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August 31, 2025, 08:07:11 am

Author Topic: Exam 2 - good or bad?  (Read 20332 times)

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#1procrastinator

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Exam 2 - good or bad?
« on: November 14, 2012, 01:44:13 pm »
Bloody disaster for me. I forgot to include a couple of formulas in my notes
and that resulted in me not being able to complete a couple of questions. Think I'd be lucky
to get 50%. That's what you get for trying to cram everything into 7 hours before the exam.
Ah well, can't do anything bout it now.

Third time's a charm, see you next year class of '13, hahahahahahahaha

InsaneMcFries

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Re: Exam 2 - good or bad?
« Reply #1 on: November 14, 2012, 01:59:50 pm »
I found it fairly easy, however it was a bit more time consuming than other years. Probably got at least 2 MC wrong and maybe some worded marks. Overall I was pretty happy.
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wanigara

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Re: Exam 2 - good or bad?
« Reply #2 on: November 14, 2012, 02:02:51 pm »
How did you go with AC motor question?
Look deep into nature, and then you will understand everything better. Albert Einstein

InsaneMcFries

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Re: Exam 2 - good or bad?
« Reply #3 on: November 14, 2012, 02:05:26 pm »
That took a bit of thought, finding the current relied on unit 3 electricity. I just found the voltage drop over the power lines using the resistances, and then divided it by the resistance to find the current. I still found it took a while.
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Mr Keshy

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Re: Exam 2 - good or bad?
« Reply #4 on: November 14, 2012, 02:10:25 pm »
I promised myself I won't read this thread... It's going to make me worry more than I want to.

Before this thread flies, I found electricity pretty good, light not so good, Detailed, alright.

I'll stop here!

Hope everyone did well!!
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ZanyCat

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Re: Exam 2 - good or bad?
« Reply #5 on: November 14, 2012, 02:19:34 pm »
For the question about two methods to reduce power loss...

First was obviously use a transformer.

But the second was confusing me. I was initially thinking to decrease the resistance by using thicker wires or a better conductor or electricity, but because I=V/R, if R decreases then I increases! And because PLoss=I^2R, any slight increase in I will be squared and actually increase the power loss by quite a bit.

So I eventually said to 'add more transmission wires in parallel', but that's effectively the same as decreasing R/increasing I, and at that point I gave up.


So which of the answers for the second method is correct (1: decrease R, or 2: add in parallel)? Are none correct, is one correct and the other isn't, or are both correct?

very confuzzled


vcestudent94

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Re: Exam 2 - good or bad?
« Reply #6 on: November 14, 2012, 02:19:44 pm »
fuck fuck fuck. I fucked up the whole transmission question, 5 marks definitely gone! I dont know what I was thinking, I went: I=V/R=900/18=50A but i know thats wrong fuck. Then I used that for the next 3 mark question. Please can anyone help me: If I used the wrong current of 50A in determining the Power loss in the next question, would I get ANY consequential marks? Please I need this answered Im dying! :(

sh00my

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Re: Exam 2 - good or bad?
« Reply #7 on: November 14, 2012, 02:22:07 pm »
Moderate exam.
There were many easy marks, but a few difficult marks. I doubt I'll be getting an A+.

My prediction for the cutoff is approximatly 89%.

The path difference question which required you to find the new wavelength will seperate students, and a lot will get caught up in finding the energy of one X-ray photon. Lastly that electron shell question was very difficult, I spoke about standing waves ect but doubt I'll be getting 3/3 for that one.

afridilegend

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Re: Exam 2 - good or bad?
« Reply #8 on: November 14, 2012, 02:22:53 pm »
#### - I couldn't finish it!! :( :( :(

Physics was my only hope... Anyone else find it a tad too long?!

what do you think the A+ cut-off score will be?

#@$%!

sh00my

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Re: Exam 2 - good or bad?
« Reply #9 on: November 14, 2012, 02:24:15 pm »
fuck fuck fuck. I fucked up the whole transmission question, 5 marks definitely gone! I dont know what I was thinking, I went: I=V/R=900/18=50A but i know thats wrong fuck. Then I used that for the next 3 mark question. Please can anyone help me: If I used the wrong current of 50A in determining the Power loss in the next question, would I get ANY consequential marks? Please I need this answered Im dying! :(

You will lose one mark max, assuming you showed working for the other questions. Relax you will be fine.

studynotes

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Re: Exam 2 - good or bad?
« Reply #10 on: November 14, 2012, 02:26:14 pm »
i forgot to put the line at NE i just put it as east fuck my lifeeeeee
im the only one who got q1 wrong :(

#1procrastinator

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Re: Exam 2 - good or bad?
« Reply #11 on: November 14, 2012, 02:30:00 pm »
I did that too. Sheeeeeeeeeeeeeeeeeeeeeeeeeeeeeit. What was the next one? I just guessed and put the unchanged one

Fayab25

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Re: Exam 2 - good or bad?
« Reply #12 on: November 14, 2012, 02:31:48 pm »


So which of the answers for the second method is correct (1: decrease R, or 2: add in parallel)? Are none correct, is one correct and the other isn't, or are both correct?

very confuzzled


For the 2nd method decrease R is correct

availn

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Re: Exam 2 - good or bad?
« Reply #13 on: November 14, 2012, 02:35:24 pm »
i forgot to put the line at NE i just put it as east fuck my lifeeeeee
im the only one who got q1 wrong :(

I'm not actually sure if it's NE, someone said it was SE, because the arrow was pointing at magnetic north. We'll need a copy of the exam to see.

Pretty neat exam, nothing out of the ordinary. Might have lost a mark on that first question, and probably dropped some in those bloody explain questions.
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vcestudent94

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Re: Exam 2 - good or bad?
« Reply #14 on: November 14, 2012, 02:37:05 pm »
fuck fuck fuck. I fucked up the whole transmission question, 5 marks definitely gone! I dont know what I was thinking, I went: I=V/R=900/18=50A but i know thats wrong fuck. Then I used that for the next 3 mark question. Please can anyone help me: If I used the wrong current of 50A in determining the Power loss in the next question, would I get ANY consequential marks? Please I need this answered Im dying! :(

You will lose one mark max, assuming you showed working for the other questions. Relax you will be fine.
Thank you for your kind words, however the way I  alculated the current was completely illogical since I just assumed that the motor would get 900V in th first place. I then went on to calculate the power loss which left the motor with less than the required power to use 9000V. Do you get what Im saying? But if I applied the rules correctly:Ploss=I^2R how many method marks am I looking at, 1 out o the total 3?