Twleve equal wires , each of resistance r ohm , are connected so as to form a frame of a cube . An electric current enters this cube at one corner and leaves from the diagnalyy opposite corner . Calculate the total resisatance between the corners .

Explanation

Let ADGCHEFS be the frame of the cube formed by tewlve equal wires each of resistance r . Let the total current entering at the corner A and leaving the diagonally opposite corner B be 6x .

Therefore , 12 wires of the cubic framework (mesh ) will have equivalent resistance
`R=(V)/(I)=(V)/(6x) " " ...(1)`
As the resistance of wire is the same , the current 6x is divied at A into three equal parts , one along AD , the other along AE and the third along AC. At points D , E and C the current is again divied into two equal parts . At point F ,G and H the currents combine If V is the potential difference across A and B , then taking the path ACHBA and applying Kirchhoff's second law we get , `2x.r+x.r+2x.r=V` or , `5xr=V` " " ...(2)
From equaltions (1) and (2) we have ,
`6xR=5xr " or , " R=(5)/(6)r`