# Mesh analysis giving different answer compared to Node Analysis

Discussion in 'Homework Help' started by Andrei Monsanto Boysen, Nov 30, 2014.

1. ### Andrei Monsanto Boysen Thread Starter New Member

Nov 30, 2014
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0
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 ?

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2. ### LDC3 Active Member

Apr 27, 2013
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It's a bit difficult to check your work when you don't tell us what V0, V1, V2, ... are.
Also, how did you determine that V0 is equal to V1? I don't see how any 2 resisters could have the same voltage drop.

3. ### Andrei Monsanto Boysen Thread Starter New Member

Nov 30, 2014
21
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Ok, I have added two additional screens showing my current convetion for the node analysis I made and another one from Tina.

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4. ### Andrei Monsanto Boysen Thread Starter New Member

Nov 30, 2014
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Besides, my mesh analysis equations were in my original post and they are pretty clear to be read.

5. ### Jony130 AAC Fanatic!

Feb 17, 2009
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If M1 and M2 loops are in Clockwise direction we have

Left loop

2*I1 + 3*I1+ (I1- I2)*1 + 3 = 0

Right loop

-3 + (I2 - I1)*1 + 5*I2 + 4*Vo =0

And Vo = -2*I1

And after we solve this we have I1 = - 5/9 = -0.555A and I2 = - 1/3 = - 0.333A

As for the nodal show as your work

6. ### Andrei Monsanto Boysen Thread Starter New Member

Nov 30, 2014
21
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I see, but I don't understand why V0=-2I1 . Why can't I ignore the polarity in the 2ohm resistor when doing mesh analysis ? Suppose we didn't know the polarity before (ok, it was given), shouldnt the result be the same ?

7. ### Jony130 AAC Fanatic!

Feb 17, 2009
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We connect ignore Vo polarity because Vo is a "control" voltage for VCVS.
And because we do first loop in Clockwise direction Vo is -2I1.

8. ### Andrei Monsanto Boysen Thread Starter New Member

Nov 30, 2014
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Thank you very much! It clarified the problem to me very well!

Feb 17, 2009
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10. ### WBahn Moderator

Mar 31, 2012
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For the same reason that you can't ignore polarity when you jump start one car with another -- because polarity counts big time!

11. ### WBahn Moderator

Mar 31, 2012
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4,800
Really?

You don't define any of your currents at all. Not which meshes they are for and not which direction they are in. You don't indicate whether you are summing voltage drops or voltage gains. You require that the reader either be able to read your mind or that they reverse engineer your work in order to guess at what you were doing, always keeping in mind that the fact that you are having trouble almost certainly means that something in there is wrong and now the reader has to try to figure out whether something reflects your approach or just a mistake you've made.

Always define your variables clearly and unambiguously. This not only helps the people reading your work (including, not to overlooked, the person grading your work!), but it also helps YOU be consistent as you work the problem.

12. ### Andrei Monsanto Boysen Thread Starter New Member

Nov 30, 2014
21
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Ok, according to the images I uploaded before:

N1: I1=I2
N2: I3 + I2 + I4 = 0
N3: I4 = I5
N4: I3=I6

I1=(0-V1)/2

I2=(V1-V2)/3

I3=(3-V2)/1

I4=(v3-V2)/5

V3=4V0
V4=3V
V1=V0

Rewriting Node equations: (I have omitted the simplification steps on purpose)

N1: -5V1 +2V2 = 0
N2: 17V1 -23V2 = 45

Thus, solving the system:

V1 = 1,111 V
V2 = 2,778 V
V3 = 4,444V
V4 = 3V
V5 (reference node) = 0 V

This concludes my node analysis.

13. ### WBahn Moderator

Mar 31, 2012
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4,800
Your mesh equations use I1 and I2.

Your figure has I1, I2, I3, I4, I5, and I6.

Not clear at all and getting less clear all the time.

14. ### Andrei Monsanto Boysen Thread Starter New Member

Nov 30, 2014
21
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Again, ok, you were right here, I missed giving this information: M1 refers to the Left Loop and M2 to the Right Loop, both in clockwise. But this is quite standard, although, as you pointed out properly, I can't assume everyone would understand this quickly, sorry for this.

15. ### Jony130 AAC Fanatic!

Feb 17, 2009
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I'm lost. I only see two "formal" nodes here (Vx and GND).

So I write

Vx/(3Ω + 2Ω) + (Vx - 3)/1Ω + (Vx - 4*V0)/5Ω =0

And Vo = (Vx/(3Ω + 2Ω)) *2Ω

After solving this I got Vx = 25/9 = 2.777V and Vo = 2/5* 25/9 = 1.111V

16. ### Andrei Monsanto Boysen Thread Starter New Member

Nov 30, 2014
21
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I posted another image in a reply with my currents and nodes on it.
Besides, b = n +l - 1 , so we have 6 branches, 2 loops, and the equation gives us that we should have 5 nodes, what is true.

17. ### Andrei Monsanto Boysen Thread Starter New Member

Nov 30, 2014
21
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You pointed out the essential nodes. An essential node connects three or more branches.

18. ### WBahn Moderator

Mar 31, 2012
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Now you have a clear basis from which to develop and communicate your work.

Nodal Analysis: One essential node -- Node B (I don't count the ground node since we get to define its value.)

$
(1) \ \ V_B $$\frac{1}{5\Omega} + \frac{1}{5\Omega} + \frac{1}{1\Omega}$$ \; = \; 4V_0 $$\frac{1}{5\Omega}$$ + 3V $$\frac{1}{1\Omega}$$
$

$
(2) \ \ V_0 \; = \; V_B $$\frac{2\Omega}{2\Omega + 3\Omega}$$
$

Multiplying both sides of (Eqn 1) by 5Ω gives us:

$
(3) \ \ 7V_B \; = \; 4V_0 + 15V
$

While solving the (Eqn 2) for Vb gives:

$
(4) \ \ V_B \; = \; \frac{5}{2}V_0
$

Substituting (Eqn 4) into (Eqn 3) yields:

$
(5) \ \ \frac{35}{2}V_0 \; = \; 4V_0 + 15V
$

Solving for Vo yields

$
(6) \ \ V_0 \; = \; $$\frac{2}{27}$$ 15V \; = \ 1.111V
$

EDIT: Hiccup in last computation. Don't know what I did, but as soon as I asked if it made sense that 30/27 was nearly 2 -- previous had an answer of 1.765V -- the mistake was obvious. Notice that by clearly developing the solution, I made it possible to track down and correct a mistake instead of having to start all over.

Last edited: Nov 30, 2014
19. ### Andrei Monsanto Boysen Thread Starter New Member

Nov 30, 2014
21
0

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!

20. ### shteii01 AAC Fanatic!

Feb 19, 2010
3,395
497
My Node Voltage Method solution. Same as Jony, same answer.