Special Note: All Plots are prepared for you. Do NOT change axis input.
|I need some asistance with the proper notation for the pre-lab question #6. I found the base units to be Newtons per meter. I entered N m-1 and the answer was wrong. Was it my use of the -1 or is my answer wrong? Thanks||Anonymous|
Fri. 20, Feb 2004, 11:46
|How about Ampere?|
Where did you find that about N/m? The Nm-1 is actually from the the following context, but you can see N/m are not the units of Current:
Although it was already obvious on the occasion of the 8th CGPM (1933) that there was a unanimous desire to replace those "international" units by so-called "absolute" units, the official decision to abolish them was only taken by the 9th CGPM (1948), which adopted the ampere for the unit of electric current, following a definition proposed by the CIPM in 1946:
"The ampere is that constant current which, if maintained in two straight parallel conductors of infinite length, of negligible circular cross section, and placed 1 meter apart in vacuum, would produce between these conductors a force equal to 2E-7 newton per meter of length."
Thanks for asking.
Fri. 20, Feb 2004, 12:00
|Are there any movies for the OHMSLAW lab?, also The lab CD I burned dosen't seem to have anything in the RC, Induction, and RLC Folders, Will There be anything to copy into these folders, or did I not get everything in my Lab CD when I Burned it?||Anonymous|
Fri. 20, Feb 2004, 13:39
|Yes, the Ohm's Law movies are on the lab computers.|
Sorry that they weren't on the CD that your were given last week.
That's right. RC, Induction, and RLC are not ready yet.
We won't be doing Induction, as previously announced via email to the phylab2 email address.
Sat. 21, Feb 2004, 08:03
|Let's talk physics translation: last semester we learned that a graph of resistance vs. length might have the trendline y = 33.282x + 0.3063, and that the physics translation of it would be Resistance [Ohms] = 33.282 [Ohms/m] * L [m] + 0.3063 [Ohms]. This semester, apparently, we need to display TWO equations: one equation with numbers (eg. y = 33.282x + 0.3063), and one with names and units (something like: Resistance [Ohms] = ??? [Ohms/m] * Length [m] + Resistance [Ohms]). If so, I'm a little confused about where the changeover happened, and what exactly the second equation is. Any thoughts? Thanks!||Anonymous|
Tue. 02, Mar 2004, 20:29
Resistance = 33.282 [Ohms/m] * Lenth + 0.3063 [Ohms]
is the only required form.
The other format ( y = 33.282x + 0.3063) is only "needed" in the sense that some programs (eg. Spreadsheets) spit out y=___x+___ information. We, of course, want it accurately translated A.S.A.P. into the physical form as the strictly mathematical form is not informative. But, do NOT edit the y=mx+b box in MSExcel. If you do, you take ownership of the box and MSExcel will never update it again, even in the event you change the graph's source data. That is why we ask that you add a text box to the graph to separately add your physics translation.
Wed. 03, Mar 2004, 04:43
|Without giving away the answer to the question, what exactally do you mean by "Ohmic" on the postlab question #2?||Morrison, Michael|
Sat. 06, Mar 2004, 11:56
|Something "Ohmic" would obey Ohm's Law. Are V and I directly proportional? (Do they have a linear relationship?)||Douglas|
Sat. 06, Mar 2004, 16:37
The % difference equation refers to dividing by the sum - this is the sum of the differences of all three resistances, correct? Why is the % difference equation using 200%?
Sat. 06, Mar 2004, 23:48
|It doesn't make any sense to compare the values of all three resistors to one another. They are all supposed to be different.|
No, the percent difference is when you have two values:
say you have 10 [Ohm] and 10.89 [Ohm],
Percent Diff = |Diff|/Average*100% = |Diff|/(Sum/2)*100%
Percent Diff = |Diff|/Sum*200%.
I hope this helps.
Sun. 07, Mar 2004, 15:21
|Question 3 and 4 on this post lab appear to be identical. Did you mean to refer to the 10 Ohm resistor on question 3??? Just wondering if there was a typo.|
Sun. 07, Mar 2004, 22:31
|No typo. Just answer it twice.|
Mon. 08, Mar 2004, 08:53
|Douglas, in the Ohmic lab postlab, Q's # 3 and 4 do not provide enough information to calculate a line slope: only one data point is given for the 33ohm line. The slope could be determined by making the ASSUMPTION that it passes through the origin, but the 100ohm line has a non-zero y-intercept suggesting that no assumption could be made about the 33ohm line. i answered N/A in Q's 3&4 and this was graded as incorrect. What answer were you looking for?||McCaskill, Douglas|
Sun. 07, Mar 2004, 18:15
|Just assume it passes through the origin.|
If you consider uncertainties, all three pass through the origin.
Mon. 08, Mar 2004, 08:54
|Hi, on question #3 and #4 on the PostLab (OhmsLaw) their is not enough information to solve the problem or to calculate the R. How do we obtain the slope without enough information?||k, timni|
Thu. 09, Jun 2005, 10:51
Fri. 02, Sep 2005, 03:34
|For question #5 on the Ohm's Law postlab it asks to describe each type of item tested in terms of its V vs. I being "Ohmic" or not. I understand that ohmic means that the graph of voltage vs. current will yeild a one to one relationship but I am not sure what each item tested are. I think all of them are ohmic because no matter where you picked a point it would have only one voltage and one current.||Anonymous|
Mon. 15, Sep 2008, 22:12
|Not all the items are ohmic; and some are ohmic only under certain conditions. By the way, 1:1 means that for each current value there is only one voltage value. Is the Capacitor Ohmic? Is the Diode Ohmic? Is the Coil Ohmic?|
Take the light bulb for instance. For a given current value, there are two voltage values at low frequencies; thus the light bulb is not ohmic. But at high frequencies, the light bulb doesn't have a chance to sporadically change temperatures, therefore resistance is constant; at high frequencies there is only one voltage value for each current value (the plot of V vs I is linear and one to one.) Thus, the light bulb is ohmic at high frequencies.
Tue. 16, Sep 2008, 07:46
|Hi Douglas, I was curious about Question #6 in the Pre-lab. It asks for the base unit for Current and I looked at the website as well as in the book and it said it was s*A. So I plugged in both s A and s*A and both were incorrect. So I'm just wondering, what exactly are you looking for? Thank you and hope you had a nice weekend!||Anonymous|
Sun. 22, Jan 2012, 23:04
|[sA] is an [ampere-second]. An [amp-sec] is synonymous with the unit [coulomb].|
The FLOW RATE of charge is defined as Current measured in [Ampere].
The QUANTITY of charge transferred is measured in [coulomb].
Ampere is a base unit.
length: meter m
mass: kilogram kg
time: second s
electric current: ampere A
thermodynamic temperature: kelvin K
amount of substance: mole mol
luminous intensity: candela cd
Mon. 23, Jan 2012, 03:41