#$&*
Phy 202
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 **
** **
3.5
** **
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.
****
4.0, 100
0, 4
10, 3.5
23, 3.0
37, 2.5
61, 2.0
86, 1.5
126, 1.0
152, .75
193, .50
263, .25
#$&*
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.
****
61
63
65
67
The graph is decreasing at decreasig rate. The amount of time between the
intervals gets bigger as time goes on. I subtracted the end time of the interval
from start time.
#$&*
Measure current vs. t when you discharge the capacitor through the resistor
You have observed and graphed the voltage across the capacitor as a function of
time.
Now repeat the same procedure, but this time measure the current. In the
process be sure to verify that the capacitor is in fact at 4.00 volts +- .02
volts.
Remember to be very sure the meter is connected in series rather than in
parallel. The safest way to do this is to first put the meter in series with
the bulb, the capacitor and an open switch, set to read volts, then close the
switch and watch for a few seconds to be sure the reading on the meter doesn't
change. If you don't have a switch, you can simply unclip one of the leads
leading to capacitor, bulb and meter in order to simulate an open switch, then
connect it to simulate a closed switch.
The reason this works is that the voltmeter has a very high resistance and will
not permit significant current to flow. However if the meter is in parallel
with the bulb and capacitor, the capacitor will discharge, resulting in a change
of voltage.
If the voltmeter shows unchanging voltage, then it's probably safe to open the
switch, then turn the dial to the 200 mA setting.
The ammeter has a very low resistance and if connected in parallel, a very large
current will flow, with the potential to damage the meter.
After setting up the circuit go ahead and close the switch (or connect the lead)
and observe current at a function of clock time, then give your table of current
vs. clock time in the box below, using the same conventions you used above in
reporting voltage vs. clock time. Starting in the following line give a short
synopsis of what your results mean and how they were obtained.
****
0, 38
9.1, 35
23.4, 30
38.7, 25
60.3, 20
85.8, 15
123.2, 10
190.5, 5
The graph shows that the current is decreasing at a decreasing rate. Using the
meter to measure the voltage then I calculate the current.
#$&*
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.
****
62s
61s
64s
63s
The graph of current vs time is decreasing at a decreasing rate. I used the
intervals in the graph to estimat how many seconds had elapsed for the current
to fall a certain percentage of the initial current to a certan value.
#$&*
Within experimental uncertainty, are the times you reported above the same? Are
they the same as the times you reports for voltages to drop from 4 v to 2 v, 3 v
to 1.5 v, etc? Is there any pattern here?
****
My times are off but not by more than a few seconds. This could be caused by me
being slow to recognize the information and record the correct time.
#$&*
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.
****
23s, 3, 30, 99
47s, 2.2, 23, 97
89s, 1.5, 15, 99
150s, .8, 7.6, 105
229, .3, 3.9, 79
#$&*
Sketch a graph of resistance vs. current. Fit the best possible straight line
to your graph. Give in the first line the slope and vertical intercept of your
graph. In the second line give the units of your slope and vertical intercept.
In the third line give the equation of your straight line, using R for
resistance and I for current. Beginning in the fourth line describe your graph,
explain what you think it means and explain how you got the slope, the vertical
intercept and the equation of the line.
****
.068, 95
Ohms/mA, ohms
R=.068ohms/mA * (I + 95ohms)
The graph is a horizontal line, which shows that resistance is constant. I
obtained the line by plugging the slope and y-intercept of the graph into the
fromula y=mx+b.
#$&*
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
19 +or- 2
I made a table and estimated the
R=.023*I+33
I reapeated the same procedure but this time with a 33ohm resistor. The data I collected was used to calculate the resistance. Risistance was proved to stay constant even when current changes.
#$&*
Charge capacitor through bulb then use 'square wave' pattern until voltage
reaches 0
Set up a circuit using the 'switch', the 6.3 volt .25 amp bulb, and the
capacitor, in series with the generator. Put the voltmeter in parallel with the
capacitor. Set the 'beeps' program to 1.5 beeps per second.
Discharge the capacitor and close the switch.
Crank the generator at this rate for 100 'beeps'. Keep an eye on the voltmeter
and the bulb.
Immediately reverse the generator for 5 'beeps'; don't miss a beat. Keep an eye
on the voltmeter and the bulb.
Again, immediately reverse the generator for 5 'beeps'. Keep an eye on the
voltmeter and the bulb.
Continue reversing the generator every 5 'beeps' until you first register a
negative voltage.
Report in the first line about how many times you had to reverse the cranking
before you first saw a negative voltage. You probably won't remember the exact
number; just give your best estimate. In the second line report whether you
think your estimate was accurate, and if not how close you think you probably
were. Starting in the third line describe the behavior of the bulb, your best
explanation for this behavior, your impressions of how the capacitor voltage
changed with time, and how you think the brightness of the bulb was affected by
the direction of cranking and the voltage across the capacitor.
****
15turns backwards
My estimate was acurate + or - a turn
When is strated cranking the generator backward the bulb became really bright immeadiately. Then it started dimming. This was caused by the capacitor and generator voltage moving back toward the generator.
#$&*
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?
****
Voltage changes most quickly when the bulb is at its brightest. When the current is at its highest the change of voltge is at its peak.
#$&*
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.
****
32cranks
+or- 3 cranks
After cranking backwards the capacitor's voltage reached its lowest point. After cranking backwards around 16 times twice the capacitors voltage read zero.
#$&*
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.
****
24, 12s
The voltage change happened more quikly and the total voltage dropped rapidly to zero.
#$&*
What voltage is produced by your generator at 1.5 cranks per second? If you
aren't sure, hook the generator up to the voltmeter and find out.
****
4.5volts
#$&*
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, .22, .80, 3.9
I used the above equations to find the amount of voltage coming from the source
#$&*
The final value reported above, that of V(t) = V_source * (1 - e^(- t / (RC) ),
should theoretically be equal to the voltage you observed at the time of
reversal. However the resistor you used is accurate only to within +-2%, and
the meter isn't completely accurate either.
If necessary in order to get an accurate result, you may repeat the charging
process, first discharging the capacitor then charging through 100 'beeps'. If
you obtained an accurate reading before, you may use that reading.
Report in the first line your reported value of V(t) = V_source * (1 - e^(- t /
(RC) ) and of the voltage observed after 100 'cranks'. In the second line give
the difference between your observations and the value of V(t) as a percent of
the value of V(t):
****
3.9, 4
2.6%
#$&*
According to the function V(t) = V_source * (1 - e^(- t / (RC) ), what should be
the voltages after 25, 50 and 75 'beeps'?
****
1.6, 2.7, 3.4
#$&*
Now, when you reverse the voltage the function becomes
V1(t) = V_previous + V1_0 * (1 - e^(- t / (RC) ),
where V_previous is the voltage at the instant of reversal, V1_0 = reversed
voltage - V_previous and t is the time since the instant of reversal.
After the first reversal what is the reversed voltage? Write this quantity
down. (e.g., if the generator voltage was +2.5 volts, the reversed voltage
would be -2.5 volts)
What are the values of V_previous and V1_0? Write these quantities down.
How long after the reversal did the voltage reach 0? This is the value of t.
Write it down.
What do you get when you use these values to find V1(t)?
Report the four numbers you have written down in the first line, in comma-
delimited format. Report the value of V1(t) in the second line. Starting in
the third line explain how you did your calculations, and what the result means.
****
-4, 4, -8, 12
1.4
I found v1(t) by plugging in the given number for the variables in the equation V1(t) = V_previous + V1_0 * (1 - e^(- t / (RC) ). I used a stopwatch to calculat t.
#$&*
When its voltage is V the charge on the capacitor is Q = C * V. A Farad is a
Coulomb per volt: Farad = Coulomb / Volt.
How many Coulombs does your capacitor store at 4 volts? Explain how you got
your result.
****
4
Q = C * V
C = Q / V
= 1 C/volt * 4volts
= 4 C
#$&*
The negative charge on the capacitor is repelled by other negative charges and
attracted by positive charges. The capacitor keeps the negative and positive
charges separate. The positive charges are in contact with one of the posts,
the negative charges with the other.
If you connect a lead between the two posts, the negative charge carriers (the
electrons) in the wire are repelled from the negative post and attracted to the
positive post, so that electrons migrate from the negative to the positive. As
electrons flow from the wire onto the positive 'plate' of the capacitor, they
are replaced by electrons from the negative 'plate'. This continues until both
the positive and negative charges on the capacitor are neutralized. If only a
lead is connected, there is no significant resistance to the flow of current and
the exchange takes place quickly. The capacitor in your kit actually releases
its charge very slowly compared to the capacitors used in most applications.
If a circuit containing resistance is connected across the posts of the
capacitor, then the flow of current is slowed and it takes longer for the
exchange to occur. This is the case for the circuits you have observed.
As the charge on the capacitor decreases, its voltage decreases.
How many Coulombs does the capacitor contain at 3.5 volts? How many Coulombs
does it therefore lose between 4 volts and 3.5 volts? Reply with two numbers,
delimited by commas, in the first line, an explanation starting in the second.
****
3.5, 0.5
As the voltage decreases so does the charge.
#$&*
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.
****
4, 0.13
The resistor took four seconds to discarge 0.5volts from 4 volts to 3.5 volts. The discharge rate was found by dividing the amount of coulomb discharged by the time it took to discharge that amount.
#$&*
You have graphs of voltage vs. clock time and current vs. clock time, and
current vs. voltage. According to your data, what was the average current
during the time the voltage dropped from 4 V to 3.5 V? Answer in the first
line. In the second line, state how this compares with the result you reported
in the previous box, how you think it should compare and why:
****
101
101mA is .101 C/sec. This is fairly close to the calculted value of0.13 C/S. This is as close as I could hope to get given variables that I did not calculat that could throw off my answer.
#$&*
*#&!
&#Your work on this lab exercise looks good. Let me know if you have any questions. &#