A and B are two points on a uniform ring of resistance R. The ∠ ACB = 0, where C is the centre of the ring. The equivalent resistance between A and B is

$\frac{R θ (2 π - θ)}{4 π^{2}}$

$R (1 - \frac{θ}{2 π})$

$\frac{R θ}{2 π}$

$\frac{R (2 π - θ)}{4 π}$

Solution

$\frac{R θ (2 π - θ)}{4 π^{2}}$

Resistance per unit length ρ = $\frac{R}{2 πr}$

Lengths of sections APB and AQB are rθ and r $(2 π - θ)$

Resistances of sections APB and AQB are

R₁ = ρr θ

= $\frac{R}{2 πr} rθ$

R₁ = $\frac{R θ}{2 π}$

and R₂ = $\frac{R}{2 πr}$ r( 2 $π -$ θ )

R₂ = $\frac{R (2 π - θ)}{2 π}$

As R₁and R₂ are in parallel between A and B , their equivalent resistance is

R_eq = $\frac{R_{1} R_{2}}{R_{1} + R_{2}}$

= $\frac{\frac{Rθ}{2 π} . \frac{R (2 π - θ)}{2 π}}{\frac{Rθ}{2 π} + \frac{R (2 π - θ)}{2 π}}$

R_eq = $\frac{\frac{R^{2} θ (2 π - θ)}{4 π^{2}}}{\frac{R}{2 π} [θ + 2 π - θ]}$

⇒ R_eq = $\frac{R (2 π - θ) θ}{4 π^{2}}$

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