EE people help me solve this problem?
 

What is current ib? (aka current accross 2 ohm resistor).

http://www.x.com/ckt.JPG

The diamond shaped thing is a dependent current source that is dependent on ia (current across the 8 ohm resistor) and has a gain of 2 ( in other words it is 2ia).

Not a EE major but as a CS major I've taken some ECE classes on circuits...
From what I can tell, that dependent current source doesn't do anything, look at this:

The voltage across the R2 and R3 must be the same and V = IR, thus:
2*ia*R2 = ia*R3
2*R2 = R3
8 = 8

Thus, if that current source weren't there, it would behave the same way

With that in mind, you can simply do this:

(R2*R3/(R2 + R3) +R1) * ib = 42V
plug it in, do the math, and get ib = 9A

Not positive if this is right.

Now, if that current supply had been different, the problem would have been a whole lot harder.

Just as an aside: If you're having trouble with problems like this, I HIGHLY recommend going to your professor (I assume you're in college?) and asking for help NOW. Make SURE you understand this stuff thoroughly because it gets much, much harder after this. Once you get to capacitors, inductors, and the dreaded op amps, if you can't do simple resistor work, then you're basically screwed.

I remember one circuits midterm I had... it was out of 100 points and the class average was a 54. I got below average because I lost 50 points on one problem due to a very simple error in resistor work at the very beginning.

the correct answer is 9A. I did it using mesh analysis (loops). You get the equation:

42 - 2*ib - 8*ib + 8*(2*ia) = 0
for the first loop (simply summing the voltages around the loop)

8*ib - 8*(2*ia) - 4*(2*ia) = 0
for the second loop.

From here you have two equations and two unknowns, should be easy to solve. I don't know if you've done mesh analysis yet, but I highly recommend becoming friends with it. You'll use it a LOT.

edit: I changed the picture a little to help visualize what I did.
http://img200.exs.cx/img200/2172/ckt7od.jpg

note: The dependent current supply didn't do anything, but if it's gain were different, this method would still give you the correct answer.

Thanks guys! I guess I wasn't thinking clearly, we haven't done much with dependent sources so it threw me off, even though it wasn't even a big deal. I had my final today, hope i did well enough....

Rangsk, I prefer opamps :) over everything you listed hehe.

Next time use Spice :) I'm kidding. But Spice can help you check your answers. In fact all that mesh current analysis and stuff like is exactly how computer simulations analyze circuits. They basic create matrices based on the equations such as yatzr posted. Except, you can have a gadzillion unknowns and a gadzillion equations.

I remember doing this stuff back in the day. The dependent current and voltage source stuff was a little weird because they are not really a concrete actual part per say. But later on you'll learn that BJT's are actually current controlled current sources and MOSFET's are voltage controlled current sources.

Why mesh analysis?

I just took the parallel circuit and found it's equivalent resistance:

1/Req = 1/R1 + 1/R2 = 1/4 + 1/8
Req = 2.667

Now you have two resistors in series, so add them up:
R = 4.667

With V = iR, i=V/R = 42/4.667 = 9

kutulu, that's what I did to.

When things get really complicated, probably mesh is needed.
So you practice using it on easy circuits first.
or so I'm thinking.