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.
** #$&* Your comment or question: **
I am sorry for the sketchy answers, i assure you i did the best i could on this lab, as i was not able to open the beeps program on my computer, so i was having to do everything by myself, including counting revs, watching voltage and time as well.
** #$&* Initial voltage and resistance, table of voltage vs. clock time: **
3.98 volts 41*10^0=41ohms
4 , 3.5
10 , 3.0
18 , 2.5
27 , 2.0
40 , 1.5
60 , 1.0
73 , .75
93 , .50
127 , .25
** #$&* Times to fall from 4 v to 2 v; 3 v to 1.5 v; 2 v to 1 v; 1 v to .5 v, based on graph. **
28 seconds
30 seconds
34 seconds
32 seconds
I drew my graph on microsoft excel and estimated the time i thought it would require for the voltage drop indicated.
** #$&* Table of current vs. clock time using same resistor as before, again starting with 4 volts +- .02 volts. **
175 , 5
150 , 11
125 , 17
100 , 29
75 , 38
50 , 61
38 , 71
25 , 94
13 , 129
My results are the clock timein seconds and the current in mA.
** #$&* Times to fall from initial current to half; 75% to half this; 50% to half this; 25% to half this, based on graph. **
29 seconds
32 seconds
35 seconds
31 seconds
I drew my graph on microsoft excel and estimated the time i thought it would require for the current drop indicated.
** #$&* 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? **
They are very close, which is surprising to me, as my numbers never match when they are supposed to!
In a physics experiment, that's always cause for celebration.
** #$&* Table of voltage, current and resistance vs. clock time: **
7 seconds ,3.2V ,160mA ,.02ohms
19 seconds ,2.4V ,120mA ,.02 ohms
38 seconds ,1.5V ,80mA ,.019 ohms
71 seconds ,.75V ,40mA ,.019 ohms
105 seconds ,.35V ,20mA ,.018 ohms
I got these values by approximating them from a graph i drew on excel. I got the resistance by dividing the current by the voltage.
160 mA is 160 milliamps, or .160 amps. The resistance in your first line is therefore 3.2 V / .160 amp = 20 ohms, not .02 ohms.
** #$&* Slope and vertical intercept of R vs. I graph; units of your slope and vertical intercept; equation of your straight line. **
m=70,000 vertical intercept=0
slope is current/resistance
The equation for the line is I=70,000R-1240
Not bad, but you could be off by a factor of a million. Your milliamps were treated as amps, and your ohms were therefore off by a factor of 1000. Then you graphed current in milliamps, which gives you another factor of 1000.
Your resistance looks like it decreases by a couple of ohms. Your current changes by about .1 amp. That gives you a slope in the neighborhood of -20 ohms/amp.
Now i am thinking i have a miscalculation, but all the math is correct to my knowledge. I used basic algebraic equations to find the slope and the line equation.
** #$&* Report for the 'other' resistor:; Resistance; half-life; explanation of half-life; equation of R vs. I; complete report. **
100 ohm
9 seconds+- 2 seconds
I looked back in my notebook and saw the 33 ohm (which I originally thought was 42 ohm) took about 27 seconds to lose half its voltage and current. So it makes sense that this resistor has three times the resistance, it should take 1/3 the time, 27/3=9 seconds.
I=23,333R-413
** #$&* Number of times you had to reverse the cranking before you first saw a negative voltage, with 6.3 V .15 A bulb; descriptions. **
probably around 20 times
I dont think this was accurate at all, the beeps program does not work on my computer so i am trying to watch a lot of different things at once.
The bulb seemed to be brighter when i reversed the crank, i believe the capacitor was charging until i reversed the crank, which seemed to dump all the stored voltage into the bulb.
** #$&* When the voltage was changing most quickly, was the bulb at it brightest, at its dimmest, or somewhere in between? **
Brightest, i think when the capacitor voltage changes rapidly, it causes the bulb to burn brighter.
Brighter bulb implies more current; the current comes from the capacitor's charge; the capacitor's charge is proportional to its voltage.
So a brighter bulb implies a quicker change in voltage, as you say.
** #$&* Number of times you had to reverse the cranking before you first saw a negative voltage, with 33 ohm resistor; descriptions. **
Again, i do not have the luxury of the beeps program so i am sort of winging this. I reversed direction probably 15 times. i do not think this is very accurate but it may be close.
The capacitor voltage was sort of dwindling down as time went by.
** #$&* How many 'beeps', and how many seconds, were required to return to 0 voltage after reversal;; was voltage changing more quickly as you approached the 'peak' voltage or as you approached 0 voltage; 'peak' voltage. **
This is very difficult to do without the beeps program, i bet it is nice!!
It probably took around 6 seconds to return to zero when i reversed the cranking.
The voltage was chaging quicker when it was approaching zero than when it was approaching peak.
My peak voltage was around 4V.
** #$&* Voltage at 1.5 cranks per second. **
around three volts
** #$&* Values of t / (RC), e^(-; t / (RC) ), 1 - e^(- t / (RC)) and V_source * (1 - e^(- t / (RC) ). **
.1212
e^.1212
2.11
3.11
I plugged my values into the given equations.
** #$&* Your reported value of V(t) = V_source * (1 - e^(- t / (RC) ) and of the voltage observed after 100 'cranks'; difference between your observations and the value of V(t) as a percent of the value of V(t): **
3.67%
** #$&* According to the function V(t) = V_source * (1 - e^(- t / (RC) ), what should be the voltages after 25, 50 and 75 'beeps'? **
.78 volts
1.56 volts
2.33 volts
** #$&* Values of reversed voltage, V_previous and V1_0, t; value of V1(t). **
-4 volt
3 volts
-3 volts
zero
The opposite values cancel out leaving behind multiplication by zero.
** #$&* How many Coulombs does the capacitor store at 4 volts? **
Q=C*V
1=C/4
1*4=C
4=C
** #$&* 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?; **
3.5 , .5
** #$&* According to your data, how long did it take for this to occur when the flow was through a 33-ohm resistor?; On the average how many Coulombs therefore flowed per second as the capacitor discharged from 4 V to 3.5 V? **
4 seconds , .125
It took 4 seconds for the voltage to drop from 4 to 3.5, so i took 4 and divided it by .5 to get .125 coulombs per second.
** #$&* According to your data, what was the average current as the voltage dropped from 4 V to 3.5 V?; How does this compare with the preceding result, how should it compare and why? **
188mA
They match up as reported above, they should match up as the current and voltage decreased together in this set up.
** #$&* How long did it take you to complete the experiment? **
5 hours
** **
i am sure i am all over the place here, i am very sorry. I am doing the best i can without the beeps program.
You did fine on this lab. See my notes.