Please see attached i need to work out the equivilant thevenin circuit to the left of AB

what is the process for thevenin when you have two voltages like this?

i have already worked out the resistance = 1.62Ω

any help would be appreciated

http://yfrog.com/5xtheveninj

 

Thevenin voltage is the open circuit output voltage, which is how it's already setup, so simply calculate the voltage from A to B. One way is to use superposition: short source #2 and calculate \(V_{\small{AB,1}}\) using only source #1, then short source #1 and calculate \(V_{\small{AB,2}}\) using source #2, and finally add the two together to get \(V_{\small{Thevenin}} = V_{\small{AB,1}}+V_{\small{AB,2}}\).

 

So therefore:
4V VAB1= 5/8*4=2.5V
6V VAB2= 5/17*6=1.764V

VT = VAB1 + VAB2
VT = 2.5V + 1.764
VT = 4.264V

Is this correct?

 

No,
For 4V Vab_2 = (3||5 ) / ( 12 + 3||5) * -4 = -0.54054054V

For 6V Vab_2 = ??

 

For 6V = (12//5) / (3 + 12//5) * 6 = 3.243

therefore

3.243 + - 0.540 = 2.703 V

is this correct? am i getting any closer to grasping this?

 

For 6V = (12//5) / (3 + 12//5) * 6 = 3.243

therefore

3.243 + - 0.540 = 2.703 V

is this correct? am i getting any closer to grasping this?

You are very close to get the right answer.
In your diagram all voltage source have positive terminal point toward point B.
So this mean that voltage on point B has higher potential than point A

So correct answer will be ??

 

im sorry im not sure, have you any more help or hints?

 

does that mean it will just be a negative result?

Therefore: revised:

For 6V = (12//5) / (3 + 12//5) * -6 = -3.243

therefore

-3.243 + - 0.540 = -3.783 V

 

If you measure voltage between to points A and B
Then VAB is potential difference when point B is reference point.
And in your circuit point B has a higher potential than point A.
Look on this picture
http://forum.allaboutcircuits.com/attachment.php?attachmentid=10090&stc=1&d=1292272813
And think about why I put "minus" sign when I calculated 4V VAB voltages
Vab_2 = (3||5 ) / ( 12 + 3||5) * -4 = -0.54054054V

 

does that mean it will just be a negative result?

Therefore: revised:

For 6V = (12//5) / (3 + 12//5) * -6 = -3.243

therefore

-3.243 + - 0.540 = -3.783 V

Yes you right.
Vth = -3.783 V