# Question about mesh and node analysis

1. Sep 7, 2015

### Frank-95

Hi, I'm trying to figure out if I haven't undestood the correct way to do a circuit analysis yet, or if I am just mistakening my calculi.

Here is a scheme of the circuit: https://drive.google.com/file/d/0B52I7EByNa_6ODJIWnVpRTRFems/view?usp=sharing [Broken]

The component in the right bottom angle is a voltage dependent voltage source controlled by V0.

I tried to make a mesh anlysis at first, and I obtained the resulting system (considering the two mesh on the right as a supermesh):

-12 + 2i1 + 4 (i1 - i2) = 0

-10i1 - 4(i1 - i2) + 8i3 + i2 = 0

i1 + i2 = i3 + 3

The results, for i1, i2 and i3, are: 14/3, 4, 17/3

With this results V0 would be 28/3, and the dependend source 140/3, which is wrong.

Can anyone help me to figure out the right process for such circuits? Thank you in advance

Last edited by a moderator: May 7, 2017
2. Sep 7, 2015

### Hesch

What is the sign of V0 across R4 ?
( Attach +/- ).

3. Sep 7, 2015

### Frank-95

It is a V0, necessary to find the voltage source on the bottom right. The + is on the left the - on the right

4. Sep 7, 2015

### Hesch

Well, I would make two nodes, V1 above R1 and V2 above R2.

Make two equations, using Kirchhoffs current law, to find V1 and V2. Afterwards you could find the currents usings ohms law, if desired.

( Pay attention to the signs ).

EDIT: Sorry, call the two nodes V3 and V4. ( V1 and V2 are used ).

Last edited: Sep 7, 2015
5. Sep 7, 2015

### Frank-95

In order to use the KVL I should consider nodes with source and resistance as a single supernode, right? It is what I thought at the beginning but I had (and still have) difficulties to immagine the incoming and outcoming current, and the supernode itself.

I mean, I can generally figure them out pretty easily, but I'm still getting used to this new argument of supernodes and supermesh

6. Sep 7, 2015

### Hesch

You should not use KVL, because you don't know the voltage across a current source ( I1 ).

Attached a sketch with currents and nodes included. As for V3 you could write an equation:

I2 - I4 - I1 - I3 = 0 =>

( V1 - V3 ) / R4 - ( V3 - V4 ) / R3 - I1 - V3 / R1 = 0 ( you know V1, R4, R3, R1, I1 ).

Now write an equation as for V4 and find V3 and V4, then I4 = ( V3 - V4 ) / R3 , etc..

I've never heard of "supernodes".

#### Attached Files:

File size:
264.1 KB
Views:
44
Last edited: Sep 7, 2015
7. Sep 7, 2015

### Jony130

Because this circuit do not have any supernode.

8. Sep 7, 2015

### Frank-95

That makes sense (and thanks for the image), but how can you know a priori which is the orientation of the current through R1? It depends from both V1 both V2, but you don't know V2 voltage, since it is dependent from V0. For this I talked about supernodes and supermesh.

The use of the supermesh analysis lets you "skip" the current generator inside the mesh for KVL application, while the supernode analysis lets you "skip" the voltage generator for KCL. I tried both kinds of analysis: as regards the first one I don't know where I mistakened, as regards the second one I cannot figure out the two (one?) supernodes correctly and the incoming and outcoming currents.

EDIT: Jony130 a supernode isn't usually defined as a resistor and a voltage source in series during the application of the KCL?

9. Sep 7, 2015

### Hesch

I cannot. I decide an orientation. Having decided a "wrong" orientation, the value of the current will just come out with a negative sign.
V2 = 5 * V0
V0 = V1 - V3.

10. Sep 7, 2015

### Jony130

Yes, you're right I missed that. But in this case we do not need to use a supernode because we know V2 value. V2 = 5*Vo = 5*(V1 - V3) and R2 current is V4 - V2. So we do not need a supernode here.

As for the mash

For I1 we have

-12V + I1*2Ω + 4Ω*(I1 - I2) = 0

for I3

I3 = -3A

and for I2 ??

Last edited: Sep 7, 2015
11. Sep 8, 2015

### Frank-95

Really thanks to both. Unluckily I don't have the right results so can you pease tell me if my steps are correct?

By nodes analysis I get (using this scheme):

i2 = (v1 - v3) / 2
i3 = v3 / 4
i4 = (v3 - v4) / 8
i5 = v4 - 5v0 = v4 - 5v1 + 5v3 (knowing that v0 = v1 - v3)

i2 = i4 + 3 + i3
i1 + i4 = i5

v1 / 2 - v3 / 2 = v3 / 8 - v4 / 8 + i1 + v3 / 4
i1 + v3 / 8 - v4/8 = v4 - 5v1 + 5v3

Which results in this system:

7v3 - v4 = 24
39v3 + 9v4 = 504

Whose results are:

v3 = 120/17 V = 7,06 V
v4 = 432/17 V = 25,41 V

By mesh analysis I get (using this scheme)

5v0 = 5(v1 - v3)
i1 = (v1 - v3) / 2 -> v3 = v1 - 2i1 -> 5v0 = 10i1

-12 + 2i1 + 4(i1 - i2)
-10i1 - 4(i1 - i2) + 8i3 + i2 = 0

I have a doubt as regard the underlined voltage. Your scheme makes me suppose that the right formula is 8(i3 + i2), but actually my book says the contrary and considers only i3

12. Sep 8, 2015

### Hesch

As for node analysis, you get:
It should be:

i5 = v4 - ( -5v0 )
Try again!

By means of ohms law, you can cross-check your results.

13. Sep 8, 2015

### Jony130

As for the super node, we do not have any supper node because we "ground" V2 and V1 source. If for example we pick a ground at V3,then we will have two supernodes.

EDIT

Why you have a minus sign here (- 4(i1- i2)) ?? And because this is a equation for loop 2 I2 current is the master, so the correct way is 4Ω * (I2 - I1)

And the correct way is 8(i2 + i3) again I2 is the master

Last edited: Sep 8, 2015