MIT 6.002x has officially kicked off...

[metalphreak]

I've had no issues finishing all the labs and homeworks up to (and including) week 2. The method of answering questions like that is similar to what I did in the equivalent course at my university. The SPICE circuit stuff is quite cool, although it is a bit... limited in its scope and usability. There's only so much you can do in a browser and only so much they can do in terms of developing a program like that from scratch.

[MikeK]

I am not even out of week "zero" and there are problems aplenty.

University isn't for babysitting.

The introductory videos have background music that threatens to swamp the speech.

It bothered me so little that I didn't even notice it.

To copy the simple circuit in Lab 0, I have to scroll my screen up and down all the time.

It is a computer based class afterall.  I've got a 17" monitor at 1280x1024 and had little trouble.

The simulator instructions state that voltage sources will show current flowing out of the (+) pin. Which it sort of does. The arrow indicates current flowing into the pin, but with a negative value. Great way to confuse newbies.

Great way to get newbies to pay attention to signs and directions.  Current flowing into the positive terminal is negative by definition.  This was explained in the text and at least one place in the introductory descriptions.

It requires knowledge of Kirchhoff's Voltage Law (KVL) more than a week before it is dealt with in lectures - or should I say a week before it was dealt with in the 2007 videos.

KVL is dealt with in the assigned readings.  Again, no babysitting here.  Read the text!

It tells us to use the sign of the first terminal of the first component we come to, to provide the basis of voltage differences. The resistors we have just added have no sign. This would be added arbitrarily in a hand sketch, the polarity being insignificant so long as it is consistent. I am using a negative voltage for a drop and a positive for the supply, but if they chain resistor polarity in the same order as they have denoted the source, my signs will be wrong.

I have zero schooling in engineering and had little trouble with it.  I've seen a couple confusing descriptions in the exercises, but nothing I didn't learn from.

Try harder, and good luck, mate!

[8086]

I wouldn't say badly worded because I think the question was intended to get the student to focus on the lesser noticed abstraction of conservation of energy of an LTI system with independent sources rather than the ability to mechanical apply P=IV.

Okay, but my issue was really with the use of the word "entering" in the question. I mean, power doesnt "enter" a battery while it's discharging. It doesn't really "enter" the resistors either, but that made more sense to me at the time. It just seems like quite a non-intuitive use of language.

But anyway, other than that I think the whole thing works pretty nicely. I'm sure the discussion area will liven up as the course goes on, too.

[IanB]

Okay, but my issue was really with the use of the word "entering" in the question. I mean, power doesnt "enter" a battery while it's discharging. It doesn't really "enter" the resistors either, but that made more sense to me at the time. It just seems like quite a non-intuitive use of language.

But anyway, other than that I think the whole thing works pretty nicely. I'm sure the discussion area will liven up as the course goes on, too.

You really have to think of it this way in engineering, otherwise energy balances will mess up your day.

Power enters the resistor as electrical energy (and leaves the resistor as heat). If you draw a control envelope around the resistor then an energy balance is satisfied: power in minus power out equals rate of accumulation of energy in the resistor.

Electrical power leaves the battery while it is discharging, which means, according to the chosen sign convention (power entering has a positive sign), that power leaving the battery has a negative sign. If you draw a control envelope around the battery, then again an energy balance is satisfied: power in minus power out equals rate of accumulation of energy in the battery. In this case power in is zero, power out is positive, meaning the rate of accumulation of energy in the battery is negative--the battery is losing energy to the circuit. Specifically, the battery has a store of chemical energy which is being depleted. In engineering you always have to remember the sign of each term, as a wrong sign is another thing that will mess up your day.

If you now draw a control envelope around the whole circuit and consider only electrical energy, then the power leaving the battery (negative sign) is exactly balanced by the power entering the resistors (positive signs). The sum of these is zero. Which means in effect that the chemical energy in the battery is being converted to heat energy in the resistors with the electricity acting as the transfer medium. This "electrical energy balance" only works because there are no other ways the circuit can generate electrical energy or dissipate electrical energy or accumulate electrical energy. If, for example, the circuit could radiate energy as electromagnetic waves then the balance would need to be extended to be complete. That's the last thing in engineering that will mess up your day: making the wrong assumptions about which terms to include and which terms can be neglected.

[metalphreak]

All of those questions were also "practise" questions. They do not contribute to your final mark, and you are given the correct answer if you get it wrong. This is so you can LEARN

The homework and lab questions, will tell you if you have a wrong answer, but will not give you the solution. You can continue trying until you understand the answer and get it correct. You can attempt the questions as many times as you like, and still achieve a 100% mark. So if you misread a question and get it wrong because of that, don't worry

[slateraptor]

All of those questions were also "practise" questions. They do not contribute to your final mark, and you are given the correct answer if you get it wrong. This is so you can LEARN

The homework and lab questions, will tell you if you have a wrong answer, but will not give you the solution. You can continue trying until you understand the answer and get it correct. You can attempt the questions as many times as you like, and still achieve a 100% mark. So if you misread a question and get it wrong because of that, don't worry

Actually, "homework" and "lab" questions count for a combined 30% of your final grade. This can be verified in the Profile section.

[metalphreak]

If you look in the Profile section you will perhaps spot the difference between "Practice Scores" and "Problem Scores" 

Many of the questions people are complaining about are practice questions.

[8086]

Well I was definitely talking about a "problem score" question. It seems that any lab or homework question counts, except Lab0.

[slateraptor]

If you look in the Profile section you will perhaps spot the difference between "Practice Scores" and "Problem Scores" 

The only question I've seen a complaint on was a "homework" problem.

[metalphreak]

I thought the sum of all powers question was one in between the lecture videos, but its not. In any case, don't feel apprehensive about trying a solution as you have nothing to lose


The lab in week 2 is a bit tricky, but the solution is quite trivial. It should help people learn about using resistor dividers for signal attenuation as well as resistor balancers for high impedance inputs (like a SPICE probe).

[MrPlacid]

8086, was your question similar to this? Is it 27mW or zero?

[8086]

8086, was your question similar to this? Is it 27mW or zero?

That's pretty similar to the question, yeah.

[MrPlacid]

That's pretty similar to the question, yeah.

Cool, since I didn't sign up I was just curious about what it looked like. Someone should have recorded the whole thing and upload it for the rest of us.

edit: Imagine the youtube hits.

[aluck]

Went onto Week 3. Kewl! 

[Zad]

I'm about 75% of the way through this week's stuff.

Videos S6V6 and S6V7 both seem to be the same, not sure if one is missing. I'm certainly struggling to solve S6E2 using the techniques described. Not least because the graph in S6E2 has different scales for the load than for the source. The intercept implies that a (i=v^3) component has a load point of somewhere around 30V, impressive for a 6A source.

[MikeK]

If you ask in their discussion forum you'll get plenty of help.  I noticed that the graph was not to scale, but it really doesn't matter since you don't use need it to solve the problem.  All you need to do is construct the equations for the two lines, set them equal to each other, and solve.  I'll point out that the technique shown in the lecture video requires that the solution value converges, which it does not in this example...You just have to try a bunch of values and see which direction it's heading and then narrow it down.