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> you can get the real resistance using:
>
> R4 + R5 = 24K => Ra
> Then
> Ra*R3/(Ra+R3) = 8 => Rb
> Then
> R1 + R2 + Rb = 32K => RE (efetive resistance)

Great, now I'm going to have to go through all the trouble of working
out the Kirchoff's Voltage Law (KVL) equations for this.

First apply an arbitrary voltage to the circuit.  It doesn't matter
what it is.

See the attached schematic for labling

The equations for the current in the loops are as follows:

V-( I1(R1+R3) - I2R1 - I3R3) = 0
- -(-I1R1 + I2(R1+R2+R5) - I3R5) = 0
- -(-I1R3 - I2R5 + I3(R3+R4+R5)) = 0

Converting this to a convenient form for matrix calculations, you get
(As with ascii art, this must be viewed in a monospaced font to look
right)
[V]   [I1]   [R1+R3, -R1, -R3]
[0] = [I2] * [-R1, R1+R2+R5, -R5]
[0]   [I3]   [-R3, -R5, R3+R4+R5]

You may follow the calculations through if you wish, but, here are my
results:

Applying an abitrary voltage of 3V to the 12k grid that was
originally shown, we get:

[3]   [I1]   [24k, -12k, -12k]
[0] = [I2] * [-12k, 36k, -12k]
[0]   [I3]   [-12k, -12k, 36k]

- From that, my result matrix is:

[I1]   [250uA]
[I2] = [125uA]
[I3]   [125uA]

I1 is the current of interest. for determining the thevenin
resistance.
In this case it is:
R = V/I1

Which is: 3V/250uA
Which is: 12k

Which is the response I originally posted.

Dinho, please review your math, I think you may have made a mistake.

- --Brendan

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