In a uniform ring of resistance `R` there are two points `A` and `B` such that `/_ ACB = theta`, where `C` is the centre of the ring. The equivalent resistance between `A` and `B` is

Options

A. `(R theta)/(2pi)`
B. `R (1 - (theta)/(2 pi))`
C. `(R theta (2 pi - theta))/(4 pi^(2))`
D. `(R (2 pi - theta))/(4 pi^(2))`

Answer

C

Explanation

(c ) For ring, the resistance per unit length `lambda = (R )/(2 pi r)`
Resistance of sections `APB` and `AQB` are
`R_(1) = rho r theta (R )/(2 pi r) r theta = (R theta)/(2 pi)`
and `R_(2) = (R )/(2 pi r) r (2 pi - theta) = (R (2pi - theta))/(2 pi)`
As `R_(1)` and `R_(2)` are in parallel between `A` and `B`, their equivalent resistance is
`R_(eq) = (R_(1) R_(2))/(R_(1) + R_(2))`.
`= ((R theta)/(2 pi) . (R (2 pi - theta))/(2 pi))/((R theta)/(2 pi) + (R (2 pi - theta))/(2 pi)) = ((R^(2) theta (2pi - theta))/(4 pi^(2)))/((R)/(2 pi) [theta + 2 pi - theta]) = (R (2 pi - theta) theta)/(4 pi^(2))`