Equivalent resistances from series-parallel combinations

Steps:

• The four configurations for three identical resistors of resistance R yield equivalents: R/3 (all parallel), 2R/3 (two series parallel with one), 3R/2 (two parallel in series with one), and 3R (all series).
• Assign 4Ω to 2R/3, so 2R/3 = 4 → R = 6Ω.
• Verify with 3R/2 = 9: 3(6)/2 = 9Ω, which matches.
• Confirm other values: R/3 = 2Ω and 3R = 18Ω, consistent with distinct configurations.

Why B is correct:

• For R = 6Ω, the formulas for mixed series-parallel equivalents exactly produce 4Ω (2R/3) and 9Ω (3R/2), per parallel resistance law 1/Req = 1/R1 + 1/R2.

Why the others are wrong:

• A: R=3Ω gives 2Ω, 4.5Ω, 6Ω, 9Ω—no 4Ω match.
• C: R=12Ω gives 4Ω, 8Ω, 18Ω, 36Ω—no 9Ω match.
• D: R=16Ω gives 16/3≈5.3Ω, 32/3≈10.7Ω, 24Ω, 48Ω—neither 4Ω nor 9Ω.

Final answer: B