Twelve straight uniform wires of length `a` and resistance R are joined to form the edges of a cube of side `a`. Current enters the system at one corner andd leaves from a point on one of the edges meeting at the opposite corner at a distance `Ka(ltKlt1)` from it.
Q. The maximum value of the equivalent resistance is

Options

A. `(9R)/(5)`
B. `(9R)/(2)`
C. `(9R)/(10)`
D. `(5R)/(12)`

Answer

C

Explanation

`Ry+R(y-q)-RP-RZ=0` (or)
`2y-Z=p+q` ..(i)
`Rx+Rr-R(z-p)-RZ=0` (or)
`2z-x=r+P` .(ii)
`Ry+Rq-R(x-R)-Rx=0` (or)
`2x-y=q+r` ..(iii)
Therefore `2p=3y+z-3x` ..(iv)
`2q=3x+y-3z` ..(v)
`2r=3z+x-3y` ..(vi)
Again `(x-r)+(x-r+q)=r+(z-p+r)`
(or) `2x-z=4r-p-q` ..(vii)
substituting values of p,q,r, we find `z=y` ..(viii)
again
`=(z-p)+(z-p+r)+K(z+x+q-p)`
`=p+(1-K)(y-q+p)`
this gives `K(2y+x)+8(x-y)=0` (or)
`V=R[x+(x-r)+(x-r+q)+K(z+x+q-p)]`
`V=R[(7x-2y)/(2)+2K(2x-y)]` ..(10)
`I=x+2y` ..(11)
`(V)/(i)=Req=R[((7x-2y)+4K(2x-y))/(2(x+2y))]`
`=(R)/(12)[10+4K-5K^(2)]`
R is maximum when `K=(2)/(5)` and `R_(max)=(9R)/(10)`