Equivalent resistance of R parallel to 2R in series

Steps:

• Identify the network: one R directly between X and Y, paralleled by two R in series (equivalent 2R).
• Calculate series branch: two resistors give 2R.
• Find parallel equivalent: $\frac{R \times 2R}{R + 2R} = \frac{2R^2}{3R} = \frac{2R}{3}$.
• Set equal to 8Ω: $\frac{2R}{3} = 8$, so $R = 12$Ω.

Why D is correct:

• Parallel combination formula yields $R = 12$Ω when total resistance is 8Ω.

Why the others are wrong:

• A: Assumes all three in parallel ($R/3 = 8$), ignores series branch.
• B: Assumes two in parallel plus one in series ($R/2 + R = 8$), mismatches configuration.
• C: Approximates two in parallel plus one in series ($1.5R \approx 8$), but exact is 5.33Ω for different setup.

Final answer: D