Mesh method problems

  • Thread starter esmeco
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  • #1
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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 [Broken]

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...
 
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Answers and Replies

  • #2
berkeman
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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
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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
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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 [tex](2)(I_3) [/tex] ?
What is [tex] (8)(I_3) [/tex] ?
What is [tex] (0.8)(V_0) [/tex] ?
What is [tex] 5 [/tex] ?
 
  • #5
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Well,I'm trying to determine the mesh current using voltages,so the 2xI3 and all those values are the voltages...
 
  • #6
berkeman
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But 0.4*Va is a current. That is a voltage dependent current source. The voltage across is is Vo.
 
  • #7
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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
berkeman
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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
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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
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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
berkeman
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esmeco said:
Va=-2xI3
Vo=-13 + 10xI1
I2 - I1=0.4va
I3 - I2=5
Is the sign correct on Vo?
 
  • #12
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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
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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
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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
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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
161
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I believe the correct equations are:
Va = 2 * I3
13 = 10 * I1 + Vo
 

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