For the circuit shown find the Thevenin’s equivalent circuit between terminals A and B
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?