Tuesday, August 2, 2011

Tentative exam 3 coverage

This is all pending a discussion with Dr. Mazumdar so we can coordinate ... but here are the minimal sections you should worry about:

22.2,4 reactance & power in ac circuits, there is certain to be a question about filters
23.1-3, 5-9 EM waves
24 all (geometric optics, but you won't have to draw ray diagrams)
28.1-3, 6-7 quantum and atomic physics, note in particular the Bohr model

Tentatively, there will not be any questions from Ch. 29 since that is almost purely qualitative stuff.

The test will be open book & notes, the only restriction is that no internets are allowed (i.e., no phones, PDAs, laptops, etc.).

And, very important, both sections will have the exam at 4pm on Wednesday in the usual lecture room. You will have at least 2 hours for the exam, and will probably not need all of it. If, for some good reason, you cannot make it from 4-6pm (e.g., job constraints, another class, etc.) we will work out an alternate time, but please contact me as soon as possible.

There will be 10 problems out of which you must solve 8, and they will be shorter problems than the previous exams. That is, don't freak out about there being more questions than before, the questions will be much less involved than usual HW or exam problems. They will be at the level of example problems in the book, for the most part find the right formula and go. Nothing really subtle.

More details tomorrow (Tues) after Dr. Mazumdar and I discuss a bit more ...

18 comments:

Anonymous said...

Is Chapter 25 included?

pleclair said...

No, since it is just applying mirrors & lenses anyway.

Anonymous said...

Dr. Leclair,

Would you consider the 2 options: opting out of the final and keep our grade as it stands if we so choose.. or take the final if we are not satisfied with our current grade to bring it up. We would definately appreciate it.

Thank You,
Students(27 signatures)

Anonymous said...

+1 (28 signatures)

Anonymous said...

+1

Anonymous said...

Dr. Leclair,

You said there would be a question regarding filters.... Are you referring to the high and low pass filters?

Thanks

Anonymous said...

My question is irrelevant if you agree to the previous post.

Thanks again

pleclair said...

I appreciate the suggestion, and it is a nice idea ... but I can't really do it. I will explain my reasoning a little more tomorrow, but we need to have a third exam for everyone. It will be much easier than the last one, for sure. And open book ... some problems will be almost like example problems from the book that you can look up.

I do mean low- and high-pass filters, indeed.

Anonymous said...

On the sum 2010 exam, are 4,5,7,and 8 relevant to the exam?

pleclair said...

5, 7, and 8 are not relevant. I would say the following questions from the summer 2010 final are relevant:

1, 2, 4, 11, 15, 16, 17

(5, 7, and 8 are on stuff we covered on exam 1 already)

Anonymous said...

Im sorry, I meant the 2010 exam 2 not the final. I did not clarify.

pleclair said...

Ah, right. Then I would say 4, 7, and 8 are relevant from the 2010 exam 2, but 5 is not (#5 is induction, so covered on the last exam).

Number 6 is relevant too, but a bit too hard.

I will try to post some answers for those problems tonight.

Anonymous said...

Will we still get a formula sheet with all the constants?

Anonymous said...

On the summer10 final, #17 I keep getting 15.1 for electric field. What am i doing wrong?

pleclair said...

The formula I have there has a typo, it should be I=E^2/(2 mu_0 c) ... that is, the E should be squared.

pleclair said...

You will have all the constants you need provided (and they are in your book).

Anonymous said...

Where can we find the answers to sum10 exam 2 at? The ones we need to know for our final? (#4,6,7,8)

pleclair said...

Number 8: see problem 3 here. http://faculty.mint.ua.edu/~pleclair/ph126/Homework/HW8_13Nov09_SOLN.pdf

Number 7:
a) p = E/c = (power)(time)/c; given power & time
b) I = (power)/(area) = E^2/(2 mu_0 c); given power, diameter to get area

Number 6: safely ignore

Number 4:
a) low pass for speech - capacitor to ground doesn't let high frequencies out, low pass easily through inductors to the output.
b) add up all the inductors up to get the equivalent, Leq = (10mH + 0.04mH + 0.22mH) = 20.5mH. Cutoff is when inductive and capacitive reactances are equal, Xc=XL, giving f=1/(2*pi*sqrt(LC))~7500Hz.