Help finding Thevanin equivalent circuit

[asdf12312]

Homework Statement

Homework Equations

a load resistor = 25ohm connecting A and B

The Attempt at a Solution

trying to find V(th) first...any help would be appreciated.

Mesh 1: 14I1-10I2-4I3=19
Mesh 2: -10I1+35I2-20I3=0
Supermesh: -4I1-20I2+24I3+8I4=0

Aux. 1: I(x)=I1-I2
Aux. 2: 2I(x)=I4-I3

 

[The Electrician]

The next step is to solve your equations. What do you get for a solution?

 

[asdf12312]

not sure if i did it right..that's what. i mean the equations for the meshes. is the dependent current source only shared by two meshes? because it seems that the mesh at the top also shares that source. so should all 3 be a supermesh ??

anyway, ended up with the matrix
|14 -10 -4 | |I1| = |19|
|-10 35 -20| |I2| = |0|
|12 -36 32 | |I3| = |0|

for the supermesh i used the auxilary equations to get rid of I4. so then i had three equations /w three unknowns. and got:
1.508A for I1.
0.302A for I2.

So Ix=I1-I2=1.206A

 

[The Electrician]

What you have so far is ok, but you need I4 to calculate Vth.

Let me show a way to set up a matrix to get everything all at once.

You can include your auxiliary equations in the matrix by establishing another unknown variable, namely Ix. Do it like this:

[  14 -10  -4    0   0] [I1] = [19 ]
[ -10  35  -20   0   0] [I2] = [ 0 ]
[ -4  -20   24   8   0] [I3] = [ 0 ]
[  1   -1    0   0  -1] [I4] = [ 0 ]
[  0    0   -1   1  -2] [Ix] = [ 0 ]

Solving this system will give you I1,I2,I3,I4 and Ix. You don't need Ix, but it's there.

Now, having I4 what is Vth?

 

[The Electrician]

not sure if i did it right..that's what. i mean the equations for the meshes. is the dependent current source only shared by two meshes? because it seems that the mesh at the top also shares that source. so should all 3 be a supermesh ??

The current source is only shared by I3 and I4. It isn't in the I2 mesh.

 

[asdf12312]

OK, so I didn't feel like doing much work with that matrix so i just plugged in the numbers into a determinant calculator http://www.bluebit.gr/matrix-calculator/

I got 1.20A for I(x), and 2.186A for I4. actually I4 is all I need to solve for V(th), i believe.

V(th)=2.186A*8ohm=17.5V

did i do this rite?

 

[The Electrician]

Yes, you've got Vth correct.

Now, if you'll look at the blue hyperlinks in the lower right of that page, you'll see one "Systems of Linear Equations". Click on that and you'll end up at this page:

http://www.bluebit.gr/matrix-calculator/linear_equations.aspx

I chose "values are delimited by commas" and 6 digit results.

This is much better than using determinants and Cramer's rule. This linear solver makes it really easy.To use that nifty linear system solver, your A matrix will be (choose comma delimited):

14,-10,-4,0,0
-10,35,-20,0,0
-4,-20,24,8,0
1,-1,0,0,-1
0,0,-1,1,-2

and the B matrix (I'd rather call it a column vector in your case):

19
0
0
0
0

You can just copy and paste into the appropriate place on that solver page.

Run the solver and get values for I1, I2, I3, I4, Ix of:

I1 = 1.507937
I2 = .301587
I3 = -.22619
I4 = 2.186508
Ix = 1.206349Now you need to find the current through the 2 ohm resistor when you short terminal a to b. You could do this by replacing the 8 ohm resistor in your matrix by the value 8||2 (that means 8 ohms in parallel with 2 ohms), a value of 8/5, or 1.6 ohms.

So, you just solve this system:

14,-10,-4,0,0
-10,35,-20,0,0
-4,-20,24,1.6,0
1,-1,0,0,-1
0,0,-1,1,-2

19
0
0
0
0

This is just the previous matrix but with the 8 changed to 1.6. This is why formulating your problem as a matrix is a good thing. You can solve the system with the 2 ohm resistor in parallel with the 8 ohm resistor by just changing one value in the matrix and solving again. The solver is doing all the hard arithmetic so it's no big deal to have a matrix as big as 5x5.

You can save the comma delimited matrix as a text file and then you don't ever have to type it all in again. Just copy and paste it into the solver.

Anyway, if you solve this latest system (with the 1.6 instead of 8), you will get a value for I4 of 4.816434 amps.

Now you can calculate the voltage across the parallel combination of 8 ohms and 2 ohms as 4.816434 amps * 1.6 ohms. Now with that voltage you can calculate the current through the 2 ohm resistor as 4.816434 amps * 1.6 ohms / 2 ohms. This is the short circuit current through the 2 ohm resistor when you short a to b; it's Ith.

Now, divide the Vth that you got earlier by Ith and you've got Rth.

But, just to show you how cool using matrix arithmetic can be for problems like this, here's a direct method to get Rth.

Look at your original circuit. If you put a voltage source where the a-b terminals are you will have another mesh with a current of I5. If the added voltage source (call it a test source) has a value of 1 volt, then with all other voltage sources (the 19 volt source) set to zero (replace the 19 volt source with a short), the current I5 which flows will allow us to calculate the impedance looking into the a-b terminals.

In other words, the 1 volt test source will cause a current I5. The ratio of 1 volt/I5 amps is a resistance--it is the very resistance we want, Rth.

So, if we can find the current in the I5 mesh with all other voltage sources reduced to zero (replaced with a short) and our test source applied at the terminals a-b having the value of 1 volt, the value of Rth will be the reciprocal of I5.

Adding one more mesh for I5 just requires adding one more row and column to our A matrix and one more row to the B column vector. We can add those to the bottom and right of the matrix like this:

[  14 -10  -4    0   0  0] [I1] = [ [COLOR="Red"]0[/COLOR] ]
[ -10  35  -20   0   0  0] [I2] = [ 0 ]
[ -4  -20   24   8   0 [COLOR="Red"]-8[/COLOR]] [I3] = [ 0 ]
[  1   -1    0   0  -1  0] [I4] = [ 0 ]
[  0    0   -1   1  -2  0] [Ix] = [ 0 ]
[  0    0    0  [COLOR="Red"]-8[/COLOR]   0 [COLOR="Red"]10[/COLOR]] [I5] = [ [COLOR="Red"]1[/COLOR] ]

The values in red come from the equation for the I5 mesh. If you look at the mesh equation you would have to write for the I5 mesh, you can see where they come from.

Notice that the 19 volt value in the right hand B vector has been set to zero, and the bottom element of the B vector is set to 1 to represent the 1 volt test source applied to the a-b terminals.
You can use the comma delimited values from the first matrix system in this post with an additional row and column like this:

14,-10,-4,0,0,0
-10,35,-20,0,0,0
-4,-20,24,8,0,-8
1,-1,0,0,-1,0
0,0,-1,1,-2,0
0,0,0,-8,0,10

0
0
0
0
0
1

See how easy this is if you have kept the original comma delimited matrix values in a text file? You only have to add one more row and column with a few values. You now have a 6x6 matrix. If you had to solve this by hand, it would be really difficult to avoid making arithmetic mistakes. But by using the linear system solver, it's no big deal.

If you solve this new matrix, you'll get a value for I5. The reciprocal of that number will be the value of Rth, because Rth = 1 volt test source / I5 amps.

Can you get it all to work? Do you get the same value for Rth using both methods I outlined in this post?

 

[asdf12312]

wow..that is pretty complex what u have there

i actually already calculated I(sc). I just wanted to find I(x) cause it seemed important with the dependent current source and all. i actually decided to use a 4x4 matrix instead of the 5x5 one you had (seemed..larger). used this to find V(th).

[ 14 -10  -4   0]  [I1]  =  [19]
[-10  35 -20   0]  [I2]  =  [ 0]
[ -4 -20  24   8]  [I3]  =  [ 0]
[  2  -2   1  -1]  [I4]  =  [ 0]

and to find I(sc)=I(th), i just added another mesh, and edited mesh #3 equation. now i had a 5x5 matrix like you had. could've still simplified it into a 4x4, didn't feel the need though.

[ 14 -10  -4   0  0]  [I1]  =  [19]
[-10  35 -20   0  0]  [I2]  =  [ 0]
[ -4 -20  24   8 -8]  [I3]  =  [ 0]
[  0   0   0  -8 10]  [I4]  =  [ 0]
[  2  -2   1  -1  0]  [I5]  =  [ 0]

and yep, got the same value as you. I(sc)=I(th)=3.853A. thanks!

 

[The Electrician]

wow..that is pretty complex what u have there

i actually already calculated I(sc). I just wanted to find I(x) cause it seemed important with the dependent current source and all. i actually decided to use a 4x4 matrix instead of the 5x5 one you had (seemed..larger). used this to find V(th).

[ 14 -10  -4   0]  [I1]  =  [19]
[-10  35 -20   0]  [I2]  =  [ 0]
[ -4 -20  24   8]  [I3]  =  [ 0]
[  2  -2   1  -1]  [I4]  =  [ 0]

and to find I(sc)=I(th), i just added another mesh, and edited mesh #3 equation. now i had a 5x5 matrix like you had. could've still simplified it into a 4x4, didn't feel the need though.

[ 14 -10  -4   0  0]  [I1]  =  [19]
[-10  35 -20   0  0]  [I2]  =  [ 0]
[ -4 -20  24   8 -8]  [I3]  =  [ 0]
[  0   0   0  -8 10]  [I4]  =  [ 0]
[  2  -2   1  -1  0]  [I5]  =  [ 0]

and yep, got the same value as you. I(sc)=I(th)=3.853A. thanks!

Wow! Did you do all this before my immediately previous post? I'm impressed. You seem to have grasped the use of matrix algebra fairly well. Did your instructor teach this or did you figure it out on your own?

I hope you are using the linear system solver on the web page I pointed out; it's much better than using determinants.

The thing to realize here is that when you use a solver, there is no need to bother simplifying a matrix even if you could; just let the solver do the work.

If you make a tiny change to your last stimulus vector above:

[ 14 -10  -4   0  0]  [I1]  =  [[COLOR="Red"]0[/COLOR]]
[-10  35 -20   0  0]  [I2]  =  [0]
[ -4 -20  24   8 -8]  [I3]  =  [0]
[  0   0   0  -8 10]  [I4]  =  [[COLOR="Red"]1[/COLOR]]
[  2  -2   1  -1  0]  [I5]  =  [0]

You can get a value for I5 with a 1 volt test source applied to the a-b terminals. Then Rth will be the reciprocal of I5

You could also have solved for both Vth and Rth directly by using nodal analysis. Here's how that would work: