Applying Kirchhoff’s second rule (ΣE + ΣRI = 0) to the closed mesh XYMLKAX,
we get.
-E + [r × 2I + r × I + r × 2I] = O
or 5rI = ε …….. (i)
where E is the emf of the battery of negligible internal resistance. If r is the resistance of the cube between the diagonally opposite corners A and M, then according to the Ohm’s law, we have
6RI = E ……. (ii)
From eqs. (i) and (ii),
6RI = 5rI
Or
R = 5/6r
Hence, the resistance of the cube between any two diagonally opposite corners is 5/6 of the resistance of each wire.