the rc circuit

Your work on the rc circuit 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:

4.02 V, 33 Ohm resistor 2.17, 3.5 6.51, 3.0 11.49, 2.5 17.58, 2.0 25.78, 1.5 35.36, 1.0 47.15, .75 61.07, .5 86.06, .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.

18 sec 20 sec 15 sec 45 sec My graph started high on the y axis and decreased and approaches zero on the x axis. I used the graph to estimate the time intervals between each of these volts.

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

17 sec 33 sec 23 sec 15 sec I used the best fit line of my graph to determine the clocktimes at the given currents.

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

1.64, 90 4.34, 80 7.89, 70 12.05, 60 17.1, 50 23.24, 40 30.57, 30 40.77, 20 56.88, 10

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?

The times were fairly close, they were also within reason for the reports of voltages

Table of voltage, current and resistance vs. clock time:

3.2V, 80mA, .04 ohms, 6 sec 2.5, 60, .042, 13 1.6, 40, .04, 23 .7, 20, .035, 41 .6, 10, .06, 57

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

Slope is .0003, vertical intercept is .0356 ohms/sec, ohms I = 0.0003R + 0.0356 My data were not very linear, I used Excel to do my graph and determine my slope and y intercept

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

100 ohm resistor 64.32 +- .50 sec

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

We reversed 15 times before we saw a negative voltage We noted this so it is very accurate When we cranked backward the bulb glowed very brightly, when we cranked forward the bulb glowed just slightly. When we were cranking backward we were taking energy out of the system.

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

The voltage was changing most quickly when the bulb was the brightest, the brighter the bulb the greater the rate at which the voltage changes

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

4 times Our estimate was very accurate because we noted the number of cranks When we cranked 125 times the voltage built up very slowly and when we reversed cranking the voltage went down very quickly

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.

15 beeps, 15 sec

15 beeps in 15 seconds is not consistent with a voltage of 4 volts.

The voltage changed more quickly as we approached 0 voltage 3.19 V

Voltage at 1.5 cranks per second.

3.5 V

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

3.03, .048, .952, 3.332 Our V_source was 3.5 V, t/(RC) was 3.03, e^(-t/(rc)) was .048 -- we used all of these values to solve the 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):

3.332 v, 3.19 v the difference was .142 V

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

after 25 beeps voltage should be 1.14 V 50 beeps 1.92 V 75 beeps 2.43 V

Remember that beeps are beeps, not seconds. I'm not sure which you intended here.

Values of reversed voltage, V_previous and V1_0, t; value of V1(t).

-3.19, +3.19, -3.5, 15 1.91

How many Coulombs does the capacitor store at 4 volts?

.030 Coulombs. If R=V/C and we know that our system used a resistor of 33 ohms and produced 4 volts then C=V/R=4/33=.12 farads, if we plug this C into Coulomb=farad/volt we get .030 coulombs

The capacitor stores 1 Coulomb per volt, so it stores 4 Coulombs with 4 volts.

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?;

.03, I didn't get any loss of coulombs between 4 and 3.5 volts with my calculations -- they may not be correct.

The capacitor stores 1 Coulomb per volt, so it stores 3.5 Coulombs with 3.5 volts. It therefore loses .5 Coulombs.

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?

I am not sure where to find this

Your graph of voltage vs. clock time will tell you how long it took for the voltage to drop from 4 V to 3.5 V.

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?

Your graphs of current vs. clock time and voltage vs. clock time will give you a very good idea of the average current between the times when the voltage was 4 V and 3.5 V.

I dont know the time interval

2.5 hours