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 currents are as shwon. From Ohm's law applied between 1 and 7 via 1487 (say)

`IR_(eq) = (1)/(3) R + (1)/(6) R + (1)/(3) R = (5)/(6) RI`
Thus, `R_(eq) = (5R)/(6)`
(b) Between 1 and 2 from the loop 14321,
`I_(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) 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) = 4 l_(1) R_(eq)`
`= l_(2) R + l_(2) R = 3l_(2) R`
Hence, `R_(eq) = (3)/(4) R`.