A 10 V battery of negligible internal resistance is connected across diagonally opposite corners of a cubical network of 12 resistors, each 1 $\Omega$. The equivalent resistance of the network is

1. By symmetry the current $3I$ entering corner A splits equally as $I$ into three edges. 2. Applying the loop rule along a body diagonal path gives $\varepsilon = (5/2)IR$. 3. The equivalent resistance is $R_{eq} = \varepsilon/(3I) = (5/6)R$. 4. For $R = 1\ \Omega$, $R_{eq} = 5/6\ \Omega$; B inverts this and the others ignore the symmetry. _Source: NCERT Class 12 Physics Ch 3 "Current Electricity", p.18_