Simplifying Resistor Combinations(EASY PROBLEM)

Problem-Solving Strategy for solving the Simplifying Resistor Combinations

STEP 1. Resistors $R_1$ and $R_2$ are in series if they are connected end to end with one common node and carry exactly the same current. They can then be combined into a single resistor $R_S$, where $R_S=R_1+R_2$

STEP 2. Resistors $R_1$ and $R_2$ are in parallel if they are connected to the same two nodes and have exactly the same voltage across their terminals. They can then be combined into a single resistor $R_P$, where $R_P=\frac{R_1R_2}{R_1+R_2}$

PROBLEM:

Note: This post is based on Irwin's Basic Engineering Circuit Analysis 11th ed. 

If there are any errors in solution, please let me know in the comments

SOLUTION:

$12k \Omega$ and $6k \Omega$ are connected in parallel. Their combined resistance is

$12k \Omega || 6k \Omega=\frac{12k \times 6k}{12k+6k} =4k \Omega$

$2k \Omega$ and $4k \Omega$ are connected in series. Their combined resistance is

$2k \Omega + 4k \Omega=6k \Omega$ 

$12k \Omega$ and $6k \Omega$ are connected in parallel. Their combined resistance is

$12k \Omega || 6k \Omega=\frac{12k \times 6k}{12k+6k} = 4k \Omega$ 

$4k \Omega$ and $4k \Omega$ are connected in parallel. Their combined resistance is

$4k \Omega || 4k \Omega=\frac{4k \times 4k}{4k+4k} =2k \Omega$ 

Therefore, the $R_{AB}$ is

$R_{AB}=2k \Omega$

Related Post: Series and Parallel Resistor Combinations(MORE DIFFICULT)