the rc circuit

#$&*

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: **

** #$&* Initial voltage and resistance, table of voltage vs. clock time: **

4.01, 33

4.01, 0

3.5, 2.04

3.0, 3.78

2.5, 6.68

2.0, 9.62

1.5, 13.72

1.0, 17.53

0.75, 20.64

0.50, 22.94

0.25, 26.00

as the clock time progressed, it is evident that it takes longer and longer for the voltage to drop smaller amounts.

** #$&* 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. **

10 seconds

13 seconds

16 seconds

3 seconds

the graph is related by a polynomial, but also exhibits aspects of exponential decay. as the voltage increases, the time to reach those voltages decreases. when read from right to left, it is evident that the time increases exponentially between voltages as the voltages decrease

** #$&* Table of current vs. clock time using same resistor as before, again starting with 4 volts +- .02 volts. **

132.33, 0

115.5, 2.04

99, 3.78

82.5, 6.68

66, 9.62

49.5, 13.72

33, 17.53

24.75, 20.64

16.5, 22.94

8.25, 26

** #$&* Times to fall from initial current to half; 75% to half this; 50% to half this; 25% to half this, based on graph. **

7 seconds

5 seconds

8 seconds

6 seconds

the values are all fairly similar, which makes sense because the distances between currents decreases, so the time decrease evens out to equal what it would have been had it been a larger current distance.

** #$&* 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? **

yes, the experimental uncertainty would definetly account for a discrepency of 3 seconds in either direction from the mean. there is no apparent pattern in the time to accompany the current drop, but there was one in the voltage drop.

** #$&* Table of voltage, current and resistance vs. clock time: **

2.06, 105.86, 3.47, 30.51

6.61, 79.4, 2.55, 31.14

12.56, 52.93, 1.53, 34.59

26.67, 26.47, 0.58, 45.64

13.23, 13.23, 0.17, 77.82

the current values could be obtained by multiplying the original current by the values 0.8, 0.6, etc. these values were plugged into the current vs. time equation (from graph) to determine the time. this time was then plugged into the voltage vs. time equation to determine the voltage. the resistance was then found by dividing the current by the voltage. as expected, the resistance increased as current decreased.

** #$&* Slope and vertical intercept of R vs. I graph; units of your slope and vertical intercept; equation of your straight line. **

-1.541, 123.3

slope= change in current/change in resistance, current

y = -1.541x +123.3

** #$&* Report for the 'other' resistor:; Resistance; half-life; explanation of half-life; equation of R vs. I; complete report. **

100

6.71 +- 0.01

the time was simply the clock time measured after t=0 that the voltage was measured to be 2V. the uncertainty is based upon the uncertainty of the time measurement tool, which is only accurate to the 1/100 place, and therefore has an uncertainty of 1/100

y = -0.257*x+185.7

for this experiment, i measured time as a function of decreasing voltage using my multimeter. i then used the multimeter to determine the current values at these time intervals. with these values i was able to divide current by voltage to determine the restistance as a function of current for the system.

** #$&* Number of times you had to reverse the cranking before you first saw a negative voltage, with 6.3 V .15 A bulb; descriptions. **

twice

the estimate wsa pretty accurate, it actually took about 1 1/2 times to get a negative number, because halfway through my second crank i saw a negative number.

the bulb immediately dimmed, almost to no glow at all when the crank was reversed, but then brightened quickly when i began to turn in the positive direction again. brightness looks to be proportional to voltage, increased voltage (positive) causes brighter lights.

** #$&* When the voltage was changing most quickly, was the bulb at it brightest, at its dimmest, or somewhere in between? **

when the voltage changed most quickly from positive to negative, the bulb was at its brightest. during the positive turns after the first negative turn, the bulb was somewehre in between brightnesses. the brightness of the bulb seems to be proportional to the rate at which the capacitor voltage changes. reasoning for this may be because as the voltage increases, the resistance decreases, leaving less interference for the electric current to reach the bulb, thus a greater brightness

** #$&* Number of times you had to reverse the cranking before you first saw a negative voltage, with 33 ohm resistor; descriptions. **

1 time, only about 5 cranks in the negative direction

this estimate was accurate, i finished my fifth crank in the negative direction and the voltage turned negative.

i made sure that my cranking was all done at the same speed, even the reverse cranking, and i just monitored the voltage to determine how quickly the numbers would turn negative. the capacitor voltage changed much more quickly in time that with the lightbulb.

** #$&* 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. **

7 beeps, 3.18 seconds

the voltage changed much more quickly as it approached zero

the peak voltage was 4.65V

** #$&* Voltage at 1.5 cranks per second. **

2.10V

** #$&* Values of t / (RC), e^(-; t / (RC) ), 1 - e^(- t / (RC)) and V_source * (1 - e^(- t / (RC) ). **

1.38, 0.252, 0.75, 1.575

R and C were known from the values of the materials used. t was the time that i reversed cranking, i know i made 100 cranks, at a rate of 2.2 cranks/second, which equals 45.5 seconds, which was plugged into the equation. this number was then taken to the e^- power, to get 0.75. the Vsource was the value measured in the step above, which was 2.1V. multiplying 0.75 to 2.1, i got 1.575

** #$&* 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): **

1.575,4.65

3.075

** #$&* According to the function V(t) = V_source * (1 - e^(- t / (RC) ), what should be the voltages after 25, 50 and 75 'beeps'? **

using my numbers (which appear to be wrong, even after retrying the charging), the voltages after these beeps are:

0.61V

1.11V

1.35V

** #$&* Values of reversed voltage, V_previous and V1_0, t; value of V1(t). **

-4.65, 2.1, 2.27, 4.34

4.34

i simply plugged in the numbers to the equation, using the values of R and C found on the materials. the voltage at a certain time was 4.34V

** #$&* How many Coulombs does the capacitor store at 4 volts? **

4 coulombs

the capacitance is 1F, and the voltage = 4, so 1*4 = 4

** #$&* 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?; **

3.5C, 0.5C

the change in Q = 4-3.5, or 0.5C. this is how many coulombs the capacitor loses during the time it takes to drop from 4 V to 3.5V

** #$&* 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? **

68.6 seconds, 7.3*10^-5C/s

i plugged the two values for charge into the voltage equation, and solved for t, taking the natural log to get rid of the exponential function. i then divided 0.5 C by this time to determine how many coulombs were lost each 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? **

7.3*10^-5

current is equal to the change in charge over a change in time, so the current is equal to (4-3.5V)/68.6 seconds. so the current is equal to 7.3*10^-5A

** #$&* How long did it take you to complete the experiment? **

3.5 hours

** **

the beeps program still wouldnt work, so i did my best to even out my generator cranks, but if there's a big discrepency in my accuracy and/or precision, that is most likely the culprit.