Thevenin theorem problem

Right its been a while since I've touched this kinda thing so I may have a mistake here, check my details as you go along.

Also check this out as it could help:
http://www.electronics-tutorials.ws/dccircuits/dcp_7.html

Right,
Firstly, the theorem essentially states that any (linear) system can be reduced down to a single voltage source with a series resistance and load resistance.

Step 1)
Remove the load from your circuit, this leaves you with a series circuit with the two sources and resistances of 1 Ohm and 2 Ohm.

Step 2)
Remove the voltage sources, this leaves just the resistances, note that they are now in || as both ends are connected together.
Calculate the equivalent resistance,

This is worked out as (R1*R2)/(R1+R2) = (2*1)/ (2+1) = 0.67 Ohms - This will be your series resistance in the Thevenin equivalent circuit.

Step 3)
Put the sources back in and calculate the current using ohms law:
It's in series so you get (10-3)V / (1+2)Ohms = 2.33A

Step 4)
Using this current in the series circuit - remember the load (5 Ohm) is still taken out, calculate the voltage drop across the 1Ohm resistor.
This drop is 2.33A * 1 Ohm = 2.33V
Using this the voltage at the node between the two resistors is (7V - 2.33V) = 4.67V

The 7V comes from 10V - 3V

Step 5)
The 4.67V is the equivalent voltage for our equivalent circuit.

Draw the circuit - this is detailed in the link above -
you get aa 4.67V source in series with the 0.67 Ohm series resistance and the 5 Ohm load.

Using Ohms law the current is 4.67 / 5.67 = 0.823
Therefore, the drop across the load is 5*0.823 = 4.12V

No bother at all

Looks to me (from the diagram) that the voltages add, 10V + 3V, going around the loop, as the 3V cell is shown inverted in the diagram.

*Edit* assuming the 3V cell is upside down, compared to the 10V Cell, then the voltages add and the current in the 5 Ohm resistor should be about 235 mA.

You can redraw the circuit as follows, which I find easier as the voltage sources are both positive on the top.

         -------[ 1R ]-------
         |        i1 ->     |
         |                  |
       ----- +              |
        3V                  |
         -   -              |
         |                  |
     A   -------[ 5R ]-------  B
         |        i2 ->     |
         |                  |
       ----- +              |
        10V                 |
         -   -              |
         |                  |
         |                  |
         -------[ 2R ]-------
                  i3 ->

Kirchhoff's Laws:
        Voltage A to B is same on all three paths:
             -3 + i1 * 1  =  i2 * 5  =  10V + i3 * 2            (1)
        
        Current at B sums to zero:
            i1 + i2 + i3 = 0                                    (2)
            
        Using (1), we can express both i2 and i3 in terms of i1:
            i2 = (i1 - 3 ) / 5                                    (3)
            i3 = (i1 - 13) / 2                                    (4)
        
        Substituting for i2 and i3 in (2), we get:

            i1 + (i1-3)/5 + (i1-13)/2 = 0
        
        Simplify by multiplying by 10:
            10i1 + 2i1 - 6 + 5i1 - 65 = 0 
            17i1 - 71 = 0
            
            i1 = 71/17 = 4.17647 A                                (5)
            
        From (3) and (4):
            
            i2 = (4.16647 - 3) / 5 =  0.23329 A   (actually   4/17 A =  0.235294118)
            i3 = (4.16647 -13) / 2 = -4.41677 A   (actually -75/17 A = -4.411764706)
        
        i2, the current in the 5 Ohm resistor:  235 mA    A to B
        
        Other currents:       1 Ohm resistor: 4.176 Amps  A to B 
                              2 Ohm resistor: 4.411 Amps  B to A

Thevenin:

        Consider 5 Ohm load on AB as "load", and calculate equivalent 
        Vs and Rs for the rest of the circuit.  (This way we can change 
        the 5 Ohm resistor to other values and easily determine the 
        new current.)

         -------[ 1R ]-------
         |                  |
         |                  |
       ----- +              |
        3V                  |
         -   -              |
         |                  |
         ----- A      B ----- 
         |                  |
         |                  |
       ----- +              |
        10V                 |
         -   -              |
         |                  |
         |                  |
         -------[ 2R ]-------
                 
        
        Step 1: Remove 5 Ohm resistor, short all Voltage sources and 
                   calculate R(AB).  Effectively we have a 1 Ohm and a 2 Ohm
                   resistor in parallel, result is 2/3 Ohm or approx 0.6667 Ohms.
                   This is equivalent series resistance Rs.
        
        Next we consider the Voltage across AB at no load.  
        The current running through our no-load circuit is 
        13V / (1+2) = 4.333 A.   If 4.333A runs through a 1 Ohm
        resistor, we get a voltage drop of 4.333V.
        The left side of the resistor is 3V above the Voltage at A, so the 
        no load voltage across AB is 4.333V - 3V = 1.333V, i.e. 
        we find that A is +ve wrt B.
 
        So Vs = +1.333V, this is our equivalent voltage value.
        
        Equivalent circuit:
        
             -------[ 0.6667 ] -------
             |                       |
             |     +                 A
           1.333 V
             |   -                   B
             |                       |
             -------------------------
        
        So what happens when we put a 5 Ohm resistor between A and B?

             -------[ 0.6667 ] -------
             |                       |
             |   +                   |      A
           1.333 V                 5.0 Ohms
             |   -                   |      B
             |                       |
             -------------------------
        
        We get a current of 1.333 V / (5 + 0.6667 Ohms) =  235 mA, 
        running from A to B.
        
        (Actually it's (4/3) / (5 + 2/3) = 4/17 = 235.294118 mA, but that's
         close enough!)

HTH