For the given mixed combination of resistors calculate the total resistance between points A and B.

Choose the correct answer from the options given below.

The correct answer is 18 Ω.

To find the total equivalent resistance between points A and B, we can simplify the resistor network step-by-step from the top of the triangular structure down to the terminals A and B.

Step 1: Simplify the top-most branch
At the very top apex of the circuit, two $4 \mathrm{\Omega }$ resistors are connected in series. Their combined resistance, $R_{\text{top\_series}}$, is calculated as:
$R_{\text{top\_series}}=4 \mathrm{\Omega }+4 \mathrm{\Omega }=8 \mathrm{\Omega }$

Step 2: Simplify the parallel combination with the horizontal 8 Ω resistor
This combined $8 \mathrm{\Omega }$ resistance is in parallel with the horizontal $8 \mathrm{\Omega }$ resistor connected directly below them. The equivalent resistance of this upper section, $R_{\text{upper}}$, is:
$R_{\text{upper}}=\frac{8\times 8}{8+8}=4 \mathrm{\Omega }$

Step 3: Add the next series resistors on the sides
This $R_{\text{upper}}=4 \mathrm{\Omega }$ combination is now in series with the two diagonal $4 \mathrm{\Omega }$ resistors located below the $8 \mathrm{\Omega }$ horizontal line. Their combined series resistance, $R_{\text{middle\_series}}$, is:
$R_{\text{middle\_series}}=4 \mathrm{\Omega }+4 \mathrm{\Omega }+4 \mathrm{\Omega }=12 \mathrm{\Omega }$

Step 4: Simplify the parallel combination with the horizontal 12 Ω resistor
This $12 \mathrm{\Omega }$ equivalent branch is in parallel with the horizontal $12 \mathrm{\Omega }$ resistor. The equivalent resistance, $R_{\text{middle\_parallel}}$, is:
$R_{\text{middle\_parallel}}=\frac{12\times 12}{12+12}=6 \mathrm{\Omega }$

Step 5: Calculate the final total resistance between points A and B
Finally, this $R_{\text{middle\_parallel}}=6 \mathrm{\Omega }$ combination is in series with the two lower-most diagonal $6 \mathrm{\Omega }$ resistors connected to terminals A and B. The total resistance, $R_{\text{AB}}$, is:
$R_{\text{AB}}=6 \mathrm{\Omega }+6 \mathrm{\Omega }+6 \mathrm{\Omega }=18 \mathrm{\Omega }$