Problem with simple Thevenin circuit

[mondo90]

Hello, I am getting headache trying to solve this simple circuit o.0


I set mesh and traditional currents as follows


Now for the left mesh current:

-Vx + 10i1 + 4Vx + 15(i1-i2) = 0 -> 10i1 - 15i2 = 0
(assuming Vx = -5i1)

and for the right mesh:
15(i2-i1) = 0 -> i1 = i2

and here is the problem, the left equation and the right is contradiction.

What's wrong ?

Thanks!

 

[albbg]

Since only dependent sources are present in you circuit, you have to apply 1A source from b to a (as probably you did, but your equations are not correct).
The following equations apply (KVL and KCL):

I = I1 + I2 (I = 1A)
I1 and I2 are that you indicated by straight arrows. I don't know why you reversed the meshes currents.

15*I2 = 10*I1+5*I1-4*Vx
Since Vx = 5*I1, then

15*I2 = -5*I1

Using, now I = I1+I2 and solving by I2:

15*I2 = -5*I + 5*I2 that means

I2 = -1/2*I

Vab = 15*I2 that is:

Vab = -7.5V

The Thevenin resistance is Rth = Vab/I = -7.5 ohm

Since no independent sources are present the Thevenin equivalent is given only by the Thevenin resistance

 

[mondo90]

I don't know why you reversed the meshes currents.

The direction of mesh current is arbitrary and can be chosen freely. So I deliberately choosen as shown in first post just to practice this method. Now can you tell me why I fail with my mesh equations ?

PS: you did this with Kirchoff laws, and it's ok, I did it also, but what abut mesh - current ?

 

[The Electrician]

You have chosen to label your branch currents I1 and I2, and you labeled the mesh currents i1 and i2.

These are very bad choices because on small computer, tablet or smart phone screens, I1 and i1 are nearly indistinguishable, as are I2 and i2. Furthermore, you have chosen the direction of I1 and i1 to be opposite directions through the 15 ohm resistor; similarly I2 and i2.

This is very confusing to the people who would help you.

 

[albbg]

The direction of mesh current is arbitrary and can be chosen freely. So I deliberately choosen as shown in first post just to practice this method. Now can you tell me why I fail with my mesh equations ?

PS: you did this with Kirchoff laws, and it's ok, I did it also, but what abut mesh - current ?

In general is better to choose branch and mesh currents flowing in the same direction. However you equations are nor correct.

The first has to be:

-Vx+10i1+4Vx = 15i2, instead of 15(i1-i2)

from which you can obtain the same result I obtained.

 

[mondo90]

Ok, I solved it finally using mesh analysis. Anyway, going back to your first post. You found I2 = -1/2*I. What does '-' sign tells us ? That the real current is reversed ? But how can it be if we have only one current source and the current should flow as my arrows on schematic indicates.

Thanks

 

[CataM]

You found I2 = -1/2*I. What does '-' sign tells us ? That the real current is reversed ? But how can it be if we have only one current source and the current should flow as my arrows on schematic indicates.

Anyone would thought like you did, but after the analysis you found that I1=1.5 A so , that means that the dependant voltage supply "manage" the circuit for whatever current you put(whatever current from B to A, let us call it "ix") from B to A. I1=1.5*ix .. this is what I was referring with "manage", the dependant voltage supply drives the circuit, and the current goes from + to - from the voltage supply...