# Mesh method problems

1. Jul 12, 2006

### esmeco

I'm trying to get the equations for this exercise but they're giving me some troubles...I think this could also be done with node voltage,but I've tried with the mesh method.Those are my equations:

Eq.1: -13 + 4xI1 + 6xI1 + 2.5xI2 + 7.5xI2=0
Eq. 2: 2.5xI2 + 7.5xI2 + 2xI3 + 8xI3 + 0.8vo=0

Va=-2xI3
Vo= -13 + 4xI1 + 6xI1
I1- I2=-0.4va
I2-I3=-5

http://i75.photobucket.com/albums/i281/esmeco/Meshmethod.jpg

I'm not sure if those equations are right(which probably aren't because the solution doesn't match) and I'm a bit in doubt with the polarities...Any help is really appreciated!Thanks in advance...

2. Jul 12, 2006

### Staff: Mentor

Sorry, I'm confused. There are three mesh circuits, but you only have two equations listed (with more terms each than I would expect). What are your beginning three equations?

3. Jul 12, 2006

### esmeco

Well,those are the 2 supermesh equations,since there is the dependent current source and the current source which are both shared by two mesh circuits...So,the 3 equations for the individual meshes would be:

Eq. 1:-13 + + 4xI1 + 6xI1 + Vo=0
Eq. 2: +0.4Va + 2.5xI2 + 7.5xI2 + 5=0
Eq. 3: 2xI3 + 8xI3 + 0.8Vo + 5=0

4. Jul 12, 2006

When you perform a mesh what values are writing down?

For example in this expression that you wrote:

Eq. 3: 2xI3 + 8xI3 + 0.8Vo + 5=0

What is $$(2)(I_3)$$ ?
What is $$(8)(I_3)$$ ?
What is $$(0.8)(V_0)$$ ?
What is $$5$$ ?

5. Jul 12, 2006

### esmeco

Well,I'm trying to determine the mesh current using voltages,so the 2xI3 and all those values are the voltages...

6. Jul 12, 2006

### Staff: Mentor

But 0.4*Va is a current. That is a voltage dependent current source. The voltage across is is Vo.

7. Jul 12, 2006

### esmeco

So re-writing the equations they would be like this?

Eq. 1:-13 + + 4xI1 + 6xI1 + Vo=0
Eq. 2: -Vo + 2.5xI2 + 7.5xI2 -V1=0
Eq. 3: 2xI3 + 8xI3 + 0.8Vo +V1=0

V1 would be the voltage on the 5A current source....

8. Jul 12, 2006

### Staff: Mentor

I think that looks better. There's a small typo in equation 1 with a ++, but other than that I think you are on the right track.

9. Jul 12, 2006

### esmeco

So,those equations would reduce to one equation since subtracting V1 from equation 2 and equation 3 would eliminate it and would also eliminate Vo from the equations since we would subtract Vo from eq. 1 and 2...Am I right?
Also we would have equations for Vo,Va,I2-I1 and I3-I2:

Va=-2xI3
Vo=-13 + 10xI1
I2 - I1=0.4va
I3 - I2=5

I hope these are right...

10. Jul 12, 2006

### esmeco

I've tried to use those other equations on my main equations to get the values of the currents on the meshes,but unfortunately with no success...Could anyone give me some hints on how to correct them?

11. Jul 12, 2006

### Staff: Mentor

Is the sign correct on Vo?

12. Jul 12, 2006

### esmeco

Well, the inicial equation for Vo,I think, it would be like this:

-13 + 4xI1 + 6xI1 - Vo=0

Since the current is flowing in a clockwise direction I think it would be -Vo...Am I right?Also,are the other equations right?

13. Jul 13, 2006

### esmeco

So, Vo woould be Vo=13 - 10xI1?It's kind of difficult for me to get Vo equation(if it wan't the one I provided)...

14. Jul 13, 2006

### doodle

Like your other problem with the nodal analysis, I am unsure if the given answer is correct. I got Vo = 56.863, I1 = -4.3863, I2 = -1.9314 and I3 = 3.0686 when I tried to work this one out.

15. Jul 13, 2006

### esmeco

Well, the solutions I had for the currents aren't even close to that...ANd also,the problem is I'm not even sure if these equations are right:

Va=-2xI3
Vo=-13 + 10xI1
I2 - I1=0.4va
I3 - I2=5

Especially Vo I don't know if it's right...Any more help is greatly welcomed...

16. Jul 13, 2006

### doodle

I believe the correct equations are:
Va = 2 * I3
13 = 10 * I1 + Vo