course Phy 202
Experiment 19: Batteries, Circuits and Measurement of Voltage and CurrentUsing a basic multimeter the relationship between voltage and the cranking rate for a hand-held generator is quantified and modeled. Measurement of current vs. cranking rate 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 are then investigated.
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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, which you will be responsible for replacing.
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).
Other than this caution, you will not the dealing with voltages and currents capable of damaging the meter.
• Also, when the meter is not in use it should be turned to the OFF position to avoid running down the battery.
Begin by observing how the cranking rate of the generator affects the voltage it produces:
Plug the probes into the meter, with the red plug in the + hole and the black plug in the - hole.
Turn the dial to the DC 15 volt setting and attach the leads of the generator to the probes.
• 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 reverse the attachment of the leads so that the meter deflect in the correct direction.
• Using the BEEPS program, determine the voltage obtained by cranking the generator at 1, 2, 3 and 4 complete cycles per second.
1 cycle/sec = 3.1 volts
2 cycles/sec = 4.1 volts
3 cycles/sec = 5.7 volts
4 cycles /sec = 7.7 volts
• Plot the voltage vs. the number of cycles per second.
The graph extends from left to right, starting at zero on the X – Y intercept and has a linear slope from left to right. This provides the evidence that the voltage output is proportional to the cranking rate. Graph is included, but it may not load into the submission form.
Now switch the meter to the 150 mA scale to measure the current created by 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.
4.25 turns / 10 seconds
4.25 cycles / 10 seconds
0.425 cycles / second
• Set the BEEPS program for this cranking rate and observe the current obtained, as accurately as possible.
Nearly all of the resistance in this circuit is in the generator itself. Determine this resistance is follows:
• From the cranking rate determine the voltage (as your graph of voltage vs. cranking rate should have showed you, voltage is proportional to cranking rate).
4.25 cycles per second = 2.25 volts per graph
• Using the current (in amps) and the voltage (in volts) determine the resistance of the circuit in Ohms (you will either divide current by voltage or voltage by current; 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).
Ohms Law: E=I*R
R = E/I = 2.25 v / 100 ma = 22.5 ohms
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 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. BE SURE THE METER IS NOT ON THE 150 mA SETTING OR YOU WILL BURN OUT THE METER. Connect the voltmeter in parallel across the two bulbs and crank the generator, starting slowly and watching to be sure you aren't going to blow the meter up, then increasing the rate until the bulbs both glow, but not too brightly. Estimate your cranking rate then set the BEEPS program to give you approximately this rate.
Rate = 44 cycles / 10 seconds
Rate = 4.4 cycles/sec
• Crank at this rate, with the bulbs glowing as before, and read the meter to determine the voltage across the bulbs.
Voltage = 6.3volts
• Reposition the probes from the meter 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.
Bulb #1 = 1.4 volts
• Repeat for the other bulb.
Bulb #2 = 4.6 volts
Answer the following questions:
• How do the voltages across the two bulbs compare to the voltage across both bulbs?
One bulb had a voltage drop that was significantly higher than the other., but the overall voltage was the sum of both voltages.
• How much voltage was produced by the generator, according to the beeping rate?
I do not have the BEEP program, but the total voltage produced, in estimation, was approximately 8.2 volts.
How much of this voltage would you therefore conclude was associated with the current through the generator itself?
6.3 volts – 1.5 volts – 4.5 volts - X = 0 volts
X = 0.3 volts
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.
Bulb #1 = 6.3 volts, 0.25 amps
I = V/R = 6.3 / 0.25 = 25.2 ohms
Bulb #2 = 14 volts, 0.2 amps
I = V/R = 14/0.2 = 70 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.
Bulb #1 = 1.5 volts, 25.2 ohms
I = 1.5 / 25.2 = 59.5 mA
Bulb #2 = 4.5 volts, 70 ohms
I = 4.5 / 70 = 65.7 mA
Are the currents you calculated approximately equal? Should the currents through the two bulbs be equal?
The currents are close in relationship, only 6 mA difference, however, this is significant since we are only dealing on the milliamp level. In answer to the second question, yes the current throughout the circuit should be the same. Each voltage drop should be able to be calculated with circuit current and the component resistance, and the voltage drops across the components and the internal generator drop should add to the total voltage of the circuit. However, I believe there is some induced error by not being able to maintain a constant voltage in the circuit. Manually cranking the generator allows the voltage to fluctuate, unlike some type of voltage regulator to maintain constant voltage. The voltage deviation will cause a resulting current deviation, due to the force to move the charge past the point in a given time. For this lab, I am only able to approximate the same cranking rate, and in turn voltage in the circuit. Additionally, I am using a digital Fluke meter, unlike the analog meters in the video, thus the resolution is better, but harder to regulate. Additionally, as the filaments begin to heat up, there could be a small resistance change that also affects the current reading. The total current provided to the circuit is dependent upon the total circuit resistance since the bulbs are in series.
actually the resistance changes in the filaments are very significant. The resistance is expected to more than double as the filament heats up.
Connect the meter in series with 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.
• How does this current compare with the current you predicted from the computed resistances of the bulbs?
The current is approximately 102 mA, which is comparably close to the total computed current, although there is a slight variance.
Using the measured current and the resistance you computed for each bulb, determine the voltage change across each bulb.
Bulb #1: V = I*R
V = 102 mA * 25.2 ohms
V = 2.6 volts
Bulb #2: V = I*R
V = 102 mA * 70 ohms
V = 7.1volts
Using the measured current and the resistance of the generator, as previously calculated, determine the voltage change across the generator.
Generator: V = I*R
V = 105 mA * 22.5
V = 2.36 volts
• Is the total voltage change around the circuit consistent with the voltage that should have been produced by the generator at the cranking rate used? Should it be?
The voltage calculated in these sections is off by approximately 50%, since the total voltage just calculated is 12.1 volts and the presumed voltage from the number of cranks per second was deemed to be approximately 6.3 volts. Understandably, there is a voltage drop in the generator, but this drop should not be that significant. Apparently, I have some error in the presumption that I am cranking the generator at the same rate, or my presumption that the bulbs are dimmer or brighter as the lab progresses, or more realistically, the resistance of the circuit is not what is calculated, since the bulbs are not being operated at rated voltage.
Under these conditions the equivalent resistance of the generator is quite significant, on the order of 20-30 volts. I say 'equivalent resistance' because in addition to the actual resistance in the generator there is significant inductive reactance.
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 lights both bulbs, but not too brightly. Use the BEEPS program to keep your cranking rate steady.
I do not have the BEEPS program, and if it is available for download, I do not know where it is located. I am performing the lab strictly on assumption that my cranking is somewhat consistent.
• Note whether this circuit requires more or less force then the previous series circuit with these bulbs.
The parallel combination requires more force to crank the generator handle.
• Connect the voltmeter, set to the 15 volt DC position, in parallel across the second bulb and a determine the voltage across this bulb.
The meter appears to read approximately 3.5 volts across the second bulb.
• Connect the voltmeter in parallel across the first bulb and determine the voltage across this bulb.
The meter appears to read approximately 3.5 volts across the second bulb.
Answer the following questions:
• How much voltage is there across each bulb?
Approximately 4 volts
• What then should be the current across each bulb?
The current across each bulb should sum to the total circuit current. Therefore, the current across each will be different.
• What should be the total current through the generator?
The current in the will be in series with the equivalent resistance of the branch circuits. So, the total current through the generator will be the sum of the branch circuits plus the current due to the internal generator resistance.
• Is the measured voltage across the bulbs equal to the voltage produced by the generator, according to the cranking rate?
The voltage is comparable, but not exact. From my graph and perception, there is about a 1.0 to 1.5 volt difference.
• 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?
The bulb with the lower resistance burns brighter. The bulb with the higher resistance burns dimmest.
• Compared to the series circuit as you investigated it, does this circuit expend more or less energy per unit of time?
The parallel circuit expends the most energy per unit of time. This circuit requires more power than the parallel circuit.
• If the two bulbs were replaced by a single bulb whose resistance was such that the same current flows for the same voltage (i.e., for the same cranking rate), what would its resistance have to be, according to the measured voltage and the total current flowing through the generator?
• This resistance is called the equivalent resistance for the parallel circuit. Ideally we should have 1 / R = 1 / R1 + 1 / R2, where R is the equivalent resistance and R1 and R2 are the resistances of the bulbs.
R(eq) = (R1*R2) / (R1+R2)
R(eq) = (25.2*70) / (25.2+70)
R(eq) = 14.1 ohms
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.
Connect ammeter in series with second bulb after first bulb branch, into the second bulb branch = 105 mA
Connect ammeter in series from at first bulb branch, with meter reading first bulb
current =55 mA
Connect ammeter from second bulb to generator, total for both branches and generator and current = 171 mA
I(r1) + I(r2) +I(gen) = 124 mA
51 mA + 105 mA + I(gen) = 169 mA
I(gen) = 171 mA – 105 mA – 55 mA
I(gen) = 11 mA
As stated above the values may contain some error, due to attempting to maintain a constant cranking rate.
In addition to the note above, with an ohm meter, just as a check, and the resistances of the bulbs were determined to be 6.5 and 2.5 ohms, thus a significant difference between the theoretical and actual. Thus, the significant error in the voltage drops in the series circuits.
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Good work. The lab should have included a link to the BEEPS program; however due to the transition in the lab component, there might be some confusion in the instructions. In any case your results were fine (better than most students get with the BEEPS program) and you clearly understand what you were doing here.