# Homework Help: Exercise about Thevenin equivalent circuit

1. Sep 14, 2015

### Frank-95

Hi all :)

I'm trying to figure out the Thevenin equivalent of this circuit:

I start by switching off every source and calculating the equivalent resistance:

RTh = RN = (6 + 14) || 5 = 4 Ω

Then I proceed with finding the voltage at 2.

I note that the nodes between 1 and 2 form a supernode:

So I set up the (super)nodal equation:

1 = v1/6 + 3 + v2/5

Then I proceed with mesh analysis. I note than the voltage source and the current source I1 form a supermesh.

And now I have some problems at setting up the equation :/

I know that:

-14 + 14*i1 + 5*i2 - 6(1 - i1) = 0

To bring this equation as a function of v1 and v2

I tried to think up about these equivalences, but I probably mistakened since the equation would become v1 = 0

14i1 = 14 - v2
5i2 = v2
6(1 - i1) = v1

2. Sep 14, 2015

### Jony130

If you have a problem with a mesh analysis why you do not proceed with the node analysis ? Also you can use a source transformation and convert a current source into a voltage source.
And your supernode equation is also wrong because V2 - V1 is not equal to 14V.

EDIT

The correct super mash look like this:

So we have
I1 = 3A
I2 = 1A

and for loop I3 we have:

-14V + 14Ω*I3 + 5Ω*(I3 - I1) + 6Ω*(I3 - I2) = 0

And please notice that I3 is always in first place in parentheses.

And the solution is
http://www.wolframalpha.com/input/?i=-14+++14*x+++5*(x+-+3)+++6*(x+-+1)+=+0

I3 = 7/5A = 1.4A and the current flowing through 5 ohm resistor is equal to:
IR1 = I3 - I1 = 1.4A - 3A = -1.6A and this means that V2 voltage is equal to :
V2 = -1.6A*5Ω = -8V

Last edited: Sep 14, 2015
3. Sep 15, 2015