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course Phy 232
batteries _ circuits and measurement of voltage and current#$&*
Phy 232
Your 'batteries _ circuits and measurement of voltage and current' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** Batteries _ Circuits and Measurement of Voltage and Current_labelMessages **
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3 hours and 15 minutes
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Experiment 19: Batteries, Circuits and Measurements of Voltage and Current
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):
20 V
200 V
2000 V
20 mA
200 mA
10 A
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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):
Digital Multimeter
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Digital, analog or both
Summarize the rules given above for using these meters:
Your answer (start in the next line):
A digital multimeter gives results of voltage and current digitally on a screen while the analog meter reports the results using a pointer which can be read using different types of scales.
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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):
Wearing protection such as safety glasses is highly recommended. This is so if a bulb were to burst, I would remain safe. Also, avoid handling hot bulbs can protect me from any types of burns. The dangers are very real although we area just dealing with low voltages. These are precautions to avoid any injury.
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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):
Be careful not to connect the meter in parallel when the setting is set to measure the current. Making sure this does not happen ensures that the fuse in the meter will not be blown out. Parallel should not mix with measuring current. Turning off the meter while it is not in use is also a rule to avoid battery loss or any other danger.
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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 rate.
Your answer (start in the next line):
voltages at 1, 2, 3, 4 cranks / sec:
slope and vertical intercept of best-fit line:
equation of straight line:
2.5, 5.1, 7, 10
2.4, 0
V = 2.4*rate_crank
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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 thing you measured voltage and current. In your third line include your brief discussion/description/explanation
Your answer (start in the next line):
100 mA, 1 V
First, I found a cranking speed that corresponded to 100 mA and then I found the voltage that corresponded to that same cranking rate. I feel like my measurements are accurate because the meter was consistent and I didn’t have to accommodate for any changes in crank rate. Inconsistent cranking would have caused skewed data.
The process above describes how I found the current and voltage and I recorded them in the first line. The current followed by the voltage.
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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):
40 mA, 0.75, 18.75
80 mA, 1.10, 13.75
120 mA, 1.45, 12.08
160 mA, 1.70, 10.63
I reported the current, in mA, the voltage, in volts, and the resistance, in ohms, in each line. The formula current equals voltage divided by resistance is used to solve for the missing resistance. I took the voltage and divided it by the current to get the resistance for each scenario.
your brief discussion/description/explanation:
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(generator and meter only) voltage and current, corresponding 'resistance', 40 mA:
80 mA:
120 mA:
160 mA:
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):
My equation for my line of best fit is y = -15.85x + 17.76. The data points are pretty much linear as a whole so the equation fits nicely. The slope of the graph is negative, which shows that while the cranking rate increases, the resistance decreases. Crank rate and resistance are inversely proportional.
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):
3.15
1.2
0.9
3.15 V for the total, 1.2 V for the first bulb, and 0.9 V for the second bulb.
your brief discussion/description/explanation:
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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 two bulbs together is significantly higher than the voltage across the single bulb circuit.
The generator produced a voltage of about 3.25 V.
The voltage used by the two bulbs was 3.15 V, therefore the voltage associated with the generator is about .1 V.
your brief discussion/description/explanation:
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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):
75, 105, 145
108
I obtained these results by recording the readings of the current when the bulbs were together and separate in a series circuit. A digital multimeter was used to make the readings.
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):
75 mA
.42 ohms
.0315 V
2.132 V
The voltage drop of the generator, when compared to the voltage drop of the whole circuit, is very small.
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):
6.3, 0.15, 42
1.2, 0.105, 11.4
6.3, 0.25, 25.2
0.9, 0.145, 6.2
your brief discussion/description/explanation:
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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):
My calculated results are along the same proportion as the voltage and current given on the bulb. However, the actual readings were a little higher than my recorded readings. I still believe that my results can still be considered good data though my data is slightly less than the expected values.
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):
1.2, 1.4
your brief discussion/description/explanation:
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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):
33.3, 48
389
All current readings above are in mA.
your brief discussion/description/explanation:
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It isn't clear how you got that result 389.
Please clarify.
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The total current across the generator is the sum of the currents through each bulb. Therefore, the total current across the generator is 33.3 ohms + 48 ohms = 81.3 ohms.
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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 second bulb, .25 A, is the brighter bulb.
The second bulb, .25 A, carries more current.
The first bulb, .15 A, has the greater resistance.
Brightness can be thought about as power, so the brighter means more power. Current is equal to volts divided by resistance, and power is equal to volts times current. Therefore, combining them we get power is equal to volts^2 divided by resistance. This shows that the greater the resistance, the less bright the bulb is. It also shows that more current means more power and a brighter light.
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 believe that the parallel circuit expends more energy per unit of time. I believe this is so because in a parallel circuit, both bulbs are using energy and power at the same time while in a series circuit the energy has to finish in one before moving on to the next one. Also, it appears that in the parallel circuit the path of the current is longer than that in a series circuit, causing more energy to be expended. Another reason is that the parallel circuit has lower resistance and therefore drains more current, which is proportional to draining more power. Testing this out also proved that more voltage was used for the parallel circuit compared to the series circuit.
your brief discussion/description/explanation:
@&
The same current flows through both bulbs in the series circuit. For a given cranking rate energy must be supplied according to the demands of the circuit.
So energy doesn't get used up in one bulb before getting to the other. The order of the bulbs makes no difference, as you can easily verify with a test, nor does the direction in which you crank, also easily verified with a test.
For a given cranking rate the parallel circuit will certainly require more energy.
For a given brightness of one of the bulbs, it is harder to determine because of the difference in the necessary cranking rates.
Please modify your answer to this question based on the information in this note.
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The series circuit will require less power than the parallel circuit because in the parallel circuit, power must be used in two different locations at one given time. In a series circuit, the power is transferred through one bulb, and then it moves on to move through the second bulb. At a given cranking rate, it takes more power for a parallel circuit to operate because it requires multiple sources of power at once since each bulb is in a difference path unlike the series circuit where the bulbs are one after the other and the power is recycled to the second bulb after passing through the first bulb.
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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):
2.4 V, 82 mA
29.27 ohms
Current is defined by the equation, current = voltage divided by resistance. Therefore, if given voltage and current, we can solve for resistance by the equation, resistance = voltage divided current.
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):
This equation indicates that the total resistance of the circuit must be less than the individual resistances. R, the total resistance, can be calculated by the equation R equals R1 times R2 divided by R1 plus R2. Plugging back in numbers into the given equation above can give us an idea of how accurate the results were. I calculated at the beginning of the tests with the parallel circuit resistances, R1 and R2, of 33.3 ohms and 48 ohms. Those were individual resistances and the recently calculated total resistance was 29.27, which is indeed less than both individual resistances. Therefore, my data seemed to be going on the right track.
your brief discussion/description/explanation:
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You have values for R1 and R2. How well does the result based on your experimental evidence reconcile the values of 1 / R, and 1 / R1 + 1 / R2?
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Using my experimental values of 33, 48, and 29 ohm resistances, I see that my results using the above formula are along the correct magnitude. Knowing this, I feel confident with my answers and readings. R1 and R2 represent the 33 ohm and 48 ohm resistances while the 29 ohms represents the final value of R. I believe that my evidence shown is good proof of accurate data.
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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):
For my test, I set up a parallel circuit and recorded the voltage and current throughout the system while cranking at a rate of 1 cycle per second. After doing so, I was able to compare the total current of the circuit to the individual currents inside of the bulbs in the circuit. The current however that I recorded was higher than the expected current calculated by the formula.
your brief discussion/description/explanation:\
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*#&!
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Good.
I've inserted two questions, and I'll ask you to copy them into a document, insert your answers and submit them.
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Self-critique (if necessary):
------------------------------------------------
Self-critique rating:
________________________________________
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&#Good responses. Let me know if you have questions. &#