Explanation

(c) Here points $B$ and $D$ are common. So $2R$ in arm $DC$ and $2\ R$ in arm $CB$ are in parallel between $C$ and $B$. Their effective resistances
$= \frac{2R \times 2R}{2R + 2R} = R$
The modified and simpler circuit will be shown in figure. The effective resistance between $A$ and $B$ is
[image:3_image182.emf]
$R_{eff} = \frac{R \times (R + R)}{R + (R + R)} = \frac{2}{3}R$