The correct option (C) (2R/3)Ω
Explanation:
The circuit can be redrawn as here in every delta. R is in series with R
∴ Req = R + R = 2R.
hence it can be reduced as
∴ 2R || R given {(2R × R) / (2R + R)} = (2R / 3)
∴ RAB = (4/3)R || (4/3)R
= [{(4/3)R × (4/3)R}/{(4/3)R + (4/3)R}]
= [{(16/9)R2}/{(8/3)R}]
= (2R/3)Ω