds and determine the cranking rate in cycles/second.
Set the BEEPS program for this cranking rate and observe, as accurately as possible, the current obtained at this rate.
In the first line of the space below report the current and the voltage you observed. In the second line describe how you set up and observed these quantities, and how accurately you think you measured voltage and current. In your third line include your brief discussion/description/explanation
Your answer (start in the next line):
101mA, 1.5V
I used a partner to count the cycles/sec while I watched the meter and kept the cranks as close to 100mA as possible. I then switched the meter over to V DC and kept the same pace to observe the voltage. I think it is pretty accurate based on how fast my reading jumped around but I think all the possible error is human with my inability to hold a steady pace. The generator is kind of old and jumps as well but, my partners older computer has the BEEPS on it now and seems to coincide with my answer, at least close to it.
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Right. The ammeter has very low resistance so current will flow freely. The voltmeter has very high resistance so very little current flows.
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(generator and meter) cycles in 10 sec for 100 mA:
current at this rate:
We're going to use the term 'resistance' to refer to a phenomenon which is in fact something else. Note the following:
While there is some resistance in the generator, most of the apparent resistance of the generator to the flow of current is actually an inductive reactance. This is a concept that hasn't been developed at this point of the course. Inductive reactance effectively acts like a resistance, except that it changes with cranking rate and with the actual current flowing through the meter.
In the following we will use the term 'resistance' to include actual resistance as well as inductive reactance. Just remember that in this situation, 'resistance' can be expected to vary with current and cranking rate.
Nearly all of the resistance in this circuit is in the generator itself. Determine this resistance as follows:
From the cranking rate determine the voltage (as your graph of voltage vs. cranking rate should have indicated, voltage is proportional to cranking rate).
Using the current (in amps) and the voltage (in volts) determine the resistance of the circuit in Ohms (to get resistance in ohms, you will either multiply or divide current in amps by voltage in volts or voltage in volts by current in amps; recalling that a smaller current implies a greater resistance, you should be able to reason out which way to divide without having to resort to a formula).
You will obtain a 'resistance' for each of the four currents.
Crank to produce a current of 40 mA and record the cranking rate.
Crank to produce a current of 80 mA and record the cranking rate.
Crank to produce a current of 120 mA and record the cranking rate.
Crank to produce a current of 160 mA and record the cranking rate.
In the first line below give the voltage and the current for 40 mA, and give the corresponding 'resistance'. In lines 2, 3 and 4, do the same for 80 mA, 120 mA and 160 mA. Starting at the 5th line, explain how you obtained your results.
Your answer (start in the next line):
40mA = .8 crank/sec, .9V = 22.5 ohms resistance
80mA = 1.5 cranks/sec, 2.1V = 26.25 ohms
120mA = 2 cranks/sec, 3.1V = 25.83 ohms
160mA = 2.8 cranks/sec, 4V = 25 ohms
Using the observed crank speed, I attempted to recreate the voltage and used ohms law to figure resistance.
I did observe that the generator itself, at rest has 23.7 ohms resistance at rest. Confusing.
your brief discussion/description/explanation:
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Sketch a graph of the generator's 'resistance' vs. cranking rate, and describe the graph below. include the equation of your estimated best-fit straight line, and indicate how well the straight line appears to fit the data:
Your answer (start in the next line):
Given my points of (cranks/sec, current in mA) my slope is 54.5 so y = 54.5x -29.1, when using the slope intercept equation. I dont think this is correct because my eye tells me that the y intercept of the best fit line is around y=2
your brief discussion/description/explanation:
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graph of generator 'resistance' vs. cranking rate
best-fit line
Now construct a circuit consisting of the bulb marked 6.3, .25 and the bulb marked 6.3, .15, connected in series to the generator (recall from Experiment 16 that in a series circuit the current does not branch but flows straight from one circuit element to the other).
Set the meter to the DC 15 volt setting, or the setting closest to this value. BE SURE THE METER IS NOT ON THE 150 mA SETTING (or on another current-measuring setting) OR YOU WILL BURN OUT THE METER. Connect the voltmeter in parallel across the two bulbs (i.e., connect the voltmeter so the current 'branches', with one branch identical to the original path through the bulbs and the other through the voltmeter) and crank the generator, starting slowly and watching to be sure you aren't going to damage the meter, then increasing the rate until the bulbs both glow, but with the dimmer bulb just barely glowing. Estimate your cranking rate then set the BEEPS program to give you approximately this rate.
Using the BEEPS program, crank at this rate, with the bulbs glowing as before, and read the meter to determine the voltage across the bulbs.
Reposition the probes in order to measure the voltage across only one of the bulbs. That is, the meter should be connected in parallel to one of the bulbs. Crank at the same rate as before and measure the voltage across this bulb.
Repeat for the other bulb.
In the space below give the voltage across the two-bulb combination, the voltage across the first bulb and the voltage across the second:
Your answer (start in the next line):
3.2V across both, 1.9V across the first inline bulb, .79V across the second inline bulb.
your brief discussion/description/explanation:
I used my Fluke micro wire tap perforators to go inline with the bulbs on either side of each bulb and then on the outside of both of them.
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voltage 2-bulb series combination, across first, across second
Answer the following questions:
How do the voltages across the two bulbs compare to the voltage across the two- bulb circuit?
How much voltage was produced by the generator, according to the beeping rate?
How much of the voltage produced by the generator would you therefore conclude was associated with the current through the generator itself?
Your answer (start in the next line):
The voltage across both of them was greater than the sum of the two individual voltages across each bulb.
4V
3.2V I would assume. I know that there will be a voltage drop with the associated components and their resistance. But without using the current probe, which I think would be cheating, I cannot tell for sure, it could be a few things. My incorrect hook up, the old slipping generator, the resistance in the line itself, not sure
your brief discussion/description/explanation:
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The meters themselves could be reducing the voltage, as they do draw some current.
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voltages across series vs. individual bulbs compared
voltage from generator
how much generator voltage associated with current thru generator itself
Connect the generator in series with the two bulbs, then place the meter in series across the two bulbs, and turn it to the 150 mA setting.
Crank at the same rate as before and determine the current through the circuit.
Connect the circuit so meter is again in series, but now between the generator and the first bulb, and repeat.
Connect the circuit so meter is again in series, but now between the second bulb and the generator, and repeat.
In the space below, give your three readings for the current, in comma-delimited format in the first line. In the second line give the mean of these three results. Starting in the third line explain how you set up and observed these three currents:
Your answer (start in the next line):
272mA, .3mA, 2.7mA
your brief discussion/description/explanation:
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three readings for current with meter between different pairs of devices
mean of three currents
The generator produces a voltage which is determined by its cranking rate. This results in the current.
There are three resistances in the circuit--that of the first bulb, that of the second, and the 'resistance' of the generator. If you multiply the current through a current element by its resistance you get the voltage drop across that element.
You have measured the voltage drop across the two bulbs.
What is the current in the circuit?
According to your graph of 'resistance' vs. cranking rate, what would be the 'resistance' of the generator at the cranking rate used here?
What would therefore be the voltage drop due to the current flowing through the generator?
What is the total of the voltage drops around the circuit?
Answer these questions in the first four lines below, one number to a line, and in the fifth line answer the question
How does the total voltage drop around the circuit compare with the voltage produced by the generator?
Your answer (start in the next line):
272mA, 200+mA not sure,?, ?
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The generator probably has a resistance in the neighborhood of 20-30 ohms.
How much voltage drop would this imply due to the resistance (actually impedance, but don't worry about that) of the generator?
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your brief discussion/description/explanation:
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total voltage drop around circuit compared with generator voltage:
Compute the resistance of each bulb:
The numbers on each bulb are the voltage (in volts) at which the bulb is designed to operate, and the current (in amps) that should flow through the bulb at this voltage.
From the voltage and current you should be able to determine the resistance of each bulb, in Ohms.
For each bulb, use the measured voltage across the bulb and the resistance to determine how much current should have been flowing through the bulb.
Report as following: Give the voltage and current marked on the first bulb and the resistance calculated from this voltage and current in the first line, in comma-delimited format. Give the voltage and current you observed and the resulting resistance in the same format in the second line. In the third and fourth lines give the same information for the second bulb.
Your answer (start in the next line):
There is glue over the V of the first bulb but I am assuming it is 6.5V with .15A =43.33 ohms
3.2V with .3mA = 10666 ohms, something is wrong.
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I suspect you're reading more like .3 amps (which would be 300 mA) or 30 mA, but not .3 mA.
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your brief discussion/description/explanation:
I am using ohms law to determine the resistance, I understand how it works so I did something wrong. I went back and started the lab over and all my readings are way off again. Changed batteries in my multi meter, did the fuse check and did the series circuit all over again, Same thing, readings are all over the place/ went back to the beginning and put the generator in series with the multimeter, and reading are way off of what I originally got. Called my partner over from his room and he is scratching his head with me
I give up on this one.
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marked voltage and current, resistance so indicated
voltage and current observed, resulting resistance:
2d bulb marked voltage and current, resulting resistance:
2d bulbs voltage and current observed, resulting resistance:
Are the resistances you obtained based on the bulb's markings consistent with the resistances you calculated from observed voltages and currrents? Does one way of calculating the resistance seem to give higher or lower results than the other?
Your answer (start in the next line):
I would go with what is printed on the bulbs to determine the resistance. My way didnt seem to work out.
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Bulb resistance changes with filament temperature, and hence with current. The posted resistance applies only at the posted voltage.
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your brief discussion/description/explanation:
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consistency of markings, observed results
Connect the two bulbs in parallel and determine the voltage across each.
To connect two bulbs in parallel, begin by connecting the first bulb to the generator so that the current flows through the generator to the bulb and back to the generator. Then, just before the first bulb, allow the circuit from the generator to branch off to the second bulb, where it passes through the bulb and then rejoins the current that has passed through the first bulb before continuing back to the generator.
Crank the generator at a rate that causes one of the bulbs to barely glow. Use the BEEPS program to keep your cranking rate steady.
Note whether this circuit requires more or less force than the previous series circuit constructed with the same bulbs. You may need to alter the circuit between the series and parallel configurations a few times to be sure of the comparison.
Connect the voltmeter, set to the 15 volt DC position (or appropriate position on your meter), in parallel across the second bulb and determine the voltage across this bulb.
Connect the voltmeter in parallel across the first bulb and determine the voltage across this bulb.
Give your two voltages, in the first line separated by a comma:
Your answer (start in the next line):
It is harder to crank now, 1.6V, 2.6V
your brief discussion/description/explanation:
This seem to go smoother.
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(bulbs in parallel) two bulb voltages
Using the voltage just determined, and the bulb resistances as determined earlier, answer the following questions:
What should be the current across each bulb?
What should be the total current through the generator?
In the space below give in the first line the current that should be flowing across the first bulb and the current that should be flowing through the second, delimited by a comma. In the second line give the total current that should be flowing through the generator:
Your answer (start in the next line):
.03692A, .06A (first, second bulbs)
your brief discussion/description/explanation:
I used the Amps off the light bulbs and the currents that I received from my multi meter
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predicted currents based on voltage, previously calculated resistance
total predicted current through generator
Answer the following:
Which bulb is clearly brighter?
Which bulb carries more current?
Which bulb has the greater resistance?
For these bulbs, how are brightness, current and resistance related?
Your answer (start in the next line):
The first bulb in line
The first has more current
The second bulb had more resistance
I am assuming that the duller bulb has more resistance and the brighter bulb is carrying more current.
your brief discussion/description/explanation:
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which bulb brighter
which more current
which greater resistance
how brightness, current, resistance related
Compared to the series circuit as you investigated it, does this circuit expend more or less energy per unit of time? Explain how you can tell.
Your answer (start in the next line):
I am guessing here, but I would think that the series circuit expends less energy. When you think about it, when dealing with electricity - you wouldnt think that there would be any difference on the load, but I think the series circuit carries less resistance.
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Two resistors in series will have a resistance equal to the sum of the two.
Two resistors in parallel have less resistance than either.
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your brief discussion/description/explanation:
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more or less energy than in series circuit:
Measure the total current flowing through the parallel circuit:
Disconnecting one of the the generator leads.
Connect this lead to one of the leads of the meter.
Connect the other meter lead to the circuit at the point where you disconnected the generator lead.
The current from the generator will therefore flow through the meter then into the circuit.
What is the voltage across the parallel combination and what is the current? Answer with two numbers in the first line, delimited by commas.
Based on this voltage and current what is the resistance of the parallel combination? Answer in the second line.
In the third line explain how you obtained your results, and what you think they tell you about the circuit:
Your answer (start in the next line):
3.8V, .3A, 12.67 Ohms
I took the voltage and current and plugged it into ohms law
your brief discussion/description/explanation:
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voltage and current across parallel combination
resistance of parallel combination
The resistance R of the parallel circuit should be related to the resistances R1and R2 of the two bulbs by
1 / R = 1 / R1 + 1 / R2.
How well do the results you have obtained support this theoretical result? Explain how you justify your answer based on your data and your results.
Your answer (start in the next line):
I believe this is discussing Kirchoffs Law, I am use to seeing it as R1*R2/R1+R2 I dont think any of my answers are accurate. I went a lot off the Voltage and Current on the bulbs. I am still stumped by my readings.
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1/R = 1/R1 + 1/R2 can be rearranged by multiplying each term by R * R1 * R2. After doing so you can solve for R and you'll get R = R1 R2 / (R1 + R2).
The former is easier to remember, the latter quicker to compute.
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your brief discussion/description/explanation:
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degree of support for parallel-resistance formula
Devise a procedure to test with the ammeter whether the total current through the generator is equal to the sum of the two currents through the bulbs, using a steady cranking rate of 1 cycle per second.
Note that at this rate, it is possible that neither bulb will dissipate enough energy to light. This does not change the fact that current is flowing through the bulbs; they just aren't getting hot enough to emit electromagnetic radiation.
Your procedure should measure the total current through the generator as well as the currents through each of the two bulbs.
Conduct your test and describe your procedure and your results in the space below:
Your answer (start in the next line):
With the parallel still set up, I connected the meter before the bulbs, took a reading, and then after the bulb. Results in about 10 this time. Individually I got 5.1 and 4.5 so 9.6. I feel like I am missing something somewhere but that is what I got after several attemps.
your brief discussion/description/explanation:\
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your test of whether generator current is sum of currents through bulbs, and results
Your instructor is trying to gauge the typical time spent by students on these experiments. Please answer the following question as accurately as you can, understanding that your answer will be used only for the stated purpose and has no bearing on your grades:
Approximately how long did it take you to complete this experiment?
You may add optional comments and/or questions in the space below.
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You're OK here, though some of your results appear questionable (as you realize).
One problem is that a digital meter doesn't work well with the generator, which doesn't produce a 'smooth' voltage.
Check my notes.
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