This doesn't seem to agree with your answer, but perhaps I'm misreading something on your schematic. My vision isn't the best anymore.

Because you made a mistake in final step
\(
(6) \ \ V_0 \; = \; \frac{2}{27}*15V \;
(7) \ \ V_0 \; = \; \frac{30}{27} \ = 1.111V;
\)

 

No problems. You were right when pointed out the missing information on my diagrams. I will avoid occurring in this error again, thanks for the advice!
In fact, you are right, according to what I have studied in Sadiku's book and in Nilsson's book, you can only put the equations for the essential nodes in order to solve the circuit. Although, you will need one more equation due to the voltage sources (you don't know what are the currents thru them before, so KCL can't be applied to voltage sources, making you to come up with KVL or a loop equation). My node analysis is correct (but I took a bit of time to come up with it . My main question was about my Mesh analysis. Jony130 answered it properly. I was making confusion with the V0=-2I1, I couldnt understand why it should be negative, after his explanation I could understand it.
Thank you too for your help!

The signs of the current come from the Initial Assumptions. When we to solve the problem, we Assume that all the mesh currents are in one direction (us Europeans assume that they are clockwise, I noticed that people from Middle East are taught to assume that they are counter clockwise). Now that we have made our assumptions, we setup the equations. Then we solve the equations. Once we found the solutions, we see that some currents are negative. What does it mean? It means that our initial assumptions were wrong. That current X, that we assumed to be positive and that is clockwise, is now negative. The negative sign tells us that our assumption about the direction is wrong. So change the direction of current X and change the sign to plus. That is all.

 

The attached image has the circuit being analyzed.

Using Node Analysis I get:
V1=V0=1.111V , V2 = 2.778V, V3=4V0=4,444V and V4=3V.
I have double checked this in Tina Pro Version 9,3,50.248

Although, performing Mesh analysis, I get a complete different result.
V0 = 2I1
M1: 2I1 + 3I1 + I1 - I2 + 3 = 0 => 6I1 - I2 = -3
M2: -3 + I2 - I1 + 5I2 + 4(2I1) = 0 => 7I1 + 6I2 = 3
Solving this, I get I1 = -0.349A and I2 = 0.907A, what gives me V0 = -0.698V , what is different from Node Analysis result and Tina Pro simulator.

What is wrong here ?

In saying that Vo = 2I1, you are implying that I1 is a current flowing downward in the 2Ω resistor, meaning that it is a counter-clockwise current. Fine.

In your first mesh equation, your first term implies that you are summing the voltage drops in the direction of the mesh current. But the 3V source is a voltage gain in the direction of the mesh current, so it should be subtracted.

This would have been very easy to spot -- and you might well not have made the mistake -- had you clearly labeled your assumed current directions. You have a similar problem in your second mesh equation with respect to the polarity of both of the voltage sources.

 

My Node Voltage Method solution. Same as Jony, same answer.

View attachment 76449

Cool, thanks!

 

Because you made a mistake in final step
\(
(6) \ \ V_0 \; = \; \frac{2}{27}*15V \;
(7) \ \ V_0 \; = \; \frac{30}{27} \ = 1.111V;
\)

Yep, and I should have caught it right at the point I made it.

 

In saying that Vo = 2I1, you are implying that I1 is a current flowing downward in the 2Ω resistor, meaning that it is a counter-clockwise current. Fine.

In your first mesh equation, your first term implies that you are summing the voltage drops in the direction of the mesh current. But the 3V source is a voltage gain in the direction of the mesh current, so it should be subtracted.

This would have been very easy to spot -- and you might well not have made the mistake -- had you clearly labeled your assumed current directions. You have a similar problem in your second mesh equation with respect to the polarity of both of the voltage sources.

Indeed, you are right. In general, when doing mesh analysis, I avoid labeling polarities for the branches, unless they are current or voltage sources (or capacitors for 1st or 2nd order circuits, or even higher order), and when I see a resistor with a polarity, it messes up my mind. I totally agree with your explanation, it made my understanding of it much clearer. Thank you!

 

Start with a clearly labeled diagram that matches the notation you will use in your analysis.
View attachment 76447

Now you have a clear basis from which to develop and communicate your work.

It's strange that all the previous posts state they were using clockwise current in the calculation, and you put in counterclockwise in the diagram. Was this intentional?

 

It's strange that all the previous posts state they were using clockwise current in the calculation, and you put in counterclockwise in the diagram. Was this intentional?

I could say that in this problem the quantity that is being sought, Vo, lends itself to using a counterclockwise current for the meshes. I could also say that the OP's original mesh equations implied a counter-clockwise current. But the truth is that I just tend to use counter-clockwise current. I don't really know why. It might be as simple as drawing a counter-clockwise arc with direction arrow just being more natural for me. I also tend to sum up voltage drops going in the direction of the current on the left-hand side and set that equal to voltage rises on the right. In addition, I tend to work my way around the loop clockwise, which seems counterintuitive even to me, but it's the way I tend to do it. Having developed those habits, I tend to stick with them even if the problem pretty obviously lends itself to using a clockwise current.

When teaching this stuff, especially initially, I will often work the first several problems using a coin to make arbitrary choices, even within the same problem. So I might choose I1 going CCW and I2 going CW and when developing the equations I might sum voltage drops around one mesh but voltage gains around another. Doing Nodal Analysis, I might pick some nodes and sum the current entering them and others sum the currents leaving them and yet other nodes I might sum the current leaving out the horizontal branches and set that equal to the currents coming in from the vertical branches, or vice versa. I find that that seems to dispel some of the cookie-cutter recipe-following mentality and get across the notion that all of these are merely different ways of coherently applying the same underlying KVL and KCL concepts. It really seems to work with most students and seems to hopelessly confuse a small handful, primarily those that have gotten by to that point by simply memorizing a bunch of formulas and procedures and have no clue as to the underlying concepts.