Twelve 1 Ω resistors are used as edges to form a cube. The R_{ eq } seen between the two diagonally opposite corners of cube is
(a) \frac{6}{5} Ω (b) \frac{5}{6} Ω (c) 6 Ω (d) \frac{5}{12} Ω
The equivalent circuits is
Applying KVL, we get
\begin{gathered} V-\frac{I}{3} R-\frac{I}{6} R-\frac{I}{3} R=0 \\ V=I\left\lgroup \frac{R}{3}+\frac{R}{6}+\frac{R}{3} \right\rgroup=I \frac{5 R}{6} \\ \Rightarrow \frac{V}{I}=R_{ eq }=\frac{5}{6} \Omega \end{gathered}