Find the resistance of a wire frame shaped as cube (Fig) when measured between points (a) 1.7, (b) 1.2, (c) 1.3.
The resitance of each edge of the frame is `R`

Explanation

(a) The current are as shown .
From Ohm's law applied between 1 and 7 via 1487 ( say )

`IR_(eq) = (l)/(3)R+(l)/(6)R+(l)/(3)R= (5)/(6)Rl`
Thus, `R _(eq)= (5R)/(6)`
(b) The curren distribution is as shown.
Between 1 and 2 from the loop 14321
`-l_(1)R = 2l_(2)R + l_(3)R` or `l_(1) = l_(3) = 2l_(2)`
From the loop 48734
`(l_(2)-l_(3))R+ 2 ( l_(2)-l_(3))R+ (l_(2)-l_(3))R = l_(3)R`
or ,`4(l_(2)-l_(3))= l_(3)` or `l_(3) = (4)/(5)l_(2)`
So, `l_(1) = (14)/(5) l_(2)`
Then, `(l_(1)+2l_(2))R_(eq)= ( 24)/(5) l_(2) R_(eq) = l_(1)R =(14)/(5)l_(2)R`
or ,` R_(eq) = (7)/(12)R`

(c) The current distribution is as shown.
Between 1 and 3
From the loop 15621
`l_(2)R = l_(1)R + (l_(1))/(2)R `or , `l_(2) = 3(l_(1))/(2)`
Then`, (l_(1)+2l_(2))R_(eq)= 4l_(1)R_(eq)`
Hence, `R_(eq) = (3)/(4)R`