P and Q are two points on a uniform ring of resistance R. O is the centre of the ring. If the part PQ of the ring subtends an angle θ at the centre O of the ring (i.e. ∠ POQ = θ), the equivalent resistance of the ring between the points P and Q will be ........ [Radius of the ring = r and resistanc…

Question

P and Q are two points on a uniform ring of resistance R. O is the centre of the ring. If the part PQ of the ring subtends an angle θ at the centre O of the ring (i.e. ∠ POQ = θ), the equivalent resistance of the ring between the points P and Q will be ........ [Radius of the ring = r and resistance per unit length of the ring = λ]

Accepted Answer

(1) (R θ(2π-θ))/(4π²). Given: a uniform ring of total resistance R; the arc PQ subtends θ at the centre. Idea: the resistance of an arc is proportional to the angle it subtends. Between P and Q the two arcs are in parallel. Minor arc: R₁ = R θ/(2π). Major arc: R₂ = R (2π-θ)/(2π). Note that R₁ + R₂ = R. Parallel: R_PQ = (R₁R₂)/(R₁+R₂) = (R² θ(2π-θ))/(4π² R). So R_PQ = (R θ(2π-θ))/(4π²). With R = 2π rλ this is also (rλ θ(2π-θ))/(2π).

Step-by-step solution

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