#$&*
Phy 122
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 **
8/30
** **
4.02 V, 33 ohms
0, 4.02
4.28125, 3.5
7.1875, 3.0
12.26563, 2.5
19.51563, 2.0
28.79688, 1.5
41.96875, 1.0
51.0625, 0.75
64.4375, 0.5
88.20313, 0.25
The first line shows the initial voltage and resistance of the circuit. Each line after that consists of the times a voltages over the period of discharging the capacitor. The times are listed first in seconds in each line, and the voltage in voltage are listed second in each line.
** **
22.1
22.1
22.1
22.1
The differences are in seconds. I made the graph and made a curved exponential line of best fit. This gave me an equation for the line of y = 3.8041 e ^ -0.0314x . I plugged each of the voltages into this equation for y and calculated the times. Then I took the difference in the times at each of the voltages listed. I got the same time difference for each of these voltage intervals depicting that this is an exponential relationship.
** **
4.0 V, 33 ohms
0, 94
5.15625, 84
10.73438, 74
14.96875, 64
20.98438, 54
29.32813, 44
37.6875, 34
52.39063, 24
72.46875, 14
124.0469, 4
The first line shows the initial voltage and the resistance in the resistor. The following lines show the current in mA and the time in seconds as the capacitor is discharged. The first number in each row is the time, and the second number in each row is the current.
** **
27
27
27
27
The differences are in seconds. I made the graph and made a curved exponential line of best fit. This gave me an equation for the line of y = 93.729e ^ -0.0257x . I plugged each of the current values into this equation for y and calculated the times. Then I took the difference in the times at each of the currents listed. I got the same time difference for each of these current intervals depicting that this is an exponential relationship.
** **
The times I reported are the same, they are not quite the same as the times for the voltage values. My voltage time was about 5 seconds less. There is a pattern though because both have an exponential decrease in value over time as the capacitor discharges at the same resistance.
** **
8.57, 2.91, 75.2, 38.696809
19.76, 2.05, 56.4, 36.347518
35.54, 1.25, 37.6, 33.244681
62.31, 0.538, 18.9, 28.465608
89.48, 0.229, 9.4, 24.361702
Each line contains the clock time in seconds, the voltage in volts, current in mA, and resistance in ohms. First I plugged in the current values to my current vs. time equation to calculate the times at these values of current. Then I plugged the calculated times into the equation for voltage vs. clock times to get the voltages. I then divided my voltages by my currents in Amps and found the resistance of each.
** **
0.2118, 23.858
The slope is in current in amps, and the y- intercept is resistance in ohms.
R = 0.2118I + 23.858
My graph has a positive slope that is slightly curved. It would fit a logarithmic line of best fit better than a linear line. I got the slope, vertical intercept, and the equation of the line through excel’s, trend line options.
** **
150 ohms
100.105 +- 3.28 seconds
I took the difference from initial voltage to half, 75% initial voltage to half of this, 50% original voltage to half of this value, and 25% initial voltage to half its value. I averaged these differences and found the standard deviation.
R = 702.8I + 134.28
30, -24.045, 4.614, 153.8
20, 39.31, 2.999, 149.95
10, 147.613, 1.436, 143.6
5, 255.917, 0.688, 137.6
3.5, 311.648, 0.471, 134.5714286
First I set up the circuit in parallel with the meter to measure the voltage. I used this to make a voltage vs. time chart. Then I set up the circuit in series to measure current and used this to make a current vs. time chart. I then took a list of currents, used my current vs. time equation to calculate the corresponding times. Then I plugged these times into my voltage vs. time graph to get the corresponding voltages. I divided the voltages by the current values to get the resistance values. Then I set up a graph of resistance vs. current. In the data listed above the values are current in mA, time in seconds, voltage in volts, and then resistance in ohms. This shows that as time goes on current and voltage decrease as the capacitor is discharged. Also this shows that resistance increases when current increases.
** **
17 crank reverses
I think my estimate was accurate because I was tallying them while I was cranking.
The bulb came on initially and went out as the capacitor stored the voltage after it reached about 2 volts. Then when I reversed the crank the bulb came back on, and when reversed direction again the bulb turned off. This continued on but eventually the bulb wasn’t as bright when I reversed directions initially and when I reversed again the bulb wouldn’t go completely out anymore. The voltage started increasing up to about 4 volts as I cranked through the initial 100 cranks. Then when I reversed directions it went down, and when I reversed again it went up a little but not as much as it was just decreased. This is because the capacitor was charged in the initial 100 cranks and then when the direction of the current was reversed the capacitor was discharging its stored voltage which powered the light causing the voltage available in the circuit to decrease because it was being used up. When I reversed directions again and the bulb went out the capacitor was trying to store more voltage but this rate was not as high as the rate of discharge when reversed in the opposite direction. This is what caused the voltage to decrease all the way down to a negative value.
** **
When the voltage was changing the most quickly in a decreasing manner the bulb was lit the brightest. The bulb is lit the brightest when the capacitor voltage rate is decreasing quickly which takes place when the generator direction is reversed. The capacitor stores voltage when the generator is cranked. When the crank is reversed the circuit is reversed which then leads the stored voltage away from the capacitor, decreasing its voltage. This is because the capacitors stored voltage is being discharged to the rest of the circuit which powers the light bulb.
** **
8 crank reverses
I think my estimate was accurate because I took a tally while I was cranking.
I followed the same procedure as before cranking at a rate of about 1.5 cranks/second for 99 cranks. Then I reversed the direction for 12 cranks and continued to reverse it for 12 cranks until the voltage reached a negative value. The voltage increased during the 99 cranks and decreased in the initial reverse while increasing at a smaller rate when reversed again. When the crank was reversed each time the capacitors voltage was released at a faster rate than it was being stored when the crank was in the other direction. This caused the overall voltage to decrease to a negative value.
** **
47 beeps, 26 seconds
The voltage was changing much more quickly as it approached 0 voltage.
3.2 volts
** **
3.2 volts
** **
0.788, 0.455, 0.545, 1.745
The first value is t/R*C which was 26 seconds/ 33 ohms * 1 Farad. The second value is e^(-t / (RC) ), so I did e ^ 0.788. The third value was 1 - e^(-t / (RC)), so I subtracted 0.455 from 1. The final number is V_source * ( 1 - e^(-t / (RC) )), so I multiplied 3.2 volts by 0.545.
** **
1.745, 3.2 V,
1.455 difference, 55 %
** **
1.27 V, 2.03 V, 2.5 V
** **
-3.2 V, 3.2 V, -6.4 V, 26 seconds
-0.289
The first number is the voltage seen at the instant of reversal, the second is V_previous, the third is V1_0, and the fourth is t. The line in the second row is V1(t) which was calculated from the following equation V1(t) = V_previous + V1_0 * (1 - e^(- t / (RC) ). In this equation R was 33 ohms and C was 1 Farad. T was plug in as 22. V1_0 and V_ previous were stated above. When all plugged in this gave -0.289 volts.
** **
Q = CV
Q = 1 Farad * 4 volts = 4 Coulomb
I calculated this as explained above. My capacitor had a capacitance of 1 Farad which was written on the side of the capacitor.
** **
3.5 C, 0.5 C
I used the formula Q = CV knowing that C was 1 Farad. Then the voltages were given above so I could just calculate Q and subtract the difference.
** **
19 seconds, 0.026 C/s
I had the time from the experiment at the beginning, and I divided the 5 C difference calculated in the last question by 19 seconds to get the rate.
** **
0.0242 A
This answer is smaller than the calculated value above but very close. These numbers should theoretically be the same but because of experimental error they are not exactly the same.
** **
3.5 hours
** **
&#This lab submission looks very good. Let me know if you have any questions. &#