Consider the following circuit: 

The Thévenin equivalent resistance ($$Omega$$) at Points A-B is most nearly:

A.   8
B.   12
C.   20
D.   26

The Thévenin equivalent resistance at Points A-B can be found by shorting all voltage sources and placing open-circuits for all current sources. The circuit in question only contains a $12 V$ independent voltage source, so we will replace this with a short circuit. 

Once the voltage source is shorted, we can see that the $6\Omega$ and $12\Omega$ resistors are in parallel with each other and in series with the $8\Omega$ resistor.

Note: Resistors in parallel are represented with two verticle lines ($\left|\right|$) and the equivalent resistance of N resistors in parallel is calculated as follows:

$R_1\left|\right|R_2\left|\right|...\left|\right|R_{N-1}\left|\right|R_N = {\left(\frac{1}{R_1}+\frac{1}{R_2}+...+\frac{1}{R_{N-1}}+\frac{1}{R_N}\right)}^{-1}$

The Thévenin equivalent resistance can now be seen as 

$$\begin{gathered} R_{TH} = 6\Omega \left|\right|12\Omega + 8\Omega \\ = {\left(\frac{1}{6}+\frac{1}{12}\right)}^{-1} +8 \\ = {\left(\frac{2}{12}+\frac{1}{12}\right)}^{-1}+ 8 \\ = {\left(\frac{3}{12}\right)}^{-1}+ 8 \\ = 4+8 \\ = 12\Omega \end{gathered}$$

Therefore, we can conclude that the answer is B $12 \Omega$.