Hi I have been struggling on this question for a few days now and have come up with V thevenin = 11/12V, but other people in my class think it is 7/8V, I used source conversion to come up with this answer.

Cheers

 

Show your work. How did you get to 11/12V?

 

Once I break it down to the section I convert it back to current source and parralel resistance, so that (1/2V)/ 2 =1/4I and 1V/3R=1/3I and 1V/3R=1/3 I, then I add the sources together to get 11/12I in parralel then change to thevenin to get 11/12V. Looking at it now it looks completely wrong but I have got completely lost somewhere and I keep getting back to this point.

Cheers

 

Just to give you an idea of at least one thing you're doing wrong, when you convert the second 1/2 amp current source which is in parallel with 2R back into a voltage source plus series resistance, you can't let the 2R resistor coming from the left be connected directly to the top of your +V voltage source. The 1/2 amp current source plus the parallel 2R resistance becomes a voltage source of V in series with 2R ohms, and the junction of the V source and its series 2R resistance is not accessible to the rest of the circuit.

It should look like your original circuit; the top of each V source is not connected to anything except its own series 2R resistance.

You may have to resort to solving the circuit with the loop equations.

The correct solution is 7/8V.

 

Thanks for that it looks neater now I will give it a go.
Cheers.

 

Chavez,

The easiest way to solve for this is to first find Vth by means of loop or mesh analysis.

Using loop current symbols I(1), I(2), and I(3), solve for I(3).
When you've solved for I3, multiply that by 2R. Looking in, the load will "see" a Vth of V-I(3)*2R.

Then you can find Rth by replacing all V sources with shorts, and solve for the resistance from the load's point of view.