The Write-Up asks you to formally derive the equation for t1/2.|
Here is a link to a pdf file that shows one derivation for t3/4.
If you do not like the viewing and printing advantages of pdf's, then here is an htmlversion.
MOVIESPlease note: If the movie is too small,
please watch via youtube using the YOUTUBE button on bottom right of each video.
And, some videos have much higher quality. Find the HD version using the gear icon.
RC: Part 1, Set Up RC: Part 2, Analyze This DMM-01- Inputs and Selector for Amprobe 37XR-A Digital Multimeter DMM-02- Measure Resistance with DMM Amprobe 37XR-A DMM-03- Measure Capacitance with DMM Amprobe 37XR-A RC: Part 4, Equation Editor
|Hi Douglas, A quick question regarding the R C Circuit pre-lab. The forward note for the quiz states that if we try the quiz we will recieve a 100% regardless of how we do on the quiz. Is this correct? And, it also states that we don't need to worry if we miss the deadline, which is 6/30/04. Is this true? It mentions that the deadline is one week later than normal, but 6/30/04 is the correct deadline. I am a little confused!?!?|
Thanks Douglas - hope all is well.
Thu. 24, Jun 2004, 19:25
|None of that is true. Those notes are no longer there.|
I am so sorry. Please let me know if the deadline creates a problem for you.
And thank you so much for alerting me of this silly error.
Fri. 25, Jun 2004, 05:25
|Hi Douglas, I am working on question 1 of the RC Circuit pre-lab. "What is the percentage of qmax, when t=4*tau*ln2." I am reading the lab manual, and I am following how the t3/4 was obtained, however, I only have one more submittal left and I keep getting 20%. Can you please tell me what Im doing wrong or how I can approach this question?|
Thanks Douglas, have a good day.
Sun. 27, Jun 2004, 13:36
What is q when t=4τln2?
q= qmax(1 - e-t/τ)
q= qmax(1 - e-4
Where the τ's cancel.
q= qmax(1 - e-4ln2)
q = 93.75% qmax
Another way to look at it is to realize that t= 4 τ ln2 is simply 4*t1/2's.
So, the way half lifes work for charging is:
After 1 t1/2's you've reached 1/2 of 100% = 50% max charge.
After 2 t1/2's you've reached 1/2 of 100% + 1/2 of (1/2 of 100% )= 75% max charge.
After 3 t1/2's you've reached 1/2 of 100% + 1/2 of (1/2 of 100% ) + 1/2 of 1/2 of (1/2 of 100% )= 87.5% max charge.
And so on and so forth.
Then... the lab manual (I think the writeup pages) show you a simplified method of the last solution.
Percent of qmax = 1 - 2-n
where n is the number of half times (t1/2's).
qmax = 1 - 2-4 = 93.75%
Mon. 28, Jun 2004, 19:13
|Hey Douglas, |
On question 7&8 i am a little confused. How are you suppose half of max vaule i just took the max value and divided it by two but that is not right. Question 8 is asking for the capiactance I was using the equation T1/2=RC*ln2 i sloved for C which was C=t1/2/RC*ln2 but that did not work i don't think i am getting the equation right
Sun. 27, Feb 2005, 18:32
|Q7 shows on offest in the "zero" point. You must add the half quantity to the offset to find the scale's value for 1/2Max.|
Q8 Are you using μF for your answer?
Tue. 01, Mar 2005, 10:42
I was trying to access the RC circuit post lab as well as some other postlabs and the deadline date is for Spring semester. Are we not allowed to access some post labs before a certain date/time?
Tue. 27, Jun 2006, 10:31
|Try the PostLab now. I don't always have time to edit the quiz for the current semester until "last minute..." RC PostLab is ready now. Thanks for asking.||Douglas|
Thu. 29, Jun 2006, 06:19
I am unsure of what to provide as the uncertainty values for Time. Are we to give values for t1/2 at n=0 and n=7, or n=2? Or, are we to give values of t (eg. t = 1*tau*ln2 and 2* or 7*tau*ln2)?
Sun. 08, Oct 2006, 15:22
|The uncertainty in time is related to the inverse of the sample rate.|
If I sample the circuit at 1000[Hz], then I am sampling 1000 times every second. Therefore the best I can measure time is
1/1000[Hz] = 0.001 [second/sample].
±0.001 [s] could be considered the uncertainty in time measurement.
I think the sample rate for RC was 200[Hz].
You can check for sample rate with the DataStudio Template. Find the "Signal Generator" window; that contains the sample rate.
I hope this helps.
Mon. 09, Oct 2006, 04:06
|Ok, I can't seem to find the uncertainty of resistance for the postlab. Furthermore, I don't know the reason for uncertainty.||Anonymous|
Tue. 27, Mar 2007, 18:46
|Every measurement has uncertainty.|
Resistance was measured with the Fluke Multimeter.
The readability of the meter in Ohms (which in this case is the best approximation of uncertainty) is ±0.1Ω
Wed. 28, Mar 2007, 09:51
|I can't find the symbol for tao, should I type in the word for the equation?||Trotter, Jennifer|
Wed. 04, Apr 2007, 13:18
|The tau is there. use |
& tau ;all together to get τ
Wed. 04, Apr 2007, 14:12
I need to derive the t(1/2) equation however I do not have equation editor on my computer. How do I get it on my computer? The downloads on my class disk don't have it.
|Zufall, Lauren |
Tue. 03, Jul 2007, 09:47
Your install disk for MSOffice should have the option to intall extra packages or extra stuff, Equation Editor is one of those things.
If you can't make it work, use one of the campus computer labs or you may hand-write the derivation sheet. Just be careful to write clearly.
Tue. 03, Jul 2007, 09:58
I have a question about postlab problem #2. Would you like us to report the range for the values in terms of a minimum value and a max value, or would you like us to report the max - min value? On the same problem, what ranges would you like us to report for time? Should we start with the time value at the beginning of the charge cycle and end with the automatic stop time for the run? Thanks!
Thu. 06, Mar 2008, 19:58
|Voltage and Time can be a range spanning the whole voltage range and whole time range. These were the measurements; all else was a calculation based upon the base measurements.||Douglas|
Fri. 07, Mar 2008, 09:56
I am confused about finding the uncertainty for the voltage.
Mon. 06, Oct 2008, 10:43
|Highlight the region where the voltage was supposed to be around 0 volts. Find the standard deviation here. That is a good representation for the uncertainty in Voltage.||Douglas|
Mon. 06, Oct 2008, 16:32
|The question 7 in the prelab is kinda confusing. Taking the maximum voltage value minus the offset below zero point then divided by two given a wrong answer. Is it needed to be the maximum value + the below zero point value then divide the total by two? Kinda confused.||Anonymous|
Mon. 28, Sep 2009, 14:32
|Think about a signal on a plot where the signal starts at 1 V and then rises to 6 V. What is the half-way point?|
Half the distance from start to finish is (6-1) [V] / 2 = 5/2 [V] = 2.5 [V].
But, that is a 2.5 [V] rise above the 1 [V] starting point.
The measurement for half-way point on the axis is therefore:
(Max - Min) / 2 + Min
This formula still works if the minimum is negative.
Start at -1, rise to +5.
(Max - Min) / 2 + Min = (5 - (-1)) / 2 + (-1) = +2
Tue. 29, Sep 2009, 15:21
|I see that you want "normal units" for #6, but that doesn't tell me how many sig figs you want. Can you please enlighten me? :)||Casey, Jie|
Fri. 08, Oct 2010, 12:20
|Use 3 sig figs.||Douglas|
Fri. 08, Oct 2010, 13:40
|Help!! I tried to do the prelab for RC and got so many wrong!! I wonder if you could recommend a website that would help me with questions 7-9. I got so many questions wrong!! Yikes!!||clark, Peggy|
Tue. 01, Mar 2011, 08:02
|All the answers are in the lab manual. Be sure to investigate the WriteUp portion too. These questions 7-9 are fairly specific to this experiment in our laboratory.||Douglas|
Wed. 02, Mar 2011, 07:51