Let ABCDKLMN be the skeleton cube formed by joining twelve equal wires each of resistance r. Let the current enter the cube at corner A and after passing through all twelve wires, let the current leave at N, a corner diagonally oppsoite to corner A.
For the sake of convenience, let us suppose that the total current is `6 I`. At A, this current is divided into three equal parts each `(2 I)` along AB,AD and AL as the resistance along these paths are equal. At into two equal parts each equal to `i` because resistance along the two paths are eaual. Thus, the distribution of current in the various arms of skelton cube is shown according to Kirchhoff's First rule. The current leaving the cube at N is again `6 I`. If `epsilon` is the emf of the battery used, then applying Kirchhoff's Second rule to the closed circuit ALKN `epsilon` A, we get
`2 Ir + Ir + 2 Ir = epsilon or 5 Ir = epsilon`
Where `epsilon` is the e.m.f. of the cell of negligible internal resistance.
If R is the resistance of the cube between the diagonally opposite corners A and N, then according to Ohm's Law, we have
` 6 I xx R = epsilon`
From (i) and (ii), ` 6IR = 5 Ir or R = 5/6 r`
Here, `r= 1 Omega`, so `R = 5/6 xx 1 = 5/6 Omega`
Current through battery `= ("voltage of battery")/("total resistance")`
` = 10/(5//6) = 12 A`
