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Free undergraduate EE course from MIT Answered

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MIT is starting up a free online courses  program called MITx.  The first one they're doing is their basic Circuits & Electronics course, 6.002x.  It is being presented in an "experimental prototype form".  It starts March 5, 2012, and runs through June 12.

What an opportunity!  I'm taking this for sure.  It requires a lot of math that I don't know, but either I will muddle through without it, or learn as much as I can and then drop out when it gets too far beyond my level.  Either way it's a huge win.  I'm also trying to organize or join a study group at my local hackerspace.

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Goodhart (author)2012-03-18

I am stuck on one of the final week one assignments. I can not for the life of me figure out what they want as an answer....I have tried just about every iteration of interpretation I can think of. I'd hate to "flunk out" because I don't understand the question, especially if I am able to get the answer once I do.

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Lithium Rain (author)Goodhart2012-03-18

I don't think cheating is any more savory with a virtual class. ;)

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Goodhart (author)Lithium Rain2012-03-19

I don't want the answer, I can determine the answer, I need to understand the Question

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caitlinsdad (author)Goodhart2012-03-19

Can you post the gist of the question? It may be one of those non-proofread questions, typo or something made up that assumes you have the other pre-requisite courses that you have not taken yet or something thrown in to just to get people's heads scratching. Did they mix it up with questions off a MENSA test?

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Goodhart (author)caitlinsdad2012-03-19

Ok, well, I ran out of time, and the problem got answered....not sure yet how they came up with it yet. Looking at the first picture (one to the left)  I figured out every one of the "variables" that was unmarked EXCEPT: The value (in Amperes) of iV is:  
 
 
 
   

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caitlinsdad (author)Goodhart2012-03-19

In a single path(series) the current is the same at any point, in parallel they sum up for all the split out paths?

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Goodhart (author)caitlinsdad2012-03-19

Yeah, I made the question harder then it need be....I tried to calculate it from the other components and voltages and "put too much into it"; like a cake with added wood chips, it didn't turn out "right". After supplying the answer, I was able to find out, in their annotated version of the coursework why that is so. Thanks. *blush*.

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CheckDavid (author)2012-03-19

Damn, you guys already started lol

I don't have all the requirements either, but I guess I will try just for the sake of it.

I wonder if they will have another course soon.

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neuropol (author)2012-03-07

So, who is doing this? I started yesterday. It is beyond any maths I've ever done, but I'm managing so far.

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Kiteman (author)2012-03-01

Is it available outside the US?

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FoolishSage (author)Kiteman2012-03-01

According to the site: This course will run, free of charge, for students worldwide from March 5, 2012 through June 8, 2012.

This is a great find Rachel! Thanks!

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rachel (author)Kiteman2012-03-01

When I registered there was a slot asking for my preferred language. I suspect it will be primarily in English but maybe there will be some translations if enough people request other languages.

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kooth (author)2012-03-01

I just enrolled, it was easy!

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kooth (author)2012-03-01

It is according to their website.

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