Here I began with Vth making mesh equations :

48-6I_{1}-12(I_{1}+I_{2})=0

12(I_{1}+I_{2})+4_{2}+8I_{2}=0

With which I get I_{1}=4A & I_{2}=-2A

Then I considered the open circuit part AB to use the 2 mesh currents there which is where I do not understand how to proceed.

For Rth also I am getting an error as well.

## Best Answer

Start with this: -

The impedance into the node marked with a red arrow is \$\dfrac{1}{\frac{1}{12}+\frac{1}{6}}\$ = 4 ohms.

And this is in series with another 4 ohms when you look at node B. That makes 8 ohms but, there is a resistor to node C\$^1\$ of 8 ohms. So the total impedance between B and A is: -

\$\dfrac{1}{\frac{1}{8}+\frac{1}{8}}\$ = 4 ohms.

\$^1\$ node C connects to node A when analysing impedances because voltage sources short to zero volts. Can you get the voltage next yourself?