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PHY 242
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 **
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1.75 hours
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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.
2.36 V, 100 ohms *I was having difficulties getting the capacitor to go higher than this.
2.36, 0
2.0, 10
1.5, 32
1.25, 45
1.0, 60
0.75, 83
0.5, 116
0.25, 170
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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.
- 32 sec from 2.36 to 1.5 V
- 50 seconds from 2.0 to 1.0 V
- 56 seconds from 1 to 0.5 V
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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.
40 mA, 0
30 mA, 3
20 mA, 10
10 mA, 45
5 mA, 92
1 mA, 135
I put the meter in series with the capacitor to measure the current. There was a little bit of difficulty with measuring initially with clipping the lead to the post of the capacitor. Also, my millimeter is digital and does not read well with quickly changing measurements.
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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.
- 0.5 current - 10 sec
- 0.25 current - 45 sec
- ½ of ½ - 35 sec
- ¼ of ½ - 47 sec
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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?
I think it takes longer and longer to reach ½ the previous value. The voltage measurements are closer too this compared to the current measurements, though I know that there is some uncertainty in current measurements due to my meter jumping around.
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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.
0, 2.4, .04, 60
15, 1.92, .018, 106.6
35, 1.44, 0.012, 120
60, 0.96, 0.0075, 128
120, 0.48, 0.0025, 192
170, 0.24, 0.0010, 240
I obtained my values by first using the voltage as the reference then obtaining the corresponding current value. I used Excel to make the graphs. It is now apparent that there are a couple current measurements that were off.
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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.
0.9629, 76.91
Ohms per amp, ohms
R = 0.9629*I + 76.9606
The graph is increasing in a linear fashion. I think the resistance and current are related inversely. The slope of the line is V/I. The y intercept is the value of the resistor. This shows that as the capacitor discharges the resistance increases greater than the initial value of the series resistor.
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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:
33
20 s +- 3s
The voltages I picked were ½ of each other and every dT is about 20 seconds apart.
R(I) = 38.92 + 27.76
I used the voltage times as reference to make a similar table as before. Then using Ohm’s Law I found the resistance from V/I. It appears as if the y intercept will be the value of the resistor then resistance will tend to increase as the capacitor discharges.
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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.
It was about on the 5th cycle approximately
Within +- 1 cycle
On the initial cranking the bulb eventually did not light. The capacitor voltage was increasing as time went on. On the reverse cranks the bulb would light up brighter and the voltage would drop. On the next reverse cycle the voltage would drop lower. I think this is due to the “back emf” of the reverse cycle.
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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 bulb was brightest when the voltage would drop quickly. I think the capacitor drops voltage and the current increases with the resistance of the capacitor playing a role in how it happens.
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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.
About 9 cycles
Within +- 2 cycles
I think the capacitor will drain at a speed determined by the resistance of the circuit. I think the forward and backward cranking in effect cancel each other out will the capacitor is draining.
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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.
18 s
It appeared to drop faster near the peak voltage
3.6 V
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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.
Approximately 3V
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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.
1.5, 0.223, 0.777, 2.80
I plugged in each value to the appropriate equation then at the end multiplied by the initial voltage.
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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):
3.6, 2.7
0.9, 25%
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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'?
1.9, 2.7, 3.23
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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.
3.0, 3.6, -3.6, 18
I plugged everything into the equation, though I think either my voltage is off a little bit or the time value. I think the equation should be yielding zero, but it is more like 1.1
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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.
Q = C*V
Q = 1 F * 4V
Q = 4 Coulombs
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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, 0.5
The capacitor will lose 0.5 Coulombs of charge when it drops from 4 to 3.5 V.
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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.
2, 0.25
My data was slightly off from 4 V but the intial drop was about 3 sec so I think it would be even quicker with a higher initial voltage. 0.5 C/2 s = 0.25 C/s
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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:
50 mA
I think drop in current will be the same percent drop as the voltage. This should compare because the current is also a function of time and resistance.
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This will be the case if the resistance is constant. However the resistance, as you have seen, decreases with decreasing current. So this will have an effect on the expected behavior.
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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).
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