Q: A and B are two points on a uniform ring of resistance R. The ∠AOB = θ, where O is the center of the ring. The equivalent resistance between A and B is:

(a) $ \displaystyle \frac{R \theta}{2 \pi}$

(b) $ \displaystyle \frac{R (2\pi – \theta)}{4 \pi}$

(c) $ \displaystyle R (1- \frac{\theta}{2 \pi} ) $

(d) $\displaystyle \frac{R}{4 \pi^2} (2\pi -\theta )\theta $

Ans: (d)

Sol: Let r be the radius of the ring.

Resistance per unit length of ring , ρ = R/(2 π r)

Lengths of section making angle θ = r θ

Resistance of this section $ \displaystyle R_1 = \rho r \theta = \frac{R}{2 \pi r} r \theta $

$ \displaystyle R_1 = \frac{R \theta}{2 \pi } $

Lengths of section making angle (2π -θ) = r (2π -θ)

Resistance of this section $ \displaystyle R_2 = \rho r (2\pi -\theta) = \frac{R}{2 \pi r} r (2\pi -\theta) $

$ \displaystyle R_2 = \frac{R (2\pi – \theta ) }{2 \pi } $

Here, R_{1} and R_{2} are in parallel between A and B,

$ \displaystyle R_{eq} = \frac{R_1 R_2 }{R_1 + R_2} $

$ \displaystyle R_{eq} = \frac{\frac{R \theta}{2 \pi }\times \frac{R (2\pi – \theta )}{2 \pi} }{\frac{R \theta}{2 \pi } + \frac{R (2\pi – \theta ) }{2 \pi }} $

$ \displaystyle = \frac{R}{4 \pi^2} (2\pi -\theta )\theta $