A ring is made of a wire having a resistance `R_(0) = 12 Omega`. Find the point `A` and `B`, as shown in the figure, at which a current carrying conductor should be connected so that the resistance `R` of the subcircuit between these points is equal to `(8)/(3) Omega`

Options

A. `(l_(1))/(l_(2)) = (5)/(6)`
B. `(l_(1))/(l_(2)) = (1)/(3)`
C. `(l_(1))/(l_(2)) = (3)/(8)`
D. `(l_(1))/(l_(2)) = (1)/(2)`

Answer

d

Explanation

Let `R_(1),R_(2)` be the resistances of a wire ring of length `l_(1)` and `l_(2)` respectively As per question
`R_(1) + R_(2)= 12 orR_(1) = 12 - R_(2)`…(i)
and `(R_(1)R_(2))/(R_(1) + R_(2)) = (8)/(3)`
or `R_(1) + R_(2)= (8)/(3)(R_(1) + R_(2)) = (8)/(3)xx 12 = 32`.....(ii)
Solving (i) and (ii) we get,
`R_(1) = 4 Omega and R_(2) = 8 Omega`
as `R prop l`, so `(R_(1))/(R_(2)) = (l_(1))/(l_(2)) = (4)/(8) = (1)/(2)`